5

Astronomical Scales

UNIT 1 ASTRONOMICAL SCALES

Structure

1.1 Introduction

Objectives

1.2 Astronomical Distance, Mass and Time Scales

1.3 Brightness, Radiant Flux and Luminosity

1.4 Measurement of Astronomical Quantities

Astronomical Distances

Stellar Radii

Masses of Stars

Stellar Temperature

1.5 Summary

1.6 Terminal Questions

1.7 Solutions and Answers

1.1 INTRODUCTION

You have studied in Units 9 to 11 of the Foundation Course in Science and

Technology (FST-1) that the universe is vast. You know that the Sun is one amongst

billions of stars situated in as many galaxies. You have also learnt that the distances

between planets and stars are huge, and so are their masses. For example, the distance

between the Sun and the Earth is of the order of 1.5×1011

m. The radius of the Sun

itself is about 7×108m, which is almost 100 times the Earth’s radius. The mass of the

Earth is of the order of 1024

kg and the Sun’s mass is a million times larger. The time

scales involved are also huge. For example, the estimated age of the Sun is about

5 billion years, compared to the lifetime of a human being, which is less than

100 years in most cases. All these numbers are very large compared to the lengths,

masses and time scales we encounter everyday. Obviously, we need special methods

to measure them and represent them.

The distances and masses of celestial objects are of fundamental interest to astronomy.

Does a star in the night sky seem bright to us because it is closer, or is it so because it

is intrinsically bright? The answer can be obtained if we know the distance to a star.

You have also learnt in Unit 10 of FST-1 that the mass of a star determines how it will

evolve.

In this unit we introduce you to some important physical quantities of interest in

astronomy, such as distance, size, mass, time, brightness, radiant flux, luminosity,

temperature and their scales. You will also learn about some simple methods of

measuring these quantities. In the next unit, you will learn about the various

coordinate systems used to locate the positions of celestial objects.

Objectives

After studying this unit, you should be able to:

• describe the distance, mass, time and temperature scales used in astronomy and

astrophysics;

• compare the brightness and luminosity of astronomical objects; and

• determine the distance, size and mass of astronomical objects from given data.

6

Basics of Astronomy 1.2 ASTRONOMICAL DISTANCE, MASS AND TIME

SCALES

In astronomy, we are interested in measuring various physical quantities, such as

mass, distance, radius, brightness and luminosity of celestial objects. You have just

learnt that the scales at which these quantities occur in astronomy are very different

from the ones we encounter in our day-to-day lives.

Therefore, we first need to understand these scales and define the units of

measurement for important astrophysical quantities.

We begin with astronomical distances.

Astronomical Distances

You have studied in your school textbooks that the Sun is at a distance of about

1.5 × 1011

m from the Earth. The mean distance between the Sun and the Earth is

called one astronomical unit. Distances in the solar system are measured in this unit.

Another unit is the light year, used for measuring distances to stars and galaxies.

The parsec is a third unit of length measurement in astronomy.

We now define them.

Units of measurement of distances

• 1 Astronomical Unit (AU) is the mean distance between the Sun and the

Earth.

1 AU = 1.496 ×××× 1011

m

• 1 Light Year (ly) is the distance travelled by light in one year.

1 ly = 9.460 ×××× 1015

m = 6.323 ×××× 104 AU

• 1 Parsec (pc) is defined as the distance at which the radius of Earth’s orbit

subtends an angle of 1″ (see Fig.1.1).

1 pc = 3.262 ly = 2.062 ×××× 105AU = 3.085 ×××× 10

16m

Fig.1.1: Schematic diagram showing the definition of 1 parsec. Note that 1°°°° ≡≡≡≡ 60′′′′ and 1′′′′ = 60″″″″.

Thus, 1″″″″ = 1/3600 degree

STAR

EARTH

distance ==== 1 parsec

angle ==== 1 arc second

SUN

1 AU

7

Astronomical Scales Dimensions of Astronomical Objects

The sizes of stars or stellar dimensions are usually measured in units of solar radius

RΘ. For example, Sirius (), the brightest star in the sky, has radius 2RΘ. The

radius of the star Aldebaran () in Taurus is 40RΘ and that of Antares ( ) in

Scorpius is 700 RΘ.

Unit of measurement of size

1 solar radius, RΘΘΘΘ = 7 ×××× 108m

Mass

Stellar masses are usually measured in units of solar mass MΘ. We know that

MΘ = 2 × 1030

kg. For example, the mass of our galaxy is ~ 1011

MΘ. The mass of a

globular cluster is of the order of 105 − 10

6 MΘ. S. Chandrasekhar showed (Unit 11)

that the mass of a white dwarf star cannot exceed 1.4 MΘ. This is called the

Chandrasekhar limit.

Unit of measurement of mass

1 solar mass MΘΘΘΘ = 2 ×××× 1030

kg

Time Scales

The present age of the Sun is about 5 billion years. It has been estimated that it would

live for another 5 billion years in its present form. The age of our galaxy may be

around 10 billion years. Various estimates of the age of the universe itself give a

figure between 12 and 16 billion years. On the other hand, if the pressure inside a star

is insufficient to support it against gravity, then it may collapse in a time, which may

be measured in seconds, rather than in millions of years.

In Table 1.1, we list the distances, sizes and masses of some astronomical objects.

Table 1.1: Distance, radii and masses of astronomical objects

Distance Radius Mass Remarks

Sun 1 AU 1 RΘ 1 MΘ −

Earth − 0.01 RΘ 10−6

MΘ −

Jupiter 4 AU

(5 AU from

the Sun)

0.1 RΘ 10−3

MΘ Largest planet

Proxima

Centauri

1.3 pc 0.15 RΘ 0.12 MΘ Nearest star

Sirius A 2.6 pc 2 RΘ 3 MΘ Brightest star

Sirius B 2.6 pc 0.02 RΘ 1 MΘ First star identified

as white dwarf

Antares 150 pc 700 RΘ 15 MΘ Super giant star

8

Basics of Astronomy You may like to express the distances and sizes of some astronomical objects in

various units introduced here.

SAQ 1

a) Express the distance between Jupiter and Sun in parsecs, and the distance between

the Earth and the Sun in light years.

b) Express the radius of the Earth in units of RΘ.

Next time when you look at the familiar stars in the night sky, you will have some

idea of how far these are from us, and also how massive they are.

An important problem in astronomy is to find out how much energy is emitted by

celestial objects. It is expressed in terms of the luminosity and is related to the radiant

flux and brightness of the object. You may have noticed that some stars in the night

sky appear bright to us, some less bright and others appear quite faint. How do we

estimate their real brightness? Let us find out.

1.3 BRIGHTNESS, RADIANT FLUX AND LUMINOSITY

It is a common experience that if we view a street lamp from nearby, it may seem

quite bright. But if we see it from afar, it appears faint. Similarly, a star might look

bright because it is closer to us. And a really brighter star might appear faint because it

is too far. We can estimate the apparent brightness of astronomical objects easily, but,

if we want to measure their real or intrinsic brightness, we must take their distance

into account. The apparent brightness of a star is defined in terms of what is called

the apparent magnitude of a star.

Apparent Magnitude

In the second century B.C., the Greek astronomer Hipparchus was the first astronomer

to catalogue stars visible to the naked eye. He divided stars into six classes, or

apparent magnitudes, by their relative brightness as seen from Earth. He numbered

the apparent magnitude (m) of a star on a scale of 1 (the brightest) to 6 (the least

bright). This is the scale on which the apparent brightness of stars, planets and other

objects is expressed as they appear from the Earth. The brightest stars are assigned

the first magnitude (m = 1) and the faintest stars visible to the naked eye are

assigned the sixth magnitude (m = 6).

Apparent Magnitude

Apparent magnitude of an astronomical object is a measure of how

bright it appears. According to the magnitude scale, a smaller

magnitude means a brighter star.

The magnitude scale is actually a non-linear scale. What this means is that a star, two

magnitudes fainter than another, is not twice as faint. Actually it is about 6.3 times

fainter. Let us explain this further.

The response of the eye to increasing brightness is nearly logarithmic. We, therefore,

need to define a logarithmic scale for magnitudes in which a difference of 5

magnitudes is equal to a factor of 100 in brightness. On this scale, the brightness

ratio corresponding to 1 magnitude difference is 1001/5

or 2.512.

Therefore, a star of magnitude 1 is 2.512 times brighter than a star of magnitude 2.

Spend

10 min.

9

Astronomical Scales It is (2.512)2 = 6.3 times brighter than a star of magnitude 3.

How bright is it compared to stars of magnitude 4 and 5?

It is (2.512)3 = 16 times brighter than a star of magnitude 4.

And (2.512)4 = 40 times brighter than a star of magnitude 5.

As expected, it is 2.5125 = 100 times brighter than a star of magnitude 6.

For example, the pole star (Polaris, Dhruva) has an apparent magnitude +2.3 and the

star Altair has apparent magnitude 0.8. Altair is about 4 times brighter than Polaris.

Mathematically, the brightness b1 and b2 of two stars with corresponding magnitudes

m1 and m2 are given by the following relations.

Relationship between brightness and apparent magnitude

=−

1

21021 log5.2

b

bmm (1.1)

( ) 5/

2

15/)(

1

2 2121 100100mmmm

b

b

b

b −−− == (1.2)

In Table 1.2, we give the brightness ratio for some magnitude differences.

Table 1.2: Brightness ratio corresponding to given magnitude difference

Magnitude Difference Brightness Ratio

0.0 1.0

0.2 1.2

1.0 2.5

1.5 4.0

2.0 6.3

2.5 10.0

3.0 16.0

4.0 40.0

5.0 100.0

7.5 1000.0

10.0 10000.0

Modern astronomers use a similar scale for apparent magnitude. With the help of

telescopes, a larger number of stars could be seen in the sky. Many stars fainter than

the 6th magnitude were also observed. Moreover, stars brighter than the first

magnitude have also been observed.

Thus a magnitude of zero or even negative magnitudes have been assigned to extend

the scale. A star of −1 magnitude is 2.512 times brighter than the star of zero

magnitude. The brightest star in the sky other than the Sun, Sirius A, has an apparent

magnitude of − 1.47.

10

Basics of Astronomy The larger magnitude on negative scale indicates higher brightness while the

larger positive magnitudes indicate the faintness of an object.

The faintest object detectable with a large modern telescope in the sky currently is of

magnitude m = 29.

Therefore, the Sun having the apparent magnitude m = − 26.81, is 1022

times brighter

than the faintest object detectable in the sky.

In the following table we list the apparent magnitudes of some objects in the night

sky.

Table 1.3: Apparent magnitudes of some celestial objects

Object Indian Name Apparent Magnitude

Sun Surya −26.81

Full Moon Chandra −12.73

Venus Shukra −4.22

Jupiter Guru −2.60

Sirius A Vyadha −1.47

Canopus Agastya −0.73

α-Centauri −0.10

Betelgeuse Ardra +0.80

Spica Chitra +0.96

Polaris Dhruva +2.3

Uranus Varuna +5.5

Sirius B +8.68

Pluto +14.9

Faintest Star

(detected by a modern

telescope)

+29

Let us now apply these ideas to a concrete example.

Example 1: Comparison of Brightness

Compare the brightness of the Sun and α-Centauri using the apparent magnitudes

listed in the Table 1.3.

Solution

From Table 1.3, mSun − mαC = − 26.81 − (− 0.10) = −26.71. Therefore, using

Eq. (1.2), we obtain

7.105/)71.26( 10100 −−α ==Sun

C

b

b

or

7.1010=αC

Sun

b

b, i.e. the Sun is about 10

11 times brighter than α-Centauri.

You may now like to solve a problem based on what you have studied so far.

11

Astronomical Scales SAQ 2

a) The apparent magnitude of the Sun is − 26.81 and that of the star Sirius is − 1.47.

Which one of them is brighter and by how much?

b) The apparent magnitudes of the stars Arcturus and Aldebaran are 0.06 and 0.86,

respectively. Calculate the ratio of their brightness.

The apparent magnitude and brightness of a star do not give us any idea of the total

energy emitted per second by the star. This is obtained from radiant flux and the

luminosity of a star.

Luminosity and Radiant Flux

The luminosity of a body is defined as the total energy radiated by it per unit

time.

Radiant flux at a given point is the total amount of energy flowing through per

unit time per unit area of a surface oriented normal to the direction of

propagation of radiation.

The unit of radiant flux is erg s−1

cm−2

and that of luminosity is erg s−1

.

In astronomy, it is common to use the cgs system of units. However, if you wish to

convert to SI units, you can use appropriate conversion factors.

Note that here the radiated energy refers to not just visible light, but includes all

wavelengths.

The radiant flux of a source depends on two factors:

(i) the radiant energy emitted by it, and

(ii) the distance of the source from the point of observation.

Suppose a star is at a distance r from us. Let us draw an imaginary sphere of radius r

round the star. The surface area of this sphere is 4π r2. Then the radiant flux F of the

star, is related to its luminosity L as follows:

24 r

LF

π=

(1.3)

The luminosity of a stellar object is a measure of the intrinsic brightness of a star. It is

expressed generally in the units of the solar luminosity, LΘ, where

13326 serg104W104 −Θ ×=×=L

For example, the luminosity of our galaxy is about 1011

LΘ.

Now, the energy from a source received at any place, determines the brightness of the

source. This implies that F is related to the brightness b of the source: the brighter the

source, the larger would be the radiant flux at a place. Therefore, the ratio of

brightness in Eq. (1.2) can be replaced by the ratio of radiant flux from two objects at

the same place and we have

5/)(

1

2 21100mm

F

F −= (1.4)

Spend

10 min.

12

Basics of Astronomy You know from Eq. (1.3) that the flux received at a place also depends on its distance

from the source. Therefore, two stars of the same apparent magnitude may not be

equally luminous, as they may be located at different distances from the observer: A

star’s apparent brightness does not tell us anything about the luminosity of the star.

We need a measure of the true or intrinsic brightness of a star. Now, we could easily

compare the true brightness of stars if we could line them all up at the same distance

from us (see Fig. 1.2). With this idea, we define the absolute magnitude of a star as

follows:

Absolute Magnitude

The absolute magnitude, M, of an astronomical object is defined as its

apparent magnitude if it were at a distance of 10 pc from us.

Fig.1.2: Absolute magnitude of astronomical objects

Let us now relate the absolute magnitude of a star to its apparent magnitude. Let us

consider a star at a distance r pc with apparent magnitude m, intrinsic brightness or

luminosity L and radiant flux F1. Now when the same star is placed at a distance of

10 pc from the place of observation, then its magnitude would be M and the

corresponding radiant flux would be F2. From Eq. (1.4), we have

5/)(

1

2 100 Mm

F

F −= (1.5)

Since the luminosity is constant for the star, we use Eq. (1.3) to write

2

1

2

pc 10

pc

=

r

F

F (1.6)

Using Eq. (1.6) in Eq. (1.5), we get the difference between the apparent magnitude

(m) and absolute magnitude (M ).

13

Astronomical Scales It is a measure of distance and is called the distance modulus (see Fig. 1.3).

Distance modulus

5log5pc10

pclog5 1010 −=

=− r

rMm (1.7)

Fig.1.3: Star cluster showing distance modulus as a measure of distance. For the star farther away,

m = 12.3, M = 2.6, r = 871 pc. For the closer star, m = 8.0, M = 5.8, r = 28 pc

We can also relate the absolute magnitudes of stars to their luminosities. From

Eq. (1.3), we know that the ratio of radiant flux of two stars at the same distance

from the point of observation is equal to the ratio of their luminosities. Thus, if M1

and M2 are the absolute magnitudes of two stars, using Eq. (1.5), we can relate their

luminosities to M1 and M2.

Relationship between Luminosity and Absolute Magnitude

5/)(

1

2 21100MM

L

L −= (1.8)

or

=−

1

21021 log5.2

L

LMM (1.9)

Thus, the absolute magnitude of a star is a measure of its luminosity, or intrinsic

brightness.

Often if we know what kind of star it is, we can estimate its absolute magnitude. We

can measure its apparent magnitude (m) directly and solve for distance using

Eq. (1.7). For example, the apparent magnitude of Polaris (pole star) is +2.3. Its

absolute magnitude is −4.6 and it is 240 pc away. The apparent magnitude of Sirius A

is −1.47, its absolute magnitude is +1.4 and it is at a distance of 2.7 pc.

You may now like to stop for a while and solve a problem to fix these ideas.

SAQ 3

a) The distance modulus of the star Vega is −0.5. At what distance is it from us?

b) If a star at 40 pc is brought closer to 10 pc, i.e., 4 times closer, how bright will it

appear in terms of the magnitude?

Spend

5 min.

Farther away

Closer

14

Basics of Astronomy We now discuss some simple methods of measuring astronomical distances, sizes,

masses and temperatures.

1.4 MEASUREMENT OF ASTRONOMICAL QUANTITIES

Since the brightness of heavenly objects depends on their distances from us, the

measurement of distance is very important in astronomy. You must have measured the

lengths of several objects in your school and college laboratories. But how do we

measure astronomical distances? Obviously, traditional devices like the metre stick or

measuring tapes are inadequate for such measurements. Other less direct ways need to

be used. We now discuss some common methods of measuring astronomical

distances. Since stars have been studied most extensively, we will focus largely on

them in our discussion.

1.4.1 Astronomical Distances

You may be familiar with the method of trigonometric parallax. To get an idea of

what it is, perform the following activity.

Activity 1: Trigonometric Parallax

Extend your arm and hold your thumb at about one foot or so in front of your eyes.

Close your right eye and look at your thumb with your left eye. Note its position

against a distant background. Now close your left eye and look at your thumb with

your right eye. Do you notice that the position of the thumb has shifted with respect to

the background? Your thumb has not moved. However, since you have looked at it

from different point (left and right eyes), it seems to have shifted. The shift in the

apparent position of the thumb can be represented by an angle θ (Fig. 1.4).

Parallax is the apparent change in the position of an object due to a change in the

location of the observer.

Fig.1.4: Parallax angle and baseline

We call θ/2, the parallax angle. The distance b between the points of observation (in

this case your eyes), is called the baseline. From simple geometry, for small

angles, ,2 d

b=

θ where d is the distance from the eyes to the thumb.

θ

b

d

15

Astronomical Scales The parallax method can be used to measure the distances of stars and other objects in

the sky. The principle of the method is similar to the one used in finding the height of

mountain peaks, tall buildings, etc.

Let us now find out how this method can be used to measure astronomical distances.

Stellar Parallax

For measuring the distance of a star, we must use a very long baseline. Even for

measuring the distance to the nearest star, we require a baseline length greater than the

Earth’s diameter. This is because the distance of the star is so large that the angle

measured from two diametrically opposite points on the Earth will differ by an

amount which cannot be measured. Therefore, we take the diameter of the Earth’s

orbit as the baseline, and make two observations at an interval of six months (see

Fig. 1.5).

One half of the maximum change in angular position (Fig. 1.5) of the star is defined as

its annual parallax. From Fig. 1.5, the distance r of the star is given by

θ= tanr

dSE (1.10a)

where dSE is the average distance between the Sun and the Earth. Since the angle θ is

very small, tan θ ≅ θ, and we can write

θ

= SEdr (1.10b)

Remember that this relation holds only when the parallax angle θθθθ is expressed in

radians.

Fig.1.5: Stellar parallax

Since, dSE = 1 AU, we have

θ

=AU

r1

(1.10c)

r

dSE

16

Basics of Astronomy If we measure θ in arc seconds, then the distance is said to be in parsecs.

One parsec is the distance of an object that has a parallax of one second of an

arc (1″″″″).

The nearest star Proxima Centauri has a parallax angle 0.77 ″. Thus its distance is

1.3 pc. Since the distance is proportional to 1/θ, the more distant a star is, the

smaller is its parallax.

In Table 1.4 we give the parallax angles and distances of some stars.

Table 1.4: Parallaxes and distances of some bright stars

Star θθθθ (in arc-seconds) distance (r pc)

α-CMa 0.375 2.67

αCMi 0.287 3.48

αAquila 0.198 5.05

αTauri 0.048 20.8

αVirginis 0.014 71.4

αScorpii 0.008 125

Note that the angle θ cannot be measured precisely when the stellar object is at a large

distance. Therefore, alternative methods are used to determine distances of stellar

objects.

You could now try an exercise to make sure you have grasped the concept of parallax.

SAQ 4

a) The parallax angles of the Sun’s neighbouring stars (in arc-seconds) are given

below. Calculate their distances.

Star Parallax

Alpha Centauri 0.745

Barnard’s star 0.552

Altair 0.197

Alpha Draco 0.176

b) A satellite measures the parallax angle of a star as 0.002 arc-second. What is the

distance of the star?

You have just learnt that the parallax method helps us in finding the distances to

nearby stars. But how can we find out which stars are nearby? We can do this by

observing the motion of stars in the sky over a period of time.

Proper Motion

All celestial objects, the Sun, the Moon, stars, galaxies and other bodies are in relative

motion with respect to one another. Part of their relative motion is also due to the

Earth’s own motion. However, the rate of change in the position of a star is very slow.

It is not appreciable in one year or even in a decade. For example, if we photograph a

small area of the sky at an interval of 10 years, we will find that some of the stars in

the photograph have moved very slightly against the background objects (Fig.1.6).

Spend

10 min.

17

Astronomical Scales

Fig.1.6: Motion of a star with respect to distant background objects

The motion of a star can be resolved along two directions:

i) Motion along the line of sight of the observer, (either towards or away from the

observer) is called the radial motion.

ii) Motion perpendicular to the line of sight of the observer is called proper motion

(see Fig. 1.7).

Fig.1.7: Radial and proper motion of a star

Radial motion causes the spectral lines of a star to shift towards red (if the motion is

away from the observer) or towards blue (if the motion is towards the observer). This

shift is the well-known Doppler shift. The proper motion is very slow. It is measured

over an interval of 20 to 30 years. It is expressed in arc seconds per year. The average

proper motion for all naked eye stars is less than 0.1 arc second/yr.

The proper motion is denoted by µ. For a star at a distance r from the Earth it is

related to its transverse velocity as follows:

proper motion = star theof distance

velocitytransverse

or r

vθ=µ , (1.11a)

where vθ is the transverse velocity.

Hence,

rv µ=θ (1.11b)

Space

motion

Radial

motion

18

Basics of Astronomy If µ is measured in units of arc-seconds per year and r in pc, the transverse velocity is

given by

rv µ=−θ 74.4)skm( 1 (1.11c)

If we add the radial velocity vector and the proper motion vector, we obtain the space

velocity of a star (Fig.1.7).

We can locate stars that are probably nearby by looking for stars with large proper

motions (see Fig. 1.8a). Proper motion of a star gives us statistical clues to its

distance. If we see a star with a small proper motion, it is most likely to be a distant

star. However, we cannot be absolutely certain since it could also be a nearby star

moving directly away from us or toward us (see Fig. 1.8b).

Fig.1.8: a) If two stars have the same space velocity and move perpendicular to the line of sight, the

one with the larger proper motion will be nearer; b) Two stars at the same distance with

the same velocity may have different proper motions, if one moves perpendicular to the

line of sight and the other is nearly parallel to the line of sight

We know that the Sun itself is not stationary. The space velocity vector of a star must

be corrected by subtracting from it the velocity vector of the Sun.

The space velocity of a star corrected for the motion of the Sun is termed as the

peculiar velocity of the star.

The peculiar velocities of stars are essentially random and their typical magnitude is

such that in a time of about 106 years the shape of the present constellations will

change completely and they would not be recognisable (Fig. 1.9).

Fig.1.9: Change in the shape of the Big Dipper due to peculiar velocities

You may now like to solve a problem based on these concepts.

(a) (b)

100,000 years ago Present 100,000 years from now

19

Astronomical Scales SAQ 5

The star η CMa is at a distance of 800 pc. If the proper motion of the star is 0.008″/yr,

calculate its transverse velocity in km s−1

.

So far you have learnt how we can find distances of stars. In astronomy, it is equally

important to know the sizes of stars. Are they all the same size, or are some of them

smaller or larger than the others? Let us now find out how stellar radii may be

measured.

1.4.2 Stellar Radii

There are several ways of measuring the radii of stars. Here we describe two methods:

• the direct method, and

• the indirect method.

Direct Method

We use this method to measure the radius of an object that is in the form of a disc. In

this method, we measure the angular diameter and the distance of the object from the

place of observation (see Fig. 1.10).

Fig.1.10: Direct method

If θ (rad) is the angular diameter and r is the distance of the object from the observer

then the diameter of the stellar object will be

D = θ × r (1.12)

This method is useful for determining the radii of the Sun, the planets and their

satellites. Since stars are so far that they cannot be seen as discs even with the largest

telescopes, this method cannot be used to find their radii. For this we use other

methods.

In Table 1.5 we give the radius of some stars.

Table 1.5: Radius of some stars

Star θθθθ (in arc seconds) Radius (in RΘ)

α Tau 0.020 48

α Ori 0.034 214

α Sco 0.028 187

The luminosity of a star can also reveal its size since it depends on the surface area

and temperature of star. This provides a basis for the indirect method of determining

stellar radii.

Spend

2 min.

r θ

D

20

Basics of Astronomy Indirect Method

To obtain stellar radii, we can also use Stefan-Boltzmann law of radiation

F = σ T4 (1.13)

where F is the radiant flux from the surface of the object, σ, Stefan’s constant and T,

the surface temperature of the star. You have learnt in Sec. 1.3 that the luminosity L of

a star is defined as the total energy radiated by the star per second. Since 4π R2 is the

surface area, we can write

FRL24π=

where R is the radius of the star. If the star’s surface temperature is T, using

Eq. (1.13), we obtain

424 TRL σπ= (1.14)

The knowledge of L and T gives R.

Now let us consider two stars of radii R1 and R2 and surface temperatures T1 and T2,

respectively. The ratio of luminosities of these two stars will be

4

2

1

2

2

1

2

1

=

T

T

R

R

L

L (1.15)

But from Eq. (1.8), we have

5/)(

1

2 21100MM

L

L −= (1.16)

where M1 and M2 are the absolute magnitudes. Therefore, from Eqs. (1.15) and

(1.16), we get

( )214.0

41

21

42

22 10

MM

TR

TR −= (1.17)

Using Eq. (1.17) let us now determine the ratio of radii of Sirius A and Sirius B.

Example 2: Determining stellar radii

The surface temperatures of Sirius A and Sirius B are found to be equal. The absolute

magnitude of Sirius B is larger than that of Sirius A by 10. Thus, M1 − M2 = −10 and

we have

R2 = 0.01 × R1

Thus the radius of Sirius A is 100 times that of Sirius B.

You may like to attempt an exercise now.

SAQ 6

The luminosity of a star is 40 times that of the Sun and its temperature is twice as

much. Determine the radius of the star.

Mass is also a fundamental property of a star, like its luminosity and its radius.

Unfortunately, mass of a single star cannot be found directly. If, however, two stars

revolve round each other, it is possible to estimate their masses by the application of

Kepler’s laws.

Spend

5 min.

21

Astronomical Scales 1.4.3 Masses of Stars

Two stars revolving around each other form a binary system. Fortunately, a large

fraction of stars are in binary systems and therefore their masses can be determined.

Binary stars can be of three kinds:

1. Visual binary stars: These stars can be seen moving around each other with the

help of a telescope. If both the stars have comparable masses, then both revolve

around their common centre of mass in elliptic orbits. If, however, one is much

more massive than the other, then the less massive star executes an elliptic orbit

around the more massive star (Fig.1.11).

Fig.1.11: Orbit of visual binary stars

2. Spectroscopic binary stars: The nature of these stars being binary is revealed by

the oscillating lines in their spectra. Consider the situation in Fig.1.12a. Here star

1 is moving towards the observer and star 2 is moving away from the observer.

The spectral lines of star 1 are, therefore, shifted towards blue region from their

original position due to Doppler Effect. The lines of star 2 are shifted towards red.

Half a period later, star 1 is moving away from the observer and star 2 is moving

towards the observer (Fig.1.12b). Now the spectral lines of the two stars are

shifted in the directions opposite to the earlier case. In this way the spectral lines

oscillate.

Fig.1.12: Spectroscopic binary stars

2 1

To Earth

1 2

To Earth

(a) (b)

22

Basics of Astronomy Observations of oscillating lines indicate that the stars are binary stars. If only one

of the stars is bright, then only one set of oscillating lines is observed. If both the

stars are bright, then two sets of oscillating lines are seen.

3. Eclipsing binary stars: If the orbits of two stars are such that the stars pass in

front of each other as seen by an observer (Fig.1.13), then the light from the group

dips periodically. The periodic dips reveal not only the binary nature of the stars,

but also give information about their luminosities and sizes.

Fig.1.13: Eclipsing binary stars

Now suppose M1 and M2 are the masses of the two stars and a is the distance between

them, then we can write Kepler’s third law as

( ) 3212

2

4aMM

GP=+

π (1.18)

where P is the period of the binary system and G is the constant of gravitation. This

relation gives us the combined mass of the two stars. However, if the motion of both

the stars around the common centre of mass can be observed, then we have

2211 aMaM = (1.19)

where 1a and 2a are distances from the centre of mass. Then both these equations

allow us to estimate the masses of both the stars.

Masses of stars are expressed in units of the solar mass, MΘ = 2 × 1030

kg. Most stars

have masses between 0.1 MΘ and 10 MΘ. A small fraction of stars may have masses of

50 MΘ or 100 MΘ.

So far we have discussed the ways of measuring stellar parameters such as distance,

luminosity, radii and mass. Stellar temperature is another important property of a star.

1.4.4 Stellar Temperature

The temperature of a star can be determined by looking at its spectrum or colour. The

radiant flux (Fλ) at various wavelengths (λ) is shown in Fig.1.14. This figure is quite

similar to the one obtained for a black body at a certain temperature. Assuming the

star to be radiating as a black body, it is possible to fit in a Planck’s curve to the

observed data at temperature T. This temperature determines the colour of the star.

c a

b

d

Due to limb

darkening

To

tal

lig

ht

ou

tpu

t

b

c

d

a

b

a

Time

23

Astronomical Scales

Fig.1.14: Total energy flux at different wavelengths.

The temperature of a star (corresponding to a black body) may be estimated using

Wien’s law:

λmaxT = 0.29 cm K (1.20)

Such a temperature is termed as surface temperature, Ts. In general it is difficult to

define the temperature of a star. For instance the temperature obtained from line

emission is indicative of temperature from a region of a star where these lines are

formed. Similarly the effective temperature of a star corresponds to the one obtained

using Stefan-Boltzman law, i.e., F = .4eTσ

In Table 1.6 we give the range of values of stellar parameters of interest in astronomy

such as mass, radius, luminosity and stellar temperature.

Table 1.6: Range of Stellar Parameters

Stellar Parameters Range

Mass 0.1 − 100 MΘ

Radius 0.01* − 1000 RΘ

Luminosity 10−5

− 105 LΘ

Surface Temperature 3000 − 50,000 K

*It is difficult to put any lower limit on the radii of stars. As you will learn later, a neutron star has a

radius of only 10 km. The radius of a black hole cannot be defined in the usual sense.

We can find various empirical relationships among different stellar parameters, e.g.,

mass, radius, luminosity, effective temperatures, etc. Observations show that the

luminosity of stars depends on their mass. We find that the larger the mass of a star,

the more luminous it is. For most stars, the mass and luminosity are related as

5.3

=

ΘΘ M

M

L

L (1.21)

In this unit we have introduced you to a number of astronomical quantities and

described some simple ways of measuring them. We now summarise the contents of

this unit.

1.5 SUMMARY

• The astronomical units of distance, size, mass and luminosity are defined as

follows:

− 1 astronomical unit (AU) is the mean distance between the Sun and the

Earth. 1 AU = 1.496 × 1011

m.

− 1 light year (ly) is the distance travelled by light in one year.

1 ly = 9.4605 × 1015

m = 6.32 × 104 AU.

T

Eλ

λmax λ

24

Basics of Astronomy − 1 parsec (pc) is defined as the distance at which the radius of Earth’s orbit

subtends an angle of 1″. 1 pc = 3.262 ly = 2.06 × 105AU = 3.084 × 10

16m.

− 1 solar radius RΘ = 7 × 108

m.

− 1 solar mass MΘ = 2 × 1030

kg.

− 1 solar luminosity LΘ = 4 × 1026

W.

• Apparent magnitude of an astronomical object is a measure of how bright it

appears. Its absolute magnitude is defined as its apparent magnitude if it were at

a distance of 10 pc from us.

• The difference in apparent and absolute magnitude is called the distance modulus

and is a measure of the distance of an astronomical object:

=−

pc10

pclog5 10

rMm

• Radiant flux is the total amount of energy flowing per unit time per unit area

oriented normal to the direction of its propagation. The luminosity of a body is

defined as the total energy radiated per unit time by it.

• Brightness and radiant flux of an object are related to its apparent magnitude as

follows:

5/)(

1

2 21100mm

b

b −=

5/)(

1

2 21100mm

F

F −=

• The absolute magnitude and luminosity are related as follows:

=−

1

21021 log5.2

L

LMM

• If θ is the parallax of an object in arc seconds, then its distance in parsecs is given

by

θ

=AU1

r

• The motion of an object can be resolved into two components: radial motion and

proper motion. The proper motion µ of a star is related to its transverse velocity

vθ as follows:

r

vθ=µ

where r is its distance.

• Stellar radii are related to the absolute magnitudes and temperatures of stars:

( ) 2141

21

42

22

4.0antilog MMTR

TR−+=

25

Astronomical Scales • The masses M1 and M2 of stars in a binary system can be estimated from the

following relations:

( ) 3212

2

4aMM

GP=+

π

2211 aMaM =

where P is the time period of the binary system, G the constant of gravitation, a

the distance between them and a1, a2, their distances from the centre of mass,

respectively.

• The temperature of a star can be estimated by fitting observed data to Planck’s

black body radiation curve or using Wein’s law: cm29.0max =λ T

The temperature T of a star can also be estimated from Stefan-Boltzmann law:

4TF σ= .

1.6 TERMINAL QUESTIONS Spend 30 min.

1. The apparent magnitude of full moon is − 12.5 and that of Venus at its brightest is

− 4.0. Which is brighter and by how much?

2. The apparent magnitude of the Sun is − 26.8. Find its absolute magnitude.

Remember that the distance between the Sun & the Earth is 1.5 × 1013

cm.

3. After about 5 billion years the Sun is expected to swell to 200 times its present

size. If its temperature becomes half of what it is today, find the change in its

absolute magnitude.

4. The mass of star Sirius is thrice that of the Sun. Find the ratio of their luminosities

and the difference in their absolute magnitudes. Taking the absolute magnitude of

the Sun as 5, find the absolute magnitude of Sirius.

1.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. a) Jupiter is 5 AU from the Sun.

1 pc = 2.06 × 105 AU

5 AU = 2.43 × 10−5

pc

Distance between Earth and the Sun = 1 AU.

1 ly = 6240 AU

1 AU = 1.6 × 10−4

ly

b) The radius of the Earth is 0.01 RΘ .

2. a) The Sun is brighter

mSun − mSi = − 26.81 − (− 1.47) = − 25.34

26

Basics of Astronomy

07.55/34.25

2

1100100 ==

b

b = 1.38 × 10

10

The Sun is about 1010

times brighter than Sirius.

b) 5/)86.006.0(

2

1100 −−=

b

b = 100

+.80/5 = (10)

0.32 = 2.09

3. a) m − M = − 0.5 = 5 log10

pc10

r

1.05

5.0

pc10log10 −=

−=

r

( ) 1.010

pc10

−=r

r = 7.9 pc

b) 1610

402

1

2 =

=

F

F

It will appear 16 times brighter, which corresponds to m = 3 from Table 1.2.

4. a) θ

=AU1

r pc ;

Alpha Centauri 1.34 pc;

Barnard’s star 1.81 pc;

Altair 5.07 pc;

Alpha Draco 5.68 pc

b) Distance = 500 pc

5. vθ = 4.74 µ r = 30.34 km s−1

6. M1 − M2 = 40

402

1.

4

1

2

42

41

21

22

×

==

L

L

T

T

R

R

( ) 5.221

22 ×= RR

Θ= RR 58.12 .

27

Astronomical Scales Terminal Questions

1. Apparent magnitude of moon is lower (larger negative number), than that of

Venus. Therefore, moon is brighter than Venus.

Moreover,

)(4.0

venus

moon venusmoon10mm

b

b −−=

)0.45.12(4.010 +−−= 4.35.84.0 1010 == −− .105.2 3×=

2. The relation between apparent magnitude m and absolute magnitude M is

M = m − 5 log r + 5

where the distance r is in parsec. Distance of the Sun in parsec is

1.5 × 1013

/3 × 1018

= 5 × 10−6

. So,

M = − 26.8 − 5 (log 5 − 6) + 5 = − 26.8 − 5 log 5 + 30 + 5

= 8.4 − 3.5 = 4.9

3. According to Eq. (1.17)

)(4.0

4

22110

)2(

)200( MM −=

where M1 is the present absolute magnitude of the Sun. Therefore,

×=−

16

200200log5.221 MM

= 2.5 log (2500) = 2.5 × 3.4 = 8.5

So, the absolute magnitude of the Sun will decrease by 8.5 and it will, therefore,

become much more luminous.

4. Using Eq. (1.21)

5.3Sirius)3(=

ΘL

L

28

Basics of Astronomy Now using Eq. (1.16),

)(4.05.3 10)3( SiriusMM −Θ=

where MΘ and MSirius are absolute magnitudes of the Sun and Sirius.

So, (MΘ − MSirius) = 2.5 log (46.8)

= 2.5 × 1.7 = 4.25.

∴ MSirius = 5 − 4.25 = 0.75.

29

Basic Concepts of

Positional Astronomy UNIT 2 BASIC CONCEPTS OF POSITIONAL

ASTRONOMY

Structure

2.1 Introduction

Objectives

2.2 Celestial Sphere

Geometry of a Sphere

Spherical Triangle

2.3 Astronomical Coordinate Systems

Geographical Coordinates

Horizon System

Equatorial System

Diurnal Motion of the Stars

Conversion of Coordinates

2.4 Measurement of Time

Sidereal Time

Apparent Solar Time

Mean Solar Time

Equation of Time

Calendar

2.5 Summary

2.6 Terminal Questions

2.7 Solutions and Answers

2.1 INTRODUCTION

In Unit 1, you have studied about the physical parameters and measurements that are

relevant in astronomy and astrophysics. You have learnt that astronomical scales are

very different from the ones that we encounter in our day-to-day lives.

In astronomy, we are also interested in the motion and structure of planets, stars,

galaxies and other celestial objects. For this purpose, it is essential that the position of

these objects is precisely defined. In this unit, we describe some coordinate systems

(horizon, local equatorial and universal equatorial) used to define the positions of

these objects. We also discuss the effect of Earth’s daily and annual motion on the

positions of these objects.

Finally, we explain how time is measured in astronomy. In the next unit, you will

learn about various techniques and instruments used to make astronomical

measurements.

Study Guide

In this unit, you will encounter the coordinate systems used in astronomy for the first

time. In order to understand them, it would do you good to draw each diagram

yourself.

Try to visualise the fundamental circles, reference points and coordinates as you make

each drawing.

You could use clay/plastecene balls or spherical potatoes to model various planes on

the celestial sphere. You could also mark the meridians and the coordinates on them

for a visual understanding of the coordinates we describe in this unit.

30

Basics of Astronomy Objectives

After studying this unit, you should be able to:

• identify the fundamental great circles such as horizon, celestial equator, ecliptic,

observer’s/local meridian;

• assign the horizon and equatorial coordinates of a celestial object; and

• calculate the apparent solar time at a given mean solar time on any day.

2.2 CELESTIAL SPHERE

When we look at the clear sky at night, the stars appear to be distributed on the inside

surface of a vast sphere centred on the observer. This sphere is called the celestial

sphere (Fig. 2.1).

Fig.2.1: The celestial sphere centred on the observer. Note that the celestial sphere is extremely

large compared with the Earth; for all practical purposes, the observer and the centre of

the Earth coincide

Actually, the stars are at different distances from us. But since the distances of the

stars are very large, they appear to us as if they are at the same distance from us.

So it is sufficient if only the directions of stars are defined on the sphere. Since the

distance is not involved, it is usual to take the radius of the celestial sphere as

unity.

You know that on the surface of a sphere, we need just two coordinates to describe

the position of a point. We shall study below some ways used to specify these two

coordinates for stars and other celestial objects. But before that you should familiarise

yourself with the geometry of a sphere.

31

Basic Concepts of

Positional Astronomy 2.2.1 Geometry of a Sphere

Let us consider a sphere of radius R = 1 unit (Fig. 2.2). We know that a plane

intersects the sphere in a circle. If the plane passes through the centre of the sphere we

get a great circle, and if it does not pass through the centre then it is a small circle.

Fig.2.2: Great and small circles on a sphere. A great circle is the intersection of a sphere and a

plane passing through its centre.

In Fig.2.3, FG is a great circle whose plane passes through the centre O, and EH is a

small circle. Points P and Q are called the poles of the great circle FG. Notice that PQ

is the diameter of the sphere perpendicular to the plane of the great circle FG. The

distance between two points on a sphere is measured by the length of the arc of the

great circle passing through them. It is the shortest path on the surface of the sphere

between these points. Thus, in Fig. 2.3, the distance between points C and D on the

great circle FG is the arc length CD. It is equal to R (=OD) times the ∠COD in

radians. Since R = 1, arc CD = ∠COD. In astronomy, it is convenient to denote

distances on a sphere in angles (radians or degrees) in this way.

An angle between two great circles is called a spherical angle. It is equal to the angle

between the tangents to the great circles at the points of their intersection. It is also the

angle between the planes of the two great circles. In Fig. 2.3, PACQ and PBDQ are

two great circles intersecting at the points P and Q. ∠APB and ∠CPD are spherical

angles between these great circles. These angles are equal. They are also equal to

∠COD, since the arc length CD is common to both great circles.

Fig.2.3: Spherical angle is the angle between two great circles

P

F G O

D C

A B

Q

E H

32

Basics of Astronomy 2.2.2 Spherical Triangle

A closed figure formed by three mutually intersecting great circles on a sphere is

called a spherical triangle. For example, in Fig. 2.3, PCD is a spherical triangle. All

the 6 elements of the triangle, namely, 3 sides and 3 angles are expressed in angular

units. The elements of a spherical triangle ABC in Fig. 2.4 are the 3 spherical angles

denoted by A, B and C and the 3 sides denoted by a, b and c.

Fig.2.4: Spherical triangle is bounded by three arcs of great circles: AB, BC and CA. The spherical

angles are A, B, C and the corresponding sides are a, b and c

Note that a spherical triangle is not just any three-cornered figure lying on a sphere.

Its sides must be arcs of great circles.

The essential properties of a spherical triangle are:

i) The sum of the three angles is greater than 180°,

ii) Any angle or side is less than 180°,

iii) Sum of any two sides/angles is greater than the third side/angle.

Note: Remember that the sides are being measured in degrees or radians.

The four basic formulae relating the angles and sides of a spherical triangle (Fig. 2.4)

are given below. These are for reference only. You need not memorise them.

Sine formula

c

C

b

B

a

A

sin

sin

sin

sin

sin

sin== (2.1)

Cosine formula

cos a = cos b cos c + sin b sin c cos A

cos b = cos c cos a + sin c sin a cos B (2.2)

cos c = cos a cos b + sin a sin b cos C

Analogous cosine formulas are:

sin a cos B = cos b sin c − sin b cos c cos A

sin a cos C = cos c sin b − sin c cos b cos A (2.3)

C

A

B

b

a

c

33

Basic Concepts of

Positional Astronomy Four parts formula

cos a cos B = sin a cot c − sin B cot C (2.4)

Now that you have become familiar with the geometry of the celestial sphere, you can

learn about the different coordinate systems used to locate a celestial object.

2.3 ASTRONOMICAL COORDINATE SYSTEMS

The method of fixing the position of a star or celestial object on the celestial sphere is

the same as that of fixing the position of a place on the surface of the Earth. We shall,

therefore, first consider geographical coordinates of a place on the surface of the

Earth.

2.3.1 Geographical Coordinates

As you know, any place on the Earth is specified by two coordinates: geographical

latitude and geographical longitude (see Fig.2.5).

Fig.2.5: The geographical coordinates (longitude and latitude) uniquely define any

position on the Earth’s surface

Note that the line joining the poles is always perpendicular to the equator.

The circles parallel to the equator are the circles of latitude. The great circles drawn

through the north and South Poles are called the circles of longitude.

How do we specify the coordinates of the place P on the Earth’s surface? The great

circle passing through P is called its meridian.

By international agreement, the geographical coordinates (longitude and latitude)

of the place P are specified with respect to the equator and the meridian through

Greenwich, which is also called the prime meridian.

P

Latitude

Longitude Prime meridian

34

Basics of Astronomy The general definitions are given in the box ahead.

Definitions of longitude and latitude

The longitude of a place is defined by the angle between Greenwich

meridian and meridian of the place. It is measured from 0 to 180°°°° east or

west of Greenwich.

The latitude is defined by the angular distance of the place from the

equator. It is measured from 0 to 90°°°° along the meridian of the place

north or south of the equator.

In Table 2.1, we give the longitudes and latitudes of some cities in India.

Table 2.1: Longitudes and latitudes of some cities in India

Cities Longitude Latitude

Allahabad 81° 50′E 25° 26′N

Ahmedabad 72° 44′E 23° 08′N

Chennai 80° 10′E 13° 06′N

Kanyakumari 77° 32′E 8° 05′N

Kolkata 88° 21′E 22° 34′N

Mumbai 72° 51′E 19° 07′N

New Delhi 77° 12′E 28° 46′N

Shillong 91° 52′E 25° 34′N

Srinagar 74° 48′E 34° 05′N

Ujjain 75° 46′E 23° 10′N

We shall now extend the concept of geographical coordinates on the Earth to define

the coordinates of a celestial body on the celestial sphere. But before that you may

like to work out a problem.

SAQ 1

Using Table 2.1, find the difference between a) the longitudes of Ahmedabad and

Shillong, and b) the latitudes of Srinagar and Kanyakumari.

The longitude-latitude system illustrates the principle of an astronomical coordinate

system. What we require for reference is:

i) A fundamental great circle (such as the equator), and

ii) A reference point, or origin on the chosen great circle.

Depending on the choice of these two references, we describe the following

coordinate systems, which are most often used in astronomy:

1. Horizon system

2. Equatorial system

Spend

3 min.

35

Basic Concepts of

Positional Astronomy 2.3.2 Horizon System

Suppose the observer is at the point O at latitude α (see Fig. 2.6). The point vertically

overhead the observer is called the zenith. It is denoted by Z in Fig. 2.6. The point

vertically below the observer is called the nadir. It is denoted by Z′.

The great circle on the celestial sphere perpendicular to the vertical line ZO Z′ is

called the celestial horizon or just the horizon. This is the fundamental great circle

chosen for reference in the horizon system.

The great circle ZNZ′SZ containing the zenith of the observer is called the observer’s

meridian or the local meridian. N and S are the points of intersection of the horizon

and the observer’s meridian. These points indicate geographical north and south

directions. Either N or S can be taken as the reference point in this system.

Fig.2.6: The celestial horizon and the observer’s meridian

Let X be the celestial body, say a star. You can now fix the coordinates of X in this

system.

Draw the great circle ZXZ′ through the star X in Fig.2.6 (see Fig.2.7a). Let it intersect

the horizon at Y.

The position of X in the horizon system is defined with respect to the horizon and the

reference point N or S.

One coordinate of X is the arc YX, called the altitude (a). Mark it on Fig. 2.6.

To fix the other coordinate, choose N as the origin. Then the arc length NY is the

second coordinate, called the azimuth (A). Mark it on the above figure.

The position of X from the zenith, the arc ZX = 90−a, is called the zenith distance and

is denoted by z. Mark z on Fig. 2.6.

Thus, in the horizon system the position of X can be specified by the coordinates

(A, a) or (A, z).

Note that the term horizon

used here does not have its

common meaning as the

line where the sky meets the

land.

horizon

Zenith Z

Nadir Z′′′′

N S

O

Observer′′′′s meridian

αααα

X

Celestial equator

36

Basics of Astronomy Compare your drawing with Fig. 2.7a.

(a) (b)

Fig.2.7: a) Horizon coordinate system; b) the azimuth gives the direction in which to look for X and

the altitude gives the angle by which a telescope must be raised from the horizon

The altitude (a) varies from 0° to 90° from the horizon to the zenith or 0° to −90°

from the horizon to the nadir. The azimuth (A) of the star is measured from N

eastward or westward from 0 to 180°. Thus, in Fig.2.7a, the great circle arc NY along

the horizon is the azimuth (east) of X. If S is chosen as origin, the azimuth is measured

from 0 to 360° through west.

Horizon coordinate system

Horizon Coordinates: (A, a) or (A, z)

The azimuth, A, is the arc length NY along the horizon if N is taken as the

origin.

The altitude, a, is the arc length YX along the great circle ZXYZ′ containing the

zenith and the star.

The zenith distance, z, is the arc length ZX from the zenith to the star on the

great circle ZXYZ′.

Fundamental great circle: Horizon

Reference point: The points of intersection N or S of the horizon and the

observer’s meridian.

The horizon system is a very convenient system and most small telescopes use it to

locate a celestial object. The azimuth gives the direction in which to look for an

object. The altitude then gives the angle by which the telescope must be raised from

the horizon to locate the object (Fig. 2.7b).

Zenith Z

horizon

Nadir Z′′′′

S N O

a altitude

A azimuth

X

Y E

W

zenith distance z

horizon

S N O

altitude

azimuth

X

Y

a

W

E

A

37

Basic Concepts of

Positional Astronomy The pole star is situated in the geographical north direction. Let us locate its

coordinates in the horizon system.

Example 1: Coordinates of the pole star in the horizon system

What are the horizon coordinates of the pole star?

Solution

Study Fig. 2.8. The pole star is at P. Notice that the angle α (∠BCO) is the latitude of

the observer. Simple geometry shows that this angle is equal to ∠PON. Do you

recognise that the arc PN or ∠PON is the altitude of P?

Fig.2.8: Horizon coordinates of the pole star. BC and OD are parallel. Hence ∠∠∠∠ZOD = αααα,

∠∠∠∠POD = 90°°°° and ∠∠∠∠ZON = 90°°°°. ∠∠∠∠POZ = 90°°°° −−−− αααα and ∠∠∠∠PON = αααα

Thus, the altitude of the pole star is equal to the latitude of the observer.

What is its azimuth? With respect to N, it is 0° and with respect to S, it is 180° west.

Thus the horizon coordinates of the pole star are (0°,α).

You may now like to attempt a problem.

SAQ 2

Suppose you wish to point a small telescope to a star whose azimuth is 30° and

altitude is 45°. Describe the procedure you will adopt.

The horizon system, though simple, has two shortcomings. Firstly, different observers

have different horizons. Therefore, at a given time, the horizon coordinates of an

object will be different for different observers. Secondly, as the Earth rotates from

west to east, celestial objects move from east to west across the sky. Looking from far

above the North Pole, say from a spacecraft, the motion of the Earth is anticlockwise.

Thus, even for the same observer, the horizon coordinates keep changing with time.

So, the horizon system of coordinates is not very useful for practical purposes. These

Spend

3 min.

Nadir Z′′′′

S N O

P

Equator

W

Zenith Z

D

αααα αααα

αααα

C

B

90°°°° −−−− αααα

North Pole

Earth

Celestial sphere

RECALL

The latitude of a point P

on the surface of the Earth

is given by the angle between

the line OP and a line in the

plane of the equator. Here O

is the centre of the Earth.

The longitude tells us our

east-west position. It is

measured from the 0º line of

the prime meridian of

Greenwich. Lines of

longitude run from the

North Pole to the South Pole.

P

O λλλλ

38

Basics of Astronomy drawbacks are removed in the equatorial system, a system that gives the same

coordinates for an object for all observers on the Earth.

2.3.3 Equatorial System

Consider Fig. 2.9 showing the celestial sphere for an observer O. The great circle

whose plane is parallel to the equatorial plane of the Earth, and contains the centre O

of the celestial sphere is called the celestial equator. P and Q are the poles of the

celestial equator: P is the north celestial pole and Q, the south celestial pole. These

poles are directly above the north and south terrestrial poles. As you know, the point P

also points to the pole star.

In Fig. 2.9, the great circle NESW is the observer’s horizon with zenith Z as its pole

and RWTE is the celestial equator for which P and Q are poles.

The celestial equator and the observer’s horizon intersect in two points, E and W,

called the East and West points.

Any semi-great circle through north and south celestial poles P and Q is called a

meridian. But as you have learnt in Sec. 2.3.2, the full great circle through the

observer’s zenith (PZRQT) is called the observer’s meridian or the local meridian.

Fig.2.9: Schematic diagram of the celestial equator and diurnal circle UVX, upper (U) and lower

(V) transits. In the segment DUC the star is above the horizon. It is below the horizon in the

segment CVD

As a result of the Earth’s rotation from west to east, the celestial bodies appear to

move across the sky in the opposite direction, i.e., from east to west. This daily or

diurnal motion is common to all celestial objects. The path of a star X will be along

the small circle XUV parallel to the celestial equator. It is called the diurnal circle

(Fig. 2.9). It intersects the observer’s meridian at the points U and V.

The star completes one circuit of its diurnal circle in 24 hours, which is the Earth’s

rotation period. As you can see from the Fig. 2.9, the star will cross (transit) the

observer’s meridian at the points U and V during its diurnal motion. These are called

the upper and lower transits. A star is said to rise when it rises above the observer’s

Notice that the observer’s

meridian is a full great

circle, unlike the meridian.

U

Diurnal

circle

Observer’s

meridian

horizon

Celestial

Equator

Zenith Z

Nadir Z′′′′

N S O

Celestial

sphere

Q

P

W

E

R

T

V

C

D

X

C

REMEMBER

THE DIURNAL

CIRCLE OF A STAR

IS ALWAYS

PARALLEL TO THE

CELESTIAL

EQUATOR.

39

Basic Concepts of

Positional Astronomy horizon. It is said to set when it goes below the horizon. In Fig. 2.9, in the segment

DUC, the star is above the horizon. It is below the horizon in the segment CVD.

Now, suppose we want to define the equatorial coordinates of a star X (see Fig. 2.10).

The celestial equator (RWTE) is the fundamental great circle in this system. The

reference point is R, the point of intersection of the observer’s meridian (ZRQP) and

the celestial equator above the horizon.

The semi-great circle PXQ is known as the hour circle of the star X. It is called hour

circle because it indicates the time elapsed since the star was on the upper transit on

the observer’s meridian. Let the hour circle of X intersect the celestial equator at J

(Fig. 2.10).

Fig.2.10: Schematic diagram showing local equatorial coordinates (H = RJ, δδδδ = JX). Note that as

the diurnal circle is parallel to the celestial equator, JX does not change due to the star’s

diurnal motion. But RJ changes as the star moves in the diurnal circle

One coordinate of the star X is given by the great circle arc RJ along the celestial

equator. It is called the hour angle (H) of the star. It is measured towards west from 0

to 24 hours.

The other coordinate required to define the position of the star is the circle arc JX

from the equator along the star’s hour circle. It is called declination (δ).

As the path of the diurnal motion of the star is parallel to the celestial equator, the

declination, δ, of a given star does not change during the diurnal motion.

The hour angle and the declination specify the position of the star in the local

equatorial system.

In astronomy, angles such as hour angle are frequently expressed in time units

according to the relation:

24 hours = 360°, lh =15°, 1m = 15′ and 1s = 15″.

Diurnal

circle

Celestial

Equator

Z

Z′′′′

N S O

Q

P

W

E

R

T

U

V

D

X

C

J

40

Basics of Astronomy The declination of a star varies from 0 to 90° from the celestial equator to the celestial

poles. It is taken as positive if the star is situated north of the equator and negative if it

is south of the equator.

Notice that in this system, one of the coordinates, the declination, does not change

during the diurnal motion of the star while the other coordinate, the hour angle, varies

from 0 to 24 hours due to the rotation of the star. That is why this system is called the

local equatorial system.

Local Equatorial System

Local Equatorial Coordinates: (H, δ)

The hour angle, H, is the arc length RJ along the celestial equator.

The declination, δ, is the arc length JX along the star’s hour circle.

Fundamental great circle: Celestial equator.

Reference point: Intersection of observer’s meridian and celestial equator.

We shall now define a coordinate system in which both the coordinates of a star

remain unchanged during its diurnal motion. For this, we choose the celestial equator

as the fundamental great circle and the vernal equinox (ϒ) as the reference point. Let

us understand what ϒ is.

Vernal equinox

You know that the Earth moves around the Sun in a nearly circular orbit. The plane of

the Earth’s motion around the Sun intersects the celestial sphere in a great circle

called the ecliptic (Fig. 2.11a).

As seen from the Earth, the Sun appears to move around the Earth in this plane. Thus

the ecliptic is the annual path of the Sun against the background stars. In Fig. 2.11a, R

ϒ T is the celestial equator and M ϒ M′ is the ecliptic. The ecliptic is inclined to the

equator at an angle of 23°27'. Points K and K′ are the poles of the ecliptic.

The ecliptic and the celestial equator intersect at two points called vernal equinox (ϒ)

and autumnal equinox (Ω).

During the annual apparent motion, the Sun is north of the equator on ϒ M Ω and

south of the equator on Ω M′ ϒ. Thus, the Sun’s declination changes from south to

north at vernal equinox (ϒ) and from north to south at autumnal equinox (Ω).

We can define the position of a star X with respect to the celestial equator and the

vernal equinox (Fig. 2.11b). Let the hour circle of X intersect the celestial equator at J.

One coordinate of X is the arc length ϒJ along the celestial equator measured from

vernal equinox eastward. It is known as the right ascension of the star and is

denoted by α.

The other coordinate is the declination δ (arc JX ) as already defined.

As the Earth rotates, the points ϒ and J on the celestial equator rotate together.

Thus, the separation between ϒ and J does not change and the right ascension of the

star remains fixed.

This system of coordinates (α, δ) is called the universal equatorial system. The α

and δ of stars do not change appreciably for centuries.

41

Basic Concepts of

Positional Astronomy

Fig.2.11: a) The ecliptic, vernal equinox and autumnal equinox; b) universal equatorial coordinates

(αααα, δδδδ)

Universal Equatorial System

Universal Equatorial Coordinates: (α, δ)

The right ascension, α, is the arc length ϒJ eastward along the celestial equator.

The declination, δ, is the arc JX along the star’s hour circle.

Fundamental great circle: Celestial equator.

Reference point: Vernal equinox.

Table 2.2 gives the universal equatorial coordinates of some prominent objects.

Table 2.2: Universal equatorial coordinates of prominent astronomical objects

Object Universal equatorial coordinates

Crab nebula RA 5: 35 (h,m) Declination 22°:01m

Andromeda RA 00:42.7 (h,m) Declination 41°:16m

Ring nebula RA 18:53:35.16 (h,m,s) Declination 33°:01m

Orion nebula RA 05:35.4 (h,m) Declination −05°:27m

(b)

Ecliptic

αααα

δδδδ

ϒϒϒϒ

(a)

Celestial

equator

ϒϒϒϒ Vernal equinox

South

celestial pole

North

celestial pole

Autumnal equinox

ΩΩΩΩ

Ecliptic

Earth

Sun R

T

M M'

K'

K P

Q

42

Basics of Astronomy Let us now find the coordinates of a star.

Example 2: Equatorial Coordinates

Calculate at latitude of 50°N, the zenith distance of a star of declination 20°N when it

is on the observer’s meridian.

Solution

Look at Fig. 2.12.

Fig.2.12

Since the latitude of the place is 50°N, the altitude of the pole star is also 50°. So, arc

PN = 50°. Therefore, arc PZ = 40°, since arc NZ is 90°. As the declination of X is

20°N, RX = 20°. Thus, arc ZX, which is the zenith distance of X, is 30°, since

RP = 90°. If the declination of the star were 20°S, it would have been at X′, then the

zenith distance ZX′ would have been 50° + 20° = 70°.

You should now practice calculating the equatorial coordinates of a few celestial

objects.

SAQ 3

a) Determine the declination (δ) of (i) celestial North Pole, (ii) celestial South Pole

(iii) zenith. (Hint: Look at Fig. 2.12 again).

b) Determine the right ascension (α) and declination (δ) of (i) vernal equinox, (ii)

ecliptic north pole, (iii) ecliptic South Pole. (Hint: Look at Fig. 2.13. Notice that

arc ϒR = 6h.)

Spend

10 min.

horizon

Celestial

equator

Z

N S

X

R

T

P

50°°°° X'

40°°°° 20°°°° N

20°°°° S

O

Q

43

Basic Concepts of

Positional Astronomy

Fig.2.13

We shall now briefly describe the diurnal motion of the stars and the Sun as seen from

different places on the surface of the Earth.

2.3.4 Diurnal Motion of the Stars

See Fig. 2.14, drawn for an observer in northern latitudes. The diurnal circles of stars

north of the equator are more than half above the horizon (arc JKL). The diurnal

circles for stars south of the equator are less than half above the horizon (arc MCM′).

Fig.2.14: a) Diurnal circles for stars north and south of the equator; b) diurnal circles on particular days

Celestial

equator

ϒϒϒϒ Vernal equinox

Ecliptic

Earth

Sun R

T

M M'

K'

K P

Q

δδδδ

δδδδ

Celestial

equator

Z

N S

Q

P R

T

K

C

M'

M O

W

E

J

L

Z'

Celestial equator

Horizon N S

Z

Dec. 21

June 21 Mar. 21 Sept. 22

(a) (b)

44

Basics of Astronomy

Z

O

Horizon

Q P

Celestial

equator

δδδδmax = 90°°°°

Z'

δδδδmax = 90°°°°

Thus, an observer in northern latitudes will see more northern stars for longer duration

above the horizon compared to the southern stars. For an observer on the equator, the

zenith is on the celestial equator. Hence, both northern and southern stars are equally

visible for 12 hours as their diurnal circles are perpendicular to the horizon (Fig. 2.15).

Fig.2.15: Diurnal circles of stars seen from the equator

If we observe stars to the north, we find that some stars never set, i.e., they never go

below the horizon. Instead they trace complete circles above it (Fig. 2.16). Such stars

are called circumpolar stars. (The Egyptians called such stars as the ones that know

no destruction.) Such stars are also there near the South Pole.

Fig.2.16: Circumpolar stars

Observer’s

meridian

Horizon

Celestial

Equator

Zenith Z

Nadir Z′′′′

N S O

Celestial

sphere

Q

P

W

E

R

T

C

D

X

C

Star

never

sets

Star

never

rises

45

Basic Concepts of

Positional Astronomy For an observer at latitude φ in the northern hemisphere, the declination of

circumpolar stars is δ ≥ (90 − φ) (Fig. 2.17). The southern stars with δ < (φ − 90) are

never seen above the horizon. They never rise. The situation is exactly the opposite in

the southern hemisphere. For an observer on the equator, φ = 0, and the maximum

value of δ is 90° at the north and South Poles. Hence, there are no circumpolar stars,

as the declination of stars can never exceed 90° (see Fig. 2.15).

Fig.2.17: Declination of circumpolar stars in the northern hemisphere for an observer at latitude φφφφ

in the northern hemisphere

At North Pole (φ = 90°), all the stars lying north of the equator, that is, δ ≥ 0 are

circumpolar, while the southern stars (δ< 0) are below the horizon. These will be

invisible (Fig. 2.18).

Fig.2.18: Circumpolar stars at the North Pole

Motion of the Sun

You have studied in Sec. 2.3.3 that because of the yearly revolution of the Earth round

the Sun, the Sun appears to move around the Earth along the ecliptic. It moves at a

Z', Q

Z, P

O

Horizon

Celestial

equator

δδδδ

Celestial

equator

Z

Z′′′′

N S O

Q

P

W

E

R

T

X

J

φφφφ

αααα

90°°°°−−−− φφφφ

46

Basics of Astronomy rate of about 1° per day. As a result, the right ascension and declination of the Sun

change continuously. The Sun is at ϒ (Fig. 2.19) on or around March 21 every year.

At that time its coordinates are α = 0, δ = 0.

At M, around June 21, the Sun is at α = 6h, δ = +23.5°. At Ω around September 23,

α = 12h, δ = 0° for the Sun. At M' around December 22, the Sun is at α = 18h,

δ = −23.5° (see Table 2.3).

The point M with δ = δmax = + 23.5° is the summer solstice which marks the

beginning of Dakshinayan, the south-ward motion of the Sun. The opposite point M'

with δ = δmin = − 23.5° is the winter solstice which is the beginning of Uttarayan,

the north-ward journey of the Sun. Note that the arc lengths ϒM and ϒM′ are equal

even though it may not seem so in Fig. 2.19.

Fig.2.19: Motion of the Sun

The Sun as a Circumpolar Star

You have learnt that for an observer at latitude φ, all stars with δ ≥ 90° − φ are

circumpolar. Thus, at North Pole φ = 90° and for a star to be circumpolar, δ ≥ 0. You

have also learnt that the declination of the Sun varies from −23.5° to +23.5° annually.

Hence, for an observer on the North Pole, the Sun is circumpolar from March 21 to

September 23 when the Sun is north of the equator since δ ≥ 0 for the Sun during this

period.

Further, from September 23 to March 21 the Sun is invisible at the North Pole, since

its declination is negative during this period. On March 21, the Sun appears above the

horizon and continues to remain above the horizon for 6 months till September 23.

~ Day (α, δ)

March 21 (0, 0)

June 21 (6h, + 23.5°)

September 23 (12h, 0°)

December 22 (18h, -23.5°)

Table 2.3

(June 21)

αααα = 6h, δ δ δ δ = +23.5°°°°

South

celestial pole

Celestial

equator

ϒϒϒϒ (Mar 21)

α α α α = 0, δ δ δ δ = 0

North

celestial pole

ΩΩΩΩ ( ( ( (Sept. 23)

αααα = 12h, δδδδ = 0°°°°

Ecliptic

Earth

Sun R

T

M M'

K'

K P

Q

(Dec. 22)

αααα = 18h, δδδδ = −−−−23.5°°°°

47

Basic Concepts of

Positional Astronomy Thus, it is seen that the Sun is above the horizon for six months and below the horizon

for 6 months for observers at the North Pole.

Can you say how an observer at the South Pole sees the Sun?

The Sun is visible from September 23 to March 21 at the South Pole and invisible for

the next six months.

During the ‘day’ of 6 months, that is, from March 21 to September 23 at the North

Pole, the Sun moves around the sky in small circles parallel to the celestial equator,

which coincides with the horizon (see Fig. 2.20).

Fig.2.20: Sun’s motion for an observer on the North Pole for whom the celestial equator coincides

with the horizon.

On a given date, the Sun is circumpolar at all those places whose latitude φ ≥ (90 − δ),

where δ is the declination of the Sun on the date under consideration. On June 21/22,

for example, when the Sun's declination is +23.5°, the Sun is circumpolar at all the

places north of 66.5° latitude.

You may now like to stop for a while and concretise these ideas. Try this SAQ.

SAQ 4

a) Find the approximate zenith distance of the Sun on June 22 and December 23 at

Delhi (φ = 28°46′N).

b) At what latitudes is the star Procyon (δ = 05°18′S) circumpolar?

2.3.5 Conversion of Coordinates

We now state the relevant formulae for conversion of coordinates from local

equatorial system to horizon system. Their mathematical derivation can be found in

any standard book dealing with positional astronomy such as Smart (1960). You

need not memorise these formulae.

Spend

10 min.

Z, P

O

Horizon

Celestial

equator

Z', Q

48

Basics of Astronomy Local Equatorial and Horizon System:

Conversion of (H, δ) to (A, z)

sin z sin A = cos δ sin H

sin z cos A = sin δ cos φ − cos δ sin φ cos H (2.5)

cos z = sin δ sin φ + cos δ cos H cos φ

Conversion of (A, z) to (H, δ):

cos δ sin H = sin z sin A

cos δ cos H = cos z cos φ − sin z sin φ cos A (2.6)

sin δ = cos z sin φ + sin z cos φ cos A

So far we have discussed how to locate the position of a celestial object in horizon,

local equatorial and universal equatorial coordinate systems. However, you know

that along with the space coordinate, we also need to specify the time to be able to

completely describe the motion of an object. Let us now learn about how time is

measured in astronomy.

2.4 MEASUREMENT OF TIME

The rotating Earth is our natural clock. The ancient people were familiar with rotation

and revolution of the Earth around the Sun and the revolution of the moon around the

Earth. It was natural for them to base time measurement on these motions. Their basic

unit of time was from one Sunrise to the next. They called it the day. Revolution of

the Earth around the Sun provided another unit called the year. A third unit, the

month, is approximately the period of revolution of moon around the Earth.

The basic unit of time, the day, is defined as the time required for the Earth to rotate

once on its axis. If this rotation is considered with respect to the Sun then the time for

one rotation relative to the Sun is the solar day, that is, the time from one Sunrise to

the next Sunrise. On the other hand, rotation period of the Earth relative to the stars

is called a sidereal day. The sidereal day is different from a solar day. This difference

arises due to the Earth’s revolution around the Sun (see Fig. 2.21).

Fig.2.21: The difference between one sidereal day and one solar day

In Fig. 2.21, the Earth is initially at A in its orbit around the Sun. Suppose at a given

place F, the Sun and a distant star are overhead. Thus, it is noon for an observer at F

(the foot of the solid arrow) and midnight at F′ (the foot of the dotted arrow). One

sidereal day later (say at time t), the Earth is at B but now the solid arrow does not

point to the Sun, though it would still point towards the distant star as it is very far

F'

C

F B

F

F

49

Basic Concepts of

Positional Astronomy away compared to the Sun. So at time t, the Sun is not overhead at F, and it is not

noon at F. The solid arrow points towards the Sun only a little later at C, when the

Earth has turned upon its axis and moved a bit in its orbit.

The time from A to C is equal to one solar day, while the time from A to B is equal to

one sidereal day.

The Earth has to rotate by about 1° more to bring the Sun overhead and complete the

solar day. It takes roughly 4 minutes for the Earth to rotate through 1° since it rotates

through 360° in 24 hours. Thus, the sidereal day is shorter than the average solar day

by about 4 minutes. In a month (30 days), the sidereal time falls behind the mean solar

time by two hours, and in six months the difference becomes 12 hours. Then at solar

midnight we have sidereal noon.

Since the Sun regulates human activity on the Earth, we use solar time in our daily

life. The sidereal time is not useful for civil purposes but is used in astronomical

observations.

2.4.1 Sidereal Time

Due to the rotation of the Earth, the vernal equinox ϒ, moves along the equator once

in 24 hours like any other star. The hour angle of ϒ for an observer is called the

sidereal time for that observer or the local sidereal time (LST).

Fig.2.22: Sidereal Time

In Fig. 2.22, NWSE is the horizon, P and Z are the north celestial pole and the zenith,

respectively. The hour angle (HA) of vernal equinox is the great circle arc Rϒ

measured towards the west from the observer’s meridian. When ϒ is at R, at upper

transit on the observer’s meridian, its HA is 0h and consequently the local sidereal

time (LST) is 0h at this instant.

The interval of time between two successive transits of vernal equinox over the same

meridian is defined as one sidereal day. One sidereal day is divided into 24 sidereal

hours, one sidereal hour into 60 sidereal minutes and one sidereal minute into 60

sidereal seconds.

Let X be the position of a star in Fig. 2.22. The arc RJ (towards west) is the HA of X

(HAX) and the arc ϒJ (towards east) is the right ascension of X (RAX). We have

Rϒ = RJ + ϒJ. Hence

Z'

Z P

Q

N S

E

W

ϒϒϒϒ

R J

X

50

Basics of Astronomy LST = HAX + RAX (2.7)

This is an important relation in the measurement of time. If X is the Sun, denoted by

, then we have the relation

LST = HA + RA (2.8)

2.4.2 Apparent Solar Time

The hour angle of the Sun at any instant is the apparent solar time. The interval

between two consecutive transits of the Sun over the same meridian is called one

apparent solar day (see Fig. 2.23).

Fig.2.23: Apparent solar day

In Fig. 2.23a, we show the Sun at noon on March 21, when the Sun and the vernal

equinox are on the local meridian. The celestial sphere rotates and some time during

the next day, the vernal equinox is again on the meridian. That interval of time is one

sidereal day; a fixed point on the celestial sphere has gone around once.

But in that time the Sun has moved eastward along the ecliptic (see Fig. 2.23b). But

the Sun is not yet on the meridian and it is not yet noon. The celestial sphere has to

rotate a bit more to bring the Sun up to the meridian. This time interval represents the

apparent solar day.

The apparent solar day is found to vary throughout the year. Thus, the Sun is an

irregular timekeeper and consequently the time based on the motion of the Sun is not

uniform. The non-uniformity in the solar time is due to two reasons: the first is that

the path of apparent motion of the Sun is elliptical and not circular. This means that

the Sun moves at a non-uniform rate. The second reason is that the Sun's apparent

motion is in the plane of the ecliptic and not in the plane of the equator; and the

measurement of time, that is, the measurement of hour angle is made along the

equator.

EAST SOUTH WEST

Ecliptic

Celestial Equator

Local celestial meridian

EAST SOUTH WEST

Ecliptic

Celestial Equator

Local celestial meridian

(a)

(b)

51

Basic Concepts of

Positional Astronomy 2.4.3 Mean Solar Time

In order to have a uniform solar time, a fictitious body called the mean Sun is

introduced. The mean Sun is supposed to move uniformly (that is, in a circular path)

along the equator completing one revolution in the same time as the actual Sun does

round the ecliptic. The hour angle of the mean Sun at any instant is defined as the

Mean Solar Time (MST), or simply, the Mean Time.

The time interval between the successive transits of the mean Sun over the same

meridian is defined as the mean solar day.

The mean solar day is subdivided into hours, minutes and seconds. The civil day

begins at mean midnight for reasons of convenience. Our clocks are designed to

display the mean solar time.

By international agreement, the time of the Greenwich meridian is called the

Greenwich mean time (GMT). It can be related to the mean time of any other place

through the longitude of that place.

Since the Earth completes one rotation on its axis, that is 360 degrees of longitude in

24 hours, each degree of longitude corresponds to 4 minutes of time. Therefore, GMT

at a given instant is related to the mean time of any other place by a simple relation:

Local Mean Time = GMT + λ (in units of time) (2.9)

where λ is the geographical longitude of the place expressed in units of time. In the

above relation, λ is to be taken as positive if the longitude of the place is east of

Greenwich, and negative if it is west of Greenwich. The above relationship holds good

for both the sidereal as well as the solar time.

It is obvious that if each town and city in a country were to use its own local time, the

life of citizens would become very difficult. To overcome this problem, a whole

country uses the local time of a place as its standard time.

For example, the Indian Standard Time is the local time of a place on longitude 82.5°

east. The IST is, therefore, ahead of the GMT by 5½ hours. Very large countries such

as United States have more than one standard time.

Example 3

What will the local time of Shillong be when the local time of Ahmedabad is 6 p.m.?

Solution

Refer to Table 2.1. The difference between longitudes of Ahmedabad and Shillong is

approximately 19°. Since each degree of longitude is equal to 4 minutes of time, the

difference in the local time of these two places is 19 × 4 = 76 minutes. Since Shillong

is east of Ahmedabad, local time of Shillong will be ahead of the local time of

Ahmedabad by one hour and sixteen minutes.

And now you can attempt an exercise.

SAQ 5

Refer to Table 2.1. If the local time of Mumbai is 8:00 p.m., what would be the local

time at Kolkata at that time?

We will now discuss the Equation of Time, which relates the apparent solar time and

the mean solar time.

Spend

5 min.

52

Basics of Astronomy 2.4.4 Equation of Time

As we have seen in Eq. (2.8)

LST = HA + RA (2.10)

Using Eq. (2.10) for the Sun and the mean Sun, we can write

RAMS − RA = HA − HAMS (2.11)

The difference in right ascensions of the mean Sun and true Sun at a given instant (see

Fig. 2.24a) is called the Equation of Time (ET).

Fig.2.24: Equation of time

non-uniform

rate

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

18

16

14

12

10

8

6

4

2

0

−−−−2

−−−−4

−−−−6

−−−−8

−−−−10

−−−−12

−−−−14

−−−−16

Su

nd

ial-

clo

ck t

ime

Real Sun

Fictitious Sun

Mean Sun

Celestial equator

Ecliptic

uniform rate

uniform rate

Equation of time

time of equinox

time of

perihelion

RAMS −−−− RA Not to scale

(a)

(b)

NOTE

HA is the symbol of

hour angle for the Sun.

RA is the symbol of

right ascension of the

Sun.

HAMS is the symbol

of hour angle for the

mean Sun.

RAMS is the symbol

of right ascension of

the mean Sun.

53

Basic Concepts of

Positional Astronomy Thus, we have

ET = HA − HAMS (2.12)

We see that the ET at any instant is the difference between the apparent solar time

given by HA and the mean solar time given by HAMS.

In Fig. 2.24b, you can see that during the year the ET varies between −14 1/4 to + 16

1/4 minutes and it vanishes 4 times during the year on or about April 16, June 14,

September 1 and December 25.

Any discussion on time in astronomy would perhaps be incomplete if we do not

mention the calendar.

2.4.5 Calendar

The civil year contains an integral number of 365 mean solar days. As the actual

period of Earth’s revolution, called a tropical year contains 365.2422 mean solar

days, a fraction 0.2422 of a day is omitted each year. This resulted in the loss of a

number of days over several centuries and the civil year would get out of step with the

seasons.

To overcome this confusion, Julius Caesar introduced Julian calendar, named after

him, in which the year was made of 365.25 mean solar days. Consequently, every 4th

year was made to contain 366 mean solar days. This was called a leap year. The extra

day was added in February. Thus, in this calendar, a year, which is divisible by 4, is a

leap year in which the month of February has 29 days.

In Julian calendar, the tropical year was assumed to be of 365.25 mean solar days, but

its actual length is 365.2422 mean solar days. The small difference between assumed

and actual lengths of the year, created serious problem by the 16th century and the

civil year was out of step in relation to the seasons. To overcome this problem, Pope

Gregory, introduced in 1582 the calendar known as Gregorian calendar, which we

are using now.

In this calendar, a leap year is defined as before, with the exception that when a year

ends in two zeros, it will be a leap year only if it is divisible by 400. Thus in a cycle of

400 years, there are 100 leap years according to Julian Calendar while there are only

97 in Gregorian Calendar, since years 100, 200 & 300 are not leap years. This makes

the average civil year to consist of 365.2425 mean solar days, which is very close to

the true length of the tropical year. No serious discrepancy can arise in the Gregorian

calendar for many centuries.

With this we come to the end of Unit 2. In this unit, we have studied various aspects

of positional astronomy. We now summarise the main points of the unit.

2.5 SUMMARY

• In positional astronomy, we are concerned with the relative directions of the

heavenly bodies, which appear to be distributed on the surface of a vast sphere

called the celestial sphere.

• Any plane passing through the centre of the sphere intersects it in a circle called

the great circle. The extremities of the diameter of the sphere perpendicular to the

plane of the great circle are called poles of the great circle. Arcs of three great

circles form a spherical triangle in which any angle / side is less than 180°. The

sum of the three angles is greater than 180°.

• Due to rotation of the Earth on its axis, the heavenly bodies appear to move across

the sky from east to west in small circles parallel to the equator.

54

Basics of Astronomy • In the horizon coordinate system, the altitude and azimuth specify a star’s

position. The altitude of the north celestial pole at any place is equal to the latitude

of the place.

• In the local equatorial system, the coordinates of a heavenly body are given by the

hour angle and the declination. In the universal equatorial system, the

declination and the right ascension specify the position of a celestial object.

• Circumpolar stars are those, which never set, i.e., they are always above the

horizon. At a given place of latitude φ, a star becomes circumpolar if its

declination δ ≥ (90 − φ). It follows that at geographical North Pole (where

φ = 90°) the Sun is circumpolar for 6 months from March 21 to September 23.

Thus, at poles the day will last for 6 months and night for 6 months during the

year. At equator (φ = 0) there is no circumpolar star and all the stars move in

circles perpendicular to the horizon. The day and night are of 12-hour duration

each.

• Hour angle of vernal equinox is defined as the sidereal time. The sidereal day is a

measure of the rotation period of the Earth with respect to the distant stars.

• The hour angle of the mean Sun is defined as mean solar time and hour angle of

the Sun is defined as apparent solar time.

• The difference between the apparent solar time and the mean solar time is known

as the equation of time. The difference between the two systems of time varies

between −14 1/4 to +16 1/4 minutes and becomes zero 4 times during the year.

2.6 TERMINAL QUESTIONS Spend 30 min.

1. What is the latitude of the place at which celestial horizon and equator coincide?

At what latitude is the celestial equator perpendicular to the horizon?

2. Show that a star attains its maximum altitude when it is on the observer’s

meridian.

3. A and B are two places in north latitude on the surface of the Earth; their latitudes

are 24° 18′N and 36° 47′N, respectively; and their longitudes are 133° 36′E and

125° 24′W, respectively. Find the difference in their local time.

4. Show that at a place of latitude φ (North) at upper transit, the zenith distance of a

star equals φ − δ .

2.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. a) Difference in the longitudes of Ahmedabad and Shillong

= 91° 52′ − 72° 44′ = 19° 08′

b) Difference in latitude between Srinagar and Kanyakumari

= 34° 05′ − 8° 05′ = 26° 00′

2. Turn the telescope 30° degrees from north to west, keeping it horizontal. Then

raise it so that it makes an angle of 45° with the horizontal.

3. a) See Fig. 2.25. δ of celestial north pole = 90°.

δ of celestial South Pole = − 90°.

δ of zenith = φ, latitude of the place.

55

Basic Concepts of

Positional Astronomy

Fig.2.25

b) See Fig. 2.26. (α,δ) of vernal equinox = (0,0). (α,δ) of ecliptic north pole

= (18h, 66½°) since ϒJ is measured eastward. (α,δ) of ecliptic South Pole

= (6h, − 66½°).

Fig.2.26

horizon

Celestial

equator

Z

N S

X

R

T

P

50°°°° X'

40°°°° 20°°°° N

20°°°° S

O

Q

Celestial

equator

ϒϒϒϒ Vernal equinox

Ecliptic

Earth

Sun R

T

M M'

K'

K P

Q

56

Basics of Astronomy 4. a) At Delhi, zenith makes an angle of 28° 46′ with the equator. The Sun on

June 22 and December 22 makes an angle of 23° 27′ with the equator.

Therefore, zenith distance of the Sun on that day = 28° 46′ − 23° 27′ = 5° 19′

b) φ ≥ 90 − δ = 84° 42′

5. The difference in longitudes of Mumbai and Kolkata is around 15 ½°. Thus, the

time difference is 62 minutes. Since Kolkata is to the east, the local time there

will be 2 minutes past 9 p.m. when it is 8 p.m. in Mumbai.

Terminal Questions

1. φ = 90°, φ = 0°.

2. See Fig. 2.27. At the transit point M on the diurnal circle, the altitude is maximum.

Fig.2.27

3. Difference in longitude = 133° 36′ E − 125° 24′ W = 259°.

∴ Time at A will be ahead by 259° × 4 min = 1036 min = 17 h 16 min.

4. See Fig. 2.28. The star is at U. In general, z = φ − δ.

Fig.2.28

Celestial

equator

Z

N S

Q

P R

T

M

O

W

E

Z'

Z'

Celestial

equator

Z

N S

U

O

Horizon

δ z

φ

57

Astronomical Techniques

UNIT 3 ASTRONOMICAL TECHNIQUES

Structure

3.1 Introduction

Objectives

3.2 Basic Optical Definitions for Astronomy

Magnification

Light Gathering Power

Resolving Power and Diffraction Limit

Atmospheric Windows

3.3 Optical Telescopes

Types of Reflecting Telescopes

Telescope Mountings

Space Telescopes

3.4 Detectors and Their Use with Telescopes

Types of Detectors

Detection Limits with Telescopes

3.5 Summary

3.6 Terminal Questions

3.7 Solutions and Answers

3.1 INTRODUCTION

In Unit 1 you have learnt about various measurable astronomical quantities and the

methods of measuring some of them such as distance, size and mass. You have also

studied about the radiant flux or brightness of different astronomical sources like the

planets, the sun, stars and galaxies. A great deal of information in astronomy is

gathered with the help of ground-based instruments. Therefore, in this unit we

describe some instruments as well as the techniques used to measure radiant flux and

to analyse the radiation emitted by astronomical objects. We shall first revisit the

basic optical definitions, viz. magnification, light gathering power and resolving

power, relevant to astronomy. You may recall having studied them in Unit 11 of the

Physics elective PHE-09 entitled Optics.

In Sections 3.3 and 3.4, we shall describe instruments like optical telescopes and

detectors that help us measure radiant flux from stars and other objects. A telescope is

like a camera. It is a light-focusing instrument, while a detector, like a photographic

film, is the medium on which the photons leave an impression, which can be

measured later.

You know the poem ‘Twinkle-twinkle little star’ from your early days. In this unit,

you will study how the Earth’s atmosphere, that makes the stars twinkle, affects the

observations made with telescopes on the ground. Finally, you will learn about

techniques to detect faint astronomical light, which originates from very distant

objects.

In the next unit we discuss the basic principles of physics applicable in astronomy and

astrophysics.

Objectives

After studying this unit, you should be able to:

• define angular magnification, light gathering power and resolving power of a

telescope;

• describe different types of reflecting telescopes and telescope mountings;

58

Basics of Astronomy • explain how atmosphere affects the incoming starlight;

• describe the different detectors and techniques used for observations; and

• estimate detectability limit of a telescope.

3.2 BASIC OPTICAL DEFINITIONS FOR ASTRONOMY

The telescope is the most common instrument used in ground-based astronomy. Most

people think of a telescope only in terms of its magnifying power. However,

magnification is only one of three powers of a telescope, the other two being its

resolving power and light gathering power. Before we describe various telescopes,

we would like to discuss these concepts as applicable in astronomy.

3.2.1 Magnification

As you know from the course in Optics, the magnifying power of a telescope refers to

its ability to make the image appear bigger. If the angle subtended by a distant object

at the objective of a telescope is β and that subtended by the virtual image at the eye is

α (Fig. 3.1), then the angular magnification (A.M.) of the telescope is given by

β

α=A.M. (3.1)

Fig.3.1: The image of a distant object is formed at the focal plane of the objective. The rays

entering the eye are rendered parallel by the eye-piece. The eye sees the image at infinity

We calculate the A.M. of a telescope by dividing the objective’s focal length by the

focal length of the eyepiece.

e

o

f

f=A.M. (3.2)

For example, if a telescope has an objective with a focal length of 60 cm and an eye-

piece of focal length 0.5 cm, its angular magnification is 60/0.5, or 120 times. We say

that the magnification is 120 X.

Further, the A.M. is also equal to the ratio of entrance pupil (e.g., the objective lens

diameter) to the human eye’s pupil diameter. If we assume the diameter of a normal

human eye pupil to be about 8 mm for the eye adapted to perfect darkness, then the

lowest possible magnification will be equal to the ratio of objective diameter (in mm)

to 8. Similarly, if we assume a normal pupil diameter of 3 mm, the highest possible

magnification will be the diameter of the objective divided by 3.

ββββ

αααα

To top’ point of

virtual image at infinity

Fo Fe

Fe

fo fe

59

Astronomical Techniques The highest possible magnification of a telescope is limited by its optics which

includes the quality of lenses and mirrors and the thermal insulation of the telescope

tube so that the exchange of heat does not disturb the air inside the tube. The

magnification is also limited by the disturbance of light rays suffered in the Earth’s

atmosphere. Also, the higher the magnification, the smaller is the field of view, i.e., the

area of the sky which can be observed by the telescope becomes smaller (see Fig. 3.2).

Fig.3.2: Increasing the magnification decreases the field of view and makes the image dimmer.

Pictures show simulated views of the Orion Nebula at magnifications of approximately 50x,

80x, and 120x

Notice that the images in Figs. 3.2 b and c are dimmer compared with the one in Fig. 3.2a.

Is there a way to make it brighter? We can do so by gathering more light from the object.

This brings us to the concept of light gathering power of a telescope.

3.2.2 Light Gathering Power

The light gathering power of a telescope refers to its ability to collect light from an

object. Most interesting celestial objects are faint sources of light and in order to get

an image we need to capture as much light as possible from them. This is somewhat

similar to catching rainwater in a bucket, the bigger the bucket, the more rainwater it

catches.

Similarly, the light gathering power of a telescope is proportional to the area (i.e.,

diameter squared) of the telescope objective. Now, the area of a circular lens or mirror

of diameter D is π (D/2)2. Thus, the ratio of light gathering powers of two telescopes

is given by

2

2

1

2

1

=

D

D

LGP

LGP (3.3)

For example, a telescope of 100mm diameter can gather (100/8)2 = 156.25 times more

light than a human eye with a typical pupil diameter of 8 mm.

Similarly, if we compare telescopes of 24 cm and 4 cm diameters, the former gathers

(24/4)2 = 36 times more light than the latter. Thus, even a small increase in telescopes

diameter produces a large increase in its light gathering power and allows astronomers

to study much fainter objects.

The ability of a telescope to reveal fine detail of an object is determined by its

resolving power, which we now discuss.

(a) (b) (c)

60

Basics of Astronomy 3.2.3 Resolving Power and Diffraction Limit

You have learnt in Sec. 11.2 of Unit 11 of Optics that the image of a point object

through an optical instrument is not a sharp point-like image but a bright circular disc

called the Airy disc. The disc is surrounded by a number of alternate bright and dark

fringes produced due to diffraction. The central bright disc represents the image of the

object.

Now suppose we wish to observe two close stars that appear equally bright. We

should be able to see two Airy discs. However, whether we see them as distinct discs

or overlapping each other, depends on the resolving power of the telescope. Fig. 3.3

shows three situations where two equally bright stars are closely placed. The pattern

of rings seen at each star image is called the Airy pattern. The two stars are said to be

just resolved when we can just infer their images as two distinct Airy discs (Fig. 3.3b).

To resolve these Airy discs, we use the Rayleigh criterion.

Rayleigh criterion

Two equally bright stars are said to be resolved when the central maximum

of one diffraction pattern coincides with the first minimum of the other.

(a) (b) (c)

Fig.3.3: Resolving a pair of two equally bright stars. In a) the stars are easily resolved; in b) the

stars are JUST resolved, and in c) the stars are too close and not resolved.

You may like to ask: What factors determine a telescope’s resolving power?

Naturally, the quality of lenses is a major factor. But even with perfect optics, the

resolving power of a telescope is limited by diffraction.

The diffraction limit of resolution (R) of a telescope is defined as

R (in radians) = (1.22 λ /D), (3.4)

where λ is the wavelength of light and D is the telescope diameter. Both λ and D have

to be expressed in the same units. [Note that 1 degree = 60 arc-minutes (60′) = 3600

arc-seconds (3600″)]. R can be expressed in arc-seconds as

R (arc-sec) = )206265(22.1

D

λ (3.5)

For human eye

R ~ 1′ for absolute sharp vision,

61

Astronomical Techniques R ~ 2′ for clear vision and

R ~ 4′ for comfortable vision.

Remember that the smaller the value of R, the better is the instrument’s ability to

resolve nearby objects.

Example 1: Diffraction limit of resolution of telescopes of different sizes for a

given wavelength

Table 3.1 shows the diffraction limit of resolution of telescopes of different sizes for

λ = 457 nm.

Table 3.1: Diffraction limit of resolution of various telescopes

Telescope Diameter (mm) R (″″″″)

50 2.3

100 1.15

200 0.58

400 0.29

500 0.23

You may like to calculate R for a telescope.

SAQ 1

Calculate the diffraction limit of resolution of Mount Palomar telescope of 200 inch

diameter for λ = 457 nm. Compare its light gathering power with a telescope of

200 mm diameter.

Based on size alone, the largest telescopes should have large resolving powers. But

the resolution of large telescopes is limited by the passage of light through the Earth’s

atmosphere. When we look through a telescope, we are looking through several

kilometres of turbulent air, which blurs the image. The Earth’s atmosphere does not

allow ground-based telescopes to resolve better than 1-2 arc-seconds in the sky (for

even the best astronomical sites).

The major limitation for ground-based astronomy is the Earth’s atmosphere and it can

affect the observations in many ways. Moreover, all wavelengths cannot pass through

the atmosphere. This brings us to the concept of atmospheric windows.

3.2.4 Atmospheric Windows

Some of the effects of the Earth’s atmosphere on electromagnetic radiation are:

Absorption, scintillation, scattering and turbulence. Atmospheric molecules such as

carbon dioxide and water vapour give rise to absorption. Thus, only certain bands of

frequencies in the electromagnetic spectrum pass through the atmosphere. These

regions of the electromagnetic spectrum are called atmospheric windows. Fig. 3.4

shows the absorption properties of the Earth’s atmosphere. On the x-axis is the

wavelength in cm and on the y-axis is the altitude in km at which the intensity of the

radiation entering the atmosphere is reduced to half.

Spend

10 min.

62

Basics of Astronomy

Fig.3.4: The spectrum of electromagnetic waves from gamma rays to long radio waves showing the

optical and radio ‘windows’ and the regions of atmospheric transparency

The atmosphere allows only visible radiation and radio waves to come through to the

surface of the Earth. For observations at other wavelengths we have to fly detecting

instruments to altitudes at which these wavelengths are not completely absorbed.

Before the advent of artificial satellites, the instruments were flown in balloons and

rockets. Now-a-days observations are carried at various wavelengths by instruments

on board the space satellites. It is important to remember that astronomers like to

observe objects in as many wavelengths as possible. This helps them to understand

astronomical objects better.

SAQ 2

For the same diameter, compare the resolving power of an optical telescope operating

at λ 457 nm and a radio telescope operating at λ 1 cm.

You have learnt in SAQ 2 that the resolving power of a radio telescope is quite poor.

Therefore, radio telescopes need to have very large apertures.

So far we have discussed some optical definitions relevant to astronomy. We now turn

our attention to the instruments and techniques used for making astronomical

observations from the Earth. Let us now learn about optical telescopes.

3.3 OPTICAL TELESCOPES

Telescopes have undergone tremendous evolution from Galileo’s times when only

refracting telescopes were used. The most recent telescopes have mirrors that can be

actively shaped according to the observer’s need. The refracting telescopes became

outdated very soon because

i) it was difficult to make good large lenses required for large gathering area,

ii) large lenses were very heavy and balancing the telescope became difficult, and

iii) lenses suffered from optical aberrations.

Reflecting telescopes with mirrors (parabolic or hyperbolic) have been around for

more than a century. In the last few decades, advanced technologies have been

developed which allow the observer to effectively control the optical system so that it

can be adapted to the needs of the observations.

Such active and adaptive optics have completely changed this basic tool of

observational astronomy. We now describe various types of reflecting telescopes.

Spend

5 min.

Micro-wave

Gamma

ray X ray UV

Visual

4 ×××× 10−−−−7

m

Infrared UHF VHF FM AM

Wavelength (metres)

Opaque

Transparent

Tra

nsp

aren

cy o

f

Ea

rth’s a

tmo

sph

ere

104 10

2 1 10

−−−−2 10

−−−−4 10

−−−−8 10

−−−−10 10

−−−−12

Visual

window Radio

window

7 ×××× 10−−−−7

m

Alt

itu

de

Gamma rays, X-rays

and UV rays are

blocked by the

atmosphere.

Infrared

and

microwaves

are blocked

by the

atmosphere.

63

Astronomical Techniques 3.3.1 Types of Reflecting Telescopes

James Gregory of Scotland (1638-75) proposed the first design for a reflecting

telescope in which he used a paraboloidal mirror as the objective (primary mirror) to

minimize chromatic and spherical aberrations (Fig. 3.5).

In this type of telescope, light from a distant object hits the primary mirror and is

reflected to a secondary mirror which reflects it down to the telescope tube again to a

secondary focus. The light emerges from the telescope through a small central hole in

the primary mirror and is observed with an eyepiece or a detector.

Fig.3.5: Gregorian reflecting telescope with equatorial mounting; ray diagram for the telescope

The performance of the Gregorian telescope was not satisfactory and Newton (1642-

1727) built a working reflector.

A Newtonian telescope (Fig. 3.6) uses a parabolic primary but a flat mirror as a

secondary, which is set at 45° to the axis of the tube. The light is brought to a focus at

the side of the telescope where the eyepiece is placed. This system is widely used

even now.

Fig.3.6: Newtonian reflecting telescope

A little later, a French optician, G. Cassegrain (who lived at the time of Newton),

designed another reflector (Fig. 3.7).

Now-a-days the most commonly used telescope is the Cassegrain type with a central

hole in the primary mirror which allows the light to come out for placing an eyepiece

or any other general detector or instruments. This design also allows the folding of the

focussed beam and thus provides a very compact telescope.

Pa

rall

el l

igh

t ra

ys

fro

m t

he

sta

r

Secondary

mirror

Paraboloidal

primary mirror

Focal

point

Secondary

mirror

Primary

mirror Eye-piece

64

Basics of Astronomy

Fig.3.7: Cassegrain reflecting telescope

All telescopes need to be pointed at the desired part of the sky, and then they have to

follow or track the objects in the sky as their direction changes due to Earth’s

rotation. For example, suppose you are viewing Jupiter and its moons with a

telescope. Due to the Earth’s rotation, you will see Jupiter moving across your view

until it is gone. You will have to move your telescope to follow Jupiter if you want to

keep viewing it. However, to track it, you will need a mounting for the telescope so

that it can be turned in the desired direction. We will now briefly describe telescope

mountings.

3.3.2 Telescope Mountings

Most of the small telescopes (less than 1 metre diameter) use the equatorial mount

(Fig. 3.8). In an equatorial mounting, the pier or the base on which the telescope is

mounted is set so that its axis points to the North Pole. This is done by raising the axis

by an angle equal to the latitude of the place. This axis is called the polar axis. A

rotation about the polar axis is used for adjustment in right ascension. The telescope is

also provided with motion about an axis perpendicular to the polar axis, called the

declination axis, for adjustment in declination. Thus, a combination of these two

motions allows the telescope to point to any object whose equatorial coordinates are

known.

Fig.3.8: The equatorial mounting

Since the telescope is fixed on the Earth, it moves with the Earth. In order that the

object remains in the field of the telescope, the mounting is made to rotate in the

direction opposite to that of the Earth with the same speed as that of the Earth. All

65

Astronomical Techniques these adjustments make this type of mounting heavier than the altitude-azimuth

mounting described ahead. That is why it is not used with very big telescopes.

All modern telescopes (larger that 2 m diameter) use Alt-Azimuth type mount shown

in Fig. 3.9. You have learnt in the last unit that altitudes and azimuths of objects

change with the location of the observer and with time for the same observer. This

was a major limitation of this type of mounting in earlier times but with advances in

the technology using computers in the past few decades, it is no more an issue.

Fig.3.9: The Alt-Azimuth type mount

In Fig. 3.10a we show the largest telescope in India set up by the Indian Institute of

Astrophysics at Kavalur Observatory in Karnataka.

Fig.3.10: a) The Vainu Bappu telescope in Kavalur Observatory; b) domes housing twin Keck

telescopes

At present, the largest operating telescopes are the twin Keck 10 m telescopes at

Mauna Kea, in Hawaii, placed at about 4200 m height above sea-level (Fig. 3.10b).

There are plans now for making very large aperture (up to 20 − 30 m) telescopes!

(a) (b)

ALTITUDE AXIS

AZIMUTH AXIS

66

Basics of Astronomy 3.3.3 Space Telescopes

You have learnt in Sec. 3.2 that the main advantages of a telescope over direct

observation with the human eye, are magnification, light collection and resolution.

For light collection, one could fabricate telescopes of increasingly large diameters.

But these will be limited by the effect of the Earth’s atmosphere. However, if we

could put a telescope in space (high above the Earth’s atmosphere, in a balloon or a

satellite), then the Earth’s atmosphere would not be a limiting factor, and we could

achieve diffraction-limited images. In such situations the size of the telescopes that

can be built for such platforms is the only limitation due to the costs involved and

limitations of available technology.

The Hubble Telescope of about 2 m diameter is the best example of this type of

telescope (Fig. 3.11). Over the past decade it has made very high quality observations

of stars, nebulae, galaxies, supernovae and other objects. Some of these objects belong

to a very early phase of the universe. These observations have led to improved

understanding of these objects.

Fig.3.11: The Hubble Space Telescope

Both the types of telescope − ground-based and space-based − have somewhat

different roles to play and complement each other. That is why there is always a

demand to build large diameter ground-based telescopes.

The active and adaptive telescopes are the latest instruments in use today. In the

active telescopes, the shape of the primary mirror can be changed (although it retains

its parabolic form) according to the observer’s need. This is done with the help of

small peizo driven pistons placed at the back of the mirror support. This is now

routinely done for large mirror based telescopes. In the adaptive telescope the

changes in the shape of the primary mirror are done in a controlled manner depending

upon the changes in the Earth’s atmosphere during night. In some sense, the adaptive

telescopes can beat the effect of the atmosphere and render images as good as the

space telescopes for some duration in the night!

Telescopes must use detectors to help us obtain useful information about the universe.

We now discuss various detectors and how these are used with telescopes.

3.4 DETECTORS AND THEIR USE WITH TELESCOPES

Detectors are used for measuring the light output from a telescope and play a major

role in obtaining information about the stars, galaxies, etc. The actual light available

from an astronomical object is very small as will be clear from the following example.

67

Astronomical Techniques

Example 2: Estimate of Available Light from a Star

Vega is one of the bright stars in the sky and is about 26 light years away. The

radiation emitted in visible band at the star’s surface is 175,000 Wcm−2

. This reduces

by 10−16

times on its arrival outside Earth’s atmosphere. Another 20% is lost in the

atmosphere due to absorption and scattering. About 30% is lost in the telescope optics.

Thus, if we use a 25 cm (10″) telescope to collect this light, then only about

0.5 × 10−9

W is available at its Cassegrain focus. Eventually only a fraction of this is

actually detected since the detector is never an ideal one and has an efficiency of at

best 80%. It is amazing that even with such a small amount of light detected, we can

study many things about a star, e.g., its light variation, brightness, composition. We

can even take a spectrum (which actually divides this small available light into smaller

bits of wavelengths)!

Detectors are used with telescopes in the following two modes of operation:

• Imaging: This involves taking direct pictures of star fields and extended objects

like gas clouds or galaxies. Since sharp images are required over a wide field

which may extend up to several square degrees, careful optical design is a natural

requirement.

• Photometry: This involves measuring total brightness, spectrum etc. of single

objects. Compared to imaging mode, poorer images are acceptable in this case but

the stellar image has still to be small enough to enter an aperture or slit of a

spectrograph.

We now briefly describe various types of detectors.

3.4.1 Types of Detectors

Detectors used in the imaging mode are mainly 2-dimensional (2D type) since we are

trying to form images of objects in a given area. Examples of such detectors are the

photographic emulsion, human eye and the most modern detector, the charge-

coupled device (CCD). Detectors used for photometry of single objects are 1D type

(one dimensional), since they receive photons from one object only. The photometer

is a 1D detector.

Photometer

Before the advent of CCDs, the measurements of light intensity and colour were made

using a photometer, a highly sensitive light meter attached to a telescope (Fig. 3.12a).

A photometer is still used in the photometry of single stars. It is used more commonly

for stars whose light output varies with time, called variable stars. The most

important component of a photometer is a photomultiplier tube that is based on the

photoelectric effect about which you have studied at +2 level. A photon when

incident on a photocathode emits an electron. The electric current thus generated is

amplified further and can be measured directly. The calibration of intensity or colour

is done by observing a comparison star. Today, however, most photometric

measurements are made on CCD images.

Charge-coupled device

A charge-coupled device (CCD) is a special computer chip of the size of a postage

stamp (Fig. 3.12b). It contains a large number (~ millions) of microscopic light

detectors arranged in an array. A CCD can be used like a small photographic plate,

though it is much more sensitive. CCDs detect both bright and faint objects in a single

exposure. The image from a CCD is stored in a digitised form in a computer.

Therefore, brightness and colour can be measured to high precision. Moreover, it is

68

Basics of Astronomy easy to manipulate the image to bring out details. At present, the only major drawback

of CCD is that its maximum size is limited (about 70 mm square) as compared to the

most basic 2D detector, i.e., photographic plates which can be as large as

300 mm square. This disadvantage of CCD is also being overcome by combining a

large number of CCDs.

(a) (b)

Fig.3.12: a) Photometer attached to a telescope; b) a CCD

Efficiency of a Detector

A basic parameter which defines the efficiency of any detector is its Quantum

Efficiency (Q.E.). It is the ratio of number of photons actually detected (or recorded)

by it to the number of photons recorded by an ideal and perfect detector. Since ideal

detector by definition would detect all photons incident on it with 100% efficiency,

this ratio is nothing but the ratio of actually detected photons by the detector versus

the number of photons incident on it. Fig. 3.13 gives the efficiency curves for various

detectors.

Fig.3.13: Quantum efficiency of various detectors in terms of wavelength of light

300 500 700 900 1100

λλλλ (nm)

1

10

100

EYE

PHOTOGRAPH

PHOTOMULTIPLIER TUBE

CCD

0.1

QU

AN

TU

M

EF

FIC

IEN

CY

(P

ER

CE

NT

)

69

Astronomical Techniques Note that the human eye and photographic emulsion are detectors with the lowest

sensitivity and photomultiplier tubes are only marginally better. The CCD works over

a large wavelength region in the visible band with a Q.E. of the order of 60-80%. You

must remember that the y-axis in Fig. 3.13 is in log scale.

As you have read in Unit 1, the limiting magnitude of the naked eye is about +6. This

means that the faintest object visible under normal observing conditions, with an eye

adapted to darkness has a magnitude of +6. It takes 25-30 minutes for the eye to get

completely dark-adapted, otherwise we can see only up to fourth or fifth magnitude.

Experiments have, however, shown that a dark-adapted eye, looking at a patch of dark

sky through a modest telescope can see stars as faint as mv = +8.5. This corresponds to

a flux of about 200 photons per second.

We would also like to find out how far the limits of detection are extended when we

use a telescope with various detectors.

3.4.2 Detection Limits with Telescopes

Modern detectors like CCDs can detect individual photons, and the limiting

magnitude is normally governed by the background noise.

For a typical photographic plate, only about 0.1% of the incident photons are recorded

depending on the type of grains on the film. On such a plate, an image is detectable

only after about 5 × 104 photons have been received.

At 200 photons per second, the photographic plate can match eye’s limit with an

exposure of about 4 minutes.

Obviously, the plate can detect much fainter objects with longer duration exposures.

The limit to the flux of visible radiation which is detectable is given by

tD

Fvis

2lim

1∝ (3.6)

where t is the exposure time and D is the telescope aperture.

For a telescope, the magnitude limit in the visible range, in a very approximate

manner without considering any efficiency factors etc., is given by

Dmv 10lim log52~ + (3.7)

where D is in mm.

Thus using the human eye as the detector along with a telescope we can have the

limiting magnitudes as mv ~ 12.9 for a 6 inch and mv ~ 15.3 for an 18 inch telescope.

SAQ 3

Find the magnitude of the faintest object that the 3.5 metre New Technology

Telescope at the European Southern Observatory in Chile can detect.

With this we come to the end of the unit. We now summarise its contents.

3.5 SUMMARY

• If a telescope has an objective of local length f0 and the eye-piece has a focal

length fe, then angular magnification is equal to the ratio f0/fe.

Spend

5 min.

In spectroscopy or

spectrophotometry, we analyze

the light in great detail by

using a spectrograph which

spreads the light into a

spectrum according to

wavelength. You are familiar

with a prism dispersing white

light into its component

colours. Nearly all modern

spectrographs use a grating

instead of a prism and the

spectrum is recorded directly

on a CCD camera. Since we

understand how light is

emitted, scattered, or absorbed

by matter, a spectrum carries a

lot of information about the

source of light and the medium

through which it has passed.

We shall discuss this in detail

in Unit 7.

70

Basics of Astronomy • The light gathering power of a telescope refers to its ability to collect light from

an object.

• The resolving power of a telescope is its ability to reveal fine detail. The

resolving power of a telescope is given by

D

6.11=α

where D is the diameter of the telescope’s objective. A point light source observed

through a telescope would not appear as a point source. Diffraction would cause

the image to appear as a round disk of light, called Airy’s disk. The diffraction

limit of resolution in radians is given by

D

Rλ

= 22.1

where λ is the wavelength of light. Both λ and D must have the same units.

• At the Earth’s surface, electromagnetic radiation can be detected only in the

radio, infrared and optical windows.

• Telescopes can be of refracting or reflecting types.

• All modern telescopes are reflecting type. There are three kinds of reflectors:

Gregorian, Newtonian and Cassegrain. The Cassegrain reflectors are the most

popular.

• Telescopes can be operated in imaging or photometry mode.

• The larger is the diameter of the objective, the fainter is the source that a telescope

can detect.

• The human eye, photographic emulsion, photometer and charge-coupled devices

are various types of detectors.

• Of these the CCDs are used most by modern astronomers. These are used for

recording images, measuring brightness and colour of celestial objects.

3.6 TERMINAL QUESTIONS Spend 30 min.

1. Why can infrared observations be made from high mountains while X-ray

observations can be made only from space?

2. Nocturnal animals have large pupils in their eyes. Can you relate that to

astronomical telescopes?

3. The moon has no atmosphere. If we had an observatory on the moon, at what

wavelengths could we observe the astronomical objects?

4. What is the resolving power of a 20 cm telescope if observations are made at λ

550nm?

5. What do two stars 1.5 arc-seconds apart, look like through a 25 cm telescope ?

6. Compare the light gathering powers of the 5 m telescope and a 0.5 m telescope.

71

Astronomical Techniques 3.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. 20626522.1

×λ

=D

R arc - sec

21054.22

06265.2457.22.1 −××

×= arc - sec

= .023 arc - sec

( ) 2.6452

10254.2

2

200

1054.2200

200mmLGP

MPLGP=×=

××=

2. Ratio of resolving powers 206265

1

222.1206265122.1

×λ

××λ

=D

D

1

21091057.4

2

1 ×−×=

λ

λ=

71057.4 −×=

The resolving power of the radio telescope is 71057.4 −× times the resolving

power of an optical telescope operating at 4.57 nm.

3. Limiting magnitude

Dm 10log52 +=ν

( )101005.310log52 ××+=

( )5.310log5152 ++=

= 19.7

Terminal Questions

1. At the top of a high mountain, we can receive only IR. X-rays are available only

much above the top of the highest mountains. Therefore, to observe in X-ray band,

we will need to observe from a space-based telescope.

2. The larger the diameter of the pupil, larger is the light gathering power. Nocturnal

animals are adapted to night vision by virtue of this. Astronomical telescopes also

require bigger and bigger objectives to see fainter and fainter objects.

3. At all wavelengths, provided we have detectors sensitive in those regions.

4. 0.58 arc-seconds.

5. Resolved.

6. The 5 m telescope has 100 times the light gathering power of a 0.5 m telescope.

72

Basics of Astronomy

UNIT 4 PHYSICAL PRINCIPLES

Structure

4.1 Introduction

Objectives

4.2 Gravitation in Astrophysics

Virial Theorem

Newton versus Einstein

4.3 Systems in Thermodynamic Equilibrium

4.4 Theory of Radiative Transfer

Radiation Field

Radiative Transfer Equation

Optical Depth; Solution of Radiative Transfer Equation

Local Thermodynamic Equilibrium

4.5 Summary

4.6 Terminal Questions

4.7 Solutions and Answers

4.1 INTRODUCTION

So far in this block we have provided the basic information which is useful in

astronomy. You have learnt about astronomical quantities of interest, various

coordinate systems, astronomical instruments and techniques. We now turn our

attention to astrophysics. The aim of astrophysics is to apply principles of physics to

understand and explain the behaviour of various astronomical systems.

There are certain physical principles and concepts which are used in astrophysics so

universally that it is worthwhile to discuss them before we begin studying specific

astronomical systems. We have decided to put together some such basic physical

principles in this Unit. You must have already learnt many of these principles in your

other physics courses. Now you will learn how these can be applied to astrophysical

systems.

Let us consider some issues which are of universal concern in astrophysics. We know

that gravitation is the dominant force in virtually any astrophysical setting. Since

gravitation is always attractive, it must be balanced in some way in a system which is

not shrinking. We shall discuss a very general and powerful principle called the virial

theorem, which helps us understand how gravitation is balanced in astrophysical

systems.

Another topic of importance in astrophysics is the interaction of radiation with

matter. Astrophysics is a very special science in which we cannot do experiments

with our systems (stars, galaxies, etc.) in our laboratories. Virtually everything we

know about these systems is learnt by analysing the radiation reaching us from these

systems. If we want to make inferences about the systems which emitted the radiation

or through which the radiation passed, then we need to understand how matter and

radiation interact with each other. Many astronomical systems like stars emit radiation

simply because they are hot. So we often need to apply various principles of thermal

physics to understand how matter in these systems behaves. Therefore, we plan to

recapitulate some of the important results of systems in thermodynamic equilibrium

and then develop the theory of transfer of radiation through matter.

Objectives

After studying this unit, you should be able to:

• apply virial theorem to simple astrophysical systems;

73

Physical Principles • identify the situations in astrophysics to which Newton’s theory of gravitation or

general theory of relativity can be applied;

• determine the specific intensity, energy density, radiant flux and radiation,

pressure for a given radiation field; and

• solve the radiative transfer equation for simple cases and interpret the results.

Study Guide

In this unit, we will be using certain concepts discussed in various units of the physics

electives PHE-01 entitled ‘Elementary Mechanics’ (Unit 10), PHE-06 entitled

‘Thermodynamics and Statistical Mechanics (Unit 9), and PHE-11 entitled ‘Modern

Physics’ (Unit 9). Please keep these units handy for ready reference.

4.2 GRAVITATION IN ASTROPHYSICS

In your elementary physics courses, you must have learnt about two important long-

range forces, whose range of influence extends to infinity. These are the gravitational

force and the electromagnetic force. The electromagnetic force can be attractive or

repulsive depending on the nature of charges. So, if a system has equal amounts of

positive and negative charges, and if there are no relative motions between these two

types of charges, then electromagnetic forces are screened off. That is, the system

does not produce a large-scale electromagnetic field. On the other hand, gravitation is

always attractive and cannot be screened off. So it is the dominant force acting over

the entire universe.

Since gravitation is always attractive, a natural question to ask is: Why do the celestial

objects not shrink in size? Obviously, it has to be balanced in a system which is not

shrinking in size (such as a star or a galaxy). We shall now use Newtonian theory of

gravitation to discuss an important theorem (virial theorem) which tells us how this

balancing takes place. Then we shall briefly consider whether Newtonian theory of

gravitation is adequate in astrophysics or whether we have to apply a more complete

theory of gravitation − the general theory of relativity due to Einstein.

4.2.1 Virial Theorem

Why is our solar system not shrinking in size? We shall analyse this problem in the

non-inertial frame of reference attached to a planet. From Unit 10 of the course

PHE-01 entitled ‘Elementary Mechanics’, you know the answer: the Sun’s

gravitational attraction on a planet is balanced by the centrifugal force due to the

orbital motion of the planet. Let M and m be the masses of the Sun and a planet in the

solar system. For simplicity, if we assume the planet to go in a circular orbit of radius

r with speed v, then the force balance equation in the given non-inertial frame of

reference is

2

2

r

GMm

r

mv=

We can rewrite this equation in the form

02

12 2 =−

r

GMmmv (4.1)

Now note that the gravitational potential energy of the system (the Sun and the planet)

is

,r

GMmEG −=

74

Basics of Astronomy whereas the kinetic energy of the system is

2

2

1mvEK =

if the Sun is assumed to be at rest. We can now rewrite Eq. (4.1) in the following form

02 =+ GK EE (4.2)

This is the virial theorem for a planet going around the Sun.

What is the significance of this result? This tells us that the gravitational potential

energy and the kinetic energy of a system will have to be of the same order if

gravitation is to be balanced by motion. We have proved virial theorem for the simple

case of a planet going around the Sun in a circular orbit. However, Eq. (4.2) can be

proved quite generally for a system in which gravitation is balanced by motions such

that the system is not shrinking in size. The motions needs not be circular, but can be

of any type. For example, inside a star, gravity is balanced by thermal motions of its

particles (atoms, electrons, ions). Even in this situation, Eq. (4.2) can be shown to

hold provided we take the total kinetic energy of all the particles in the star for EK. If a

galaxy or a star cluster is not shrinking in size, the total kinetic energy EK of the stars

in it should be related to the total gravitational potential energy EG by Eq. (4.2).

Fig.4.1: Spiral galaxies

In a type of galaxy known as spiral galaxy, stars seem to be moving in nearly circular

orbits. However, in typical star clusters and in galaxies known as elliptical galaxies,

stars move randomly. Due to these random motions, the stars do not fall to the centre

due to gravitation. Do you feel puzzled by the idea that random motions can balance

gravitation? To understand this concept, consider the air around you. Earth’s gravity is

pulling all the molecules of the air. Then why are not all molecules settling on the

floor of the room due to this attraction? It is the random motion of the molecules

which prevents this from happening.

Deriving the virial theorem Eq. (4.2) for a completely general situation is a very

mathematically involved problem. It is beyond the scope of this elementary course. So

let us apply the virial theorem in a simple situation so that you feel comfortable with

it.

Example 1: Estimating the average temperature in stellar interior

A star (we have the Sun in mind) has a mass of 1033

g and a radius of 1011

cm. Make an

order-of-magnitude estimate of the average temperature in the interior of the star.

Solution

Since this is an order of magnitude estimate, we shall take the virial theorem to imply

that the potential and the kinetic energies are approximately equal. In c.g.s. units, we

Fig.4.2: The Sun

75

Physical Principles take the following approximate values of various physical constants: gravitational

constant G ≈ 10−7

, Boltzmann constant kB ≈ 10−16

, mass of hydrogen atom mH ≈ 10−24

.

The gravitational potential at the surface of the star has the magnitude GM/R. The

approximate gravitational potential energy of the star is given by multiplying this by

M, which is

( )

erg1010

1010 48

11

23372

≈×

=−

R

GM

Now we need to equal the total kinetic energy to this. Now, suppose that the star is

made up of hydrogen. Then, the star consists of about M/mH particles. Each of them

has kinetic energy of order kBT. Therefore, the total kinetic energy of the star is

TT

Tkm

MB

H

41

24

1633

1010

1010≈

×≈

−

−

If this is equated to 1048

erg, then we obtain the temperature of the star as

T ≈ 107 K

Note that detailed calculations suggest a temperature of about 15 million degrees at

the centre of the Sun. The above order of magnitude estimate thus provides a fairly

good estimate of the average temperature inside a star.

You may now like to solve a problem to fix these ideas.

SAQ 1

A globular cluster of stars has about a million stars. The stars inside such a cluster of

radius 1020

cm have random velocities of order 106 cm s

−1. Estimate the mass of the

star cluster and the number of stars in it. Take the mass of a star to be about 1033

g.

Hint: The mass of the cluster is Nm, where N is the number of stars in the cluster.

Now, equate the total K.E. of the cluster to its P.E.

After Newton formulated his theory of universal gravitation, for more than two

centuries it was regarded as a supreme example of a successful physical theory.

However, in 1915, Einstein showed that this theory was incomplete and formulated

his new theory of gravitation known as general relativity. In this section, we

investigate the situations in which Newton’s theory may have to be replaced by the

general theory of relativity.

4.2.2 Newton versus Einstein

We now know that Newton’s theory is only an approximation. But it is such an

exceptionally good approximation in most circumstances that we do not need general

relativity at all. Only when the gravitational field is sufficiently strong, we have to

apply general relativity. Although we shall not discuss general relativity in this

elementary course, we would like to point out when you can safely use Newtonian

theory and when general relativity is needed.

Even people without any technical knowledge of general relativity now-a-days have

heard of black holes. These are objects with gravitational fields so strong that even

light cannot escape. Let us try to find out when this happens. Newtonian theory does

not tell us how to calculate the effect of gravitation on light. So let us figure out when

a particle moving with speed of light c will get trapped, according to Newtonian

theory. Suppose we have a spherical mass M of radius r and a particle of mass m is

Spend

10 min.

76

Basics of Astronomy ejected from its surface with speed c. The gravitational potential energy of the particle

is

r

GMm−

If we use the non-relativistic expression for kinetic energy for a crude estimate (we

should actually use special relativity for a particle moving with c!), then the total

energy of the particle is

r

GMmmcE −= 2

2

1

Newtonian theory tells us that the particle will escape from the gravitational field if E

is positive and will get trapped if E is negative. In other words, the condition of

trapping is

02

1 2 <−r

GMmmc

or

12

2>

rc

GM (4.3)

It turns out that more accurate calculations using general relativity give exactly the

same condition (4.3) for light trapping, which we have obtained here by crude

assumptions.

General relativity is needed when the factor

rc

2GMf

2= (4.4)

is of the order unity.

Newtonian theory is quite adequate if f is much smaller than 1.

Let us investigate the case of the Sun.

Example 2

Let us examine if the Newtonian theory is adequate for the Sun.

Solution

The Sun has mass 1.99 × 1033

g and radius 6.96 × 1010

cm. Substituting these values in

Eq. (4.4), we get

f = 4.24 × 10−6

<< 1

Hence, Newtonian theory is quite adequate for all phenomena in the solar system.

Only if we want to calculate very accurate orbits of planets close to the Sun (such as

Mercury), we have to take into consideration general relativity.

When is general relativity applicable in the case of the Sun? We can use Eq. (4.3) to

calculate the radius to which the solar mass has to be shrunk such that f is of order

unity. Then the light emitted at its surface gets trapped. Why don’t you do this

calculation yourself?

77

Physical Principles SAQ 2

Calculate r for the Sun such that f ~ 1.

You would have calculated the radius to be 3 km. Therefore, general relativity will

apply to the Sun, once it shrinks to this size. As we shall discuss in more detail in

Block 3, when the energy source of a star is exhausted, the star can collapse to very

compact configurations like neutron stars or black holes. General relativity is needed

to study such objects.

If matter is distributed uniformly with density ρ inside radius r, then we can write

3

4 3rM =

and Eq. (4.4) becomes

2

2

3

8

c

Grf

ρπ= (4.5)

We note that f is large when ρ is large or r is large (for given ρ). The density ρ is very

high inside objects like neutron stars. You may ask: Can there be situations where

general relativity is important due to large r? We know of one object with very large

size − our universe itself. The distance to farthest galaxies is of the order 1028

cm. It is

very difficult to estimate the average density of the universe accurately. Probably it is

of the order 10−29

g cm−3

. You may like to substitute these values in Eq. (4.5), and

calculate f.

SAQ 3

Calculate the value of f from Eq. (4.5) using the data given above for the Universe.

The result of SAQ 3 tells us that we should use general relativity to study the

dynamics of the whole universe. This subject is called cosmology.

Thus, in astrophysics, we have two clear situations in which general relativity is

very important:

• the study of collapsed stars and

• the study of the whole universe (or cosmology).

In most other circumstances, we can get good results by applying Newtonian theory of

gravitation.

As we mentioned in the introduction, basic principles of thermal physics apply to

many astronomical systems with high temperatures. Let us briefly revisit these

principles.

4.3 SYSTEMS IN THERMODYNAMIC EQUILIBRIUM

You have learnt about thermodynamic equilibrium in the course PHE-06 entitled

Thermodynamics and Statistical Mechanics. If a system is in thermodynamic

equilibrium, then several important principles of physics can be applied to that

system. Let us first recapitulate some important laws and equations, relevant for such

systems, namely, Maxwellian velocity distribution, Boltzmann distribution law,

Saha’s equation and Planck’s law of black body radiation. Afterwards, we shall

discuss whether we can assume astrophysical systems to be in thermodynamic

equilibrium and whether these principles can be applied to them.

Spend

2 min.

Spend

2 min.

78

Basics of Astronomy Maxwellian velocity distribution

You have studied about Maxwellian velocity distribution in Unit 9 of PHE-06.

Different particles in a gas move around with different velocities. Recall that if the gas

is in thermodynamic equilibrium at temperature T, the number of particles per unit

volume having speeds between v and v + dv is given by

dvTk

mvv

Tk

mNdN

BBv

−

ππ=

2exp

24

22

2/3

(4.6)

where N is the number of particles per unit volume and m is the mass of each particle.

Boltzmann and Saha Equations

You have studied in Unit 9 of the physics elective PHE-11 entitled Modern Physics

that a hydrogen atom has several different energy levels. It is also possible to break

the hydrogen atom into a proton and an electron. This process of removing an electron

from the atom is called ionisation. If a gas of hydrogen atoms is kept in

thermodynamic equilibrium, then we shall find that a certain fraction of the atoms will

occupy a particular energy state and also a certain fraction will be ionised. The same

considerations hold for other gases.

If N0 is the number density of atoms in the ground state, then the number density Ne of

atoms in an excited state with energy E above the ground state is given by

−∝

Tk

E

N

N

B

eexp

0

(4.7)

This is the Boltzmann distribution law.

In 1919, the famous Indian physicist M.N. Saha derived an equation which tells us

what fraction of a gas will be ionised at a certain temperature T and pressure p. The

derivation of this equation involves some statistical mechanics. Here we merely quote

the result without derivation. If nI is the number of hydrogen atoms out of which nII

are ionised at temperature T, then Saha’s equation gives

( )

−

=

Tk

ITk

h

mp

n

n

BB

ee exp

2 5/23/2

2I

II (4.8)

where I is the ionisation potential of hydrogen, pe is the partial pressure of electrons,

h is Planck’s constant and me is the mass of electron. A form convenient for

calculation is

( ) 0.48I)(5040/2.5log/log III −−= TTnpn e

SAQ 4

Assuming pe to be 100 dyne/cm2, calculate the fraction of hydrogen atoms ionised at

T = 10,000 K. The ionisation potential of hydrogen is 13.6 eV.

Planck’s law of blackbody radiation

You have studied this law and its consequences in Unit 15 of PHE-06. You know that

when radiation is in thermodynamic equilibrium with matter, it is called blackbody

radiation. The spectral distribution of energy in blackbody radiation is given by the

famous law derived by Planck in 1900 (see Fig. 4.3). The energy density uνdν lying in

the frequency range between ν and ν + dν is given by

Spend

5 min.

79

Physical Principles

1exp

8 3

3

−

ν

νν=ν

Tk

h

d

c

hdu

B

(4.9)

Fig.4.3: Blackbody radiation curve

We can now use these results to understand the interaction of matter with radiation.

4.4 THEORY OF RADIATIVE TRANSFER

Matter can both emit and absorb radiation. It is possible to use quantum mechanics to

calculate the rates at which atoms emit or absorb energy. Here, however, we shall not

do it. We would study the processes of emission and absorption by matter, by

introducing suitable coefficients of emission and absorption. Radiative transfer is the

name of the subject in which we study the interaction of radiation with matter having

prescribed emission and absorption coefficients.

Let us first consider how we can provide the mathematical description of radiation at a

given point in space.

4.4.1 Radiation Field

You know that it is particularly easy to give a mathematical description of blackbody

radiation, which is homogeneous and isotropic inside a container. Specifying the

energy density uν associated with the frequency ν, which is given by Planck’s law

(Eq. 4.9), more or less provides us complete information about blackbody radiation. In

general, however, the radiation is not isotropic. When we have sunlight streaming into

a room, we obviously have a non-isotropic situation involving the flow of radiation

from a preferred direction.

We now define the radiation field for a non-isotropic situation. Let us consider a small

area dA at a point in space (Fig. 4.4). Let dEν dν be the energy of radiation passing

through this area in time dt from the solid angle dΩ centred at θ and lying in the

frequency range ν, ν + dν. The energy dEνdν is proportional to the area dA cosθ

projected perpendicular to the direction of radiation, time interval dt, solid angle dΩ

and frequency range dν. Hence we can write

ν= dddtdAtrIddE )cosˆ,,( n (4.10)

0 2,000 4,000 6,000 8,000 10,000 12,000 14,000 16,000 18,000 20,000

wavelength (angstrom)

En

erg

y i

nte

nsi

ty (

arb

itra

ry u

nit

s)

0

1.0

0.5

T3

T2

T1

T3 >>>> T2 >>>> T1

80

Basics of Astronomy where n is the unit vector indicating the direction from which the radiation is coming

and N is the unit vector normal to the area dA. The quantity ( )n,, trI is called the

specific intensity. As you can see, it is a function of position r, time t and direction

n .

Fig.4.4: Illustration of specific intensity

Radiation field

If ( )n,, trI is specified for all directions at every point of a region at a

time, then we say that the radiation field in that region is completely

specified.

In this elementary treatment, we shall restrict ourselves only to radiation fields which

are independent of time.

If we know the radiation field at a point in space we can calculate various quantities

like radiant flux, energy density and pressure of radiation. For example, radiant

flux is simply the total energy of radiation coming from all directions at a point per

unit area per unit time. Hence, we simply have to divide Eq. (4.10) by dA dt and

integrate over all solid angles to get the flux. Thus, we can define the radiant flux

associated with frequency ν, and the total radiant flux as follows:

Radiant flux in terms of specific intensity

The radiant flux associated with frequency ν is given by

Ω= dIF cos (4.11)

The total radiant flux is obtained by integrating over all frequencies

ν= dFF (4.12)

The pressure of the radiation field over a surface is given by the momentum

exchanged per unit area per unit time, or momentum flux, perpendicular to that

surface. Let us obtain an expression for momentum flux.

You know from Unit 3 of PHE-11 that the momentum associated with a photon of

energy dEν is dEν /c.

Its component normal to the surface dA is dEν cosθ/c. On dividing this by dA dt, we

get the momentum flux associated with dEν ;

θ

n

dΩ

dA

81

Physical Principles

Momentum flux = dtdAc

dE θcos

Using Eq. (4.10), we get the expression for momentum flux in terms of specific

intensity:

cos1cos 2

dc

I

dAdtc

dE=

θ (4.13)

The radiation pressure pν is obtained by integrating the momentum flux over all

directions.

Radiation pressure

.cos1 2

= dIc

p (4.14)

If the radiation field is isotropic, i.e., it is independent of θ and φ, then

4andsin =ϕθ=Ω dddd . Hence, we get,

c

Id

c

Ip

2

3

4cos == (4.15)

We now apply these results to calculate energy density and radiation pressure.

SAQ 5

Perform the integration in Eq. (4.15) and verify the result.

Example 3: Calculating energy density, specific intensity and radiation pressure

Calculate the energy density uν of a radiation field at a point and use that expression

of energy density to write down the specific intensity of a blackbody radiation. Show

that the pressure due to isotropic radiation is given by 1/3 of the energy density.

Solution

Let us consider energy dEν of radiation associated with frequency ν as given by

Eq. (4.10). This energy passes through area dA in time dt in the direction n . Since the

radiation traverses a distance cdt in time dt, we expect this radiation dEν to fill up a

cylinder with base dA and length cdt in the direction n during this time. Now the

volume of this cylinder is cosθ dAcdt. Therefore, from Eq. (4.10), the energy density

of this radiation in the solid angle dΩ is

cos

d

c

I

dAcdt

dE=

θ

To get the total energy density of radiation at a point associated with frequency ν, we

have to integrate over all directions, so that

Ω= dc

Iu

For isotropic radiation c

Iu

ν=4

Spend

5 min.

82

Basics of Astronomy Since blackbody radiation is isotropic, the specific intensity of blackbody radiation

usually denoted by Bν (T) should be independent of direction. Hence the energy

density of blackbody radiation is simply given by

Ω= dc

TBu

)(

Since Bv(T) is independent of direction, integration over all solid angles gives 4π.

∴ c

TBu

)(4

π=

Therefore, making use of Eq. (4.9), we now get the specific intensity of blackbody

radiation as

1

exp

12)(

2

3

−

=

Tk

hc

hTB

B

Using Eq. (4.15), we get the radiation pressure for isotropic radiation as

3

1up =

For black body radiation .)(

3

4

c

TBp =

In astrophysics, we need to understand the interaction of matter and radiation to

explain spectra of objects such as stars, interstellar gas clouds and galaxies. We now

discuss the effect of matter on radiation field.

4.4.2 Radiative Transfer Equation

If matter is present, then in general the specific intensity of the radiation field keeps

changing as we move along a ray path. Before we consider the effect of matter, first

let us find out what happens to specific intensity in empty space as we move along a

ray path.

See Fig. 4.5. Let dA1 and dA2 be two area elements separated by a distance R and

perpendicular to a ray path. Let Iν1 and Iν2 be the specific intensities of radiation in the

direction of the ray path at dA1 and dA2.

Fig.4.5: Two area elements perpendicular to a ray path

We want to determine the amount of radiation passing through both dA1 and dA2 in

time dt in the frequency range ν, ν + dν. If dΩ2 is the solid angle subtended by dA1 at

R

dA1 dA2

83

Physical Principles dA2, then according to Eq. (4.10), the radiation falling on dA2 in time dt after passing

through dA1 is

222 ddtddAI

From considerations of symmetry, this should also be equal to

111 ddtddAI

where dΩ1 is the solid angle subtended by dA2 at dA1. Equating these two expressions

and noting that

2

122

21 ,

R

dAd

R

dAd =Ω=Ω

we get

Iν1 = Iν2 (4.16)

In other words, in empty space the specific intensity along a ray path does not change.

If s is the distance measured along the ray path, then we can write

0 =

ds

dI (4.17)

in empty space.

At first sight, this may appear like a surprising result. We know that the intensity falls

off as we move further and further away from a source of radiation. Can the specific

intensity remain constant? The mystery is cleared when we keep in mind that the

specific intensity due to a source is essentially its intensity divided by the solid angle

it subtends, a quantity called the surface brightness of an object. This means that the

specific intensity is a measure of the surface brightness. As we move further away

from a source of radiation, both its intensity and angular size falls as (distance)2.

Hence the surface brightness, which is the ratio of these two, does not change.

Let us now consider what happens if matter is present along the ray path. If matter

emits, we expect that it will add to the specific intensity. This can be taken care of by

adding an emission coefficient jν on the right hand side of Eq. (4.17). On the other

hand, absorption by matter would lead to a diminution of specific intensity and the

diminution rate must be proportional to the specific intensity itself. In other words, the

stronger the beam, the more energy there is for absorption. Hence the absorption term

on the right hand side of Eq. (4.17) should be negative and proportional to Iν. Thus,

we obtain the radiative transfer equation which gives the value of specific intensity in

the presence of matter:

Radiative transfer equation

Ijds

dI−= (4.18)

where αν is the absorption coefficient

The radiative transfer equation provides the basis for our understanding of interaction

between radiation and matter.

It is fairly trivial to solve this equation if either the emission coefficient or the

absorption coefficient is zero. Let us consider the case of jν = 0, i.e., matter is assumed

to absorb only but not to emit. Then Eq. (4.18) becomes

84

Basics of Astronomy

Ids

dI−= (4.19)

On integrating this equation over the ray path from s0 to s, we get

( )

′′−=

s

s

sdssIsI

0

0 exp)()( (4.20)

We will discuss below more general solutions of the radiative transfer equation. These

solutions will provide us answers to questions such as: Why is the radiation emitted

from nebula usually in spectral lines? Why do we see absorption lines in stellar

spectra?

4.4.3 Optical depth; Solution of Radiative Transfer Equation

To obtain a general solution of the radiative transfer equation, we need to define two

quantities, namely, the optical depth and the source function. Let us first define the

optical depth τν through the following relation:

dsd = (4.21)

such that the optical depth along the ray path between s0 and s becomes

( ) ′′=s

s

sds

0

(4.22)

If matter does not emit radiation, i.e., jν = 0, it follows from Eqs. (4.20) to (4.22) that

the specific intensity along the ray path falls as

)0()(

−=τ eII (4.23)

Based on the values of τν we can define objects as optically thick or optically thin.

Optically thick and optically thin objects

If the optical depth τν >> 1 along a ray path through an object, then the

object is known as optically thick.

An object is known as optically thin if τν << 1 for a ray path through it.

It follows from Eq. (4.23) that for an optically thick object Iν (τν) = 0 and it

extinguishes the light of a source behind it. What about an optically thin object? It

does not decrease the light much. Hence the terms optically thick and optically thin

roughly mean opaque and transparent at the frequency of electromagnetic radiation

we are considering.

We now define the source function

jS = (4.24)

Dividing the radiative transfer equation Eq. (4.18) by αν and using Eqs. (4.21) and

(4.24), we get

SI

d

dI+−= (4.25)

85

Physical Principles Multiplying this equation by ,

νe we can write it as:

( )

eSeI

d

d=

Integrating this equation from optical path 0 to τν (i.e., from s0 to s along the ray

path), we get the general solution of the radiative transfer equation:

General solution of the radiative transfer equation

( ) ( ) ′′+=

′−−−

0

(0))( dSeeII (4.26)

If matter through which the radiation is passing has constant properties, then we can

take Sν to be constant and solve the integral in Eq. (4.26). This gives

)(1(0))(

−− −+= eSeII

We are now interested in studying the emission and absorption properties of an object

itself without a source behind it. Then we take Iν(0) = 0 and write

)(1)(

−−= eSI (4.27)

Let us consider the cases of optically thin and thick objects.

• Optically thin object

For an optically thin object, τν << 1, and ντ−e may be approximated to 1 − τν.

Thus Eq. (4.27) becomes

)( SI =

For matter with constant properties, we take τν = ανL, where L is the total length

of the ray path. Making use of Eq. (4.24), we get the following result

Iν = jνL (4.28)

• Optically thick object

If the object is optically thick, we can neglect ντ−e compared to 1 in Eq. (4.27).

Then we get the result

Iν = Sν (4.29)

Let us put these results together for ready reference.

Specific Intensity of

Optically thin object: Iν = jνL

Optically thick object: Iν = Sν

We have derived two tremendously important results in Eqs. (4.28) and (4.29). To

understand their physical significance, we have to look at some thermodynamic

considerations.

86

Basics of Astronomy 4.4.4 Local Thermodynamic Equilibrium

Suppose we have a box kept in thermodynamic equilibrium. If we make a small hole

on its side, we know that the radiation coming out of the hole will be blackbody

radiation. We have already derived the specific intensity of blackbody radiation as

1

exp

12)(

2

3

−

=

Tk

hc

hTB

B

(4.30)

The specific intensity of radiation coming out of the hole is simply

)( TBI = (4.31)

We now keep an optically thick object behind the hole inside the box as shown in

Fig. 4.6. If this object is in thermodynamic equilibrium with the surroundings, then it

will not disturb the environment and the radiation coming out of the hole will still be

blackbody radiation, with specific intensity given by Eq. (4.31). On the other hand, we

have seen in Eq. (4.29) that the radiation coming out of an optically thick object has

the specific intensity equal to the source function. From Eq. (4.29) and (4.31), we

conclude

)( TBS = (4.32)

when matter is in thermodynamic equilibrium.

Fig.4.6: Blackbody radiation coming out of a box with an optically thick obstacle placed behind

the hole inside the box

On using Eq. (4.24), we finally obtain the famous result known as Kirchoff’s law.

Kirchoff’s law

)( TBj = (4.33)

Let us now stop and try to understand what we have derived. Very often matter tends

to emit and absorb more at specific frequencies corresponding to spectral lines. Hence

both jν and αν are expected to have peaks at spectral lines. But, according to

Eq. (4.33), the ratio of these coefficients should be the smooth blackbody function

Bν(T).

We now look at the results of Eq. (4.28) and (4.29).

Eq. (4.28) tells us that the radiation from an optically thin source is essentially

determined by its emission coefficient. Since the emission coefficient is expected to

have peaks at spectral lines, we find that the emission from an optically thin system

like a hot transparent gas is mainly in spectral lines.

Bν (T)

87

Physical Principles On the other hand, from Eq. (4.29), the specific intensity of radiation from an

optically thick source is its source function. This has been shown to be equal to the

blackbody function Bν(T) in Eq. (4.31). Hence, we expect an optically thick object

like a hot piece of iron to emit roughly like a blackbody.

The nature of radiation from an astrophysical source crucially depends on whether the

source is optically thin or optically thick. Emission from a tenuous nebula is usually in

spectral lines. On the other hand, a star emits almost like a blackbody.

Why is the radiation from a star not exactly blackbody radiation? Why do we see

absorption lines? Recall that we have derived Eq. (4.29) by assuming the source to

have constant properties. This is certainly not true for a star. As we go down from the

star’s surface, temperature keeps increasing. Hence Eq. (4.29) should be only

approximately true. It is the temperature gradient near the star’s surface which gives

rise to the absorption lines.

By assuming thermodynamic equilibrium, we have derived the tremendously

important result Eq. (4.32) that the source function should be equal to the blackbody

function Bν(T). In a realistic situation, we rarely have strict thermodynamic

equilibrium. The temperature inside a star is not constant, but varies with its radius. In

such a situation, will Eq. (4.32) hold?

We have already mentioned in section 4.3 that the Maxwellian velocity distribution,

the Boltzmann law and the Saha equation hold if the system is in thermodynamic

equilibrium. This generally means that the temperature does not vary much over it.

For Planck’s law also to hold, the radiation has to interact with matter efficiently.

We note from the radiative transfer equation Eq. (4.18) that αν has the dimension of

inverse length. Its inverse 1−να gives the distance over which a significant part of a

beam of radiation would get absorbed by matter. Often this distance 1−να is referred to

as the mean free path of photons, since this is the typical distance a photon is

expected to traverse freely before interacting with an atom.

The smaller the value of 1−να , the more efficient is the interaction between matter and

radiation. If 1−να is sufficiently small such that the temperature can be taken as

constant over such distances, then we expect Planck’s law of blackbody radiation to

hold. In other words, if both 1−να and the mean free path of particles are small

compared to the length over which the temperature varies appreciably, then all the

important laws of thermodynamic equilibrium hold.

Such a situation is known as Local Thermodynamic Equilibrium, which is abbreviated

as LTE. Inside a star, we expect LTE to be a very good approximation and we can

assume Eq. (4.32) to hold when we solve radiative transfer equation inside the star. In

the outer atmosphere of a star, LTE may fail and it often becomes necessary to

consider departures from LTE when studying the transfer of radiation there.

In this unit, we have discussed some basic principles of physics applicable to

astrophysical systems. We now summarise the contents of this unit.

4.5 SUMMARY

• The virial theorem, states that the gravitational potential energy EG and kinetic

energy EK of a system are of the same order

02 =+ GK EE

88

Basics of Astronomy • Newton’s theory of gravitation is generally adequate. However, if the quantity

rc

GMf

2

2= becomes of order unity in a system, then the general theory of

relativity has to be used instead of Newton’s theory.

• For astrophysical systems in thermodynamic equilibrium, the following results

hold:

Maxwellian velocity distribution

dvTk

mvv

Tk

mNdN

BB

−

π=

2exp

24

22

3/2

Boltzmann’s law

−=

Tk

E

N

N

B

eexp

0

Saha’s equation

( )

−

π=

Tk

ITk

h

mp

n

n

BB

ee exp

2 5/23/2

2I

II

Planck’s law of black body radiation

1

exp

8

3

3

−

π=

Tk

h

d

c

hdu

B

• The energy of radiation lying in the frequency range ν and ν + dν passing

through an area dA cosθ in time dt from the solid angle dΩ centred at θ is given

by

)cosˆ,,( dddtdAtrIddE n=

where Iν is called the specific intensity.

• The radiant flux of a time independent radiation field is defined in terms of

specific intensity Iν as

Ω= dIF cos

• The radiation pressure is given by

= cos1 2

dIc

p

• The interaction of matter with radiation is given by the radiative transfer

equation

Ijds

dI−=

The general solution of this equation is given as

( ) ( ) ′′+=

′−−−

0

(0))(

t

dSeeII

89

Physical Principles • The optical depth of an object along a ray path between points s0 and s is given

by

( ) ′′=s

s

sds

0

• Kirchoff’s law for a system in thermodynamic equilibrium is given as

)( TBj =

4.6 TERMINAL QUESTIONS Spend 30 min.

1. Suppose the Sun contracts at a uniform rate to half its present size in 107 yr.

Suppose that all its energy is radiated from its surface. Calculate the luminosity of

the Sun during the contracting phase.

2. Determine the size to which the Earth must shrink so that the use of Einstein’s

theory of gravitation becomes necessary.

3. Calculate the optical depth at which the specific intensity reduces to one-

hundredth of its original value in a system in which no emission of radiation is

taking place. Perform your calculation at a given frequency. Would the system be

optically thick or thin?

4.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. We need to equate the total kinetic energy of the globular star cluster with its

gravitational potential energy. If the cluster has N stars with masses of individual

stars of order m, and velocities of order v, the total kinetic of the star cluster is

about

Nmv2.

The total gravitational potential energy is

R

NmG2)(

,

On equating these two, we have

,2

R

GNmv ≈

from which

( )

g1010

1010 397

20262

≈×

≈≈−G

RvNm

This is the mass of the cluster. Taking the masses of stars to be of order 1033

g

(~ mass of the Sun), the cluster has about million stars in it.

90

Basics of Astronomy 2. 1

2

2==

rc

GMf

( )m

103

1099.110673.622

28

3011

2×

××××==

−

c

GMr

m109

99.1673.62 3×××

=

= 2951 m

3. 1010

2928288

2

2

103103

101010106736

3

8

3

8

×××

××××π=

ρπ=

−−.

.c

Grf

11027

67368 −××

=.

~ 0.6, which is of the order 1.

4. 480.5040

log2.5logI

II.I

TT

n

npe −−=

0.48.5040

log2.5loglogI

II −−=+ IT

Tn

npe

48.0.5040

log5.22logI

II −−+−= IT

Tn

n

( ) 6.135040

10log5.248.2 4 ×−+−=T

= 0.67

5. = sin.2.cos 2 d

c

Ip

c

I

c

I 3

3

4

03

cos.2 =

π−=

Terminal Questions

1. Gravitational potential energy of a star of radius r

GMr

2

=

For the Sun of radius 2

ΘR , gravitational P. E. Θ

=R

GM 22

Present gravitational potential energy of the Sun = ΘR

GM 2

∴ Energy radiated from its surface ΘΘΘ

=−=R

GM

R

GM

R

GM 2222

(Remember that when the star contacts, gravitational energy is released)

91

Physical Principles

∴ TR

GML

Θ

=2

(T is the time of contraction)

36002436510. 7

2

×××=

ΘR

GM

( )

32710

2338

106.3104.21065.310107

10210673.6

××××××××

×××=

−

6.34.265.37

4673.6

×××

×=

33

3535

104

1.

6.34.265.37

104673.610

××××

××=

(in terms of present solar luminosity)

2106.34.265.37

673.6×

×××=

= 3.02 (in terms of present solar luminosity)

2. rc

GMf

Earth

2

21=≈

∴ 2

2

c

GMr

Earth=

( )( )UnitsSIm

103

10610673.62

28

2411

×

××××=

−

= m109

6673.62 3−×××

= 8.9 × 10−3

m

3. Eq. (4.23) ( )

0

−=eI

I

100

1 −=e ( ) 614100log .e ==

Since 1 >>τ , the system is optically thick.

5

The Sun

UNIT 5 THE SUN

Structure

5.1 Introduction

Objectives

5.2 Solar Parameters

5.3 Solar Photosphere

5.4 Solar Atmosphere

Chromosphere

Corona

5.5 Solar Activity

5.6 Basics of Solar Magnetohydrodynamics

5.7 Helioseismology

5.8 Summary

5.9 Terminal Questions

5.10 Solutions and Answers

5.1 INTRODUCTION

From Unit 4, you know that the principles of different branches of physics such as

mechanics, thermodynamics and quantum mechanics are used in astronomy and

astrophysics. In the present and subsequent Units, you will use these principles to

investigate the behaviour and properties of the universe and its constituents.

On the cosmic scale, the Sun is just another star; there are bigger and brighter stars in

the universe. The Sun is, however, very important to us because i) it is the nearest

and the only star in our planetary system and ii) it provides almost all of our

energy. Do you know that a slight variation in the energy received from the Sun can

threaten life on the earth! Further, the Sun being the nearest star, we can study its

structure, atmosphere, and other physical characteristics in greater detail. The

information/data so obtained can be used to test the theories of stellar structure and

evolution. In this way we can improve our theories and have a better understanding of

other stars. In the present Unit, you will study about the Sun.

Due to the efforts of astronomers, today we have detailed information regarding the

Sun. In Sec. 5.2, you will learn to arrive at the estimates of the basic solar parameters

such as mass, radius and effective surface temperature. As far as we are concerned, all

the visible radiation from the Sun comes from its surface layer called the photosphere.

Above the photosphere is the atmosphere of the Sun consisting of two distinct layers

namely the chromosphere and the corona. In Sec. 5.3, you will study the characteristic

features of these layers. Interaction of the Sun’s magnetic field with highly mobile

charged particles in it gives rise to a variety of observable events. These events,

collectively known as solar activity, have been discussed in Sec. 5.4. The theoretical

analysis of the interaction of magnetic field with conducting matter in motion is

known as magnetohydrodynamics and it provides a basis to understand solar activity

and related features of the Sun. Solar magnetohydrodynamics has been discussed in

Sec. 5.5. In Sec. 5.6, you will learn, in brief, about helioseismology which provides

valuable information about the Sun’s internal structure.

Objectives

After studying this unit, you should be able to:

• estimate values of the basic parameters of the Sun;

6

The Solar System

and Stars • describe different layers of the Sun’s atmosphere;

• describe some of the observed features such as sunspot, prominence and solar

flare associated with solar activity;

• explain the role of the Sun’s magnetic field in solar activity;

• derive the basic results of solar magnetohydrodynamics; and

• explain the seismology of the Sun.

5.2 SOLAR PARAMETERS

The basic parameters of the Sun are its mass (MΘ), radius (RΘ), luminosity (LΘ) and

effective surface temperature (Teff). In the following discussion, you will learn to

estimate these parameters.

Mass: You know that the mean distance of the Sun from the Earth is 1.5 × 1011

m. It is

called the mean solar distance, a. In astronomy we measure mean distances in terms

of a and it defines the Astronomical Unit (1 AU = mean solar distance). To obtain an

expression for the mass of the Sun, we may use Kepler’s third law under the

assumption that the mass of the planet can be neglected in comparison with the mass

of the Sun. This assumption is valid for the Sun-Earth system and we can write:

Θ=π

GMP

a

2

324 (5.1)

where a, P, G and MΘ are the mean solar distance, orbital period (~ 365 days) of the

Earth, gravitational constant and mass of the Sun, respectively. With the values of the

orbital period and the mean solar distance available at present, the value of GMΘ is

estimated to be 132712438 × 1012

m3s

−2. Since the laboratory measurements for G

gives a value equal to 6.672 × 10−11

m3kg

−1s

−2, we obtain:

MΘ ≈ 2 × 1030

kg.

This value is taken as the mass of the Sun as it exists today. In fact, solar mass

decreases continuously since the Sun continuously emits radiation and particles which

carry with them some mass. However, the total mass loss during the Sun’s estimated

life time (~ 1010

yrs) is found to be less than 1027

kg. This value is much less than the

error in measurement of the solar mass and is, therefore, negligible. This method can

also be used to estimate the masses of the satellites/Moons of the planets in our solar

system. How about solving an SAQ of this nature?

SAQ 1

One of the four Galilean satellites of the planet Jupiter is Io. Its orbital period is

1.77 days. The semi-major axis of its orbit is 4.22 × 1010

cm. Calculate the mass of

Jupiter under the assumption that the Jupiter is too massive in comparison to Io.

Radius: The radius of the Sun can be estimated if we know the values of its angular

diameter, θ and the mean solar distance, a (Fig. 5.1). The angular diameter of the Sun

is 32′ and mean solar distance is 1.5 × 1011

m. Thus, with the help of Fig. 5.1, we

obtain the value of the solar radius RΘ as:

[ ]rad)109.232(m)105.1(2

1 411 −Θ ××××=R

= 6.7 × 108m.

Spend

5 min.

You may recall from Unit 6 of

the physics elective course

PHE-01 entitled Elementary

Mechanics that Kepler’s third

law is given by:

GM

aT

322 4π

=

where T, a, G and M are

respectively the time period,

semi-major axis, gravitational

constant and mass of the Sun.

In astronomy, we encounter

typically small angles (of the

order of minute (′) and second

(″)), for which simple

approximations are used for

trigonometric functions. Thus,

for small values of the angle θ,

tanθ = θ = sinθ

and

cosθ = 1

However, as you know, angles

must be measured in radian if

these formulae are to be valid.

7

The Sun Astronomical observations indicate that the solar radius is not constant; rather, its

value changes slowly. Over a period of ~ 109 years, the average change is about

2.4 cm per yr. Further, radius of the Earth is 6.4 × 106m. Thus, the Sun’s radius is

Fig.5.1: When viewed from the Earth, the angular diameter of the Sun is approximately 32′′′′

almost 100 times larger than that of the Earth. To get an idea of the relative sizes of

the Sun and the Earth, refer to Fig. 5.2.

Fig.5.2: The size of the Sun is so big that it can contain the Earth as well as the orbit of the Moon!

Luminosity: The solar luminosity, LΘ, is defined as the total energy radiated by the

Sun per unit time in the form of electromagnetic radiation. To estimate the value of

luminosity of the Sun, let us imagine a sphere with the Sun at its centre (Fig. 5.3). The

radius of this imaginary sphere is a, the mean distance between the Sun and the Earth.

Now, each unit area A of the sphere receives energy equal to S, called the solar

constant. Therefore, luminosity can be expressed as:

LΘ = 4πa2S (5.2)

Fig.5.3: Imaginary sphere of radius a surrounding the Sun where a is its mean solar distance from

the Earth

Earth

Moon’s orbit

Sun

Earth

Sun 1 AU

32´

Sun

1 AU

Earth

1m

1m

a

A

8

The Solar System

and Stars Since the solar radiation is absorbed in the Earth’s atmosphere, it is obvious that S

should be measured above the atmosphere. S has now been measured accurately using

satellites and its value is 1370 Wm−2

. Substituting the value of S and the mean solar

distance, a = 1.5 × 1011

m in Eq. (5.2), we get:

LΘ = 3.86 × 1026

W.

Temperature: The temperature of the Sun at its surface and its interior regions are

different. The surface temperature can be estimated using Stephan-Boltzmann law

which you studied in our course on Thermodynamics and Statistical Mechanics

(PHE-06). We leave this as an exercise for you in the form of an SAQ.

SAQ 2

Assume that the Sun radiates like a black body at temperature T. Calculate T using

Stephan-Boltzmann law. Take Stephan constant σ = 5.67 × 10−8

Wm−2

K−4

.

On solving SAQ 2, you would have found that the temperature of the Sun is 6000 K.

This estimated temperature is called the effective surface temperature because it is

the temperature of a black body whose surface emits the same flux as the Sun. This is

the temperature of the surface layer of the Sun called the photosphere from which all

the radiation is emitted.

To appreciate the validity of the approximation that the Sun radiates like a black body,

refer to Fig. 5.4. It shows the observed solar radiation in the ultraviolet, visible and

infrared regions of electromagnetic spectrum. Also plotted in this figure is the energy

curve of a black body at 6000 K. In view of the similarity of the two curves, it is fair

to assume that the Sun radiates like a black body at temperature 6000 K.

Fig.5.4: Solar energy curve

The Sun is a hot, bright gaseous ball and it does not have a well defined surface like

the Earth. The visible surface of the Sun is called photosphere. Let us learn about it

now.

5.3 SOLAR PHOTOSPHERE

The photosphere (Fig. 5.5) is the visible surface of the Sun. All the light received from

the Sun, in fact, comes from the photosphere. You may ask: Why do not we receive

the radiation in the same form as generated in the interior of the Sun? At the

centre of the Sun, the energy is generated in the form of high energy photons called γ-

rays. As these photons travel outwards, they collide with particles of matter and lose

energy continuously. By the time these photons reach the surface − the photosphere −

they are reduced to photons of visible region of electromagnetic spectrum. So, visible

radiation is emitted from the photosphere.

Spend

5 min.

Fig.5.5: The Sun’s photosphere

You learnt that the Sun

radiates like a black body at

approximately 6000 K. You

may think that if all the

energy is radiated outward

from the Sun, its surface

should gradually cool down.

This does not happen

because there is a constant

flow of energy from the

interior of the Sun towards

its surface.

9

The Sun The density of photosphere is 3400 times less than the density of the air we breathe.

The thickness of the photosphere is about 500 km and the temperature at its base is

~ 6500 K. The temperature decreases upward and reaches a minimum value of

~ 4400 K at the top. This assumption is corroborated by the Sun’s absorption

spectrum which indicates that the light we receive must be passing through a cool gas

in which photons get absorbed.

The photosphere is not a quiet region (see Fig. 5.6). It shows a granular structure. If

you look at Fig. 5.6a carefully, you can see that the photosphere consists of bright and

irregularly shaped granules; each granule surrounded by dark edges. It has been found

that these granules are very hot and their typical size is ~ 1500 km. The hot gas in the

granules rises up with a speed of the order of 500 ms−1

and bursts apart by releasing

energy. The cool material subsequently sinks downward along the dark edges or lanes

between granules. The rising hot granules are seen only for a very short time (~ 10

minutes) before they dissolve.

(a) (b)

Fig.5.6: a) Photograph of the photosphere showing granular structure; and b) schematic diagram showing granules and their boundaries

The question is: What causes granulation of the photosphere? It is caused due to

convection (a mode of energy transport by matter, about which you will study in

Unit 8). The granulation can be visualised (Fig. 5.6b) as the top layer of a region

where, due to convection, hot gas from below the photosphere moves upward. Thus,

the centre of the granule is hotter and it emits more radiation and looks brighter in

comparison to the edges which are relatively cooler and emit less radiation.

Convection based explanation seems valid because the spectra of granules indicate

that their centres are much hotter than the edges. Further, the solar granulation

provides observable evidence supporting the idea that there exists a convection zone

below the photosphere.

You may now like to know: What is the chemical composition of the photosphere?

It consists of 79 percent hydrogen and the remaining 21 percent consists of nearly 60

other chemical elements. Interestingly, all the elements of the photosphere are known

elements and their proportion in the earth is more or less the same as that in the

photosphere. This similarity in the chemical compositions of the photosphere and the

earth is of utmost importance for understanding the formation of the solar system.

Though the photographs of the Sun give the impression that it has a clear edge, such

clear and distinct edge does not exist. Outside the apparent edge are the Sun’s outer

layers, collectively known as the Sun’s atmosphere. These layers can be seen and

probed and valuable information about their physical characteristics can be obtained.

Let us learn about the various layers of the solar atmosphere.

Granule Granule

Granule boundary

As they look inwards into the

solar atmosphere,

astronomers have discovered

that, within a distance of

about 500 km, the solar

atmosphere changes from

being optically thin to

optically thick. This distance

is only ~ 0.07% of the Sun’s

radius. This region gives the

impression that the Sun has a

sharp edge as viewed from

the earth.

As you have learnt in Unit 4,

optically thin/thick medium

refers to the medium

characterised by low/high

absorption of electromagnetic

radiation.

10

The Solar System

and Stars 5.4 SOLAR ATMOSPHERE

The Sun’s atmosphere is divided into two layers namely, the chromosphere and the

corona. A schematic diagram of these layers is shown in Fig. 5.7.

Fig.5.7: Schematic diagram showing the layers of solar atmosphere

5.4.1 Chromosphere

Chromosphere lies above the photosphere (Fig. 5.7) and extends up to ~ 2000 km.

This layer of the solar atmosphere is normally not visible from the Earth because of its

faintness. However, it can be seen during a solar eclipse. The name chromosphere is

derived from the fact that a few seconds before and after a total solar eclipse, a bright,

pink flash appears above the photosphere (Fig. 5.8). The spectrum obtained at that

time is called a flash spectrum (Fig. 5.9).

Fig.5.9: Emission lines observed at the time of total solar eclipse

The appearance of pink colour is due to the emission of the first Balmer line (Hα)

which occurs in the red region. The temperature, density and pressure in the

chromosphere determine the intensities of various emission lines. In the

chromosphere, the density decreases by a factor of ~ 104 from that of the photosphere

Fig.5.8: Chromosphere just

before a total solar

eclipse

You may be aware that each

chemical element

emits/absorbs electromagnetic

radiations of characteristic

wavelengths. These radiations

of different wavelengths (also

called lines) constitute the

spectrum of the element.

Balmer lines, which fall in the

visible region, are the spectral

lines of hydrogen atom. The

emission/absorption of the

spectral lines is predominantly

dependent on the temperature

and density of the material.

You will study about the origin

of spectral lines in Unit 7.

Photosphere

Sun’s interior

Transition zone

Chromosphere

0 km

500 km

2300 km

2600 km

Corona

11

The Sun while the temperature rises to ~ 25000 K within a short distance of ~ 2000 km.

Therefore, the spectral lines that are not produced at relatively higher density and

lower temperature of the photosphere are formed in the chromosphere as emission

lines (see margin remarks).

At this point, it is logical to ask: Why does the temperature in the chromosphere

increase with height? The clue to the answer of this question lies in observing the

chromosphere just before the total solar eclipse. Hot gas, in the form of jets called

spicules, is observed throughout the chromosphere (Fig. 5.10). These spicules extend

upward in the chromosphere up to a height of ~ 10000 km and last for as long as 15

minutes. This implies that the lower part of the chromosphere is highly turbulent and

the spicules transport energy and matter from the photosphere to the chromosphere.

This causes heating of the chromosphere.

Fig.5.10: Spicules in the Sun’s chromosphere

Now, your next logical question could be: What causes spicules? The origin of

spicules is not yet understood completely. However, it appears to be caused by the

Sun’s magnetic field. Further, just above the chromosphere, there exists a transition region extending up to ~ 3000 km. In this region, the temperature rises sharply

to ~ 106

K (Fig. 5.11). The transition region links the chromosphere with corona, the

outermost part of the solar atmosphere.

Fig.5.11: The variation of temperature and density in the Sun’s atmosphere with distance

T

emp

era

ture

(K

)

107

106

105

104

103

1014

1012

1010

108

106

104

102 104 106

Den

sity

(cm

3)

Height from the Sun's surface (km)

12

The Solar System

and Stars 5.4.2 Corona

Corona, the outermost layer of the Sun’s atmosphere is named after the Greek word

for Crown. Like the chromosphere, the corona can be observed only during total solar

eclipse − when the Moon completely covers the solar disc (Fig. 5.12). You may

wonder why we cannot see corona at normal times! The fact is that the density of

matter in both the chromosphere and corona is very low (see Fig. 5.11). They emit

very little light and, as a result, they are very faint. In the bright light of the

photosphere, they are not visible.

Fig.5.12: Two photographs of the solar corona

The spectrum of corona consists of bright lines superimposed on a continuous

spectrum. When these lines were first discovered, they were thought to be due to a

new element, coronium, not found on the earth. Later, it was realised that these lines

were due to highly ionised atoms and not due to the so-called ‘new’ element

coronium. Fig. 5.13 shows the temperature and height in the corona at which emission

lines of various ionised elements are formed. You may note here that to excite the

emission lines from highly ionised elements, say spectral line of SiX ( read margin

remarks), a temperature greater than 2 × 107K is required. The observed emission

lines of highly ionised atoms of iron, nickel, neon, calcium etc., in the spectrum of

corona clearly indicate that the temperature prevailing in corona is very high (more

than 106

K). Now, before proceeding further, how about testing yourself?

SAQ 3

a) The temperature of chromosphere and corona is very, very high in comparison to

that of the photosphere. Still, we observe that the photosphere is the brightest of

the three. Why?

b) Calculate the temperature at which a particle will have sufficient energy to ionise

a hydrogen atom.

Due to high temperature, electrons in the corona region have high energies. These

electrons interact with ionised atoms and give rise to emission of X-rays. The coronal

X-ray emission is much larger than that of the photosphere. Remember that the

temperature of the photosphere is only 6000 K. So, it emits very little energy in the

X-ray region. The Sun, as observed in X-rays is shown in Fig. 5.14 which clearly

indicates the existence of very high (~ 106 K or more) temperature in the corona.

You have already learnt that the temperature of the photosphere is lower than that of

the chromosphere and as one goes further up in the corona, temperature rises to more

than a million degree K. This gives rise to a very simple but important question:

Despite being closer to solar interior, why is the photosphere far cooler than the

corona? You know that the second law of thermodynamics precludes such a scenario

Spend

5 min.

In astronomy, it is common

to denote a neutral atom,

such as silicon, as SiI. By

this convention, singly

ionised silicon is denoted by

SiII. Hence, SiX denotes a

silicon atom whose nine

electrons have been

removed due to repeated

ionisation.

Fig.5.13: Height versus temperature plot showing emission

lines of different

ionised atoms

1 2 4 10 20 40

40

10

200

10000

20000

40000

Si II

O II

Si IV O II

O V

O VI

Si VII

Mg X

Si X

Height above the surface(103 km)

Tem

per

atu

re (

10

3 K

)

Corona

Ch

rom

osp

her

e

13

The Sun as heat cannot flow from a cooler region to a hotter region on its own. We also know

that the radiation from the photosphere passes through corona almost freely because

of its (corona’s) low density. Since hardly any absorption of radiation takes place in

the corona, the existence of such high (~ million degree) temperature in the corona

presents a paradoxical situation. Several mechanisms have been proposed to resolve

the paradox. It is now generally believed that the magnetic field of the Sun might, in

some way, be responsible for coronal heating. You will study the basics of this

mechanism in Sec. 5.5 of this Unit. The observed overlapping of regions of intense X-

ray emission and strong magnetic fields lend support to this idea.

Solar Wind

Unlike its visual appearance, the solar corona extends much beyond into the space.

The outer layer of the solar atmosphere, in fact, continuously emits charged particles

which fill the entire solar system. This emission is called the solar wind. It comprises

streams of charged particles (mainly protons and electrons) and causes continuous loss

of mass from the Sun. The phenomenon of solar wind was predicted much before its

detection. Its characteristics can be investigated using rockets and satellites. For

instance, the solar wind velocities range from 200-700 km s−1

at the distance of the

Earth from the Sun. The number density of the solar wind at this distance

is ~ 7 particles per cm3.

You may like to know: What gives rise to the solar wind? In view of the high

temperature prevailing in the corona, the gas contained therein exerts tremendous

pressure outward. In fact, the pressure is much higher than the inward pressure due to

the Sun’s gravity. The gas, therefore, streams outward from the Sun and fills the

interplanetary space. In 1962, Mariner II spacecraft detected the solar wind by its on-

board instruments.

Fig.5.15: van Allen radiation belts

The electrically charged particles carried by the solar wind cannot cross the lines of

force of the Earth’s magnetic field. These particles are deflected by the Earth’s

magnetic field, spiral around the field lines and move back and forth between the

magnetic poles of the Earth. As a result, two doughnut-shaped zones of highly

energetic charged particles are created around the Earth and they are collectively

called the van Allen radiation belts. These radiation belts are shown in Fig. 5.15.

Astronomers have observed a variety of short-lived events, collectively known as

solar activity, occurring on or near the surface of the Sun. The root cause of all these

One of the manifestations of

solar wind is observed in the

shape of comets. You know

that the tail of a comet points

away from the Sun. It is

because the solar wind sweeps

along the material of the comet.

You know that when a charged

particle passes through a

magnetic field, it experiences a

force which changes its

direction of motion. The force

experienced by a moving

charged particle is known as

Lorentz force.

Fig.5.14: X-ray picture of the Sun

As you know, satellites and

space missions comprise

very sophisticated integrated

circuits, solar cells and other

electronic gadgets.

Therefore, care is taken to

minimise the damaging

effects of the radiation belts

on the satellites and space

missions.

Magnetic axis of

the Earth

Outer radiation belt

Inner radiation belt Rotational axis of the Earth

14

The Solar System

and Stars activities is the existence of strong and localised magnetic field in the photosphere.

Studies of these events/activities provide valuable information about the Sun and the

nature of its magnetic field. You will now learn about some of these short-lived

events.

5.5 SOLAR ACTIVITY

Sunspots

If you look at the photographs (Fig. 5.16) of the Sun, you see dark spots on its visible

surface. These dark spots are called sunspots. Sunspots can be seen sometimes even

with unaided eye at sunrise or sunset. (But you should not attempt to see the Sun with

unaided eye as it may cause irrepairable damage to your eyes because of its intense

brightness.) Naked eye observations of sunspots date back to ~ 2000 years in China. It

was in the seventeenth century that Galileo, using the telescope which he himself had

fabricated, observed sunspots and found that these dark spots were in motion. This led

him to suggest that the Sun was spinning in space. Galileo also observed that the sizes

and shapes of the sunspots kept changing as they rotated with the Sun.

Fig.5.16: Photographs of a sunspot group

The sunspot temperature is ~ 4000 K. With such high temperature, you may

wonder, why they appear dark! Sunspots appear darker because they are cooler

than their surrounding areas in the photosphere that have an effective temperature of

6000 K. A typical white light picture of a large sunspot is shown in Fig. 5.17. Note

that it consists of a dark central region, called umbra, surrounded by a less dark

region, called penumbra. We do not see such details in the picture of smaller

sunspots.

At this stage, a logical question is: Why is the temperature of the sunspots lower

than their surroundings? It is due to the existence of strong magnetic fields in the

sunspots. In the presence of a magnetic field, a spectral line emitted by an atom at a

Fig.5.17: Sunspot structure

15

The Sun single wavelength is split into three lines. This is called the Zeeman Effect. Such

Zeeman splitting is observed in the spectrum of sunspots. Since the line separation, ∆λ

is proportional to the applied magnetic field, a magnetic field up to ~ 3000 Gauss has

been estimated in sunspots. Fig. 5.18 shows the mechanism of Zeeman splitting of a

spectral line and the Zeeman splitting of a spectral line of a sunspot.

(a) (b)

Fig.5.18: Zeeman splitting of a) a spectral line; and b) a spectral line of the sunspot

The presence of strong magnetic fields in sunspots restrains the flow of hot material

from layers below the photosphere. Therefore, within a sunspot, less heat comes up

and they (sunspots) are cooler/darker than the surrounding region. Within a sunspot,

the umbral magnetic field is quite intense ~ 3000 Gauss. It spreads like an umbrella

and weakens in the penumbral region. The field strength in the penumbra is estimated

to be ~ 1000 Gauss.

Sunspots can last for weeks. The question is: How do these cooler regions survive

for so long amidst the hotter regions? This could happen due to the magnetic field.

You know that magnetic field exerts pressure (equal to B2/2µ) across the lines of

force. This pressure, along with the pressure of matter inside a sunspot balances the

material pressure outside and the sunspots can exist in equilibrium.

Sunspot Cycle

The observed motion of the sunspots indicates that the Sun is spinning in space. In

1843, Heinrich Schwabe, a German who observed the sky for fun, discovered a

periodic variation in the numbers of visible sunspots. He found an interval of 5.5 yrs

between the time when maximum number of sunspots (sunspot maxima) were

observed and the time when the minimum number of sunspots (sunspot minima) were

observed. Over the last two centuries, sunspot observations clearly suggest a periodic

variation of about 11 years between two successive sunspot maxima (Fig. 5.19a).

Another important observation pertaining to sunspot is that the sunspot zones migrate

along solar latitude. It is observed that the first sunspot zone appears at latitude

of ~ 35° in, say, the northern hemisphere and it migrates to lower latitudes. It lasts

till it reaches a latitude of ~ 10°. The latitude migration of sunspot zones is shown in

Fig. 5.19b. This is the famous butterfly diagram which shows a period of ~ 11 years

between the successive occurrences of a sunspot at a given latitude. It is believed that

the sunspot cycle is caused due to differential rotation of the Sun; it rotates faster at

the equator compared to higher latitudes.

Energy levels

Transition

Spectrum

16

The Solar System

and Stars

Fig.5.19: a) Sunspot cycle; and b) butterfly diagram which shows the migration of sunspots from

higher to lower latitudes

Solar Prominences

Refer to Fig. 5.20 which depicts loop like structures surging up into the corona when

the Sun is viewed along the edge of the solar disc. These structures are called

prominences which are intimately connected with and formed due to the Sun’s

magnetic field. The material in prominences comprises hot ionised gases trapped in

magnetic fields associated with the active regions. Since these gases came from

deeper layers of the solar atmosphere, they are cooler and therefore denser than the

coronal gas. It is for this reason that prominences appear as bright structures.

However, when viewed against the photosphere (solar disk), prominences appear as

dark snake like objects, called filament.

Solar Flares

Yet another form of solar activity is called solar flare. Solar flares are sudden eruptive

events which occur on the Sun (Fig. 5.21). Each event may involve energy in the

Fig.5.20: Solar prominences

(a)

(b)

Year

Year

Number of

Sunspots

17

The Sun

range of 10

22 to 10

25 Joules. Usually the flares last anywhere between a few minutes

to more than an hour. A large flare may have linear dimension as large as 105 km and

may be seen as a short-lived storm on the Sun. Such energetic eruptions are usually

linked to sunspots because these quite often occur at the top of magnetic loops that

have their feet in sunspots. Thus, the most likely places of occurrence of solar flare are

the regions of closely packed sunspots. The tremendous amount of energy carried in

solar flare is released in the form of X-rays, ultraviolet and visible radiation, high

speed electrons and protons. You may ask: What is the source of energy in solar

flares? This question can be answered on the basis of a model for solar flare shown in

Fig. 5.22.

Fig.5.22: A solar flare model

Fig.5.21: Solar flare

Approximate size of the Earth

18

The Solar System

and Stars A Model of Solar Flare

All kinds of solar activities i.e., the sunspots, prominences, flares etc., are possibly

linked to the release of stored magnetic energy. It is believed that the energetic solar

eruptions are caused due to coming together and merging of magnetic fields in the

active regions (the phenomenon is known as magnetic field reconnection) and

thereby releasing the stored magnetic energy. To appreciate this phenomenon,

magnetic lines of forces can be considered as stretched springs with certain amount of

energy associated with each unit length. If the length of the lines of forces gets

reduced by certain mechanism, energy is released. This is what happens when lines of

force pointing in opposite directions meet and merge with each other (Fig. 5.22).

In order to fix these ideas, you should answer the following SAQ.

SAQ 4 a) What is the basis to conclude that the Sun is rotating in space?

b) What is the difference between spicules and solar prominences?

So far, you have studied about the photosphere, solar atmosphere and solar activity.

You must have noted that the Sun’s magnetic field plays an important role in solar

activity. Further, gaseous matter in the Sun is in the ionised form, that is, it is a

conducting matter. We will now try to understand the nature of interaction between

the Sun’s magnetic field and the conducting matter (fluid) in motion. This

understanding is of utmost importance in astronomy because, everywhere in the

universe, we find conducting matter moving in the presence of magnetic fields. The

study of the motion of conducting fluid in the presence of magnetic field is called

magnetohydrodynamics. Let us now turn to this subject.

5.6 BASICS OF SOLAR MAGNETOHYDRODYNAMICS

We begin our discussion of solar magnetohydrodynamics with Maxwell’s equations.

You may recall from the physics course entitled Electric and Magnetic Phenomena

(PHE-07) that three of the Maxwell’s equations can be written as:

∇∇∇∇ . B = 0, (5.3)

∇∇∇∇ × B = µj, (5.4)

and

∇∇∇∇ × E = t∂

∂−

B (5.5)

where µ, B, E and j are the magnetic permeability of the medium, magnetic field

intensity, electric field intensity and electric current density respectively. Note that in

Eq. (5.4), we have not written the displacement current term. This is because,

magneto-hydrodynamic phenomena in the Sun are usually slow whereas the

displacement current gives rise to fast phenomena such as electromagnetic radiation.

Further, if the fluid (conducting matter) velocity is v and its electrical conductivity is

σ, then Ohm’s law gives (see the margin remark):

j = σ (E + v × B) (5.6)

You are familiar with the

Ohm’s law expressed

mathematically as:

j = σ E

However, when a conducting

fluid is moving with velocity

v in a magnetic field B, there

is an induced electric field

given as v × B. Thus, in the

presence of two kinds of

electric fields − one due to

charged particles and another

due to their motion − the

Ohm’s law takes the form:

J = σ (E + v × B)

Spend

3 min.

19

The Sun where (v × B) term represents the electric field induced due to the motion of

conducting fluid in the presence of magnetic field. Using Eqs. (5.3) to (5.6), you can

readily obtain:

BBvB 2)( ∇η+××∇=

∂

∂

t (5.7)

where η = (µσ)−1

is called the magnetic diffusivity. Its value is generally constant in

solar conditions.

SAQ 5

Derive Eq. (5.7).

The first term on the right hand side of Eq. (5.7) gives the change in B due to the fluid

motion. The second term represents the change in B due to conductivity σ. It is

generally called the Ohmic decay of the field. Note that for v = 0, Eq. (5.7) reduces to:

BB 2∇η=

∂

∂

t (5.8)

Eq. (5.8) is the diffusion equation. It gives the rate at which magnetic field diffuses

out due to conductivity. In the limit of infinite conductivity, we have η → 0, and

Eq. (5.7) becomes

)( BvB

××∇=∂

∂

t (5.9)

Without going into the mathematical details, it is possible to understand the relative

importance of the two terms on the right hand side of Eq. (5.7). To do so, let us obtain

the orders of magnitude, of the values of the two terms. The order of magnitude of the

first term is ,L

VB where V, B and L are the typical values of the fluid velocity,

magnetic field and the dimension of the system. Similarly, the magnitude of the

second term is .2

L

Bη The ratio of the two terms is called magnetic Reynold number,

Rm, and is given by:

η

=VL

Rm (5.10)

If Rm >> 1, the second term in Eq. (5.7), i.e. the diffusion term, is negligible. The

condition Rm >>1 is obtained in two situations: when the conductivity is very high

because η appears in the denominator, and secondly when the dimension of the

system, L is very large. In astrophysical systems, we have Rm >> 1 because of the

second situation as their dimensions are very large. In any case, Rm >> 1 implies that

there is no decay of the magnetic field as if the conductivity of the medium is infinite.

Actually, the conductivity is finite, but the large dimensions ensure that Rm >> 1 and

so there is no decay of the field.

If we drop the diffusion term in Eq. (5.7), the remaining equation (Eq. (5.9)), implies

that the magnetic flux linked to a cross-section of the fluid remains unchanged as the

fluid moves about. In other words, the magnetic field is frozen in the fluid. The

idea of frozen field means that the magnetic flux is transported along with the material

motion. We can show this formally in the following manner:

Spend

3 min.

20

The Solar System

and Stars Let us consider a cross-sectional area A placed in a magnetic field B (Fig. 5.23). The

magnetic flux linked with area A may be written as:

=Φ

A

d aB . (5.11)

where da is an element of area on the surface A. Let l be the curve enclosing the area

A. Let us further assume that, in the time interval dt the area changes from A to A′ as

the fluid moves around. The magnetic flux Φ′ linked with A′ may be different from Φ

because (i) magnetic field B may have changed and/or (ii) some flux might have been

exchanged through the surface of the volume generated between A and A′. Now, the

rate of change of magnetic field is ∂B/∂t. The area of the curved surface surrounding

the volume between A and A′ is v × dl dt where dl is an element of length of the

contour of A. Therefore, the difference ∆Φ between fluxes through A′ and A is given

by:

.)(.o. dtddt

Al

×+

∂

∂=∆Φ lv.Ba

B (5.12)

Further, using Stoke’s theorem, we can write the second term on the right hand side of

Eq. (5.12) as:

××∇−=×

Al

dd aBvlv.B .)()(.o (5.13)

So, we can write the rate of change of flux (Eq. (5.12)) as:

aBvB

dtdt

d

A

.)(

××∇−

∂

∂=

Φ (5.14)

From Eq. (5.9), we have )( BvB

××∇=∂

∂

t. Thus, from Eq. (5.14), we find that the

rate of change of magnetic flux is zero. It implies that Φ remains constant. We may,

therefore, conclude that in astrophysical systems the lines of forces are

completely attached or glued to the moving fluid when Rm >> 1. Now, let us pause for a moment and think about the significance of the above

conclusion for solar activity. Recall that the active regions consisting of sunspots have

strong magnetic fields and solar activities such as prominence and flare occur in these

regions. The structures associated with solar prominences and solar flares are very

similar to the magnetic lines of force. It is, therefore, believed that these activities are

caused due to frozen magnetic lines of force in the conducting fluid.

You have learnt earlier in this unit that, in almost all the events associated with the

Sun’s magnetic field, energy is also transported. The energy transported by the

magnetic field glued to the conducting matter is responsible for heating the

chromosphere and corona. Would not you like to know how it happens? To

understand this process, we begin with the fact that an electric current exists in the

solar atmosphere due to the drift of electrons with respect to ions carrying opposite

charges. If a magnetic field B is also present in the plasma, then a volume force, also

called the Lorentz force (= j × B), acts on the material. Since ( )Bj ×∇µ

=1

(from

Eq. (5.4)) we may write the expression for Lorentz force as:

Fig.5.23: Motion of a closed curve l with fluid with

velocity v

You may recall that area is a

vector quantity having direction

along the normal to the area,

whose sense is defined by the

right hand rule.

dl v ×××× dldt

A

l

A'

21

The Sun

( ) BBBj ××∇µ

=×1

( )

µ∇−∇

µ=

2

2BBB.

1 (5.15)

The first term on the RHS of Eq. (5.15) denotes magnetic tension and the second term

denotes the gradient of magnetic pressure. Therefore, in the presence of a magnetic

field, a lateral pressure (due to the second term in Eq. (5.15)) acts on a conducting gas.

To maintain equilibrium, the lateral pressure due to magnetic field is balanced by the

gas pressure. The magnetic tension term is similar to the tension of a stretched string.

When a stretched string is distorted, its tension provides the restoring force. Therefore,

when the magnetic field lines (frozen in the conducting fluid) are disturbed we have a

similar restoring force (due to the magnetic tension). Thus, a disturbance may

propagate as a transverse wave along the magnetic field lines. Such waves are called

Alfven waves. The speed of these waves is given by:

.4πρ

=B

vm

where ρ is the density of the plasma. Since Alfven waves propagate along the

magnetic field lines, it is possible to transport energy outward along magnetic field

lines which are threading the outer solar atmosphere. This energy is believed to be

responsible for heating the chromosphere and corona.

SAQ 6

A pressure of 103Pa (Pascal) prevails in the solar atmosphere. What should be the

strength of the magnetic field required to balance such a pressure?

Now, before we close our discussion about the Sun, we will briefly discuss a new area

of solar research called helioseismology.

5.7 HELIOSEISMOLOGY

In 1962, Leighton, Noyes and Simon noticed wiggling back and forth of some of the

absorption lines in the solar spectrum with periods ~ 5 min. They conjectured that the

movement of the surface of the Sun is responsible for such observations.

Subsequently, careful observations of the solar surface confirmed the idea of waves

rising up and down on the surface of the Sun (Fig. 5.24). Astronomers now use these

waves to probe the solar interior much the same way as seismologists probe the

earth’s interior using vibrations caused by an earthquake.

So far, millions of different vibrational modes also called acoustic modes or p-modes

have been observed on the Sun’s surface. All of them have different frequencies and

surface patterns. Thus, the Sun appears to have a rhythmic surface motion similar to

that of a beating heart. The observed rise and fall of the surface of the Sun is, in fact,

due to superposition of many different acoustic modes. These modes are now believed

to be driven by irregular motions in the convective envelope under the solar surface.

Spend

3 min.

The name helioseismology

derives from the fact that, in

Greek, helios means the Sun

and the word seismos is

used for an earthquake.

As you know, seismology

refers to the study of seismic

waves moving inside the

Earth. Their arrival at

various points on the Earth’s

surface enables us to find

the point of origin of these

waves. This way we are able

to construct the internal

structure of the Earth.

22

The Solar System

and Stars

Fig.5.24: Waves on the surface of the Sun

The combined effect, or the superposition of millions of these acoustic waves, results

in the observed up and down motion of the photosphere with a period of the order of 5

minutes. The extent or size of a wave is called its horizontal wavelength. The relation

between sizes and periods of waves, obtained theoretically, suggested that only

specific combination of periods and sizes can resonate inside the Sun.

Fossat and Grec observed the solar oscillation from the South pole for around 120

hours. The analysis of this continuous record showed that the entire surface of the Sun

is ringing like a bell with periods in the range of 5 minutes and the vibrations may last

for days and weeks.

The natural frequencies of oscillations can be computed for any solar model. In view

of the precision now possible for determining the frequencies, we may compare these

with those computed for a given solar model. In case there is lack of agreement

between the predicted and observed frequencies, the model is slightly modified to

improve agreement. The improvement in the model brings it closer to reality. It has

been found that the depth of the solar convection zone is at a radius of 71.3 percent of

radius of the Sun. It has also been confirmed now that the proportion of helium in the

Sun lies between 0.23 − 0.26. This is quite consistent with the value of 0.25 for

helium believed to have been formed in the early phase of the universe after the big-

bang.

Now, let us summarise what you have learnt in this unit.

5.8 SUMMARY

• The mass, radius and effective surface temperature of the Sun are 2 × 1030

kg,

7 × 108m and 6000 K, respectively.

• The Sun’s atmosphere comprises of the photosphere, chromosphere and

corona.

• The photosphere is the visible surface of the Sun and all the radiation we receive

from the Sun is emitted by this layer.

• The photosphere has a granular structure which is caused by convection of hot

gas from below the photosphere.

• The chromosphere, which lies above the photosphere, extends up to ~ 2000 km

and is normally not observable from the earth except during total solar eclipse.

The temperature of the chromosphere is much higher than the photosphere.

• Relatively higher temperature of the chromosphere is perhaps caused due to

spicules − jets of hot gas − which extend upward in the chromosphere.

Since we cannot go inside a

star, we build theoretical

models which tabulate

pressure, temperature and

density at various depths

inside the star.

Time

Disk of Sun

Po

siti

on

5 min

23

The Sun • The outermost layer of the solar atmosphere, corona is also not visible at normal

times due to its low material density. The temperature of the corona is of the order

of 106K, much more higher than that of the photosphere.

• The sunspots are the dark spots on the solar disk. They appear dark because they

are cooler than their surrounding areas in the photosphere. The movement of

sunspots indicate that the Sun is spinning in space.

• Solar prominences are the loop like structures surging up into the corona. They

are caused due to the Sun’s magnetic field.

• Solar flares are sudden eruptive events involving energy in the range of 1022

to

1025

joules. The most likely places of occurrence of solar flares are the regions of

closely placed sunspots and they are possible caused due to the release of stoned

magnetic energy.

• The study of the motion of conducting matter in the presence of magnetic field is

called magnetohydrodynamics.

• The magnetic Reynold number is given by

η

=VL

Rm

which indicates that for infinite conductivity (i.e. η → 0), there is no decay of the

magnetic field. In other words, magnetic field is frozen in the conducting matter

in motion.

• Transportation of magnetic flux with conducting matter explains some of the solar

activities like prominence and solar flare.

• Helioseismology has its origin in the observed wiggling back and forth of some of

the absorption lines in the solar spectrum. The movement of the surface of the

Sun, causing these back and forth motion of absorption lines, gives rise to waves

on it. Investigation of the nature of these waves provides valuable information

regarding the internal structure of the Sun.

5.9 TERMINAL QUESTIONS Spend 20 min.

1. In a sunspot, magnetic diffusivity, linear dimension and velocity of conducting

fluid respectively is 103m

2s

−1, 10

4km, and 10

3ms

−1. Estimate the magnetic

Reynold number, Rm. Is it possible to assume that the conductivity in the sunspot

is virtually infinite?

2. The temperature inside a sunspot is 4000 K and that of its surface is 6000 K.

Calculate the strength of the magnetic field inside the sunspot which will balance

the pressure inside and outside.

[Hint: Remember that the magnetic pressure is B2/2µ where µ is magnetic

permeability of the medium and its value for the present case can be taken as

4π × 10−7

NA−2

].

3. The number density of particles (assume hydrogen) in the photosphere is 1020

particles per cm−3

and the strength of the magnetic field of the Sun is 1 G.

Calculate the velocity of the Alfven waves in the photosphere.

24

The Solar System

and Stars 5.10 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. Since the mass of the planet Jupiter is very large compared to its satellite Io, we

can use Kepler’s third law for Jupiter and Io system. Thus, we can write:

JGMP

a=

π2

324 (i)

where MJ is the mass of Jupiter.

From the problem, we have

a = 4.22 × 1010

cm = 4.22 × 108 m

P = 1.77 days = 1.77 × 24 × 60 × 60 s

G = 6.672 × 10−11

m3 kg

−1 s

−2

Substituting these values in Eq. (i) above, we get

2

324

GP

aM J

π=

221311

382

s)60602477.1()skgm10672.6(

m)1022.4()14.3(4

×××××

×××=

−−−

= 1.97 × 1027

kg

2. According to the Stephan-Boltzmann law, the amount of energy radiated by a

black-body per unit time per unit area at temperature T is given by

E = σ T 4

(i)

where σ is Stephan constant.

The energy radiated by the Sun can also be expressed in terms of its luminosity LΘ

as

2

4 R

LE

π= Θ (ii)

where R is the radius of the Sun.

Comparing the above two expressions, we can write

4

24

TR

Lσ=

π

Θ

Rearranging the terms and substituting the values of R (= 6.7 × 108m),

σ (= 5.67 × 10−8

W m−2

K−4

) and L = (3.86 × 1026

W), we get,

4/1

24

σπ= Θ

R

LT

25

The Sun

4/1

42828

26

)KWm1067.5()m107.6()14.3(4

W1086.3

×××××

×=

−−−

4/1

4162

2

K1067.5)7.6(14.34

1086.3

×

×××

×=

= 0.5895 × 104K

≈ 6000 K.

3. a) It is because the density of matter in photosphere is much higher than the

density of matter in chromosphere and corona.

b) We know that the ionisation energy of the hydrogen atom is 13.6 eV. Further,

energy acquired by a particle at temperature T is kBT. If this energy is equal to

the ionisation energy of the hydrogen atom, it will be ionised. Thus, we must

have

J)106.16.13( 19B

−××=Tk

)JK1038.1(

J)106.16.13(

123

19

−−

−

×

××=T

≈ 13.6 × 104 K.

4. a) Motion of sunspots.

b) Spicules are the jet like structures comprising hot gas and is observed

throughout the chromosphere. Prominences look like structures surging up

into the corona.

5. Ohm’s law can be written as:

J = σ (E + v × B)

Taking curl of both sides, we get:

∇ × J = σ (∇ × E + ∇ × (v × B))

Using Eqs. (5.4) and Eq. (5.5) we can write:

)()(1

BvB

B ××∇+∂

∂−=×∇×∇

σµ t

[ ] )().().(1

BvB

BB ××∇+∂

∂−=∇∇−∇∇

σµ t

From Eq. (5.3), we have ∇ . B = 0. Thus, we get

BBvB 21

)( ∇σµ

+××∇=∂

∂

t

26

The Solar System

and Stars BBv 2 )( ∇η+××∇=

which is Eq. (5.7).

6. We know that the pressure generated by magnetic field B is equal to B2/2µ. Thus,

for generating pressure equal to 103 Pa, we must have

Pa102

32

≈µ

B

Substituting µ = 4π × 10−7

NA−2

in the above expression, we get:

G. 105 T105 22 ×=×≈ −B

Terminal Questions

1. The magnetic Reynold number is given by (Eq. (5.10)):

η

=VL

Rm

Substituting the values of V, L and η from the problem, we get

123

713

sm10

)m10()ms10(

−

− ×=mR

≈ 107

Yes, conductivity can be taken to be virtually infinite because Rm >> 1.

2. On the basis of the equation of state for the sunspot and using the fact that

magnetic pressure is equal to B2/2µ, we can write:

)(2

12

2

TTNkB

B −=µ

where N is number density. Since µ = 4π × 10−7

NA−2

, T2 = 6000 K and

T1 = 4000 K, we can write:

B2 = 2 × (4π × 10

−7 NA

−2) × (10

23 m

−3) × (1.38 × 10

−23J K

−1) × (2000 K)

B = 0.08 T = 800

G

3. Velocity of Alfven waves is given by

πρ=

4

Bv

m

where ρ is density. Substituting the values of B and ρ, we get:

)cm g 106.110(4

G1

32420 −−×××π=

mv

25 cm s−1

.

27

The Solar Family

UNIT 6 THE SOLAR FAMILY

Structure

6.1 Introduction

Objectives

6.2 Solar System: Facts and Figures

6.3 Origin of the Solar System: The Nebular Model

6.4 Tidal Forces and Planetary Rings

6.5 Extra-Solar Planets

6.6 Summary

6.7 Terminal Questions

6.8 Solutions and Answers

6.1 INTRODUCTION

In Unit 5, you have learnt about various features of the Sun such as solar atmosphere,

solar activity and energy transportation. You know that the Sun is the only star in our

solar system. It is the most massive object in the system and all the planets revolve

around it. You have already studied about the solar system in your school science

course as well as in Foundation in Science and Technology (FST-1) course.

Therefore, you know most of the facts and figures such as mass, density, distance

from the Sun, and surface temperature of all the nine planets of the solar system.

However, being a student of physics and astronomy, you may not be satisfied only

with the facts and figures and would like to know: How did the solar system come

into being? Is it possible to explain the observations pertaining to the solar system on

the basis of the principles of physics? Do all the stars have planetary systems similar

to our solar system? Does life exist on any other planet? In the present unit, we shall

address some of these issues.

In Sec. 6.2, you will recapitulate some facts and figures about the solar system. This is

necessary because any model for the formation of the solar system must be consistent

with these observations. In Sec. 6.3, you will learn the nebular model which is at the

core of all the contemporary theories of formation and evolution of the solar system.

In this section you will also discover that the nebular model explains most of the

dynamic properties of the solar system. Further, it has been argued that the

gravitational force gives rise to tides in the oceans on the Earth and planetary rings

around the outer planets such as Jupiter and Saturn. The genesis of tidal forces and

planetary rings has been discussed in Sec. 6.4. And, in Sec. 6.5, you will learn about

the efforts made by astronomers to investigate the existence of extra-solar planets.

Objectives

After studying this unit, you should be able to:

• describe the planets of the solar system;

• understand how the terrestrial planets differ from the jovian planets;

• explain the nebular model of the solar system;

• discuss the role of gravitational forces in generating tides and in the formation of

Earth’s tidal bulge and planetary rings; and

• discuss the possible existence of extra-solar planets.

28

The Solar System

and Stars 6.2 SOLAR SYSTEM: FACTS AND FIGURES

The solar system consists of the Sun, nine planets (see the margin remarks), satellites

of planets, asteroids and comets. The nine planets, arranged according to their

increasing distances from the Sun, are: Mercury (Buddha), Venus (Shukra), Earth

(Prithvi), Mars (Mangal), Jupiter (Brihaspati), Saturn (Shani), Uranus (Arun),

Neptune (Varun) and Pluto (Yama). The sizes of these planets with respect to the

Earth are shown in Fig. (6.1).

Note in Fig. 6.1a that the sizes of the first four planets are similar to that of the Earth

and they are called terrestrial planets. On the other hand, from Fig. 6.1b, it is

obvious that the sizes of the next four planets are bigger than the Earth. They are

called jovian planets. The status of the ninth planet, Pluto, is somewhere in-between.

Fig.6.1: Sizes of a) terrestrial; and b) jovian planets relative to the Earth

The properties of the terrestrial and jovian planets are different. For example,

terrestrial planets are mainly made of rocks and metals having an average density of

4 or 5 g cm−3

whereas jovian planets consist mainly of gas and ice with an average

density of 1 or 2 g cm−3

.

All these planets revolve around the Sun in elliptical orbits. The planetary orbits are

almost in the same plane except that of Pluto which is inclined at an angle of ~ 17° to

the common plane (Fig.6.2). Asteroids are believed to be the captured objects which

were wandering in the solar system. Their orbits are mostly located in between the

orbits of Mars and Jupiter. Since all the planets revolve round the Sun, it is considered

the ‘head’ of the solar family. In addition, the Sun contains almost 99.87% of the total

mass of the solar system. Among the planets, Jupiter and Saturn are the most massive,

accounting for 92% of the mass of all the planets.

Uranus

Earth

Jupiter

Neptune Saturn

Mercury

Relative sizes of the jovian planets

(Earth = 1)

Relative sizes of the terrestrial planets

(Earth = 1)

Venus

Mars

Earth

Moon

R = 4.0

R = 11.3

R = 1.0

R = 0.38

R = 0.95 R = 1.0

R = 0.53

R = 0.26

R = 3.9

R = 9.4

Pluto

R = 0.18

(a) (b)

In a recent development,

the International

Astronomical Union

(IAU) has decided to

remove Pluto from the list

of planets in the solar

system. So, the solar

system now has only

eight planets. Pluto has

now been categorised as

an object of the Kuiper

belt, found in the outer

region of the solar

system. Kuiper belt

contains many objects of

the size of Pluto. In fact,

it was the realisation that

many more objects of the

size of Pluto could be

discovered in future that

prompted IAU not to

consider Pluto as planet.

29

The Solar Family

Fig.6.2: Orbits of the a) terrestrial; and b) jovian planets in the solar system

Table 6.1 summarizes some basic data about these planets and the Moon. Many of

these features could directly be attributed to the distance of a planet from the Sun.

Table 6.1: Some basic data of planets and the Moon

Planets Mass [kg] Density

[g cm−−−−3]

Mean

distance

from the

Sun (km)

Rotation Period

(hours or days)

Revolution

Period (days)

Inclination of

orbit to the

plane of the

Solar System

(deg)

Percentage of

light reflected

Mercury 3.30×1023 5.4 5.8×10

7 58 d 88 7.0 7

Venus 4.87×1024

5.2 1.1×108 243 d 245 3.4 76

Earth 5.97×1024

5.5 1.5×108 23 h 56 min 365.25 0.0 39

Mars 6.42×1023

3.9 2.3×108 24 h 37 min 687 1.8 18

Jupiter 1.90×1027

1.3 7.7×108 9 h 55 min 4333 1.3 51

Saturn 5.69×1026

0.7 1.4×109 10 h 30 min 10743 2.5 50

Uranus 8.70×1025

1.1 2.8×109 17 h 14 min 30700 0.8 66

Neptune 1.03×1026

1.7 4.5×109 18 h 60280 1.8 62

Pluto 1.00×1022

2.1 5.9×109 6 d 09 h 17 min 90130 17.1 15

Moon 7.35×1022

3.3 3.8×105* 27.33 d 27.33 5.0

* This is Moon’s mean distance from the Earth.

SAQ 1

Using the data given in Table 6.1, verify Kepler’s third law:

.r

TConst

3

2

=

where T is the orbital period of a planet and r is its mean distance from the Sun.

Further, as mentioned above, the division of planets into two groups namely,

terrestrial and jovian, is based on the similarity of some major characteristics of the

planets in a particular group. Table 6.2 lists these characteristics of terrestrial and

Spend

5 min.

Mars

Earth Mercury

Sun Venus

Pluto

Uranus

Saturn

Jupiter

Neptune

(a)

(b)

30

The Solar System

and Stars jovian planets. Such classification helps in developing theoretical models for their

formation and evolution.

Table 6.2: Characteristics of the terrestrial and jovian planets

Characteristics Terrestrial Jovian

Basic form Rocky Gas/Liquid/Ice

Mean orbital distance (AU) 0.39−1.52 5.2−30.1

Mean surface temperature (K) 200−700 75−170

Mass (relative to the Earth) 0.055−1.0 14.5−318

Equatorial radius (relative to the Earth) 0.38−1.0 3.88−11.2

Mean density (g cm−3) 3.96−5.42 0.68−2.30

Sidereal rotation period (equator) 23.9 h − 243 h 9.8 h − 19.2 h

Number of known Moons 0−2 8−20

Ring systems No Yes

We present some salient features of the individual planets in Table 6.3.

Table 6.3a: Salient features of the terrestrial planets

Mercury Venus Earth Mars

• It can be seen near

the horizon at

sunset or sunrise

with unaided eyes.

• Like the Moon, it

has no atmosphere

and its surface is

full of craters.

• It is, at the same

time, the coldest

and the hottest

planet because its

periods of

revolution and

rotation are almost

equal which keeps

its same surface

face the Sun all

the time.

• Its surface

temperature varies

between +

340°C to −

270°C.

• It appears as the third

brightest object in the

sky after the Sun and

the Moon, as it is

nearest to the Earth.

• Its surface is dry, hot

and volcanic. Its

atmosphere contains

about 96 percent carbon

dioxide, 3.5 percent

nitrogen and remaining

half percent is water

vapours, argon,

sulphuric acid,

hydrochloric acid etc.

• The planet is covered

by a thick cloud mainly

consisting of sulphuric

acid droplets.

• Its surface temperature

is very high (~ 470°C)

which is perhaps

caused due to the

greenhouse effect: the

infrared radiation

emitted by the planet is

not allowed to escape

due to the presence of

carbon dioxide in its

atmosphere thus

causing the heat

received from the Sun

to be trapped and raise

its temperature.

• Its crust, extending to

10 km deep under the

oceans and up to 40

km under the

continents, consists

mainly of silicon

(27.7%), and oxygen

(47.3%). Elements

like aluminium, iron,

calcium, sodium etc.

make the bulk of its

matter and less than

2% is made of all

other elements.

• Its rotation axis is

tipped by ~ 23°

causing various

seasons and polar

caps.

• Its atmosphere

consists of distinct

layers called

troposphere,

stratosphere and

ionosphere.

Troposphere

comprises mainly of

78% nitrogen, 21%

oxygen; stratosphere

contains ozone which

absorbs the harmful

ultraviolet radiation

from the Sun.

• Though half in size

of the Earth, this

planet has various

similarities with the

Earth

i) A day on Mars is

24h 40 minutes

and a year lasts for

1.88 Earth year,

ii) its rotation axis is

tipped at 25° and

thus has seasons

and polar caps.

• This is the most

extensively probed

planet and a few

automated

laboratories have

also been landed.

The atmosphere and

geology of this

planet has many

features similar to

the Earth.

• Martian surface has

craters of all sizes

and enormous

volcanoes.

• Martian soil, like the

Earth, is mostly

made of silicates.

However, due to the

presence of 16

percent iron oxide in

its soil, it has the

characteristic red

colour. It is also

known as the Red

Planet.

31

The Solar Family Table 6.3b: Salient features of the Jovian planets

Jupiter Saturn Uranus Neptune Pluto

• It is the largest

(having diameter

11.2 times that

of the Earth) and

the most massive

planet (contains

71% of all the

planetary mass).

• It is like a

spinning ball of

gas and liquids

with no solid

surface. In this

regard, it is

similar to the

Sun. It has a

large number of

satellites.

• It is covered by a

turbulent,

gaseous

atmosphere

comprising of

hydrogen,

helium and small

traces of water

vapour,

ammonia,

methane etc.

• It has a Great

Red Spot on its

surface which

has an oval

shape. It is big

enough to

accommodate

two Earths! It is

presumed to be

due to huge

cyclonic

disturbance in its

atmosphere.

• In view of its

composition,

size and the

number of

Moons, Jupiter

looks like a star

having its own

‘solar system’!

• It has rings and

it emits radio

waves.

• It is the second

largest planet

exceeded in

size and mass

only by Jupiter.

Like Jupiter, it

consists mainly

of hydrogen

and helium.

• It is the last

(sixth) planet

visible from

the Earth. It

has beautiful

rings (which

can be seen

through a

telescope). The

rings are, in

fact, a

thousand

tightly packed

individual

ringlets.

• The

temperature at

its cloud tops is

− 180°C. It is

colder than

Jupiter.

• This planet is

less dense than

water and

being mostly

liquid and

rotating

rapidly, its

shape is

flattened.

• It has a large

number of

satellites

orbiting at the

edge of the

rings. These

satellites

(Moons) are

composed of

rock and ice

and have

craters.

• Its main

satellite, Titan,

is very large

(diameter

5800 km) and

has atmosphere

of its own as

dense as ours.

• This planet is

smaller than

Jupiter and four

times farther

from the Sun. In

a telescope, it

appears as a

green disc with

vague markings.

• Its axis of

rotation is tipped

97.9° from the

perpendicular to

its orbit. This

causes its poles

to nearly point

towards the Sun

and it would

seem that the

planet is rolling

along the orbit

like a wheel.

• The rings of this

planet were

discovered as

late as 1977

(Voyager 2) and

they comprise of

very dark

material, as black

as coal.

• It has a large

number of

Moons.

• This planet was

discovered in

1846 and it is so

far away that,

seen from this

planet, the Sun

would look like

a bright spot!

• Its colour is

faint blue which

is caused due to

larger

percentage of

methane present

in it.

• Its cloud

temperature is

about − 237°C.

• Like other

jovian planets,

this planet also

has rings.

• Two Moons of

this planet are

visible from the

Earth.

• Triton is the

largest Moon of

this planet

which is

orbiting it in the

clockwise

direction i.e.

opposite to the

rotation of the

planet. Triton

also has

atmosphere of

its own

comprising of

nitrogen and

methane.

• This planet was

predicted to exist

theoretically to

account for the

observed

irregularities in

the orbits of

Uranus and

Neptune.

• It is a very small

(about one-fifth

the size of the

Earth), cold and

dark planet.

• Unlike most

planetary orbits,

Pluto’s orbit is

quite elliptical

and therefore it

can come closer

to the Sun than

Neptune.

• Being so far away

from the Sun, this

planet is cold

enough to freeze

most compounds.

• Its mass is only

0.002 times the

Earth mass!

32

The Solar System

and Stars At this stage, you may pause for a moment and think about what have you learnt till

now. You have basically recapitulated what you learnt in school about the solar

system and have acquainted yourself with the characteristic features of each of its

planets. In the next section, you will learn about the origin of the solar system. A

model for the origin of the solar system must explain its characteristic features listed

below:

i) Most of the mass of the solar system is contained in the Sun.

ii) Except for Mercury and Pluto, the orbital planes of all the planets are more or less

in the same plane.

iii) When viewed from above, the planets are found to revolve around the Sun in the

anticlockwise direction; the direction of rotation of the Sun is also the same.

iv) Except for Venus, Uranus and Pluto, the direction of rotation of planets is the

same as their direction of revolution.

v) The direction of revolution of the satellites of each planet is the same as the

direction of rotation of the planet itself.

vi) Total angular momentum of all the planets is more than the angular momentum of

the Sun.

vii) Terrestrial planets comprise mainly of rocky material whereas jovian planets

comprise mainly of gaseous material.

SAQ 2

List some common features of terrestrial and jovian planets.

6.3 ORIGIN OF THE SOLAR SYSTEM: THE NEBULAR

MODEL

The formation of the Earth or, in fact, the entire solar system has been of considerable

interest to human being for ages. A variety of speculative ideas were proposed which

gave rise to two kinds of theories: catastrophic and gradualistic. According to the

catastrophic theories, the planets were formed out of the material ejected from the Sun

when a giant comet collided with it or the planets came into being due to ripping-off

of material from the Sun caused by the tides generated by a close by passing star.

However, these theories do not explain many features of the solar system. Further,

during the 20th century, astronomical evidences have been found which support the

gradualistic theories. According to these theories, formation of planets is a gradual

process and is a natural by-product of the formation of stars like the Sun.

Contemporary gradualistic theories of the solar system are based on the nebular

hypothesis. According to this hypothesis, the Sun as well as the planets formed from

an interstellar cloud of gas and dust. A model of the formation of the solar system

based on this hypothesis is called a nebular model. The basic principle of physics

underlying the nebular model is the Newton’s law of gravitation. According to the

nebular model, formation of stars including the Sun, begins when the interstellar

cloud with enough mass and low internal pressure, contracts due to its own gravity.

Can you guess the consequences of gravitational contraction of an interstellar

cloud? You are on the right track if you think that it leads to increase in density of the

cloud. Simultaneously, the kinetic energy of the particles in the cloud increases which

Spend

3 min.

33

The Solar Family causes increase in its temperature. As a result, the internal pressure of the collapsing

cloud increases. Eventually, the gravitational contraction is balanced by the internal

pressure of the cloud. The contracting cloud is called a protostar and when its

internal temperature is high enough to initiate thermonuclear reactions, a full-fledged

star (such as the Sun) is born. This is the general picture of the formation of a star

according to the nebular model.

Well, you may ask: How did our solar system come into existence? According to

the nebular model, our solar system formed due to gravitational contraction of the

rotating interstellar cloud called the solar nebula. Due to rotation, the solar nebula

takes the form of a disc (Fig. 6.3). When the Sun formed at the centre of this disc and

became luminous enough, the remaining gas and dust was pushed away due to the

Sun’s radiation pressure. The blown away gas and dust condensed into various planets

orbiting the Sun. One of the natural consequences of the solar nebula model is that

most stars in the galaxies should have planetary systems!

Fig.6.3: Schematic diagram showing different stages (a to f) of the formation of the solar system

Formation of Solar Nebular Disk

You must have noted that one of the significant features of the solar nebular model is

the formation of a disk of interstellar cloud around the Sun. You may ask: Is there

any evidence supporting the formation of the disk? Two facts support this

assumption. Firstly, as you know, the orbits of planets, except that of Mercury and

Pluto, lie in nearly the same plane. Such coplanar planetary orbits are consistent with

the formation of nebular disk. Secondly, the rotation of the Sun and revolution of the

planets is along the same direction and their equatorial planes are very close to the

plane of the solar system. Thus, the motion of the Sun and the planets are consistent

with the disk hypothesis.

Your next logical question could be: Can we explain the formation of nebular disk

on the basis of physical principles? It can indeed be done provided we assume that

the interstellar cloud, giving rise to the solar system, had some initial rotational

motion. Having assumed this, we can invoke the principle of conservation of angular

momentum to explain the formation of a disk. As the cloud contracted due to gravity,

each gas and dust particle would come closer to the axis of rotation. To conserve the

angular momentum during the contraction, the particles coming nearer to the axis of

rotation must revolve faster. At some point, due to the increased speed of revolution,

(a)

(b)

(c)

(e)

(d)

(f) Pluto

Earth

Mercury Uranus

Saturn

Venus Jupiter

Mars

Neptune Asteroids

34

The Solar System

and Stars the centripetal acceleration of the particles of the cloud balance the gravitational

contraction and equilibrium is attained.

To understand the flattening of the rotating cloud, refer to Fig. 6.4. Particles at points

A (near the pole) and B (near the equator) have the same angular momentum because

points A and B are equidistant from the axis of rotation. Thus, it is energetically more

favourable that particle at A falls under gravity along the line AB and reaches the point

B (closer to the centre) than say a particle at point C to come to point B. It is so

because, in the first situation, no change in angular momentum is involved. Thus,

rotating cloud near the poles contracts more than those near the equator giving rise to

the formation of disk.

(a) (b) (c)

Fig.6.4: a) Interstellar cloud with initial rotational motion; b) due to gravitational attraction,

material at the poles contracts along the rotation axis; and c) the final disk shaped solar

nebula

The principle of the conservation of angular momentum also explains why most of the

planets rotate along the same direction. It is because each planet has retained some of

the angular momentum of the solar nebula. Let us discuss the angular momentum of

the solar system to appreciate this point better.

Angular Momentum of the Solar System

The present distribution of angular momentum in the solar system seems inconsistent

with the basics of the nebular model. The argument goes like this. Since the Sun and

the planets are formed from the same spinning interstellar cloud, the angular

momentum per unit mass of each of them must be same. The facts are otherwise: the

Sun possesses approximately 99.9% of the total mass of the solar system but only 1%

of its total angular momentum; orbital angular momentum of Jupiter exceeds the

rotational angular momentum of the Sun by a factor of 20; among the planets having

99% of the total angular momentum of the solar system, Jupiter possesses the most of

it! To appreciate these facts and figures, you should solve an SAQ.

SAQ 3

Calculate the total angular momentum of the Sun-Jupiter system assuming that Jupiter

has a circular orbit of radius 5.2 AU, and its orbital period is 11.86 yr. Assume that the

Sun interacts only with Jupiter.

You may ask: How does the nebular model explain the inconsistency in the

distribution of angular momentum? It is proposed that during the formation of the

solar system, the angular momentum is transported from the central part of the nebula

Spend

5 min.

35

The Solar Family to the outer regions. Two possible mechanisms for such a transfer have been

suggested. According to the first mechanism, the interaction of the charged particles

and the magnetic field of the evolving Sun causes transfer of angular momentum. You

know that charged particles spiral along the magnetic field. Thus, the charged

particles created by ionisation of solar nebula by the Sun are dragged along by the

magnetic field of the rotating Sun (Fig. 6.5). The magnetic field links the outer

nebular matter (charged particles) with Sun’s rotation. As a result, the nebular

material in the outer region gains angular momentum at the expense of the Sun. The

other mechanism suggested for transfer of angular momentum is based on viscosity.

You know that due to viscosity, motion of fluid in one part is affected by the motion

of the fluid in the adjoining part. Thus, it is quite possible that the slow moving

particles located at outer edges of the nebula gain in velocity due to their interaction

with the fast moving particles in the smaller orbits of the nebula and vice-versa. This

may cause transfer of angular momentum from the Sun to the outer planets of the

solar system.

Now, you will learn about formation and evolution of planets according to the nebular

model.

Formation of Planets

You know that the solar nebula consisted of gas and dust. As the solar nebula

contracted, it became hot enough and most of the dust particles evaporated. So, the

solar nebula consisted mainly of gaseous matter. The question is: How did the

planets form from the solar nebula? The formation of planets is a two-stage

process: firstly, small solid particles are formed from the gaseous matter and then

these particles stick together and grow into planets.

Depending upon the temperature of the nebular region, the nebular gas condensed into

solid matter of different types. In the inner region, the temperature was very high and

materials with very high melting points were formed. The sequence of condensation

of gas into different types of materials, from the centre of the nebula (the Sun) to its

periphery, is called the condensation sequence and is given in Table 6.4.

Table 6.4: The condensation sequence

Temperature

(K)

Material(s) formed Planet

(Temperature of formation)

1500 Metal oxides Mercury (1400 K)

1300 Metallic Iron and

Nickel

1200 Silicates

1000 Feldspars Venus (900 K)

680 Troilites (FeS) Earth (600 K)

Mars (450 K)

175 H2O, Ice Jovian (175 K)

65 Argon-neon Ice Pluto (65 K)

From Table 6.4, it is evident that terrestrial planets formed from high density materials

and jovian planets formed from low density materials.

Fig.6.5: Magnetic lines of force

of the rotating Sun

With the help of computers, it

is possible to study the

behaviour of a large number of

particles when they interact in

the manner as in the process of

accretion. Such computer

experiments are called

simulation. The results of such

simulation studies suggest that

there could be as many as 100

planetesimals of the size of the

Moon along with a large

number of smaller objects in

the region of inner planets. In

addition, there could be about

10 bodies with masses

comparable with the planet

Mercury and many other

objects of the size of the planet

Mars.

36

The Solar System

and Stars Having discussed the general chemical composition of planets, we now focus on the

evolution of planets. The evolution of planets involves three stages:

a) The first stage involves the growth of macroscopic grains of solid matter from the

interstellar cloud. The size of these grains range from a few cm to a few km and

they are called planetesimals. Planetesimals can grow through two processes:

condensation and accretion. In the condensation process, grains grow by adding

one atom at a time to a ‘nucleus’ atom, from the surrounding gaseous cloud. This

is similar to the growth of snowflakes in the Earth’s atmosphere. In the accretion

process, solid particles stick together. Further, the planetesimals would tend to

rotate in the plane of the solar nebula.

b) In the second stage, planetesimals coalesce and form protoplanets − objects

having planetary sizes and masses. You may ask: How do the protoplanets

form? Since all the planetesimals are moving along the same direction in the

nebula, they collide with each other at a low relative velocity and stick together to

form protoplanets. Further, growth of protoplanets is helped by gravity because

the nebular matter is attracted by the protoplanets.

c) At the third stage, when a protoplanet grows into a stable planet, a large amount

of heat is generated in its core due to the decay of short-lived radioactive

elements. Heat is also generated due to collision of these planets with other

objects. Due to high temperature, the planets melt and facilitate the process of

gravitational separation in which materials in the planet segregate themselves

according to their density. Therefore, the inner regions of the planets hold heavier

elements and compounds and lighter elements are pushed to the surface.

This, in the nutshell, is the ‘story’ of planet building! You must have noted that

there are many unanswered questions and many ifs and buts in the mechanism

described above. To understand the details of formation of planets is still an active

area of research in astronomy and astrophysics. We will, however, not go into

those details and close our discussion with a few words about the success of the

nebular model.

The solar nebular model successfully accounts for the following three important

features of the solar system:

i) Disk shape of the solar system: The model suggests that the rotating solar

nebula ultimately evolves into disk shape due to gravitational contraction and

conservation of angular momentum.

ii) The orbits of most of the planets are coplanar: As per the model, this situation

exists because of the disk shape of the solar nebula from which planets are

formed.

iii) The direction of rotation of the Sun and the directions of revolution of most of

the planets are the same: It is so because the Sun and the planets formed from

the same rotating nebula.

You have also learnt that the condensation sequence of the solar nebula explains why

terrestrial planets comprise of compounds having high melting point and jovian

planets comprise of ices and gases.

The nebular model has some very obvious limitations. You know that the Moon’s

surface is not smooth; it has craters of varied sizes, small and large. Also, its surface

composition is extremely poor in hydrogen, helium etc. These observations suggest

the continued collisions of bigger planetesimals with the Moon’s surface even after its

formation. It is quite likely that other objects (planets) in the solar system experienced

similar collisions with planetesimals. Thus, the theory of solar system formation must

account for massive encounters endured by planets in the early stages of their

The creation of a massive

object, like Jupiter,

influences the orbits of

nearby planetesimals. Most

of the objects present in the

asteroid belt today had their

orbits changed gradually into

more eccentric orbits till they

were sucked in by Jupiter. It

is also likely that they either

left the solar system or

crashed into the Sun. This

process might in fact have

resulted in the smaller mass

left in the asteroid belt and

also a smaller planet Mars.

As we go further from Jupiter

and beyond in the nebula, the

material density becomes

very low. So, the accretion

process for the formation of a

planet-like object takes much

longer time. Saturn perhaps

took twice as long to form as

Jupiter, while the planet

Uranus took an even longer

time. The planet Neptune is

estimated to have taken twice

the time it took to form

Uranus.

37

The Solar Family formation. These massive collisions or encounters are possibly responsible for

different orientation of the spin axis of the planets in the solar system. It is now known

that Venus, Uranus and Pluto have retrograde rotations.

The cause for tipping of rotation axis, retrograde rotation etc. of planets cannot be

understood as such by gradualistic models like the nebular model. We need to

consider catastrophic events. During the formation of the solar system, the wandering

planetesimals did perhaps collide with other planets such as Mercury, Venus, Earth

etc. The collision with Mercury ripped off its low density mantle while such a

collision with Venus flipped its rotation axis. A collision with Earth led to the

formation of the Earth-Moon system. Further, massive planetesimals crashed into

Mars as well as other outer planets changing orientation of their rotation axes. It is

also possible that some of these planetesimals were captured by planets as their

Moons.

Now, before you proceed further, how about answering an SAQ?

SAQ 4

a) List two evidences supporting the assumption that a disk shaped solar nebula

existed during the evolution of the solar system.

b) Why do terrestrial planets comprise mainly of materials having high melting

points?

c) What is the difference between the condensation and accretion processes?

6.4 TIDAL FORCES AND PLANETARY RINGS

You must be aware of tides occurring on the Earth’s surface. In fact, those of you who

live in the coastal areas must be quite familiar with tides. Two high tides occur once in

every ~ 24 hours depending on the local features of the coastal area and its latitude.

However, generally people are not familiar with the tidal bulge of the Earth along the

equator which measures around 10 cm. Do you know what causes tides or tidal

bulge of the Earth? It is caused due to the difference between gravitational force of

the Moon at different locations, say points A and B, on the Earth (Fig. 6.6). Due to

similar reasons, tidal bulge (as large as 20 m) caused by the Earth has also been

observed on the Moon. The larger value of the bulge of the Moon is because the Earth

is more massive than the Moon.

Tidal effects play an important role in astronomy, particularly in understanding the

creation of large number of satellites and ring systems in the jovian planets. We now

derive an expression for the tidal force for the Earth-Moon system. Then we shall use

this concept for a qualitative explanation of the formation of planetary rings.

Fig.6.6: Tidal bulge of the Earth is due to the difference in gravitational force of the Moon experienced by mass elements at, say points A and B, on the Earth.

Spend

7 min.

Moon Earth

A B

38

The Solar System

and Stars Let a mass m on the Earth be located at a distance r from the centre of mass of the

Moon (Fig. 6.7). If the mass of the Moon be M, the magnitude of the gravitational

force acting on m, in the direction shown, is given by:

,2

r

MmGF = (6.1)

where G is the gravitational constant.

Fig.6.7: Gravitational force on an element of mass m located on the Earth at a distance r from the

centre of mass of the Moon

Now consider a similar mass located at a distance dr from the earlier mass element

along the same line. The difference between the gravitational forces experienced by

these two mass elements can be obtained by differentiating F in Eq. (6.1):

drr

MmGdF

32−= (6.2)

Note that the difference in the gravitational forces on the two equal masses separated

by a distance dr will cause them to move with different accelerations. The differential

force given by Eq. (6.2) is called the tidal force. The r3 term in the denominator of

Eq. (6.2) clearly shows that the tidal force has a stronger dependence on distance

compared to the gravitational force (which varies as 1/r2). Further, in case of the

Earth-Moon system, the tidal force becomes more pronounced if the mass element is

closer to the Moon.

Now to appreciate the effect of the tidal force of the Moon on mass elements located

at different points on the Earth, e.g., at the equator and at the poles, let us obtain a

general expression for the tidal force considering the Earth as a two-dimensional

object. Let us consider an element of mass m located at the centre (C) of the Earth and

another element of mass m located at an arbitrary point (P) at latitude φ on the surface

of the Earth (Fig. 6.8). By assuming that the Moon lies along the x-direction, the

components of gravitational force at points C and P can be written as:

,2,

r

MmGF xC =

,0, =yCF

and

,cos2, θ=

s

MmGF xP

.sin2, θ−=

s

MmGF yP (6.3)

Earth

m

Moon

M m

r dr

39

The Solar Family where s is the distance between points P and the centre of mass of the Moon.

Fig.6.8: The schematic diagram of the tidal force on an arbitrary point on the Earth due to Moon

Let the unit vectors along the x- and y-directions be i and j , respectively. Thus, from

Eq. (6.3), the difference between the magnitudes of gravitational forces at the points P

and C can be written as:

∆F = FP − FC

ji ˆsinˆ1cos222

θ−

−

θ=

sGMm

rsGMm (6.4)

Further, s can be expressed in terms of r, R and φ as:

φ+φ−= 2222 sin)cos( RRrs

φ−≈ cos

212

r

Rr

Substituting for s2 in Eq. (6.4) using the small angle approximations, cosθ ≈ 1 and

sinθ ≈ ,sin φr

R and using binomial expansion we obtain:

)ˆsinˆcos2(3

jiF φ−φ≈∆r

GMmR (6.5)

SAQ 5

Derive Eq. (6.5).

You may ask: What does Eq. (6.5) signify physically? This expression clearly

indicates that the magnitude of the tidal force is dependent on the latitude. For mass

element located at the equator of the Earth, that is, for φ = 0, the magnitude of tidal

force is maximum. And, at the poles, where φ = π/2, the magnitude of the tidal force is

minimum. This is the cause of tidal bulge around the equator and also causes tides in

the oceans.

Spend

6 min.

Earth

Moon

C

P

φ

R

x

y

s

r

40

The Solar System

and Stars In the above discussion, we have considered the Earth-Moon system to be an isolated

one. In fact, we must also consider the tidal forces due to the Sun. When the Sun,

Earth and the Moon are all aligned in a straight line, the differential forces discussed

above add up to produce large tides on the Earth. These tides are called spring tides.

When the Sun, Earth and the Moon form a right angle, the differential forces due to

the Sun and the Moon are directed opposite to each other and the tides on the Earth

(known as neap tides) are very small.

Yet another effect of tidal force is that the Earth’s speed of rotation is slowed down.

This results in longer days at present compared to many years ago. Further, just as the

Moon causes tides on the Earth, the Earth also gives rise to tides on the Moon. You

know that on the Earth we see the same side of the Moon which implies that its

rotation and revolution periods are the same. It is quite likely that earlier the Moon’s

rotation period was shorter than its orbital period. As time progressed, because of the

tidal friction, the rotation period of the Moon increased and has become equal to its

orbital period (also called one-to-one synchronous rotation). We find such

synchronous motion to be quite a common phenomenon in the solar system. For

instance, the two Moons of Mars and four Moons of Jupiter and majority of Saturn’s

Moons are in synchronous rotation. Many moons of the outer planets behave in a

similar manner.

You know that most outer planets of the solar system have rings around them

comprising of small (~ 10 µm to 10−5

m) particles. The concept of tidal forces can be

invoked to understand these ring systems. There are two possible scenarios. Since the

jovian planets are massive, tidal gravitational force of the planets on their satellites

must be very strong. Thus, if a satellite comes very close to, say, Jupiter, it

experiences a strong tidal force and breaks up into pieces. The particles formed during

such a process revolve around the planet in ring formation. According to the second

scenario, during the formation of jovian planets, the tidal forces restrained the

particles from condensing into a satellite.

You may argue: The tidal forces must ultimately disrupt the ring system; what

maintains the ring system? It is suggested that the Moons associated with these

planets play an important role in preserving the rings. The particles in the rings are

restrained from moving out of their orbit due to the combined gravitational forces of

these Moons.

Human beings have always wondered about the existence of planets around other

stars. Only in recent times there has been a confirmation that planets do exist outside

the solar system. These planets are called extra-solar planets. You will learn about

them now.

6.5 EXTRA SOLAR PLANETS

On August 4, 1997, the Hubble Space telescope took the first image of an extra-solar

planet around another star (Fig. 6.9). The picture shows a double star located at about

450 light years away towards the constellation of Taurus. It was the first direct look at

a planet outside our solar system. It is now believed that this planet is a “runaway”

object, thrown out of the binary system, as indicated by the filament tracing. The

planet is located at about 1500 times the Earth-Sun distance from its parent. The

binary system is believed to be 300,000 years old and the planet seems to be quite

similar to Jupiter with a mass of around 2-3 times that of Jupiter. This discovery led to

careful investigation by several groups of astrophysicists and by now more than 100

such extra solar planets have been found.

Young planets in the new systems are difficult to detect as the parent stars

overshadow their feeble glow. Since it is difficult to detect a planet orbiting a distant

Fig.6.9: Photograph of an

extra solar planet

taken by Hubble

telescope

41

The Solar Family star, we look for alternative methods of detection, such as the influence of the planets

on their parent stars. You may like to know: What are these influences? It is easy to

think of gravitational influence. As the planet orbits the star, it will tug at it from

different sides causing it to wobble back and forth. The gravitational influence gives

rise to the following methods for detection of the planet:

a) Astrometric Detection

b) Radial Velocity Detection

The first method is based on measuring the position of a star relative to its background

stars. If the star is accompanied by an orbiting planet, it gets a tug from the planet and

its position changes a little. This change in the position of the star is measured which

would show a periodic change (back and forth) indicating the presence of an orbiting

object.

The second method is based on Doppler effect. As an orbiting planet tugs on to its

companion star, the light from the star experiences a Doppler shift. If the planet pulls

the star slightly away from us, the light emitted by star would shift towards red end of

the electromagnetic spectrum while if it tends to pull the star towards the Earth, the

light would be shifted towards the blue end of the spectrum. To measure the Doppler

shift, we choose a particular spectral line and observe its shift from red to blue and

back.

In 1995, Michel Mayor and Didier Qucloz of the Geneva Observatory observed that

the Sun-like star, 51 Pegasi is wobbling back and forth at the rate of 56 ms−1

. The only

valid explanation for this observation was the presence of a planet like object orbiting

the parent star. The mass of the planet was estimated to be half the mass of Jupiter

with a radius of about 0.05 AU. Subsequently, Mayor also detected a planet of about

0.16 times the mass of Jupiter orbiting the star HD 83443 in the constellation of Vela

about 141 light years away from the Earth.

Recently David Charbonneau, Timothy Brown and Robert Gilliland used the Hubble

spectrometer and made the first direct detection and chemical analysis of the

atmosphere of the planet HD 209458b orbiting a yellow, Sun-like star, HD 209458, in

the constellation of Pegasus. The mass of the planet is estimated to be 70% of the

mass of Jupiter. It passes in front of its star every 3.5 days and contains sodium in its

outer layers. The extremely short period of the planet suggests its very close proximity

to the star and therefore its atmosphere getting heated to around 1100 degrees Celsius.

Many new results are pouring in and are compelling astronomers to have a re-look on

the theories of solar system formation. The search for extra-solar planets is going hand

in hand with the search for life in the Universe. Exploration of the solar system has so

far not revealed any signs of even primitive life on any of the planets or their

satellites. Scientists strongly believe that there must be a large number of extra-solar

planets where conditions are suitable for the existence of life. Indeed, some of these

planets could be hosting life more advanced than our own. It is possible that some of

these intelligent beings are trying to contact us just as we are looking for them. It must

be remembered that any message from any of these beings would be coded in radio

waves. That is why SETI, an organisation searching for Extra Terrestrial Intelligence,

is looking for such signals in the radiation at radio frequencies coming from outside

the solar system. Scientists are hopeful that someday they would be able to detect

extra-solar intelligence.

Let us now summarise what you have learnt in this Unit:

42

The Solar System

and Stars

6.6 SUMMARY

• The first four planets of the solar system namely Mercury, Venus, Earth and Mars

are called terrestrial planets and they mainly comprise of solid matter. The next

five planets, namely Jupiter, Saturn, Uranus, Neptune and Pluto are called jovian

planets and they mainly comprise of ices and gases.

• One of the significant features of jovian planets is the ring system.

• Some of the characteristic properties of the solar system are:

i) Most of the mass of the solar system is contained in the Sun.

ii) Orbital planes of all the planets except Mercury and Pluto are coplanar.

iii) The direction of rotation of the Sun and direction of revolution of the planets

is the same.

iv) Total angular momentum of all the planets is more than the angular

momentum of the Sun.

• According to the nebular hypothesis of the origin of solar system, the Sun as

well as the planets formed from an interstellar cloud of gas and dust. A model

based on this hypothesis is called nebular model.

• According to the nebular model, formation of stars including the Sun takes place

when the interstellar cloud contracts due to its own gravity.

• The contracting interstellar cloud takes a disk shape due to its rotational motion.

Formation of nebular disk causes coplanar planetary orbits.

• The present distribution of angular momentum in the solar system seems

inconsistent with the basics of the nebular model.

• The angular momentum problem is addressed by proposing that there is a transfer

of angular momentum from the inner to outer regions of the solar system during

its evolution.

• The formation of planets results due to condensation of nebular gas as per the

condensation sequence.

• The formation of planets involves three stages:

i) Formation of planetesimals,

ii) Formation of protoplanets, and

iii) Stabilisation of the planet.

• Tidal forces result due to the difference in gravitational force at two points on the

Earth. For the Earth, tidal forces due to the Moon have a significant effect. Tidal

force at the equator is maximum and at the poles, its value is minimum. The

general expression for the tidal force is given by

)ˆsinˆcos2(3

jiF φ−φ≅∆r

GMmR

• In the recent past, a few extra-solar planets have been detected.

6.7 TERMINAL QUESTIONS Spend 20 min.

43

The Solar Family 1. Explain, in your own words, the theory of the solar system formation based on

nebular hypothesis.

2. Explain how angular momentum can be transferred from the Sun to the outer

planets of the solar system.

3. Is it possible that the Earth also suffered collisions with other bodies of the solar

system and its surface also had craters like those on the surface of Mercury? Can

you guess what happened to those craters?

4. Explain how extra-solar planets can be detected. Why can they not be seen

directly?

6.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. To verify Kepler’s third law, let us calculate )( 32rT for three representative

planets, say the Mercury, the Earth and the Saturn. From Table 6.1, we have

MercuryT = 88 days = 60602488 ××× s = 7.6 × 106s

and

Mercuryr = 7108.5 × km = 10108.5 × m

So,

3

2

)(

)(

Mercury

Mercury

r

T =

310

26

m)108.5(

s)106.7(

×

×

= 2319 sm109.2 −×

Similarly, using data from Table 6.1, you can easily show that:

3

2

)(

)(

Earth

Earth

r

T = 2319 sm109.2 −×

and

3

2

)(

)(

Saturn

Saturn

r

T = 2319 sm109.2 −×

Since the ratio )( 32rT for all the three planets has the same value

2319 sm109.2 −× , Kepler’s third law is true.

2. See text.

3. For two bodies of mass M1 and M2 moving around their centre of mass, the

reduced mass is given by:

21

21

MM

MM

+=µ

So, for the Sun-Jupiter system, we can write:

44

The Solar System

and Stars

≅

J

J

MM

MM

+=µ

Θ

Θ

JM≈

Using Kepler’s third law, it can be readily shown that the angular momentum L

and the period T of the orbiting mass are related as:

.2

T

abL

µπ=

where a, and b respectively are the semi-major and semi-minor axes of the

elliptical orbit of the planet. As per the problem, the orbit of Jupiter is to be

considered circular. Thus, we can write:

AU2.5== ba

= 5.2 × 2.279 × 1011

m

Substituting the values of MJ = 2 × 1027

kg and T = 11.86 yr, we get:

( )s)36002436586.11(

kg102m)10279.22.5(14.32 27211

×××

××××××=L

s1044.4

mkg1063.179

251

×

×=

1242smkg109.3

−×=

4. See text.

5. Refer to Fig. 6.8. From Eq. (6.4), we have the difference between the magnitudes

of gravitational force at points P and C as:

CP FFF −=∆

ji ˆsinˆ1

s

cos

222

θ−

−

θ=

sGMm

rGMm

222 Rrs ′+=

φ+= 222 sinRr

φ+φ−= 222 sin)cos( RRr

φ+φ+φ− 22222 sincoscos2 RRRrr

22 cos2 RRrr +φ−=

φ−≈ cos

212

r

Rr (neglecting higher order terms in R)

Substituting for s

2 and making small angle approximations, 1cos ≈θ and

r

R≈θsin sin φ, we can write:

45

The Solar Family

F∆ ji ˆ

cos2

1

sinˆ1

cos2

1

1

22

2

φ−

φ

−

−

φ−

=

r

Rr

r

R

GMmr

r

Rr

GMm

φ−

φ−

−

φ−= ji ˆ

cos2

sinˆ1cos22 Rr

R

Rr

r

r

GMm

φ−

φ−

φ−

φ= ji ˆ

cos2

sinˆcos2

cos22 Rr

R

Rr

R

r

GMm

( )ji ˆsinˆcos23

φ−φ≅r

RGMm (because r>> 2 R cos φ).

Terminal Questions

1. See text.

2. See text.

3. See text.

4. See text.

46

The Solar System and Stars UNIT 7 STELLAR SPECTRA AND

CLASSIFICATION

Structure

7.1 Introduction

Objectives

7.2 Atomic Spectra Revisited

7.3 Stellar Spectra

Spectral Types and Their Temperature Dependence

Black Body Approximation

7.4 H-R Diagram

7.5 Luminosity Classification

7.6 Summary

7.7 Terminal Questions

7.8 Solutions and Answers

7.1 INTRODUCTION

In Unit 5, you have studied about the Sun and in Unit 6, you have learnt about the

solar system. You also learnt the characteristic features of the solar atmosphere and

solar activity. You now know that the Sun is the nearest and the only star in our solar

system. All the stars, including the Sun are located at very large distances from the

Earth. Thus, a logical question is: How do we obtain information about the stars? All

the information about stars is obtained by analysing their spectra and this is the

subject matter of the present Unit.

You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course

that the radiation emitted by an object at a given temperature covers a range

(spectrum) of wavelengths with a characteristic peak wavelength. The value of the

characteristic wavelength depends on temperature of the object. Therefore, by a

careful analysis of the radiation emitted by a star, we can estimate its temperature. In

addition, we can also obtain useful information regarding composition, pressure,

density, age etc. of a star on the basis of its spectrum. In Sec. 7.2, we briefly

recapitulate the atomic origin of emission and absorption spectra and explain how the

temperatures and luminosities of stars can be inferred from their spectra.

One of the earliest uses of stellar spectra was to classify stars on the basis of strength

of certain spectral lines. Later on, it was discovered that the relative strengths of

spectral lines depend basically on the star’s temperature. The spectral classification

has been discussed in Sec. 7.3. The most comprehensive classification of stars was

done on the basis of the correlation between their luminosities (an observable

parameter of stars) and temperatures. This classification gave rise to Hertzsprung-

Russell (H-R) diagram about which you will learn in Sec. 7.4. This diagram is of

utmost importance in astronomy. In Sec. 7.5, you will learn yet another classification

of stars, called luminosity classification, which tells us about the size of a star.

Objectives

After studying this unit, you should be able to:

• explain the atomic origin of emission and absorption spectra;

• discuss the correlation between the strength of spectral lines of an element and the

temperature of the region containing the elements;

• list the spectral types of stars and their characteristic features;

• discuss the salient features of different groups of stars on the H-R diagram;

• describe the need for luminosity classification of stars; and

• estimate the size of a star on the basis of its luminosity class.

47

Stellar Spectra and Classification 7.2 ATOMIC SPECTRA REVISITED

To appreciate the relation between the physical parameters of an object and the

radiation emitted by it, let us imagine heating an iron bar. First, the iron bar begins to

glow with a dull red colour. As the bar is heated further, the dull red colour changes to

bright red and then to yellowish-white. If we can prevent the bar from melting and

vaporising with further increase of temperature, it would glow with a brilliant bluish

colour. This simple experiment reveals how the intensity and colour of light emitted

by a hot object varies with temperature. Similar is the situation with stars. You must

have noticed, while looking at night sky, that all stars do not look the same in terms of

their brightness or colour: the differences are determined by their surface

temperatures.

From Unit 3, you know that spectroscopy refers to the analysis of light in terms of its

wavelength and it is extensively used to obtain information regarding temperature,

composition etc. of stars. Such analyses provide valuable information. But, the

question is: How do we analyse the light from a star? To answer this, we need to

recapitulate the atomic origin of emission and absorption of light. Recall from the

Modern Physics (PHE-11) course that:

i) Electrons in an atom exist in certain allowed energy states. Each

element has a characteristic set of energy levels (see Fig. 7.2).

(a) (b)

Fig.7.2: Energy level diagram of a) hydrogen atom; and b) helium atom

If you happen to look at Orion

nebula (Fig. 7.1), it has not only

blue stars but also some red

stars. Similarly the brightest

visible star to the naked eye −

Sirius (Vyadh) − looks whitish;

Canopus (Bramhahrudaya) is

yellowish and Aldebran

(Rohini) is reddish.

Fig.7.1: The Orion nebula

−25eV

−0.85 eV

−1.50 eV

−3.40 eV

−13.60 eV

n = 4

n = 3

n = 2

n = 1

0

−25eV

0 0

48

The Solar System and Stars

ii) When an electron makes a transition from a higher energy level to a lower energy level, radiation is emitted. An electron goes from a lower energy level

to a higher energy level when it absorbs energy and vice-versa (Fig. 7.3). Such

transitions follow certain selection rules and are always accompanied by the

emission/absorption of radiation. You know from Unit 9 of the PHE-11 course

that the wavelength (or frequency) of the emitted/absorbed radiation is determined by the difference in the energies of the two atomic energy levels:

∆ E = E2 − E1 = hv (7.1)

where E1 and E2 are the energies of the levels involved in transition, h is Planck’s

constant , v is the frequency of the emitted/absorbed radiation and ∆ E is the

energy of the corresponding photon.

(a)

(b)

Fig.7.3: The atomic electron a) emits radiation when it makes a transition from a higher energy level to a lower energy level giving rise to emission spectrum; and b) makes a transition to a higher energy level on absorbing radiation giving rise to absorption spectrum

To understand the origin of atomic spectra, refer once again to Fig. 7.3 which shows

the energy level diagram and transition of electron between these levels for a

hydrogen atom. In Fig. 7.3(a), a hydrogen atom makes a transition from the 2nd

energy level to the 1st, giving off light with an energy equal to the difference of

energy between levels 2 and 1. This energy corresponds to a specific colour or

wavelength of light -- and thus we see a bright line at that exact wavelength! This is

an emission spectrum.

On the other hand, what would happen if we tried to reverse this process? That is,

what would happen if a photon of the same energy was incident on a hydrogen atom

in the ground state? The atom could absorb this photon and make a transition from the

ground state to a higher energy level. This process gives rise to a dark absorption line in the spectrum as shown in the figure. This is an absorption spectrum.

Bright emission line Level 1

Level 2

Level 3

Level 4 etc…

Energy

Dark absorption line Level 1

Level 2

Level 3

Level 4 etc…

Energy

E4

E3

E2

E1

E4

E3

E2

E1

49

Stellar Spectra and Classification

If light from a star with a continuous spectrum is incident upon the atoms in its

surrounding atmosphere, the wavelengths corresponding to possible energy transitions

within the atoms are absorbed. Thus, an observer will see an absorption spectrum.

(a)

(b)

Fig.7.4: Emission of radiation by hydrogen atoms corresponding to Lyman and Balmer series: a) Transitions to level n=1 gives Lyman series; and b) Transitions to n=2 level gives Balmer series

On the basis of above, we can understand the genesis of atomic spectra. Let us take

the simplest example of the spectrum of hydrogen atom. In Fig. 7.4 we show the

various spectral series for the hydrogen atom. The transition of electrons from higher

energy levels to the lowest (n = 1) energy level gives rise to emission of a series of

characteristic wavelengths known as Lyman series. Similarly, transition of electrons

from higher energy levels to the first excited (n = 2) energy level results into the

emission of another series of characteristic wavelengths called the Balmer series.

You may note that some of the spectral lines in Balmer series fall in the visible region

of the electromagnetic spectrum (Fig. 7.4b). This makes them very useful spectral

lines for spectroscopic analysis. Further, transitions to the second excited (n = 3)

energy level give rise to the Paschen series.

Emission Spectrum

Now, let us consider the nature of emission of radiation by hydrogen gas (a collection

of many hydrogen atoms) kept at a high temperature. Each atom in the gas can

n = 1

λλλλ (Å) 4000 5000 7000 6000

Balmer series

n = 4 n = 3

n = 2

n = 1

n = 2

n = 3 n = 4

n = ∞ 0 eV

−13.6eV

−3.39eV

−1.51eV −0.85eV

−0.54eV

972 950 937 1025 1215

λλλλ (Å)

Lyman series

50

The Solar System and Stars

absorb thermal energy as well as the energy due to collisions with other atoms. As a

result, each atom would absorb a different amount of energy. Therefore, the electrons

in the lowest energy level of various atoms get excited to different higher energy

levels (Fig. 7.5). The question is: What kind of emission spectrum would we obtain in such a situation?

Fig.7.5: At high temperatures, electrons in hydrogen atoms get excited to various higher energy levels; transitions to levels 2 and 3 are shown here as examples

To answer this, let us consider a hydrogen atom in this hot gas whose electron has

been excited to the second excited energy level (Fig. 7.6). This electron can return to

the lowest energy level either directly (Fig. 7.6a) or in steps. For example, it can go to

the first excited energy level and then to the lowest level (Fig.7.6b). In the two cases,

the energies of the emitted photons would be different. Thus, in a gas of hydrogen

atoms at high temperature, electrons can be excited to many possible levels and can

make transitions to any of the lower levels emitting radiations of the corresponding

frequencies. This implies that photons of different frequencies will be emitted and we

will observe more than one emission line in the emission spectrum of a hot gas.

(a) (b) Fig.7.6: Different ways of transition of electron in a hydrogen atom from an excited energy level to

the lowest energy level

However, you must remember that the spectrum of hydrogen gas is not continuous,

that is, not all frequencies will be emitted; radiations of only those frequencies will be

emitted whose energies are equal to the difference in energy between the allowed

energy levels. Therefore, the emission spectrum of hydrogen gas contains bright lines

of certain definite frequencies.

The frequencies contained in the emission spectra of each element are unique:

no two elements have the same type of emission spectra because each one of

them has a characteristic set of allowed energy levels.

Level 1

Level 2

Level 3

Level 4 etc…

Energy

Level 1

Level 2

Level 3

Level 4 etc…

Energy

Level 1

Level 2

Level 3

Level 4 etc…

Energy

Level 1

Level 2

Level 3

Level 4 etc…

Energy

51

Stellar Spectra and Classification

Absorption Spectrum

Let us assume now that the hydrogen gas is not at a high enough temperature to excite

the electrons to higher energy levels. In such a situation, there is no emission of

radiation. If white light (continuous spectrum) is passed through this cool hydrogen

gas, it will absorb light of only those frequencies which give electrons just enough

energy to make a transition to one of the excited energy levels (see Fig. 7.3b).

Therefore, just as photons of certain energies are emitted by hydrogen gas at a high

temperature, photons of the same energies are absorbed by the cool hydrogen gas.

You can argue that atoms absorbing photons and giving rise to absorption spectrum

will eventually re-emit photons of the same frequencies. Thus, it should cancel out the

effect of absorption and we could not observe any absorption spectrum! The question,

therefore, is: How to explain the observed absorption spectrum? The re-emitted

photons are repeatedly absorbed by the atoms of the cool gas before they finally

escape it. These re-emitted photons escaping from the gas travel along different

directions. Thus, the photons of certain frequencies, which were originally coming

towards the spectroscope as a part of white light, are scattered by the atoms of the

cool gas. As a result, the intensities of the light corresponding to these frequencies are

very low and we observe an absorption spectrum.

Thus when white light passes through a cool hydrogen gas, we obtain a series of dark

lines and such a spectrum is called the absorption spectrum.

Do remember that for any element, the dark lines of absorption spectrum occur

at exactly the same frequencies (or wavelengths) where we observe bright lines

in its emission spectrum.

Now, before proceeding further, you should answer an SAQ to fix these ideas.

SAQ 1

a) Why do the spectra of different elements have different sets of lines?

b) What do you understand by a continuous spectrum? Under what condition do we

observe it?

Let us pause for a moment and think: How does the knowledge about the origin of spectral lines help in analysing stellar spectra? Firstly, you have learnt that each

element produces a unique set of spectral lines. Therefore, observing the spectrum of a

star, we can tell which element(s) are present in its atmosphere. Secondly, the nature

of spectrum emission or absorption depends on the temperature prevailing in the

atmosphere through which star light passes. Thus, the nature of spectrum gives us an

idea about the temperature of the star’s atmosphere.

In addition, stellar spectra also provide information about the density of the star’s

atmosphere and the motion of stars with respect to the Earth. Whether a star is moving

towards or away from the Earth can be inferred on the basis of Doppler effect:

According to the Doppler effect, there is an apparent change in the wavelength

of radiation emitted by a star due to its (star’s) motion; if the star is moving away

from the Earth, the wavelengths are shifted towards longer wavelengths (the

phenomenon is known as red shift) and if the star is moving towards the Earth,

the wavelengths are shifted towards shorter wavelengths (the phenomenon is

known as blue shift).

With this background information, you are now ready to study stellar spectra. In this

context, it is useful to remember a set of empirical rules of spectroscopic analysis

given below:

Spend

5 min.

Strictly speaking, the stellar

spectra tell us about the

composition of the star’s

atmosphere. However, it is

generally believed that the

composition of the

atmosphere reflects the

composition of the star itself.

52

The Solar System and Stars

i) A hot and opaque solid, liquid or highly dense gas emits a continuous spectrum;

Fig. 7.7a (you have learnt about optically thin and thick media in Unit 4).

ii) A hot gas produces emission spectrum and the number and position of the

emission lines depends on the composition of the gas (Fig. 7.7b).

iii) If light having continuous spectrum is passed through a gas at low temperature, an

absorption spectrum consisting of dark lines is produced (Fig. 7.7c). Again, the

position and number of dark lines are characteristic of the elements present in the

gas.

iv) When light with continuous spectrum passes through a very hot, transparent gas, a

continuous spectrum with additional bright lines is produced.

Fig. 7.7 shows some of these situations.

(a) (b)

Fig.7.7: Diagrammatic representation of empirical rules for spectroscopic analysis of stellar spectra

7.3 STELLAR SPECTRA

Before discussing stellar spectra, you may like to know how they are obtained using a

spectroscope. Refer to Fig. 7.8 which shows how to obtain a spectrum using a

spectroscope. Light from the telescope (pointing towards the star) is passed through a

slit, a collimating lens, through the prism of the spectroscope to obtain the spectrum.

The resulting spectrum is focussed by another lens and can either be viewed directly

or photographed.

Fig.7.8: Schematic diagram of a spectroscope

For a casual observer, all stars look alike. But, if observed carefully, some stars appear

bright and some faint. It is usually difficult to make out the colours of stars if they are

fainter than the second magnitude. This is due to poor colour sensitivity of the human

(c)

Bright-line spectrum Continuous spectrum

Spectroscope Hot light source

Cloud of gas

Spectroscope Spectroscope

Continuous spectrum with dark lines

Light from telescope

Collimating lenses

Prism

Red

Violet Source slit

53

Stellar Spectra and Classification

eye to fainter light. Stellar spectra, however, show that individual stars differ widely

from one another in brightness and spectral details. While spectra of some stars

contain lines due to gases like hydrogen and helium, others show lines produced by

metals. Some stars have spectra dominated by broad bands of molecules such as

titanium oxide.

A typical spectrum of a star consists of a continuum, on which are superposed dark

absorption lines. Sometimes emission lines are also present. An example of a typical

stellar spectrum is the solar spectrum (Fig. 7.9). Solar spectrum can be easily obtained

by passing a narrow beam of sun light through a prism. It consists of a continuum

background superposed by dark lines.

Fig.7.9: Solar spectrum

The dark lines in the solar spectrum are called Fraunhofer lines. You know that the

intensities of spectral lines indicate the abundance of various elements to which the

lines belong. The important lines in stellar spectra are due to hydrogen, helium,

carbon, oxygen, neutral and ionised metal atoms. Bands in the spectra are caused by

the molecules such as titanium oxide, zirconium oxide, CH, CN, C3 and SiS2.

Similarities in stellar spectra provided the basis for classification of stars into certain

categories. The earliest classification was done by Annie J. Cannon. She classified

more than 2, 50,000 stars by observing the strength of absorption lines, particularly,

the hydrogen Balmer lines. In this way stars have been classified into seven major

spectral types, namely, O, B, A, F, G, K and M.

For greater precision, astronomers have divided each of the main spectral types into

10 sub-spectral types. For example, spectral type A consists of sub-spectral types A0,

A1, A2.... A8, A9. Next come F0, F1, and so on. Thus there are 70 sub-spectral types

possible. However, in practice, all 70 types have not been observed. The advantage of

the finer division is to estimate the star's temperature to accuracy within about

5 percent. The Sun, for example, is not just a G star, but a G2 star, with a surface

temperature of about 5800 K.

7.3.1 Spectral Types and Their Temperature Dependence

Fig. 7.10 shows a set of spectra of seven main types of stars: O, B, A, F, G, K and M

(hottest O-type to coolest M-type). It was shown by M.N. Saha, an Indian scientist,

that Cannon’s spectral classification can be explained primarily on the basis of the

temperatures of stars because the intensities of spectral lines depend on the surface

temperature of a star. To elaborate this argument, let us take the case of spectral lines

of hydrogen Balmer series as an example.

You know that 50 to 80 percent of the mass of a typical star is made of hydrogen.

However, only a small fraction of the stars show Balmer lines in their spectra! It is so

because most of the stars do not have sufficient surface temperature to raise the

electrons in hydrogen atoms to the second and higher energy levels. And, as you

know, for production of Balmer lines, electrons in the hydrogen atoms make

transitions from the higher energy levels to the first excited energy level. For example,

a M type star whose surface temperature is about 3000 K, does not have enough

energy for production of Balmer lines. Therefore, even though there is an abundance

of hydrogen in the stars, Balmer lines may not be observed in the spectra of all of

them: the surface temperature prohibits the formation of Balmer lines of hydrogen.

You have learnt in Unit 1 that

the brightness of stars is

expressed in magnitudes.

In earlier times, the spectrum was

exposed on a photographic film.

The film was then developed in

the usual way and scanned for

intensity at each point of the film.

The data so obtained was plotted

to obtain the spectrum. Now-a-

days, with the advent of modern

detectors like the CCDs (about

which you have studied in Unit

3), the spectrum can be directly

stored in a computer and plotted.

M.N. Saha’s conclusion that

Cannon’s spectral classification

is based on surface temperature

of stars is based on the fact that

ionisation of atoms is a process

similar to chemical reaction.

This means that at a given

temperature, relative number of

atoms in various stages of

ionisation which are in

equilibrium, is fixed. Further,

the intensity of an absorption

line is a function of the number

of atoms which can absorb

radiation corresponding to this

line and this number itself is a

function of temperature.

54

The Solar System and Stars

Fig.7.10: Examples of spectra of various types of stars (O to M type)

On the other hand, very hot stars with surface temperatures higher than 25000 K also

do not contain Balmer lines in their spectra. Can you guess why is it so? It is because

such stars are so hot that electrons of the hydrogen atoms are ripped off and the atoms

are ionised. Such ionised atoms cannot produce spectral lines. It is only when the

surface temperature of a star is between 7500 K and 11000 K that the conditions are

favorable for the production of Balmer lines. Therefore, surface temperature of a star

determines which spectral lines would be formed and what their intensities would be.

After hydrogen, the second most abundant element in stars is helium. A helium atom

has two electrons which are held very tightly together in their lowest states. Hence,

helium absorption lines are seen in the spectra of relatively hot stars with surface

temperatures in the range 11000 K to 25000 K. In stars hotter than 25000 K, one of

the two electrons in helium atoms is torn away. The question is: Are the spectral lines produced by singly ionised helium atoms similar to those produced by un-ionised helium atom?

The spectral lines produced by singly ionised helium are different from those

produced by neutral or un-ionised helium. Further, stars with temperatures in excess

of 40000 K are so hot that helium is completely ionized which cannot produce any

spectral lines.

In some stars, conditions are favourable for a molecule to produce spectral lines. Very

cool stars with temperatures less than 3500 K show very strong, broad bands of

titanium oxide (TiO). Table 7.1 shows the values of the temperature and some other

important parameters of seven main spectral types.

Table 7.1: Spectral types and their parameters

Spectral class

Approx. temp. (K)

Hydrogen Balmer lines

Other spectral features

Naked-eye example (Star)

Color

O 40,000 Weak Ionised He Meissa (O8) Blue

B 20,000 Medium Ionised and

Neutral He

Achenar (B3) Blue/White

A 10,000 Strong Ionised Ca weak Sirius (A1) White

F 7,500 Medium Ionised Ca weak Canopus (F0) Yellow/White

G 5,500 Weak Ionised Ca

medium

Sun (G2) Yellow

K 4,500 Very weak Ionised Ca

strong

Arcturus (K2) Orange

M 3,000 Very weak TiO strong Betelgeus (M2) Orange/Red

Hydrogen Balmer series H

lines H H H

O

B

A

F

G

K

M

350 nm

K H lonized Ca

400 nm

Calcium

Various metals

450 nm 500 nm

55

Stellar Spectra and Classification

There is an alternative method to determine the temperature of a star on the basis of its

spectrum. This method is based on the principle of black body radiation and you will

learn it now.

7.3.2 Black Body Approximation

As you know, the colour and brightness of a star is different from other stars. You also

know that these parameters depend on temperature. Does it, therefore, mean that we can estimate the temperature of stars on the basis of their observed colour and brightness?

It can indeed be done if we consider a star as an ideal object called black body. Refer

to Fig. 7.11 which shows a set of black body spectra at various temperatures. Note

that hotter bodies radiate most of their total energy in the shorter wavelength part of

the spectrum. On the other hand, the cooler bodies have the peak of their radiation at

the longer wavelength side of the spectrum and the total energy radiated by them is

relatively low.

It has been observed that the outer envelope of a star’s spectrum is quite similar to a

black body spectrum at a certain temperature. Thus, a stellar spectrum can be

approximated to a black body spectrum. You should, however, note that the

absorption features of stellar spectra distinguish it from the black body spectrum.

Therefore, the main spectral types, namely O to M, can be considered as referring to

different temperatures. In other words, we can say that spectral type directly refers to

the effective temperature of the star (see Table 7.1).

Fig.7.11: Black body radiation curves for various temperatures

To estimate the temperature of an astronomical object such as a star on the basis of

black body approximation, go through the following Example carefully.

-4.5 -4 -3.5

log10 (cm)

2

0

-2

-4

-5

log 1

0 F

56

The Solar System and Stars Example 1

An astronomical object named Cygnus X-1, a strong X-ray source, is found to radiate

like a black body with peak wavelength at 1.45 nm. Calculate its temperature. Assume

that the constant for Wien’s displacement law is equal to 2.9×10−3

mK.

Solution

You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course

that the Wein’s displacement law is given by:

λpeak T = constant

where λpeak is the value of wavelength corresponding to the peak of the black body

spectrum and T is the temperature of the black body. For Cygnus X-1, we have

λpeak = 1.45 × 10−9

m

and

value of the constant = 2.9 × 10−3

mK.

Thus, temperature of Cygnus X-1,

K102m104.1

mK109.2 6

9

3

×≅×

×=

−

−

T .

As you know, the temperature of a star is not a directly measured quantity; rather, we

infer this parameter on the basis of black body approximation. Would it not be better

to classify stars on the basis of a parameter, say luminosity, which is more easily

measurable? This is what was attempted by Ejnar Hertzsprung and Henry Norris

Russell and their classification of stars resulted in a graph known as H-R diagram.

You will study about it now.

7.4 H-R DIAGRAM

H-R diagram is a graph which enables us to simultaneously classify stars on the basis

of their temperatures and their luminosities. Since H-R diagram involves parameters

such as temperature, luminosity and radius, it would be advisable to first recapitulate

these terms and their interdependence in the context of stars.

The luminosity (L) of a star is defined as the total energy radiated by it, in one second,

consisting of radiations of all wavelengths. On the other hand, you may recall from

Unit 1 that the absolute visual magnitude refers only to the visible range of the

electromagnetic radiation. Therefore, corrections need to be applied for the radiations

emitted at other wavelengths. The required extent of correction depends upon the

temperature of the star. This correction is called Bolometric Correction (B.C.) and

for medium temperature stars like our Sun, its value is small. After making necessary

correction, we obtain the absolute bolometric magnitude: for the Sun it is + 4.7 and

for Arcturus, it is − 0.3.

Thus, the expressions for the apparent magnitude as well as the absolute magnitudes

of a star are written as:

B.C.+= vbol mm (7.2a)

and

.MM vbol C.B+= (7.2b)

where mbol, mv, bolM and vM are respectively the apparent bolometric magnitude

apparent visual magnitude, absolute bolometric magnitude and absolute visual

magnitude of a star. Further, the absolute bolometric magnitude of a star can be

calculated from the following relation:

57

Stellar Spectra and Classification

−=−

log2.5L

LMM starbolbol

star (7.3)

You have already learnt that a difference of 5 in absolute magnitude implies a ratio of

100 in luminosity. So, the luminosity of Arcturus is 100 ΘL . The most luminous stars

can have luminosities of the order of 105

ΘL , whereas the least luminous ones have

luminosities ~ 10−4

ΘL .

Thus, if we can measure the parallax of a star, we can find its distance, calculate its

absolute visual magnitude, calculate the bolometric correction and obtain absolute

bolometric magnitude. As a result, we can estimate the luminosity of a star in terms of

the Sun’s luminosity. How about solving an SAQ to check your understanding of the

terms discussed above?

SAQ 2

The absolute visual magnitude of a star is 8.7 and for its temperature, the bolometric

correction is − 0.5. Calculate the absolute bolometric magnitude and the luminosity of

the star.

You know that one of the fundamental parameters of a star is its diameter and it is

related to the star’s luminosity and temperature. To appreciate the relation amongst

these parameters, consider a normal candle flame which has a low surface area. The

candle flame, despite being very hot, cannot radiate much heat and its luminosity is

low. However, if the candle flame were 25 cm long, it would have larger surface area.

In this case, despite being at the same temperature, it would radiate more heat and

would have high value of luminosity.

In the same way, a star’s luminosity is affected by its surface area and temperature. To

obtain a relation between the luminosity and diameter of a star, we can assume the star

to be spherical in shape having surface area 4πR2, where R is the radius. Further, from

basic thermodynamics (PHE-06 course), you know that, the total energy radiated by a

black body per second per unit area is σT 4 (Stefan’s law). Assuming that the star

radiates like a black body, we can express its luminosity (L) as the product of its

surface area and the energy radiated by it per unit area per second, i.e.,

L = 4πR 2σT

4 (7.4)

Thus, by comparing the luminosity L of a star, with ΘL , the luminosity of the Sun, we

get:

42

=

ΘΘΘ T

T

R

R

L

L (7.5)

where ΘR and ΘT are the radius and temperature of the Sun.

Let us now refer to Fig. 7.12 which shows a H-R diagram for all known stars in our

solar neighbourhood. Note that the H-R diagram is a plot between absolute magnitude

or luminosity (along y-axis) and temperature (along x-axis).

The H-R diagram contains quite a lot of information about stars. Since the absolute

magnitude or luminosity refers to the intrinsic brightness of a star, H-R diagram

relates the intrinsic brightness of a star with its temperature. Moreover, the H-R

diagram separates the effects of temperature and surface area on the luminosity of the

stars because the brightness of two stars at the same temperature is proportional to

their radii. This feature enables us to classify stars in terms of their diameters.

Further, you must remember that the location of a star on the H-R diagram is in no

way related to its location in space: a star located near the bottom of the diagram

simply means that its luminosity is low and similarly, a star in the right indicates that

its temperature is low (because the temperature decreases away from the origin) and

so on. Another interesting feature of the H-R diagram is that the position of a star on

Spend

5 min.

You have learnt in Unit 1

how the parallax of a star is

measured.

58

The Solar System and Stars

it changes with time. This implies that the star’s luminosity and temperature change

with its age. Again, this change in position of a star has nothing to do with the star’s

actual motion.

The stars have been divided into different types/groups on the basis of their location

on the H-R diagram. In Fig. 7.12, you may note that the distribution of stars follows a

pattern such that a majority of stars fall along a central diagonal called the main sequence. The main sequence stars account for nearly 90 per cent of all stars. The

other types of stars such as giants, supergiants, white dwarfs populate other regions.

The giant stars (named so because of their big size) located at the top right of the H-R

diagram have low temperature but high luminosity.

Fig.7.12: a) A schematic H-R diagram (Note that on the main sequence, masses of the stars are indicated in units of solar mass.); and b) magnified version of the H-R diagram

-20

-10

10

5

O B A F G K M

20000 10000 7000 6000 5000 3000

White dwarfs

Main sequence

Supergiants

Red giants

17

6

3.2

1.8

1.5 1.3

1

0.7

0.5

0.3

Spectral class

Temperature (K)

Abs

olut

e m

agni

tude

20,000 14,000 10,000 7000 5000 3500 2500 +20

+15

+10

0

−−−5

−10

+5

Abs

olut

e M

agni

tude

O B A F G K M Spectral Class

(b)

(a)

Temperature (K)

59

Stellar Spectra and Classification

When we move further up in the H-R diagram (Fig. 7.12), we find super giants.

These high luminosity and low temperature stars must be extraordinarily big in size. It

has been estimated that the diameters of super giants are roughly 100 to 1000 times

the diameter of the Sun! Further, when we come to the bottom left (below the main

sequence), we come across stars known as white dwarfs stars which are very hot

but their luminosity is very low. Obviously, white dwarf stars must be very small in

size compared to the Sun to have such low luminosities.

Example 2

A bright star in Orion constellation, Betelgeuse, has a surface temperature of 3500K

and is 105 times more luminous than the Sun. Calculate its radius in terms of ΘR , the

radius of the Sun. What kind of star could it be on the basis of the H-R diagram?

Solution

We have

TB = 3500K; ΘT = 5800K; ΘL

LB = 105

From Eq. (7.5), we have

ΘL

LB = 42

42

ΘΘTR

TR BB

or,

[ ] ( )245 5800350010 Θ= RRB /

or,

Θ≈ RRB 1000

The luminosity and surface area of Betelgeuse is much higher than the Sun whereas

its temperature is lower than the Sun. These characteristics, as per the H-R diagram,

indicate that Betelgeuse is a super giant.

To fix the ideas expressed above, you should answer the following SAQ.

SAQ 3

a) Suppose that the surface temperature of two stars A and B is the same and the

luminosity of A is higher than B. Which of the two stars is bigger in size? Why?

b) Choose a typical giant star in Fig. 7.12 and estimate its radius.

Although the basic H-R diagram depicts stars in terms of their temperatures and

absolute magnitudes, we can have its many other representations as well depending

upon the extent of information we wish to incorporate in it. Fig. 7.13 shows a

composite H-R diagram incorporating several parameters such as luminosity, absolute

magnitude, temperature, and spectral type.

Spend

8 min.

60

The Solar System and Stars

Fig.7.13: A composite H-R diagram showing various parameters of the stars SAQ 4

a) In which part of the main sequence are the less massive stars located?

b) A main sequence star has a luminosity of 400 LΘ. What is its spectral type?

Thus, on the basis of above discussion, you will agree that the H-R diagram provides

information about the following parameters of stars:

a) size

b) luminosity

c) mass

d) spectral type, and

e) absolute magnitude

It is because of this reason that the H-R diagram is so important in astronomy. So far,

you have studied about stellar spectra and classification of stars on the basis of their

temperature. You have also learnt that, on the basis of H-R diagram, one can get an

idea about the size of a star. However, to get a better idea about the size of a star, we

can use the correlation between the sharpness of spectral lines and luminosity. This

gives rise to luminosity classification. You will learn about it now.

7.5 LUMINOSITY CLASSIFICATION

We have discussed earlier in this unit that the luminosity of a star predominantly

depends on its size: the bigger the star is, the higher is the value of its luminosity.

Detailed study of the stellar spectra helps in ascertaining the density and size of a star

which give a fairly good idea of its luminosity. You may ask: Which features of the stellar spectra provide such information? Recall that the spectrum of an isolated

atom differs from that of the gas of such atoms. The difference is manifested in the

form of broadening of spectral lines of the atom in the latter case due to collision

between atoms.

The next logical question is: Under what conditions can collisions readily take place? For frequent collisions, a denser gas provides a better environment than a rarer

gas. Therefore, if the spectral lines in stellar spectra are broadened, we can safely

conclude that the density of the star is high. Examples of such stars are the main

sequence stars. On the other hand, giant stars have very low densities. As a result,

their spectral lines are fairly narrow compared to the main sequence stars.

Spend

5min.

61

Stellar Spectra and Classification

Thus, sharp lines in the stellar spectra clearly indicate that the size of the star is large

and hence its luminosity is high. The converse is also true. Thus, looking at the width

of spectral (absorption) lines, we can obtain a fairly good idea about its luminosity.

As such, stars have more or less continuous range of luminosities. Still, on the basis of

the width of their spectral lines, they are categorized into various luminosity classes.

Table 7.2 gives the various luminosity classes denoted by Roman numerals I to V

with the supergiants further subdivided into classes Ia and Ib.

The star Rigel (β Orionis) is a bright supergiant (class Ia) and Polaris, the North star is

a regular supergiant (class Ib).The star Adhara ( Canis Majoris) is a bright giant

(Class II); Capella (α Aurigae) is a Giant (III) and Altair (α Aquilae) is a subgiant

(IV). The Sun is a main-sequence star (V). Thus, the complete spectral classification

for the Sun is G2V. This complete classification is also called the spectro-luminosity classification. The spectro-luminosity class of star Vega is A0V.

Refer to Fig. 7.14 which shows the position of luminosity classes on the H-R diagram.

You must remember that the lines corresponding to each class is just an

approximation; star of a particular class may lie just above or below the line

corresponding to that class.

Fig.7.14: H-R diagram depicting the location of luminosity classes

Now, let us sum up what you have learnt in this unit.

7.6 SUMMARY

• Emission and absorption of radiation is caused due to transition of electrons in

atoms between allowed energy levels. Analysis of stellar spectra provides

information about the temperature and size of a star.

• On the basis of the strength of spectral lines, particularly Balmer lines, in stellar

spectra, stars were classified into seven main spectral types namely O, B, A, F,

G, K and M. It was shown by M N Saha that this classification essentially refers

to the temperatures of stars.

• H-R diagram enables us to classify stars on the basis of their temperatures and

luminosities.

Table 7.2: Luminosity classes

Ia Bright supergiant

Ib Supergiant

II Bright giant

III Giant

IV Subgiant

V Main-sequence star

62

The Solar System and Stars

• Luminosity of a star is defined as the total energy rediated by it in one second

consisting of radiations of all wavelengths. The relation between luminosity,

radius and temperature of a star is given by:

42

=

ΘΘΘ T

T

R

R

L

L

• On the basis of H-R diagram, stars are grouped into four categories namely main sequence, giants, super giants and white dwarfs. Stars of each group have

characteristic temperatures, sizes and luminosities.

• The density of a star affects the sharpness of its spectral lines. This fact is used for

luminosity classification of stars.

• According to the luminosity classification, stars are classified into five classes: I,

II, III, IV and V with class I further sub-divided into Ia and Ib.

7.7 TERMINAL QUESTIONS Spend 25 min.

1. Assign spectral class to each of the objects whose characteristics are given below:

a) Temperature ~ 40,000K

b) Weak Balmer lines but moderately strong Ca lines

c) Strongest hydrogen lines

d) Molecular bands of TiO

e) Neutral helium lines

2. A star has luminosity ~ 100 ΘL and apparent bolometric magnitude, 9.7.=bolstarm

If Sun has 4.7,

+=bolM calculate the distance of the star.

3. Calculate the radius of a star which has the same effective temperature as the Sun

but luminosity 10,000 times larger.

4. In the following table, which star is a) the brightest; b) the most luminous; c) the

largest; and d) the smallest?

Star Spectro-luminosity type mv

1. G2V 5

2. B1V 8

3. G2Ib 10

4. M5III 19

5. White Dwarf 15

7.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. a) Spectra of different elements have different lines because their atomic energy

levels are different.

b) A continuous spectrum contains all wavelengths. We get such a spectrum

when the density of a system is high, as in a solid.

2. From Eq. (7.2b), we know that

.MM vbol C.B+=

Substituting the values of vM and B.C., we get, 5.07.8 −=bolM = 8.2

63

Stellar Spectra and Classification

Further, from Eq. (7.3), we have:

( )

)7.42.8(4.0

4

10

10

−−

−−

Θ

=

= Θ

bolbol

starMMstar

L

L

25

1=

Thus, we get, 25

Θ=L

Lstar .

3. a) Since LA > LB, the surface area of A is greater than the surface area of B. So, A

is bigger.

b) In Fig. 7.12, we find that the temperature of a typical giant star can be taken as

5000 K. So, we can write for a typical giant, T = 5000 K. Further, the

luminosity of a typical giant can be written as, L = 100 ΘL

Thus, using Eq. (7.5), we get:

42

=

ΘΘΘ T

T

R

R

L

L

or,

42

=

Θ

ΘΘ T

T

L

L

R

R 4)5/6(100 ×=

or,

R = 14.4 ΘR .

4. a) Lower part of the main sequence.

b) From Eq. (7.3), we have

−=−

ΘΘ L

LMM starbolbol

star log5.2 400log5.2−= = 6.5

Thus,

bol

starM = 6.5 + 4.7 = 1.8

If we look at the H.R. diagram to identify the class of the star having this

value of absolute magnitude, we find that its class should be B7,

approximately.

Terminal Questions

1. (a) O (b) G (c) A (d) M (e) B

2. We have from Eq. (7.3):

−=−

ΘL

LMM starbolbol

star log2.5

or,

(100)log2.574. −=bolstarM 30.−=

64

The Solar System and Stars

Further, we have:

=−

10log5

dMm

bolstar

bolstar

( )10log5)30.(79. d/=−−

or,

.d pc1000=

3. From Eq. (7.5), we have:

42

22

41

2121 TRTRLL =

or,

;10 41

2

41

21

4TRTR=

or,

4

1 10010 RRR ==

4. [Hint: Use Fig. 7.12 to determine the absolute magnitudes.]

The answers are: a) 1; b) 3; c) 3; d) 5.

65

Stellar Structure

UNIT 8 STELLAR STRUCTURE

Structure

8.1 Introduction

Objectives

8.2 Hydrostatic Equilibrium of a Star

8.3 Some Insight into a Star: Virial Theorem

8.4 Sources of Stellar Energy

8.5 Modes of Energy Transport

8.6 Simple Stellar Model

Polytropic Stellar Model 8.7 Summary

8.8 Terminal Questions

8.9 Solutions and Answers

8.1 INTRODUCTION

In Unit 7, you have learnt about the classification of stars on the basis of their spectra.

You know that the H-R diagram is obtained on the basis of the luminosity, and

effective temperature of stars and enables us to classify them in the most

comprehensive manner. A careful look at this diagram reveals that there are gaps

between families of stars, e.g., between the main sequence and giants. You may

wonder why such gaps should exist when we have such a large number of observable

stars! Further, you may like to know: What causes some ordinary stars to become

giants and others to become dwarfs? These and similar other questions cannot be

answered on the basis of observations alone. We require the knowledge of the

physical conditions in the interior of the stars. In other words, we need to know: How

the temperature, pressure and density of a star vary in its interior? In the present unit,

you will study the physical principles which form the basis for understanding the

internal structure of stars.

You know that the Sun is emitting radiation at a constant rate and its diameter shows

no significant variation with time. This implies that the Sun as well as other stars are

in mechanical and thermal equilibria. In Sec. 8.2, you will study about hydrostatic

equilibrium and its consequences for the variation of density and pressure inside a

star. In Sec. 8.3, you will learn how to estimate the internal temperature of a star on

the basis of the virial theorem: statement of relation between the kinetic and potential

energies of a system in equilibrium. In addition to the considerations of hydrostatic

and thermal equilibria, the mechanism of energy generation and transport play an

important role in deciding stellar structures. In Sec. 8.4, you will learn why only the

energy generated due to nuclear reaction needs to be considered as the source of

stellar energy. You will also learn some important mechanisms of nuclear energy

generation in stars. In Sec. 8.5, various modes of transportation of energy from the

interior to the surface of stars have been discussed. You will discover the conditions

for the formations of convective and radiative zones in the stellar interior. Finally, in Sec. 8.6, we discuss the computation of a simple stellar model. We also

discuss how the results of this model compare with the observations.

Objectives

After studying this unit, you should be able to:

• list the basic assumptions for the theoretical study of stellar structure;

66

The Solar System

and Stars • show that the values of interior pressure and temperature of a star are higher than

the values at the surface by several order of magnitudes;

• explain that the nuclear energy generation is the only important energy generation

process in stars;

• predict when a radiative or a convective zone will be formed in the stellar interior;

• compute polytropic stellar model and compare the theoretical results with

observations; and

• solve numerical problems based on these concepts.

8.2 HYDROSTATIC EQUILIBRIUM OF A STAR

You know that stars, including the Sun, are made of hot gas. We cannot probe the

interior of stars to determine their physical parameters and their variation with time

and distance because of their high temperatures and enormous distances from the

Earth. The question, therefore, is: How do we determine the internal structure of a

star? Astrophysicists construct theoretical models of stars and compare their

predictions with observations. To keep the theoretical analysis simple, the following

assumptions are made:

i) The star is spherically symmetric: You know that stars have rotational motion

which alters their spherical shape. Since the rotational motion is slow in most

cases, it does not have appreciable effect on the shape of the stars. Spherical

symmetry is, therefore, a valid assumption.

ii) The star is in dynamic equilibrium: Dynamic equilibrium means that the energy

radiated by a star is equal to the energy supplied from its core. This assumption

seems valid because luminosities of stars have been observed to be constant over

a considerable period of time.

iii) The star is in thermally steady state: This implies that the temperature at each

point within a star is constant over a considerable period of time. Note that this

assumption does not mean that the entire interior of a star is at the same

temperature.

Under these assumptions, the theoretical understanding of stellar structure is based on

four equations based on certain fundamental principles of physics. First of all, let us

consider the principle of hydrostatic equilibrium.

You know that the observable stellar parameters such as luminosity change very

slowly. We can, therefore, safely assume that a star is in hydrostatic equilibrium, that

is, it is neither expanding nor contracting (at least not very rapidly). This equilibrium

is maintained by a balance between the force of gravity acting inwards and that due to

the gradient of pressure of the gas acting outwards. To find out the consequences of

the hydrostatic equilibrium, let us consider an element of volume dV at a point A

inside a star at a distance r from its centre (Fig. 8.1). If ρ(r) is the density of matter

inside dV, that is, at a distance r from the centre, the mass enclosed in the volume

element dV is ρ(r) dV. Further, if M(r) is the mass inside the sphere of radius r, the

gravitational force acting on the mass inside dV is given by:

dVrr

rGM)(

)(

2ρ (8.1)

67

Stellar Structure

Fig.8.1: An element of volume dV at a distance r inside a star in hydrostatic equilibrium

Now, in view of the spherical symmetry of the star, the pressure, density and

temperature may be taken as identical at all points over the spherical surface of radius

r. Therefore, the net hydrostatic force acting on the volume element dV and pushing it

outward can be written as:

dP . dA, (8.2)

where dA is the area of the volume element dV perpendicular to r and dP is the

pressure difference between two sides of the volume element along the radius. For

hydrostatic equilibrium, the gravitational force must be equal and opposite to the

hydrostatic force. Thus, on the basis of Eqs. (8.1) and (8.2), we can write:

dAdPdVrr

rGM.)(

)(

2−=ρ

dVdr

dP.−=

This yields the equation of hydrostatic equilibrium.

Equation of Hydrostatic Equilibrium

)()(

2r

r

rGM

dr

dPρ−= (8.3)

We can also write Eq. (8.3) as

)()( rgrdr

dPρ−= (8.4)

where g(r) is the acceleration due to gravity given by g(r) = .)(

2r

rGM From Eq. (8.3),

it is obvious that it is the pressure gradient that supports the star and not the pressure.

If we denote the mass of the whole star by M and its radius by R, the mean density,

>ρ< of the whole star can be expressed as:

π

=>ρ<3

3

4R

M (8.5)

dr

dV (=dA.dr)

dP.dA

r

R

68

The Solar System

and Stars Let us now consider a spherical shell of the star between radii r and r + dr. The

volume of the matter enclosed in this shell is 4πr2dr. Since ρ(r) is the density of

stellar matter at distance r, the mass of this spherical shell is:

drrrrdM )(4)( 2ρπ=

Thus, we can express the total mass inside the sphere of radius r as:

πρ=r

drrrrM0

24)()(

Differentiating both sides of the above equation with respect to r, we get the mass

continuity equation for the star:

Mass continuity equation

)(4)( 2

rrdr

rdMρπ= (8.6)

Eqs. (8.3) and (8.6) constitute two basic equations of stellar structure. Further, the

state of hydrostatic equilibrium in a star enables us to obtain a relation between its

gravitational potential energy and the kinetic energy of its constituent particles. This

relation is known as the virial theorem. You will learn about it now.

8.3 SOME INSIGHT INTO A STAR: VIRIAL THEOREM

You may be aware that the virial theorem is applicable for a system of perfect gas

particles. However, this theorem can also be applied to a star because it (star) can be

considered as a system of free particles. To obtain the relation between the potential

and kinetic energies of a star, let us consider the equation of hydrostatic equilibrium

(Eq. (8.3)). On multiplying Eq. (8.3) by 4πr3 and integrating over the radius of the

star, we get:

πρ−=πRR

drrr

rGMrdrr

dr

dP

0

3

20

3 4)(

)(4 (8.7)

On integrating by parts, the left hand side of Eq. (8.7) gives:

π−π=πRRR

drrPrPdrrdr

dP

0

2

0

3

0

3 4344

Since P = 0 at r = R, the first integral on the right hand side vanishes and the above

equation reduces to:

π−=πRR

drrPdrrdr

dP

0

2

0

3 434

Thus, Eq. (8.7) can be written as:

πρ−=π−RR

drrr

rGMrdrrP

0

2

0

2 4)(

)(43 (8.8)

Now, assuming that the star comprises of monoatomic gas, its total thermal (or

internal) energy U can be written as:

Actually, the pressure on

the surface of a star is not

zero. Its value is, however,

much smaller than the

pressure in the interior.

Therefore, it can be

assumed to be zero. This is

also the case with density.

69

Stellar Structure

π=R

drrPU0

2432 (8.9)

SAQ 1

Derive Eq. (8.9).

Substituting Eq. (8.9) in Eq. (8.8), we obtain:

πρ=R

drrr

rGMrU

0

24)(

)(2 (8.10)

The right hand side of Eq. (8.10) can be expressed in terms of the gravitational

potential energy. The gravitational potential due to the mass M(r) inside the sphere of

radius r is .)(

r

rGM− Therefore, total potential energy due to all mass elements dM (=

4πr2ρ(r) dr) of the star is:

ρπ−=−=ΩRM

drrrr

rGMdM

r

rGM

0

2

0)(4

)()( (8.11)

Thus, on the basis of Eqs. (8.10) and (8.11) we get the virial theorem.

Virial theorem

2U + Ω = 0 (8.12)

You can see from Eq. (8.11) that the gravitational potential energy Ω of the whole star

can be determined only if we know the variation of density ρ(r) inside. Assuming that

ρ(r) ≈ >ρ< , the mean density of stellar matter, we may write:

>ρ<π≈ 3

3

4)( rrM (8.13)

Substituting >ρ< for )(rρ and M(r) from Eq. (8.13) in Eq. (8.11) and integrating, we

get:

R

GM2

5

3−≈Ω (8.14)

Now, to find an expression for the internal energy of a star, we make use of the

equation of state of a gaseous system:

Tkm

P Bµ

ρ= (8.15)

where kB (= 1.38 × 10−23

J K−1

) is the Boltzman constant, T is the temperature and

µm is the mean mass of a gas particle. For pure hydrogen gas, µ = 1. Substituting

Eq. (8.15) in Eq. (8.9), we get:

=π=M

BR

dMm

TkdrrPU

00

2

2

34

2

3

Spend

3 min.

>ρ<π= 3

3

4)( rrM

rdrdM >ρ<π=2

3.3

4

drrdM 2

4π=>ρ<

70

The Solar System

and Stars since . and4 2

dMdVdVdrr =ρ=π If we define the mean temperature of the star,

=><M

TdMM

T0

,1

then the above expression can be written as:

><µ

= TMm

kU B

2

3 (8.16)

Substituting the expressions for potential energy (Eq. (8.14)) and internal energy

(Eq. (8.16)) in the virial theorem (Eq. (8.12)), we get:

R

M

k

GmT

B

µ=><

5

1

Thus

3/13/2 >ρ<∝>< MT (8.17)

where R has been expressed in terms of the mean density.

SAQ 2

a) Verify the results contained in Eqs. (8.14) and (8.17).

b) Assume that the Sun is made of pure hydrogen (µ = 1). Show that the mean

temperature of the Sun is >< T ≅ 4 × 106K.

On solving SAQ 2, you must have appreciated that the internal temperature of the Sun

can be estimated without making any detailed calculations. Further, it is obvious from

Eq. (8.17) that if two stars have the same mass, the denser one will be hotter. For a

sun-like star, the effective surface temperature is Te ≈ 5780 K. And from the solution

of SAQ 2, we find that the mean solar temperature, >< T ≈ 4 × 106

K. This means

that the internal temperature must be much higher. You may ask: What causes such

high internal temperature in stars? To answer this, we must investigate the sources

of energy generation in stars. The stars can have possibly three kinds of energy

sources: gravitational, chemical and nuclear. You will study about them now.

8.4 SOURCES OF STELLAR ENERGY

Let us first consider the gravitational energy as the source of energy for stars. From

Eq. (8.14), note that when the whole matter of the star is scattered at infinity, its

gravitational potential energy is zero. When the stellar matter is assembled to make a

star of radius R, its potential energy becomesR

GM2

5

3− . This means that during the

gravitational contraction of a star, that is, during the formation of a star, an energy

equal to R

GM2

5

3 is released.

It can be shown (TQ 2) that the Sun’s energy would last only for about 107 yrs if

gravitational energy was its only source. But, the results obtained on the basis of

radioactive dating of different types of meteorites, deep terrestrial oceanic sediments

and lunar rocks suggest an age of ≈ 5 × 109 years for the Sun. Thus, the gravitational

potential energy cannot be the source for solar luminosity. We must look for some

other source of energy for stars like the Sun.

We are now left with two other possible processes, namely, chemical and nuclear.

Spend

10 min.

71

Stellar Structure The possibility of a chemical process as the source of energy in stellar interior is also

ruled out on the basis of results obtained in the following example.

Example 1

Assume that the Sun consists of hydrogen and oxygen and the proportion of these

elements is such that the entire solar material could be burned and transformed to

water vapour. Show that the total energy available from this process would last only

for ~ 104 yr given that 10 eV is liberated in the formation of each water molecule.

Solution

The molecular weight of water (H2O) is 18 u = 18 × 1.6 × 10−27

kg = 28.8 × 10−27

kg.

The mass of the Sun is 2 × 1030

kg. Therefore, the total number of water molecules

present in the Sun is given by:

.106kg108.28

kg102 55

27

30

×≈×

×−

Thus, total energy liberated due to the formation of water vapours in the Sun is:

6 × 1055

× 10 eV × 1.6 × 10−19

J (eV)−1

≈ 9.6 × 1037

J.

Since the Sun radiates energy at the rate ~ 4 × 1026

Js−1

, the duration over which the

Sun would radiate all its energy generated due to formation of water molecules is:

.yr10~s104.2Js104

J106.9 411

126

37

×≈×

×≈

−

In view of the fact that the Sun has an estimated age of ~ 5 × 109 yr, the result of the

above example clearly shows that chemical process cannot be responsible for

generation of energy in stars.

Let us now look at the possibility of nuclear processes for generating energy in a sun-

like star. You may recall from your school physics that nuclear reactions are of two

types: fission reactions and fusion reactions. In nuclear fission, large unstable nuclei

like 238

U break into smaller nuclei and energy is released. Since the abundance of

such nuclei is negligible in stars, such a process is also ruled out as a source of stellar

energy. We are, therefore, left with only fusion process to be considered as possible

energy source.

In nuclear fusion process, two lighter nuclei combine and form a new nucleus and

energy is released. The amount of energy released depends on the binding energy per

nucleon of the elements involved in the reaction. You may recall that the binding

energy of a nucleus is the energy required to separate its constituent nucleons by a

large distance. Refer to Fig. 8.2 which depicts binding energy curve. Note that as the

mass number increases from zero, the value of binding energy per nucleon increases.

This implies that if two or more, lighter nuclei (such as hydrogen) are fused together

to create a relatively heavier nuclei (such as 3He or

4He), we will have surplus of

energy. This is precisely what happens in fusion reactions: enormous amount of

energy would be released due to fusion of hydrogen nuclei and consequent production

of helium. Further, Fig. 8.2 also shows that the value of binding energy per nucleon

saturates at around mass number 50 and shows a very slow decrease beyond. The

nature of the binding energy curve, therefore, indicates that small-mass as well as

large-mass nuclei are less tightly bound than medium mass (such as Fe) nuclei.

72

The Solar System

and Stars Before proceeding further, you may like to convince yourself whether or not the

energy generated due to nuclear fusion can account for the observed luminosity of the

Sun. To do so, solve the following SAQ.

Fig.8.2: Variation of the binding energy per nucleon with mass number

SAQ 3

Assume that originally the Sun comprised only of hydrogen and that the inner 10

percent of the Sun’s mass could be converted into helium. For how long would the

Sun be able to radiate at the rate given by L = 4×1026

Js−1

. The mass of the Sun is

MΘ = 2×1030

kg.

On solving SAQ 3 you find that the Sun would indeed be able to radiate energy at the

present rate for another 5 billion years by fusing hydrogen into helium.

Now the next logical question is: How is the fusion of hydrogen nuclei into helium

nuclei made possible? In other words, you may like to know what the pre-conditions

for nuclear fusion to take place are and how these conditions are obtained in the stellar

interiors. Nuclear fusion involves fusion of two positively charged nuclei against

Coulomb repulsion.

The possibility of such a process can be understood on the basis of the potential

energy curve that an atomic nucleus would experience when it approaches another

atomic nucleus (Fig. 8.3). Note that the potential energy curve consists of two distinct

regions. Region I corresponds to the situation when separation between the two nuclei

is such that their potential energy is due to Coulomb repulsion. Region II represents

the situation when two nuclei are very close to each other and the curve is in the form

of a potential well. The potential well illustrates the strong nuclear forces that bind the

nuclei. From Fig. 8.3, it is obvious that the two nuclei can fuse only when the

approaching nuclei are able to overcome the repulsive Coulomb barrier. This means

that the approaching nuclei have sufficient energy to overcome the barrier.

The question is: At what temperature, the approaching nuclei will have sufficient

energy? Using classical dynamics, we find that the temperature required to provide

sufficient energy to two nuclei so that they overcome Coulomb’s barrier is much

higher than the core temperature Tc (~ 1.5 × 107 K) of the Sun. Thus, classical physics

To know the amount of energy

released due to fusion of four

hydrogen nuclei and formation

of a helium nucleus, note that

the total mass of four hydrogen

atoms is 4.031280 u while that

of a helium atom is 4.002603 u.

The mass defect is 0.02867u

and in accordance with the

Einstein equation, E = mc2, the

energy released due to fusion

of four hydrogen nuclei into a

helium nucleus is equal to

(0.02867 u) × c2 where c is the

velocity of light.

Spend

5 min.

73

Stellar Structure fails to explain the possibility of fusion reactions in stars. You must convince yourself

about the inadequacy of classical physics to explain nuclear fusion by solving the

following SAQ.

Fig.8.3: Schematic representation of the potential energy barrier experienced by an atomic nucleus

approaching another atomic nucleus

SAQ 4

Suppose that two nuclei have charges Z1e and Z2e and in order to interact, they must

be separated by a distance ~ 10−13

m. Calculate their mutual potential energy. If their

relative kinetic energy is 3 kBT, calculate the temperature required by two hydrogen

nuclei to overcome this potential barrier.

The temperature in the core of the Sun is ~ 1.5 × 107K. At this temperature, only a

few proton-proton fusion can take place. There is, however, a finite probability that

particles with insufficient energy can tunnel thorough the potential barrier and react.

Now, let us look at some thermonuclear fusion reactions through which lighter nuclei

fuse and release energy.

In stars like the Sun, one of the prominent thermonuclear reactions is the so-called

proton-proton chain or p −−−− p chain. This reaction proceeds sequentially through

three steps as given below:

MeV)19.1(,HHH 211 ve ++→+

MeV)49.5(,HeHH 321 γ+→+

Spend

5 min.

Region II Region I

Fe Si

O

C

He

H

Distance

Coulomb barrier

En

erg

y

Nuclear potential well

(binding energy)

← Ekin ∼ kT

74

The Solar System

and Stars MeV)85.12(,HHHeHeHe 11433 ++→+

Note that in these reactions, four hydrogen nuclei combine to form a helium nucleus.

The amount of energy liberated at each step is given within brackets. The p − p chain

produces most of the energy in Sun and other such stars.

Further, a relatively smaller amount of energy in the Sun is also generated by the

thermonuclear reaction known as carbon-nitrogen-oxygen (CNO) cycle as given

below:

MeV)45.1(,NCH 13121 γ+→+

MeV)22.2(,CN 1313 ve ++→

MeV)54.7(,NCH 14131 γ+→+

MeV)35.7(,ONH 15141 γ+→+

MeV)71.2(,NO 1515 ve ++→

MeV).(4.96,HeCNH 412151 +→+

You may note that in the CNO cycle, the carbon destroyed in the first step gets

regenerated in the last step. Further, similar to the p − p chain, CNO cycle also

produces a helium nucleus from four hydrogen nuclei. A CNO cycle, however, cannot

begin unless carbon is present.

The thermonuclear energy generated in the stars depends on the abundance of

fusionable matter and the interior temperature of the stars. This provides us a basis to

link the luminosity of stars with its mass. Let ε denote the rate of energy generated per

unit mass in thermonuclear reactions. Then, the luminosity dL caused by an element

of mass dM can be written as:

dL = ε dM (8.18)

Since dM = 4πr2ρ(r) dr, Eq. (8.18) can be expressed as:

)(4 2rr

dr

dLρπε= (8.19)

Eq. (8.19) is one of the basic equations of stellar structure.

The energy generated at the core of a star must flow towards its surface because the

temperature of the core is very high compared to the star’s surface. Further, since the

star’s surface continuously radiates energy, it will cool off unless the radiated energy

is replaced. Energy transport in the star has important consequences for its structure

because the transport process determines the temperature (and pressure) of different

layers of the star’s interior. Would you not like to know what the different

mechanisms of energy transport in a star are? This is the subject matter of the next

section.

8.5 MODES OF ENERGY TRANSPORT

There are three basic energy transport processes: conduction, radiation and

convection. Fig. 8.4 shows a schematic diagram for these three processes.

75

Stellar Structure

Fig.8.4: Three modes of energy transport: conduction, radiation and convection

Conduction is the most familiar form of heat flow and this process works through

vibrating atoms. The energy of the vibrating atoms is transferred to the nearby cooler

atoms by collisions and energy transport through conduction works better in solids

and not in gases. You will learn later in this Unit that conduction is responsible for

energy transfer in stars like white dwarfs whose interior is in a crystallised form

having density ~ 106g cm

−3. For ordinary gaseous stars, conduction process is not

important.

Radiation is the next most familiar mode of energy transport and is responsible for

energy transport in some layers of the interiors of almost all the stars. The process of

energy generation in the central region of a star produces very high energy photons

which are γ-rays. As these photons travel outwards, they collide with matter. At each

collision, γ-ray photon loses energy and when it reaches the surface, its frequency lies

in the visible range. The progress of photons is extremely slow as they travel outward

and this, in fact, regulates the solar luminosity at the level of ~ 1026

Js−1

. The

absorption of the energy of γ-rays by the stellar gas, is characterized by the

absorption coefficient, also known as the opacity, kλ of the gas. The subscript λ

indicates that the absorption depends on the wavelength. The important sources of

opacity at high temperatures inside a star are:

i) Electron scattering: The scattering of photons by free electrons.

ii) Photoionisation: The energy of photon is used for successive ionisation of

atoms/ions.

We can obtain an expression for the opacity on the basis of qualitative arguments.

Consider a slab of the stellar gas of thickness dx. Let Fλ denote the flux of radiation

that strikes at one end. If the mass density of the gas is ρ, the amount of radiation

absorbed by the slab is proportional to i) the density of matter in the slab ii) the

incident flux and iii) the thickness of the slab. Thus, the amount of flux absorbed is:

dxFdF λλ ρ∝

,dxFk λλρ−= (8.20)

where the minus sign indicates absorption. Photons generated inside the Sun do not

reach the solar surface directly; they are scattered by the electrons and nuclei. This

scattering is isotropic and thus their forward and backward scattering is equally likely.

The travel of photons inside a star is, therefore, like that of a drunken person. It is also

called the random walk. It takes thousands of years for these photons to reach the

star’s surface. To have an idea about the time taken by a photon to reach star’s surface

from the core, go through the following example carefully.

Conduction

Radiation

Convection

76

The Solar System

and Stars Example 2

Suppose that the energy transport due to radiation process is analogous to the

random walk. Compute the time taken by a photon, generated in the core of the Sun,

to reach the solar surface. Given that for the Sun, the mean free path is ~ 0.5 cm for

photon at an average density and temperature of 1.4 g cm−3

and 4.5 × 106 K,

respectively.

Solution

You may recall from Thermodynamics and Statistical Mechanics (PHE-06) course

that, according to the theory of random walk, we can write the mean square distance,

>< 2D moved in N steps as:

>< 2D

2N=

This gives N = 2

2

>< D, the number of steps required to travel a distance >< 2

D in

steps of size in one dimension. For three dimensional space, N ~ 32

2

>< D. Now,

for the photon at the core which has to reach the surface, we have .22RD >=<

Hence, the time taken for the photon to reach the solar surface is given by

.

3 2

c

Rt =

since, in each step, the time taken is /c where c is the velocity of light. Substituting

the values of R, c and , we get

t ≅ 30,000 yr.

Now, you may like to know: How does the radiative transport of energy give rise

to temperature gradient in stars? To find out, let us consider a thin spherical shell

around a spherical surface of radius r as shown in Fig. 8.5. Let the temperature of the

sphere be T. Since the spherical region within radius r acts as a source of black body

radiation, the radiative flux incident on the inner side of the shell can be expressed by

(Stefan Law):

F(r) = σT

4, (8.21)

where σ is the Stefan’s constant. Similarly, the radiative flux emerging outward from

the shell surface at r + dr is:

F (r + dr) = σ (T + dT )

4 (8.22)

where T + dT is the temperature at the surface of radius r + dr. It is important to

mention here that dT is negative because the surface at r + dr is cooler than the inner

surface of the shell located at r. Since dT is very small compared to T, we can expand

(T + dT)4 using binomial expansion. Doing so, we get:

F (r + dr) = σ T 4 + 4σ T

3 dT

Fig.8.5: Spherical shell of thickness dr around

a surface of radius r

r

r + dr

77

Stellar Structure Therefore, the flux absorbed by the shell can be written as:

dF = F (r + dr) − F(r) = 4σT 3 dT (8.23)

Combining Eqs. (8.20) and (8.23), we get:

4σT 3dT = − k(r) ρ(r) F(r) dr (8.24)

Since luminosity, L(r) = 4πr2F(r), we can rewrite Eq. (8.24) as:

)(4

)()(

3rF

T

rrk

dr

dT

σ

ρ−=

π

ρ−=

23 4

)()()(

r

rL

acT

rrk (8.25)

since σ = ,4

ac where a is a constant and c is the velocity of light. Eq. (8.25) gives the

temperature gradient within a star due to radiative transport of energy. This expression

needs to be multiplied by an extra factor of 4

3 on the right hand side so that it

becomes consistent with the one obtained by incorporating the details of such a

process. We, therefore, write the temperature gradient as:

π

ρ−=

234

)()()(

4

3

r

rL

acT

rrk

dr

dT (8.26)

Eq. (8.26) is yet another basic equation of stellar structure.

Now, let us consider the third mode of energy transport, namely, convection which

plays an important role in stars. Convection refers to the process in which heat energy

is transported by mass motion, i.e., by transport of the hot/cool matter itself. You are

familiar with convection currents or bubbles moving up and down when water in a

beaker is gradually heated from below. Such a motion also takes place in certain

regions in stars with hot fluid masses rising outward releasing their heat energy and

the cooler matter sinking downward to receive more energy. Fig. 8.6 depicts the

photospheric granulation which is a strong evidence supporting convective transport

of energy at the base of Sun’s photosphere. Convection causes mixing of the stellar

Fig.8.6: Photospheric granulation of the Sun which is caused due to convective transport of energy

in which the hot matter comes out to the solar surface from layers below the surface

constituents in certain regions inside a star and produces homogeneity of chemical

composition by transferring heavy elements from interior to the surface. You may ask:

Radiative

zone

Convective

zone

78

The Solar System

and Stars Under what condition(s) does convection become the dominant mode of energy

transport? The process is dominant when the temperature gradient becomes too

steep. Steep temperature gradients are generally created in regions with high opacity

which restrains the flow of energy through radiative process.

To determine the temperature gradient in a convective region of a star, we consider a

situation where hot bubbles of gas rise up and expand adiabatically. After rising

through a characteristic distance, the bubbles lose extra heat and get mixed up with the

surroundings. For such a process, the bubble’s adiabatic temperature gradient is given

by:

dr

dP

P

T

dr

dT

ad

γ

−γ=

1 (8.27)

Eq. (8.27) can be obtained by using the adiabatic relation P = K ργ and the equation of

state of the gas, .B Tm

kP ρ

µ= Also, from the equation of state, we have P = NkBT,

where N is the number of particles per unit volume. In astrophysics, it is usual to take

the mass of a particle as µm, where µ is called the mean molecular weight and m is the

mass of a proton. Further, for dr

dP we use Eq. (8.3) in Eq. (8.27). With these

substitutions, it is possible to write Eq. (8.27) as:

2

)(1

r

rGM

k

m

dr

dT

Bad

µ

γ

−γ−=

(8.28)

Eq. (8.28) is one of the basic equations of stellar structure. You should note that only

if the actual temperature gradient in a star is steeper than the adiabatic gradient given

by Eq. (8.27), convective transport of energy can take place. We call the actual

temperature gradient in such a case as superadiabatic. Therefore, for convection to

take place, we must have:

adactualdr

dT

dr

dT

>

(8.29)

In fact, it can be shown that convection dominates the radiative transport of energy in

a region if actualdr

dT

is slightly superadiabatic. In any case, the actual mode of

energy transport in a region inside a star depends on the temperature gradient existing

there.

Now, you should pause for a moment and think what you have learnt so far. You have

learnt to derive certain equations on the basis of the principles of physics connecting

various parameters of a star. These equations are known as the basic equations of

stellar structure. You may ask: Why did we do all this? How do equations of stellar

structure help enhance our understanding of stars? These equations are used to

develop theoretical models of stars. If the predictions of these models are in

agreement with observations, we can conclude that the assumptions made about the

parameters of stellar interior are valid. In case of disagreement between theoretical

prediction and observations, the models are ‘fine-tuned’ by modifying the initial

assumptions. You will indeed appreciate that this is the only way to investigate the

interior of stars because we simply cannot look into those interiors. In the next

section, you will learn to develop a stellar model on the basis of equations of stellar

structure.

79

Stellar Structure 8.6 SIMPLE STELLAR MODEL

Developing a stellar model essentially involves solving the equations of stellar

structure for a star. As such, it is a very complex task because large numbers of

equations with several unknowns need to be solved. This does not mean that we

cannot get a physical picture of a star. We find that, with some valid approximations,

simple stellar models are easier to calculate. Such models help understand the basis of

some of the empirical laws, such as, mass luminosity relation. Before discussing any

stellar model, let us first list the basic equations of stellar structure.

Basic Equations of Stellar Structure

In the previous sections, we used the following basic physical principles to obtain the

equations of stellar structures:

• Hydrostatic equilibrium,

• Equation of state for stellar matter,

• Mechanism of stellar energy generation, and

• Modes of energy transport in stellar interior.

These basic equations are used to compute theoretical stellar models. This is

equivalent to “constructing a theoretical star”! Once different models are computed,

their location in H-R diagram is found out. It is so because the H-R diagram sets a

detailed standard to be met by any theory of stellar structure and evolution. A

theoretical model is wrong if the physical characteristics of the computed “star” are

such that its location falls in the gap or empty regions of the H-R diagram. The

physical parameters of a star at any point in its interior are temperature, T(r), pressure

P(r), density ρ(r), and luminosity L(r). The basic equations of stellar structure are:

Hydrostatic equilibrium: )()(

2r

r

rGM

dr

dPρ−= (8.30)

Mass continuity : )(4 2rr

dr

dMρπ= (8.31)

Energy transport: 23 4

)(

4

3

r

rL

T

k

acdr

dT

π

ρ

−= (8.32)

Energy transport: 2

)(11

r

rGM

k

m

dr

dT

B

γ−−= (8.33)

Energy generation: )()(4 2rrr

dr

dLερπ= (8.34)

Equation of state: )()()( rTrRrP ρ= (8.35)

In the above equations, ε(r) is the thermonuclear energy production rate per unit mass.

You have learnt about it earlier in relation with luminosity. Further, the opacity k

occurring in these equations depends on temperature and density of the gas. In fact,

the exact form of the opacity relation depends on the process responsible for it.

Computation of stellar opacity is, however, a complex process and is usually

approximated by Kramer’s opacity relation given as

(radiative)

(convective)

(Thermal equilibrium)

80

The Solar System

and Stars 5.3

)1(.T

XZconstkρ

+= (8.36)

where X denotes the amount of hydrogen in a gram of stellar matter (it is also called

the abundance of hydrogen), Z denotes the abundance of heavier elements. (In

astrophysics, elements heavier than helium are called heavier elements.) The opacity

relation given by Eq. (8.36) is valid for stars on the main sequence.

To obtain the values of physical parameters by integrating the stellar structure

equations, we invoke the following boundary conditions:

M(r) = 0 and L(r) = 0 at r = 0 (the centre of star), (8.37)

and

M(r) = M; L(r) = L and T(r) = Teff at r = R (surface of a star)

(8.38)

With this background information, we are now in a position to discuss a stellar model.

8.6.1 Polytropic Stellar Model

In such a model, there is no need to know the actual source of energy generation in the

star. Further, we assume that any change in the equilibrium structure of a star takes

place in such a way that the specific heat remains constant, i.e.,

== CdT

dQ constant (8.39)

where C denotes the heat capacity when neither pressure (P) nor volume (V) is

constant. We call such a change as a polytropic change. An adiabatic or an

isothermal change, therefore, represents a polytropic change of zero and infinite

heat capacities, respectively. Instead of the adiabatic relation dQ = 0, we now have

dQ = CdT. In such a situation, the first law of thermodynamics

dQ = CV dT + PdV

takes the form

CdT = CV dT + PdV

Now, using the equation of state for a perfect gas, PV = RT and the fact that

R = CP − CV, we can write the above expression as:

( ) ( )V

dVCC

T

dTCC VPV −=− (8.40)

where CP and CV are the specific heats at constant pressure and constant volume,

respectively. Let us now define an exponent γ′ similar to the adiabatic exponent γ:

CC

CC

V

P

−

−=γ′ (8.41)

Thus, we can write from Eq. (8.40) that

.constconst,1 == γ′−γ′PVTV (8.42)

81

Stellar Structure SAQ 5

Derive Eq. (8.42).

It is usual to express the physical variables, e.g., density, pressure and temperature in

terms of the polytropic index n defined as:

1

1

−γ′=n (8.43)

Since .1constPV =−γ′

, we can write:

nKP

11+

ρ= (8.44)

where K is a constant. Further, the density is expressed in terms of a non-dimensional

parameter θ defined in terms of the central density ρc as

ρ = ρc θ n

. (8.45)

We, therefore, get the following expressions for pressure and temperature:

P = Pc θ n+1

(8.46a)

T = Tc θ (8.46b)

where .and /1

1n

cB

cn

n

cc Kk

mTKP ρ

µ=ρ=

+

We shall see below that Pc and Tc are the

central pressure and temperature.

SAQ 6

Derive Eq. (8.46a) and (8.46b).

With this formal introduction to polytropic changes, let us consider the following

stellar structure equations:

Hydrostatic Equilibrium: )()(

2r

r

rGM

dr

dPρ−= (8.47)

Mass continuity: )(4 2rr

dr

dMρπ= (8.48)

Substituting Eq. (8.48) in Eq. (8.47) and rearranging terms we can write:

)(41

2

2rG

dr

dPr

dr

d

rρπ−=

ρ (8.49)

Substituting for P and ρ from Eqs. (8.44) and (8.45) in Eq. (8.49), we get:

nn

c dr

dPr

dr

d

rG

Knθ−=

ρπ

+ −2

2

11

1

4

)1( (8.50)

Spend

5 min.

Spend

5 min.

82

The Solar System

and Stars To write Eq. (8.50) in a simpler form, let us introduce a dimensionless variable ξ as

follows:

;αξ=r (8.51)

where .4

)1(2

1

11

ρπ

+=α

−n

cG

Kn Substituting Eq. (8.51) in Eq. (8.50) and rearranging

terms we get:

Lane-Emden equation

n

d

d

d

dθ−=

ξ

θξ

ξξ

2

2

1 (8.52)

Eq. (8.52) is known as Lane-Emden equation. Solution of this equation, for a given n,

provides the density and pressure profile inside a star. The boundary conditions

under which this equation must be solved are:

.0at 0 and 1 =ξ=ξ

θ=θ

d

d (8.53)

θ is also known as Lane-Emden’s function.

Analytical solutions of Eq. (8.52) with the specified boundary conditions are possible

only for n = 0, 1 and 5. The analytical expressions for the Lane-Emden functions for

these values of n are:

n = 0; 6

12

0ξ

−=θ

n = 1; ξ

ξ=θ

)(sin1 (8.54)

n = 5; 2

1

2

53

1

−

ξ+=θ

Fig.8.7 shows the density profile inside a polytrope for n = 1.5 and 3.

SAQ 7

Verify that the Lane-Emden equation (Eq. (8.52)) is satisfied by the solutions given by

Eq. (8.54).

The general solution of the Lane-Emden equation is in the form of a series for θn as

given below:

...1206

1 42

−ξ+ξ

−=θn

n (8.55)

Spend

10 min.

83

Stellar Structure

Fig.8.7: Density profile of a polytropic star

Current stellar models calculated for the Sun that fit the observations indicate that a

large fraction of hydrogen at the centre of the Sun has been converted to helium

(estimated 40 percent hydrogen and 60 percent helium). Energy transport is still

radiative in the solar interior upto ~ 0.73 RΘ and beyond this distance, the temperature

gradient reaches a value which sets up convection. Fig. 8.8 illustrates the internal

structure of the present Sun.

Fig.8.8: The internal structure of the Sun as per a stellar model which shows the main regions of

the Sun and values of its important physical parameters

0.8

0.6

0.4

0.2

0

1

1 0.8 0.6 0.4 0.2 0

Polytropes

n = 1.5

n = 3

r/R

ρρ ρρ/ ρρ ρρ

c

Corona

Chromosphere

Photosphere

Convection

Zone

Radiative

Zone

1.6 ×××× 107 K

160 ×××× 103 kg m−−−−3

1/4R

0.80R

R

2000 km

330 km

Core

Density (kg m−−−−3)

8 ×××× 10−−−−5

20 ×××× 103

4 ×××× 10−−−−4

Temperature (K)

8 ×××× 104

5 ×××× 105

6 ×××× 103 K

106

84

The Solar System

and Stars

Computation of actual stellar models involves complex mathematics and needs

extensive computer resources. It is primarily due to the complexities of the equations

of stellar structure.

Now, let us summarise what you have learnt in this Unit.

8.7 SUMMARY

• To understand the internal structure of stars, theoretical models of stars are

developed and the predictions of these models are compared with observations.

• The principle of hydrostatic equilibrium is one of the fundamental principles of

physics invoked for investigating stellar structure.

• The equation of hydrostatic equilibrium is given as:

)()(

2r

r

rGM

dr

dPρ−=

• The mass continuity equation is given as:

).(4)( 2

rrdr

rdMρπ=

• The relation between total thermal energy and the gravitational potential energy of

a system of perfect gas particles such as a star is given by the so called virial

theorem:

2U + Ω = 0.

• Of the three possible sources namely gravitational, chemical and nuclear of

energy generation in stars, only nuclear processes can give rise to such high

internal temperatures. Energy is generated in stars due to fusion reactions,

particularly due to fusion of hydrogen nuclei and consequent formation of helium.

• One of the basic equations of stellar structure, as given below, relates the

luminosity (an observable parameter) of a star with its density:

)(4 2rr

dr

dLρπε=

where ε is the rate of energy generated per unit mass in thermonuclear reactions.

• Energy generated at the core of a star is transported to its surface through one or

more than one of the three basic energy transport processes namely conduction,

radiation and convection.

• The temperature gradient in a star is produced due to radiative transport of

energy and is given as:

π

ρ−=

234

)()()(

4

3

r

rL

acT

rrk

dr

dT

• Equations of stellar structure are used to develop theoretical models of stars.

If the predictions of a model are in agreement with observations, we conclude that

the assumptions made about the parameters of stellar interior are valid.

85

Stellar Structure • Developing a stellar model basically involves solving equations of stellar

structure. This is quite a complex process because a large number of equations

with several unknowns are to be solved.

• In polytropic stellar model – a relatively simple stellar model – we do not need to

know the source of energy generation. In this model, it is assumed that any change

in the equilibrium structure of a star do not alter its heat capacity:

== CdT

dQ constant

• The Lane-Emden equation is given as:

n

d

d

d

dθ−=

ξ

θξ

ξξ

2

2

1

Solution of this equation provides the density and pressure profile inside a star.

8.8 TERMINAL QUESTIONS Spend 25 min.

1. Show that when four protons combine to form helium, the energy released is

~ 26.7 MeV.

2. At present, the Sun radiates energy at the rate of ~ 4 × 1026

W. Assuming that the

gravitational contraction is the only source of the Sun’s radiant energy, how long,

since its creation, it would have radiated energy at the present rate? Take

.m107 and kg102 830 ×=×= ΘΘ RM

3. Show that Eq. (8.55) is consistent with Eq. (8.54).

8.9 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. The left hand side of Eq. (8.8) is:

===πR RR

UPdVPdVdrrP0 00

2 22

32343

since the internal or thermal energy can be expressed as PdV2

3.

2.a) From Eq. (8.11), we can write the total potential energy as:

ρπ=Ω−R

drrrr

rGM

0

2 )(4)(

Assuming that ρ(r) ≈ >ρ< , the mean density of stellar matter, we can write:

>ρ<π>ρ<π

=>ρ<π=Ω−RR

drrrr

Gdrr

r

rGM

0

23

0

2 43

44

)(

86

The Solar System

and Stars =>ρ<ππ

=RR

drR

MrGdrrG

06

24

0

24 .33

4

3

43

(using Eq. (8.13))

=R

drrR

GM

0

4

6

23

or,

R

GM2

5

3−=Ω

This is Eq. (8.14). Further, from the virial theorem (Eq. (8.12)), we can write:

2

Ω−=U

R

GM2

10

3= (using Eq. (8.14))

Also, from Eq. (8.16), we have:

><µ

= TMm

kU

B

2

3

Comparing the above expressions for U, we can write:

><µ

= TMm

k

R

GM B

2

3

10

3 2

or,

Bk

m

R

GMT

µ=><

5

1

Further,

ρπ= 3

3

4RM

or,

3

1

ρ∝

MR

Substituting this value of R, we get

3

1

3

2

>ρ<∝>< MT

which is Eq. (8.17).

b) Eq. (8.17) can be written as:

Bk

m

R

GMT

µ=><

5

1

87

Stellar Structure Substituting the values of G, M, R and kB, we get:

cm)107()Kerg104.1(

g)106.1(g)102()sgcm107.6(.

5

1

10116

24332138

×××

×××××=><

−−

−−−−

T

≈ 4 × 106 K.

3. Mass of four hydrogen atoms = 4.031280 u

Mass of a helium atom = 4.002603 u

So, the mass defect = 0.02867 u

Thus, the energy released per gram when four protons combine to form helium

is

= 0328.4

)scm109(u)02867.0( 2220 −××

≈ 6 × 1018

erg.

Thus, we can write:

lifetime of the Sun = )serg104(

)gerg106(g)102(

133

11833

−

−

×

×××

s103 17×=

≈ 1010

yr.

4. We can write the potential energy of two nuclei Z1e and Z1e separated by a

distance r as:

P.E. = r

eZZ2

21

r

e2

= if Z1 = Z2 = 1

Since the potential barrier will be overcome by the relative kinetic energy of

the two nuclei, we can write:

r

eTkB

2

3 =

or,

rk

eT

B3

2

=

1116

20

10104.13

108.48.4

−−

−

×××

××=

≈ 5 × 107 K.

88

The Solar System

and Stars

5. From Eq. (8.40), we have:

( ) ( )V

dVCC

T

dTCC VPV −=−

If we define an exponent γ′ as

CC

CC

v

p

−

−=γ′

we can write:

11 −−

−=−γ′

CC

CC

V

P

CC

CCCC

V

VP

−

+−−=

CC

CC

V

VP

−

−=

Further, Eq. (8.40) can be written as:

( ) ( ) 0=−+−V

dVCC

T

dTCC VPV

or,

0)1( =−γ′+V

dV

T

dT

=−γ′ 1TV Const.

Further, using the equation of state of a perfect gas:

PV = RT

we can write:

.T

dT

V

dV

P

dP=+

Substituting for T

dT, we get:

0=γ′+V

dV

P

dP

or,

Const.=γ′PV

6. From Eq. (8.44), we have:

n

n

KP

1+

ρ=

89

Stellar Structure

( ) n

nn

cK

1+

θρ= (using Eq. (8.45))

1

1

+θρ=

+

nc

n

n

K

1+θ= ncP

where n

n

cc KP

1+

ρ=

Further, we can write pressure as:

Tm

kP B ρ

µ=

or,

ρ

µ=

P

k

mT

B

ρ

ρµ=

+

n

n

B

Kk

m

1

(substituting Eq. (8.44))

nnc

B

n

B

Kk

mK

k

m /1/1 )( θρµ

=ρµ

= (substituting Eq. (8.45))

θρµ

= nc

B

Kk

m /1

nc

Bcc K

k

mTT

/1where ρµ

=θ=

7. Lane-Emden equation (Eq. (8.52)) is:

n

d

d

d

dθ−=

ξ

θξ

ξξ

2

2

1

For n = 0, this equation reduces to:

11 2

2−=

ξ

θξ

ξξ d

d

d

d

And, for n = 0, we have 6

12ξ

−=θ

Substituting this value of θ in the left hand side of the above equation, we get:

.13

31

3

1

3

1 2

2

3

2

2

2−=

ξ−

ξ=

ξ−

ξξ=

ξ−ξ

ξξ d

d

d

d

90

The Solar System

and Stars Further, for n = 1, we have ξ

ξ=θ

sin from Eq. (8.54). So, we get:

2

sincos

ξ

ξ−

ξ

ξ=

ξ

θ

d

d

Again, substituting this value of ξ

θ

d

d in the left hand side of Lane-Emden

equation, we get:

( )ξ−ξξξξ

sincos.1

2 d

d

[ ] θ−=ξ

ξ−=ξ−ξ+ξξ−

ξ=

sincoscossin

1

2

as required.

For n = 5, we have,

2/12

31

−

ξ+=θ

or,

3

2.

31

2

12/3

2 ξ

ξ+−=

ξ

θ−

d

d

Substituting the value of ξ

θ

d

d in the left hand side of Lane-Emden equation, we

get:

ξ+

ξ−

ξξ

− 2/323

2 31

3

1

d

d

ξ

ξ+

ξ+

ξ+

ξ−

ξ=

−−

3

2.

31

2

3.

331

3

312/5

232/3

22

2

ξ−

ξ+

ξ+−=

ξ+

ξ+

ξ+−=

−−−

331

31

31

331

222/5

22/5

222/3

2

2/52

31

−

ξ+−=

5θ−= , as required.

Terminal Questions

1. When four protons combine to form a helium atom, the mass defect is 0.02867 u.

Thus, energy released in this process:

91

Stellar Structure 2mcE =

)sm109(kg)106.102867.0( 221627 −− ××××=

J10128.4 12−×=

eV106.1

10128.4

19

12

−

−

×

×=

= 26 MeV.

2. Assuming the Sun to be a sphere, its gravitational potential energy can be written

as:

R

GM2

5

3=Ω

So, the time τ for which the Sun will radiate with its luminosity, L can be written

as:

LR

GM 1.

5

3 2

=τ

.LR

GMyr

103

1.

1.

5

3

7

2

×=

. yr)103(

1

)smkg104(m)107(

)kg104()skgm107.6(

5

3

712268

26011311

××××

××××=

−

−−−

yr5

1108 ×=

.yr107×≈ 2

3. Eq. (8.55) is:

...1206

1 42

−ξ+ξ

−=θn

n

For n = 0, this equation reduces to:

6

12

0ξ

−=θ → a constant

for ,1=n

−

ξ+

ξ−ξ

ξ=ξ

ξ=θ ...

!5!3

1sin

1 53

1

92

The Solar System

and Stars 1206

1...1206

1 4253 ξ+

ξ−=

−

ξ+

ξ−ξ

ξ=

...1206

1 42

−ξ+ξ

−=n

for ,5=n

9

.2

1.

2

3.

2

1

3.

2

11

31

422/1

2

5ξ−−

+ξ

−=

ξ+=θ

−

...120

5

61

2461 4

242

−ξ+ξ

−=ξ

+ξ

−=

...1206

1 42

−ξ+ξ

−=n

Thus, we find that Eq. (8.55) is consistent with Eq. (8.54).

5

Star Formation

UNIT 9 STAR FORMATION

Structure

9.1 Introduction

Objectives

9.2 Basic Composition of Interstellar Medium

Interstellar Gas

Interstellar Dust

9.3 Formation of Protostar

Jeans Criterion

Fragmentation of Collapsing Clouds

9.4 From Protostar to Pre-Main Sequence

Hayashi Line

9.5 Summary

9.6 Terminal Questions

9.7 Solutions and Answers

9.1 INTRODUCTION

In Unit 8, you have learnt the basic physical principles governing the processes in the

interior of stars. The existence of stars was taken for granted and their coming into

being was not discussed. It is, however, logical to ask: Where do the stars come from?

How are the stars formed? In general terms, you did study about the formation of the

solar system including the Sun (the only star in our solar system) in Unit 6. But the

focus of that unit was to understand the formation and characteristics of the planets. In

the present unit, you will learn about the formation of stars.

You must have observed stars in the night sky. On a clear dark night away from city

lights, you can observe that the space surrounding the bright stars is fairly luminous.

This is due to the scattering of light from the star by the gas and dust in its

surroundings. Further, astronomical observations suggest that the more luminous and

massive stars are younger and have been formed recently. Since these stars are

completely surrounded by gas and dust, it is believed that the stars are formed from it.

The gas and dust in the interstellar space is called the Inter Stellar Medium (ISM) and

it is the major constituent of our galaxy − the Milky Way Galaxy. You will learn the

basic composition of ISM and methods of its detection in Sec. 9.2. Under suitable

conditions, a cloud of gas and dust condenses or collapses due to its own gravity and

forms protostars. In Sec. 9.3, you will learn the Jeans criterion which is a rough

measure of the mass and size of the interstellar cloud which may collapse and give

birth to a star. You will discover that the interstellar cloud must fragment repeatedly to

form stars. In Sec. 9.4, you will learn how a protostar evolves into a full fledged star

and becomes a member of the main sequence (in the H-R diagram discussed in

Unit 7).

The various stages from the initial collapse of a cloud of ISM upto the pre-main

sequence are collectively considered as birth of a star, that is, the process of its

formation. You may ask: What happens afterwards? This issue is addressed in the

next Unit on Nucleosynthesis and Stellar Evolution where we discuss the life of stars.

Objectives

After studying this unit, you should be able to:

• describe the basic composition of ISM;

• derive Jeans criterion for the stability of a gas cloud;

• explain the necessity of repeated fragmentation of a collapsing cloud; and

• discuss the evolution of an interstellar cloud into a pre-main sequence star.

It is usual to write our

galaxy with the upper case

G, that is, as Galaxy.

6

From Stars to Our

Galaxy 9.2 BASIC COMPOSITION OF INTERSTELLAR

MEDIUM

The interstellar medium (ISM) makes up only 10 to 15 percent of the visible mass of

the Milky Way Galaxy. It comprises matter in the form of gas and dust (very tiny

solid particles). About 99 percent of ISM is gas and the rest is dust. You may like to

know: Which elements are present in the ISM? An important clue for investigating

the composition of the ISM is the fact that the birth and death of stars is a cyclic

process. This is so because a star is born out of ISM, and during its life, much of the

material of the star is returned back to the ISM by the process of stellar wind (in case

of the Sun, it is solar wind; see Unit 5) and other explosive events such as Nova and

Supernova. The material thrown back into the ISM may form the constituents of the

next generation of stars and so on. To know the basic composition of ISM,

astronomers use photographs and spectra. In the following discussion, we shall confine

ourselves to the composition of the ISM of the Milky Way Galaxy in the

neighbourhood of the Sun. Further, for simplicity, we will first discuss the

composition of interstellar gas and then come to interstellar dust.

9.2.1 Interstellar Gas

Hydrogen and helium are the two major constituents of interstellar gas; hydrogen

constitutes about 70 percent and the rest is helium. Indeed, other elements are also

present but in very small quantities. The analysis of the radiation received from ISM

has enabled astronomers to classify the gaseous matter filling the interstellar space

into the following four types:

i) H II region,

ii) H I region,

iii) Inter-cloud medium, and

iv) Molecular cloud.

We now briefly describe these regions and their possible roles in the formation of

stars.

i) H II region: As the name suggests, these regions of ISM mainly consist of

(singly) ionised hydrogen. In addition, these regions contain ions of other

elements such as oxygen and nitrogen and free electrons. Since the ionisation

energy of hydrogen atom is very high, such regions can exist only in the

vicinity of very hot stars. H II region can be viewed even with naked eye in the

constellation of Orion. If you look carefully at Orion’s sword, you will see that

one of the objects is a hazy cloud of gas (Fig. 9.1). This bright cloud is called

Orion nebula. The spectrum of this nebula shows lines of hydrogen and some

other elements in its bright line spectrum. Such bright nebulae are also called

emission nebulae.

It has been discovered that emission nebulae do not shine by their own light. They

absorb high energy ultraviolet photons from hot stars. These photons ionise the

gas in the nebulae and subsequently, low energy photons are radiated. Since the

spectrum of the nebula consists of many emission lines of hydrogen, it indicates

that the light must have been emitted by a low-density gas. Further, the emission

lines of hydrogen are very strong and the red, blue and violet Balmer lines blend

together resulting in the characteristic pink-red colour of the nebula (Fig. 9.1).

7

Star Formation

Fig.9.1: Orion nebula −−−− a typical bright and diffuse nebula

At this stage, you may ask: Why are the H II regions always found in the

emission nebulae? Note that only those photons which have wavelengths shorter

than 91.2 nm have sufficient energy to ionise hydrogen. Such high energy photons

can be produced in sufficiently large numbers by hot (~ 25,000 K) stars only. And

stars (such as hot O or B star) having temperature of this order are located in or

near the emission nebulae. Further, H II regions have very low density

(~ 109 particles m

−3). They provide observable evidence supporting the existence

of matter in interstellar space.

ii) H I region: Although it has been generally believed by astronomers that hydrogen

atoms populate interstellar space, they could not detect H I gas till 1951. The

reasons are obvious: it is not possible for the neutral hydrogen in ISM to produce

emission line as it is in the ground state. However, with the development of radio

telescopes, it is now possible to detect H I region. The detection of H I in the

ISM is based on the detection of a unique radiation of wavelength

21 cm.

SAQ 1

Calculate the energy of electromagnetic radiation having wavelength 21-cm.

On solving SAQ 1, you must have found that the value of the energy of radiation

corresponding to wavelength 21-cm is very small. You may ask: What kind of

transition produces such a low energy photon? In a hydrogen atom, an electron

revolves around the proton. Since the proton and the electron possess spin, there

are two possible ways for their spins to align with respect to each other. The two

spins may be parallel (aligned) or anti-parallel (anti-aligned) (Fig. 9.2). It is known

that the parallel spin state of hydrogen atom has slightly more energy than the

anti-parallel spin state. Therefore, if there is a flip from the parallel to the anti-

parallel state, there is a loss in energy of the hydrogen atom and it results in the

Spend

5 min.

8

From Stars to Our

Galaxy emission of a photon. The frequency and the wavelength of such emissions are

1420 MHz and 21 cm, respectively.

Fig.9.2: Parallel and anti-parallel spin alignments of the electron and proton in a

hydrogen atom

Further, the question is: Can we obtain 21-cm radiation in laboratory

conditions? It is not possible because the best vacuum that can be produced in a

laboratory has a much higher density of atoms than that found in the ISM. So, in

laboratory conditions, practically all the hydrogen atoms get de-excited due to

higher collision rate and we cannot obtain 21-cm line. Thus, the only place where

favourable conditions for emission of such radiations can exist is outer space. The

21-cm line was first detected in the year 1951 using a radio telescope even though

it was predicted as early as in the 1940s. Since its detection, it has become a

common tool in astronomy to map the location and densities of H I regions. This

helps in determining the structure of galaxies including our own. Investigations of

diffuse interstellar H I clouds suggest that their temperatures are in the range of 30

– 80 K, masses in the range of 1 – 100 solar masses and the number densities in

the range of 108 – 10

9 cm

−3.

iii) Inter-cloud Medium: Having read about clouds of ionised and neutral hydrogen

in ISM, you would like to know: Is the space between the interstellar clouds

empty? It is not so; the inter-cloud space consists of

a) neutral hydrogen gas with a density of 105 atoms m

−3, and

b) hot (~ 8000 K) ionised gas with very low density (~ 104 ions m

−3).

You would further like to know: Is there any interaction between the H I

clouds and the inter-cloud medium? The H I clouds are very cool and have high

densities whereas inter-cloud medium has very low density and high temperature.

The pressure of a region, being a function of its density and temperature, in the

H I clouds and in the inter-cloud medium is about the same and they are in

equilibrium.

iv) Molecular Cloud: Analysis of optical spectra of interstellar medium reveals that

matter exists in molecular form in ISM. Since hydrogen is the most abundant

matter in ISM, it mainly consists of the hydrogen molecules (H2). However,

molecules of hydrogen do not emit photons of radio wavelength and vast clouds

of hydrogen molecules remain undetected by radio spectroscopy. Other

molecules, such as CO (carbon monoxide), capable of emitting in radio

wavelength, can indeed be detected. In fact, nearly 100 such molecules have been

detected. But the basic question is: How do these molecules form in ISM? It is

believed that the atoms come in the vicinity of each other and bond to form

molecules on the surfaces of the dust grains (about which you will learn later in

this Section). These molecules are very weakly bonded and can be easily broken

by high energy photons. Thus, they can exist only deep inside dense clouds. Also,

efficient release of energy by molecules makes these dense clouds very cool.

These dense, cold clouds are called molecular clouds.

Hydrogen molecules have

been detected in the ISM by

infrared spectroscopy.

Proton Electron

Parallel spins Anti-parallel spins

21-cm

radiation

9

Star Formation In our Galaxy, the largest of these cool, dense molecular clouds are called giant

molecular clouds (GMC). They are 15 to 60 pc across and may contain 100 to

106 solar masses! The internal temperature of GMC is very, very low (~ 10 K).

The question is: Do the giant molecular clouds have any role in the formation

of stars? You know that young stars are surrounded by H II regions. And H II

regions are invariably found near giant molecular clouds. This proximity indicates

that the GMC plays an important role in the formation of stars. Thousands of

GMC exist in the spiral arm of our Galaxy. The association of O and B main

sequence stars with GMC suggests that star formation takes place in these regions.

We will talk more about it later in this Section.

9.2.2 Interstellar Dust

Interstellar medium also contains dust grains, which constitute about one percent of

the interstellar mass. Although the temperature of interstellar dust region is very low

(~ 100 K), it can be detected by infrared telescopes. A typical dust grain is made of

thin, highly flattened flakes of silicates and graphites coated with ice. Fig. 9.3 shows a

typical dust grain, which is of the size of the wavelength of blue light.

Now, you may ask: What is the evidence supporting the existence of dust in ISM?

The two observable effects due to dust are extinction and reddening. Refer to

Fig. 9.4. Note that, besides the brighter gas and dust regions surrounding the stars,

there are darker regions as well. You may think that these regions are devoid of stars.

It is not so. In fact, these darker regions are so dense that they completely stop the

light emitted by stars behind them and therefore no light is able to pass through. This

phenomenon is known as interstellar extinction. The extent to which light is

scattered or absorbed in a dust cloud depends on the number density of particles and

on its thickness.

Fig.9.3: A typical dust grain

Core

(Silicates,

Graphite)

Mantle (ice)

0.1 m

10

From Stars to Our

Galaxy

Fig.9.4: Dark interstellar space caused due to extinction

To obtain an expression for the apparent magnitude of a star, located behind the dense

cloud of dust, recall (from Unit 1) that the relation between the apparent magnitude

(mλ), the absolute magnitude (Mλ) and the distance d in parsec of a star at wavelength

is written as:

5log510 −+= dMm (9.1)

since the absorption and scattering of light is dependent on the wavelength. For stars

suffering extinction, we can write Eq. (9.1) as:

λλλ +−+= adMm 5log5 10 (9.2)

where aλ is the magnitude of light scattered or absorbed along the line of sight.

Eq. (9.2) indicates that absorption increases the magnitude of a star. A star which

would be visible to the naked eye, for instance, may be invisible due to the large

extinction i.e., sufficiently large aλ. Further, the extinction may be expressed in terms

of the optical depth as:

aλ = 1.086 τλ (9.3)

SAQ 2

The relation between optical depth, τλ and intensity Iλ for a star is given by

0

τ−= eII .

Show that aλ = 1.086 τλ.

(Hint: Remember that apparent magnitude may be written as m = K− 2.5 log I, where

K is a constant.)

To appreciate the fact that extinction depends on the density of dust grains, we can

express the optical depth τλ in terms of the number density of particles, n and

scattering cross section σλ as:

λλ σ=τs

dssn0

.)( (9.4)

Spend

10 min.

Dark region

11

Star Formation Assuming that the scattering cross-section σλ is constant along the line of sight, we

obtain from Eq. (9.4):

τλ = σλ Nd (9.5)

where Nd is the column density of particles, i.e., the number of particles in a cylinder

of unit cross section stretching from the star to the observer. Eq. (9.5) shows that

extinction depends on the amount of interstellar dust present in the path of light from

the star.

Fig. 9.5 shows the variation of νa

a with

1. Here, aν is the amount of extinction in

the visual band of wavelengths centered at 5500

A . In Fig. 9.5 we observe a peak in

the ultraviolet region which indicates that radiations of corresponding wavelengths are

strongly absorbed by ISM. Such a peak, therefore, provides a basis for determining

the composition of ISM. It is now known that graphite interacts strongly with

electromagnetic radiation of wavelength around 2175

A . Therefore, the occurrence of

peak at λ = 2175

A in Fig. 9.5 suggests the presence of graphite as one of the

constituents of ISM. Further, the presence of absorption bands at 9.7 µm and 18 µm in

the observed spectrum (not shown in Fig. 9.5) indicate the presence of silicate grains

in the ISM.

Yet another manifestation of interstellar dust is in the form of interstellar reddening.

You know that an O star should be blue in colour. But, it has been observed that some

stars with the spectrum of an O star look much redder. This is caused due to scattering

of light from stars by interstellar dust.

Fig.9.5: Schematic diagram showing the variation of ν

λ

a

a with

λ

1

BD + 56524

HD 48099

Herschel 36

l/ (m-1)

a /

a νν νν

2 4 6 8 0

12

From Stars to Our

Galaxy

Fig.9.6: Interstellar reddening caused due to the scattering of blue light coming from star

Refer to Fig. 9.6. The light coming from a star behind the dust cloud is scattered.

Since the typical size of dust grains is of the order of the wavelength of blue light, the

blue light from star is scattered more than the red light. As a result, some of the blue

light from the star is lost and it appears redder.

You might wonder as to why we discussed ISM in such detail. It is because

astronomers suspect that stars are formed from ISM. The most important observation

supporting this hypothesis is the association between young stars and clouds of gas:

wherever we find the youngest group of stars, we also find large clouds of gas

illuminated by the hottest new stars. In the following section, you will learn about

the processes and principles involved in the formation of stars from ISM.

9.3 FORMATION OF PROTOSTAR

In the previous Section, you learnt that the giant molecular clouds (GMCs) are the

likely sites where star formation can take place. However, GMCs are very unlike

stars: their typical diameter is 50 pc, typical mass exceeds 105 solar masses and their

density is 1020

times less than a typical star. Thus, the question is: How do stars

form from such molecular clouds? The simplest answer to this question is that the

stars form due to gravity: a cloud of gas contracts due to self-gravity. This contraction

increases density and temperature of the core, initiating generation of nuclear energy.

You may ask : How do we reconcile the typical mass of a star with the mass of a

GMC (~ 105 solar mass)? This problem was addressed by Jeans who proposed a

criterion for the mass of a contracting cloud which may evolve into a star.

9.3.1 Jeans Criterion

Jeans proposed that there are two competing processes in the gravitational collapse of

a molecular cloud. On the one hand, the gravitational contraction increases the

internal pressure of the cloud which tends to expand the cloud. On the other hand,

gravity acts on the cloud and tends to further contract it. Which of these two processes

will dominate is determined by the mass of the cloud. If the internal pressure is more

than the gravitational force, the cloud will break up. A clump of cloud must have a

minimum mass to continue collapsing and give birth to a star. This minimum mass is

called the Jeans mass. It is a function of density and temperature.

To obtain an expression for the Jeans mass, we make the following simplifying

assumptions:

i) the cloud is uniform and non-rotating;

ii) the cloud is non-magnetic; and

iii) the gas and dust is confined to a certain region of space by the gravitational force

and is in hydrostatic equilibrium.

Blue

light

Interstellar

dust

Red

light

Blue

light

Telescope

Star

13

Star Formation For such a system, we may write the relation between kinetic and potential energies as

(see virial theorem, Unit 8):

2U + Ω = 0, (9.6)

where U is the internal kinetic energy and Ω is the gravitational potential energy of

the cloud. If M and R are the mass and radius of the cloud, respectively, the potential

energy of the system can be written as (Eq. (8.14), Unit 8):

.5

3 2

R

GM−=Ω (9.7)

If the number of particles in the cloud is N and its temperature is T, the internal kinetic

energy of the cloud can be written as:

,2

3TNkU B= (9.8)

where kB is the Boltzmann constant. Further, the number of particles m

MN

= ,

where µ m is the mean molecular weight and m is the mass of a hydrogen atom. If the

total internal kinetic energy is less than the gravitational potential energy, the cloud

will collapse. This condition reduces Eq. (9.6) into

Ω<U2

Substituting for Ω and U from Eqs. (9.7) and (9.8), respectively, we get:

.5

3

3 2

R

GM

m

TMkB <

or,

Gm

TRkM B

5> (9.9)

On substituting3

1

4

3

πρ=

MR in Eq. (9.9), we find that the minimum mass that will

initiate a collapse is given by:

2

1

2

3

4

3

5

πρ

=≈

Gm

TkMM B

J (9.10)

Here MJ is called the Jeans mass and Eq. (9.9) is known as Jeans criterion. The

Jeans mass is the minimum mass needed for a cloud to balance its internal pressure

with self-gravity; clouds with greater mass will collapse.

Jeans criterion can also be expressed in terms of the Jeans length, RJ given by:

2

1

4

15

π≈

mG

TkR B

J (9.11)

14

From Stars to Our

Galaxy which can be obtained by putting ρπ

= 3

3

4RM in Eq. (9.9). For pure hydrogen,

µ = 1.

SAQ 3

A collapsing cloud is made of neutral hydrogen (H I) only. If the temperature of the

cloud is 50 K and its number density is 105 m

−3, calculate its Jeans mass.

In a situation where Jeans criterion is satisfied, the cloud must collapse

gravitationally. When the collapsing could attain high density, it becomes

gravitationally unstable and breaks up into smaller pieces. Thus, the general picture is

that stars form in groups due to fragmentation. You will learn more about it later in

this section.

At this stage, you may ask: How long does it take for a cloud to collapse? It is a

good idea to estimate the minimum time which will be taken if we assume that the

cloud collapses only under the influence of self-gravity and there is no other process

taking place to slow down the collapse. This assumption is called free-fall collapse,

and it implies that the pressure gradient in the interior of the clump is negligibly small.

To obtain a rough estimate of the free-fall time, you may recall that a particle on the

surface of a star of mass M and radius R experiences an acceleration g given by:

2

R

GMg = . (9.12)

If the time taken by this particle to fall through a distance R is tff we can write:

2

2

1fftgR =

or

2

1

2

=

g

Rt ff (9.13)

Substituting Eq. (9.12) in Eq. (9.13), we get:

tff 2

132

=

GM

R

2

1

3

3

3

4

2

ρπ

=

RG

R

2

1

2

3

ρπ=

G (9.14)

Eq. (9.14) shows that the free-fall time, tff is a function of the cloud’s initial density

and it does not depend on the initial radius or the initial mass of the cloud. If some

part of a collapsing cloud, say in the central region, is denser than its surroundings, the

collapse of such a region is likely to be faster than that of the surrounding region.

Spend

5 min.

15

Star Formation SAQ 4

Calculate the free-fall time for a molecular cloud whose initial density is

10−17

g cm−3

.

Now you know that the larger gas clouds collapse if their masses exceed the Jeans

mass. In a typical situation of an H I cloud with T = 100 K, ρ = 10−24

g cm−3

and

µ = 1, we find that for gravitational collapse, the mass of the cloud must be greater

than 105 times the solar mass. So, you may be led to believe that the stars could be

formed with masses of this order. However, observations suggest that the stars are

formed in groups and the masses of the stars are in the range of 0.1 − 120 MΘ. Thus,

the range of stellar mass is much smaller than 105 MΘ. This has led astronomers to

propose the fragmentation of interstellar clouds during collapse. You will now learn

about it.

9.3.2 Fragmentation of Collapsing Clouds

As mentioned earlier, Jeans criterion provides the theoretical justification for

fragmentation of a collapsing cloud. Refer to Fig. 9.7 which shows the fragmentation

of interstellar cloud of mass M and Jeans mass MJ such that M > MJ. Since M is

greater than MJ, the interstellar cloud collapses. This results into the increase of the

density and temperature of the cloud which may, in turn, change the Jeans mass for

the cloud. Let the changed Jeans mass be JM ′ (≠ MJ ).

Fig.9.7: Fragmentation of interstellar cloud

Spend

5 min.

Interstellar

cloud

M > MJ

Collapse

M'J MJ

M'J > MJ M > MJ > M'J

M, M'J

M > M'J

M, M'J

M < M'J

Stable

M'J < M1 M'J < M2

M'J < M3

M'J < M3

M'J < M2 M'J < M1

Fragmentation

Stage I

Stage

II

Stage

III

Stage

IV

Stage

V

16

From Stars to Our

Galaxy Further, from Eq. (9.10) it is evident that, if the cooling of this cloud is not efficient,

the Jeans mass will increase (stage II in Fig. 9.7). If the mass of the cloud is less than

the new (increased) value of the Jeans mass, the collapse will stop and if M > JM ′ ,

the cloud will collapse further. On the other hand, if the cooling of the collapsing

cloud is efficient, the temperature of the cloud will fall and the Jeans mass would

decrease (stage III in Fig. 9.7). In such a situation, it is possible that the mass in

certain regions of the cloud is more than the reduced Jeans mass (stage IV in Fig.

9.7). This may trigger further collapse of such regions resulting into fragmentation of

the cloud (stage V in Fig.9.7). Fragmentation of clouds continues until clouds of still

smaller masses, which are gravitationally stable, are created. These fragments of

molecular clouds are the birth places of the stars in the mass range of 0.1 − 120 MΘ.

You must note that for the scenario discussed above to be true, the Jeans mass should

not be a constant during collapse. It must decrease with increase in density in a local

region during collapse. This would lead to further gravitational instabilities which

may lead to separate individual collapsing regions.

Further, while discussing the collapse of interstellar clouds, we assumed that the

collapse process is isothermal. This can be considered to be a valid assumption as far

as the initial stages of collapse are concerned. As the collapse begins, the cloud is

likely to be optically thin due to its low density. Therefore, the gravitational energy

released during collapse is radiated away completely, keeping the temperature of the

cloud unchanged.

You may ask: What happens if the collapse is adiabatic instead of isothermal? In

that case, the energy released during the collapse is used up in raising the internal

energy of the cloud and its temperature increases. The increase in temperature thus

affects the Jeans mass. Let us now obtain an expression for the Jeans mass for

adiabatic collapse.

You may recall from the course entitled Thermodynamics and Statistical Mechanics

(PHE-06) that, for an adiabatic process, the relation between temperature T and

density ρ of a system is given by:

T = K1 ργ−1

(9.15)

where K1 is a constant and γ is the ratio of heat capacities. Using this relation in the

expression for the Jeans mass (Eq. (9.10)) we can write:

2

1

2

3

4

35

πρ

µ=

mG

TkM B

J

2/)43(2

1

2

3

1

4

35 −γρ

π

µ=

mG

KkB

Thus

2/)43( −γρ∝JM (9.16)

For the cloud comprising only atomic hydrogen, we have γ = 5/3. Therefore, for

adiabatic collapse of the cloud, we get:

2/1ρ∝JM (9.17)

17

Star Formation Eq. (9.17) shows that Jeans mass increases with increase in density. However, the

relation between the Jeans mass and density for the isothermal collapse is given as

MJ ∝ ρ−1/2

. Comparison of these two results indicates that in a switchover from

isothermal to adiabatic collapse, the Jeans mass reaches a minimum for the fragments

of molecular clouds.

The gravitational collapse is a complex process. It is difficult to ascertain exactly

when a switchover from isothermal to adiabatic collapse takes place. Moreover, the

transition is neither instantaneous nor complete. The minimum Jeans mass has been

estimated to be ~ 0.5 MΘ. This is the right order of magnitude for the mass of the

fragments of clouds which become stars.

The fragmentation of collapsing clouds leads to the formation of protostars − pre-

stellar objects which are hot enough to radiate infrared radiation but not hot

enough to generate energy by nuclear fusion. Due to gravitational collapse, the

fragmented clumps of cloud contract and at the core of each of them, high-density

region develops which is surrounded by a low-density envelope. Matter continues to

flow inward from the outer parts of the clumps. The protostar begins to take shape

deep inside the enveloping cloud of cold, dusty gas. These clouds are called cocoons

because they hide the forming protostar from our view. So, these are the initial stages

of the formation of stars. Let us now discuss this further.

9.4 FROM PROTOSTAR TO PRE-MAIN SEQUENCE

We have seen above that initially a cloud collapses, perhaps isothermally, and with the

passage of time the cloud collapse tends to become adiabatic. As a result, increased

pressure inside tends to slow the time scale of collapse particularly near the centre of

the cloud. Further, fragmentation of collapsing cloud leads to a stage when a

protostar is formed and its interior attains hydrostatic equilibrium. The interior of the

protostar at this stage becomes almost convective, that is, the temperature gradient

inside becomes larger than the adiabatic gradient. The stellar object so formed is

called a pre-main sequence star. To understand the evolution of a pre-main sequence

star, its position on the HR diagram and changes in its interior during evolution, you

must study the concept of a Hayashi line.

9.4.1 Hayashi Line

In the HR diagram, Hayashi line or track runs almost vertically in the temperature

range of 3000 to 5000 K as shown in Fig. 9.8. This line is important in the discussion

of pre-main sequence evolution of stars because of the following features:

i) When a protostar is formed, the interior of the cloud attains hydrostatic

equilibrium. At this stage, it is fully convective. The position of such an object

must fall on this line.

ii) For a given mass and chemical composition, this line represents a boundary in the

HR diagram. It separates the HR diagram into allowed and forbidden regions. The

forbidden region occurs to the right of this line and the stars in this region cannot

attain hydrostatic equilibrium. For stars falling to the left of this line, energy

transport due to convection/radiation or both is possible.

18

From Stars to Our

Galaxy

Fig.9.8: The Hayashi line

The internal temperature of pre-main sequence stars is quite low and cannot ignite

thermonuclear reactions. To compensate for the loss of energy through radiation, a

protostar must contract and this leads to increase in its thermal energy. Because of

increase in temperature and pressure in the interior of the protostar, its collapse slows

down. In other words, we expect longer evolution time in the pre-main sequence

stage for smaller mass protostars. For more massive stars, the pre-main sequence

evolution is faster and their pre-main sequence lifetime is shorter.

Now, let us sum up what you have learnt in this unit.

9.5 SUMMARY

• The space between the stars, called the interstellar medium (ISM), is not empty.

It contains gas and dust, and the gas is mostly hydrogen.

• The interstellar gas is not uniformly distributed. At places, it is highly

concentrated. These regions are called gas clouds. It is in these gas clouds that

new stars are formed.

• The gaseous matter in the interstellar space is classified into four types, namely,

HI region, H II region, inter-cloud medium and molecular clouds.

• The interstellar dust, which constitutes about one percent of the interstellar mass,

gives rise to extinction and reddening. Extinction depends on the density of dust

grains and reddening is caused due to scattering of light from stars by interstellar

dust.

• Jeans proposed that a molecular cloud must have certain minimum mass, called

Jeans mass, for its collapse due to self-gravity. The expression for the Jeans

mass is:

2

1

2

3

4

35

πρ

µ=

mG

TkM B

J

• In terms of the size, the Jeans criterion implies that if a gas cloud becomes larger

than a certain size, called Jeans length, it collapses under the gravitational force.

The expression for the Jeans length is:

3.8 3.7 3.6 3.5

Main

Sequence

Hayashi line

Lu

min

osi

ty (

Lo

g s

cale

) Log Teff

19

Star Formation

2

1

4

15

ρµπ=

mG

TkR B

J

• The expression for the free-fall time taken by a cloud to collapse under the

assumption of free-fall collapse is:

2

1

2

3

ρπ=

Gt ff

• Repeated fragmentation of collapsing clouds leads to the formation of protostars.

A protostar contracts due to gravitational force and its temperature and density

increases to such an extent that the nuclear reaction at its core becomes feasible.

A star is then said to be born. Such stars are called pre-main sequence stars.

9.6 TERMINAL QUESTIONS Spend 25 min.

1. What is the origin of the 21 cm line of hydrogen? Why can we not obtain this line

in a terrestrial laboratory? Explain the importance of this line in astronomy.

2. What is interstellar reddening? What does it tell us about the composition of

interstellar matter?

3. Derive Eq. (9.11). For the data given in SAQ 3, calculate the Jeans length.

9.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. You know that the energy of electromagnetic radiation of frequency ν and

wavelength λ is given by:

hE =

λ

=c

h

where h is the Planck’s constant and c is the velocity of light. Substituting the

values of h, c and λwe can write.

E =)m21.0(

)ms103(Js)1063.6( 1834 −− ××

J1021

363.6 24−××

=

eV106.1

110

19

24

−

−

××≈

eV1025.6 6−×≈

20

From Stars to Our

Galaxy 2. As per the problem,

λτ−λλ = eII 0

Taking logarithm on both sides and multiplying by 2.5, we get :

eII 100 log5.2log5.2log5.2 λλλ τ−=

If m0 is the original magnitude and m is the increased magnitude because of

extinction, we have:

λτ×−−=− 4343.05.20mKmK

or,

0 086.1 τ=− mm

because we can write the apparent magnitude as:

IKm log5.2−= .

Further, the magnitude of light scattered or absorbed along the line of sight can

be written as:

λλ τ=−= 086.10mma

3. From Eq. (9.10), we have the expression for Jeans mass:

2

3

5

µ=

mG

TkM B

J

2

1

4

3

ρπ

For pure hydrogen, 1=µ . So, we have:

( ) ( )

( ) ( )2

3

2131127

123

skgm1067.6kg1067.1

K50JK1038.15

×××

×××=

−−−−

−−

JM

( )

2

1

3275 kgm1067.1104

3

×××π×

−−

kg67.14

30010

67.667.1

538.1510

2

1

102

3

24

×π×

×

××=

Θ

×

×××= M

33

334

102

1075.345.510

Θ≈ M

510

4. From Eq. (9.14), we have the expression for the free-fall time for a collapsing

cloud as:

2

1

2

3

ρπ=

Gt ff

21

Star Formation Substituting the values of G and ρ, we get:

2

1

31421311 kgm10skgm1067.62

3

×

××π

=−−−−−

fft

122

1

1067.62

30×

×π= s

yr103

10

7

12

×≈ yr103 4×≈

Terminal Questions

1. See text.

2. See text.

3. The expression for Jeans mass is given by (Eq. (9.10)):

2

1

2

3

4

35

ρπ

µ=

Gm

TkM B

J

Now, for a spherical cloud of mass M and radius R, we have:

ρ×

π= 3

3

4RM

So, we can write:

ρ×

π= 3

3

4JJ RM

Substituting the above expression for MJ in Eq. (9.10), we can write:

2

1

2

3

3

4

35

3

4

πρ

µ=ρ×

π

Gm

TkR B

J

or,

2

1

4

15

ρµπ=

mG

TkR B

J

As per the data given in SAQ 3, we have T = 50K and the number density

105 m

−3. Further, for pure hydrogen, 1=µ . Substituting these values in the

expression for RJ, we get:

22

From Stars to Our

Galaxy

2

1

327327521311

123

)mkg1067.1()mkg1067.110()skgm1067.6(4

K)50()JK1038.1(15

×××××××π

×××=

−−−−−−−

−−

JR

pc 300 ~ m 1019≈

23

Nucleosynthesis and

Stellar Evolution UNIT 10 NUCLEOSYNTHESIS AND

STELLAR EVOLUTION

Structure

10.1 Introduction

Objectives

10.2 Cosmic Abundances

10.3 Stellar Nucleosynthesis

10.4 Evolution of Stars

Evolution on the Main Sequence

Evolution beyond the Main Sequence

10.5 Supernovae

10.6 Summary

10.7 Terminal Questions

10.8 Solutions and Answers

10.1 INTRODUCTION

In Unit 9, you have learnt that the gravitational collapse of interstellar cloud leads to

the formation of stars. You may recall that a stable star is formed when an equilibrium

between gravitational force in the collapsing cloud and the radiation pressure due to

nuclear energy generated in its core is attained. In this regard, some logical questions

which might come to your mind are: Since nuclear fuel burning at the core of a star

like the Sun is finite, what happens when the fuel is exhausted? Does the same type of

nuclear energy generation process take place in all stars? What happens to the material

produced in the nuclear reactions in the stars? If all the new born stars find place on

the main-sequence of the H-R diagram, how do we explain the existence of red giants,

supergiants and white dwarfs? To answer these questions, you need to learn about the

life of a star (that is, stellar evolution) after it has been formed and found its place on

the main-sequence. This is the subject matter of the present Unit.

An interesting approach to understand stellar evolution is to ask ourselves: How did

the ninety-three natural (chemical) elements, which are the building blocks of all

living and non-living matter around us, come into existence? All the elements,

except hydrogen and most of helium, were made inside stars through the process of

nucleosynthesis. The stars make elements during their life-time. Interestingly, all the

stars produce the same elements which we find on the Earth. Further, we find the

same elements everywhere in the universe and more or less in the same proportion.

The relative proportion of these elements in the Universe is called cosmic

abundances about which you will learn in Sec. 10.2. In Sec. 10.3, you will learn

about various nucleosynthesis processes taking place inside the stars at different

stages of their life. You will discover that the conditions for different processes are

different. This provides a basis to track the evolution of stars along the main-sequence

and afterwards. The low-mass and high mass stars evolve differently leading to

different end products such as white dwarfs, neutron stars and black holes. You will

learn about this in Sec. 10.4. In Sec. 10.5, you will learn that, when the entire nuclear

fuel at the core of a star is exhausted, rapid gravitational collapse takes place which

may result in a violent explosion, called supernova, of the stellar envelope.

Objectives

After studying this unit, you should be able to:

• list the various methods of determining cosmic abundances;

• list and explain various nucleosynthesis processes in the context of stellar

evolution;

24

From Stars to Our

Galaxy • describe where and how chemical elements are formed;

• explain the significance of stellar mass in respect of evolution of stars;

• estimate the lifetime of stars on the main sequence;

• describe the conditions for the formation of red giants, supergiants and white

dwarfs; and

• discuss conditions for the supernova explosions and the fate of left over stellar

cores.

10.2 COSMIC ABUNDANCES

In school chemistry, you must have learnt about elements, atoms and molecules.

Elements are the simplest substances of ordinary matter. Scientists recognise

93 elements as natural. Besides these, there are 20 elements which can be

manufactured in the laboratory by bombarding nuclei with α-particles or other

high energy particles. These 20 artificial elements are not found in nature because

they are unstable and live for extremely short times.

Almost all the natural elements are found on the Earth. A few elements which are not

found on the Earth are found in other bodies of the solar system. You may be

surprised to know that the same 93 elements are also found elsewhere in the universe

and their proportion is more or less the same as their proportion in the solar system.

The chemical composition of the universe refers to the presence of different types of

elements and their proportions in the universe. Since the chemical composition of the

universe and that of the solar system are similar, it is one and the same whether we

talk of the chemical composition of the universe, or the chemical composition of the

solar system. Further, the relative proportions of elements in the universe are called

cosmic abundances.

Determination of Cosmic Abundances

Now, the question is: How do we determine abundances in the solar system, that

is, how do we determine cosmic abundances? Some of the important methods to

obtain information about abundances are as follows:

i) In the solar system, the immediate source for obtaining such information is

obviously the Earth. Samples from many locations on the Earth are analysed in

the laboratory. Care is taken that these locations are as diverse as possible.

ii) The next obvious source is the Sun. You may recall from Unit 5 that the dark lines

in the solar spectrum, called Fraunhofer lines, are actually absorption lines due

to elements present in a slightly cooler layer above the photosphere. Each line in

the spectrum is checked against the sample spectra of elements and the elements

are identified. The intensity of a particular line gives the abundance of the

corresponding element. You may also recall from Unit 5 that the higher layers of

the solar atmosphere, the chromosphere and the corona, are at relatively higher

temperatures than the solar surface. The spectra of these layers show emission

lines due to elements present in these layers. Analysis of these lines also helps in

determining solar system abundances.

iii) The Sun also emits streams of particles in the form of solar wind. Occasionally,

the Sun emits very high energy particles, called the solar cosmic rays. The

compositions of the solar wind and the solar cosmic rays are directly determined

by instruments on-board many spacecrafts orbiting the Earth.

iv) The spectrum of other objects in the solar system such as the moon and planets

are other sources of information about abundances in the solar system. Samples of

dust brought from the moon and chemical analysis of the Martian surface has

added significantly to this information.

You know that the atom is

the smallest unit of an

element and atoms of each

element are unique. Atoms

may be said to be the basic

building blocks of matter.

Molecules are formed when

atoms combine due to

electrical forces between

them. More complex

substances are formed when

there is chemical reaction

between elements.

25

Nucleosynthesis and

Stellar Evolution v) You must be aware that small rocky pieces wandering in the solar system, called

meteors, occasionally enter the atmosphere of the Earth. If meteors are not burnt

completely by the heat generated due to atmospheric friction, they reach the Earth.

These pieces are called meteorites. Analysis of their composition provides

valuable information about abundances. The spectra of comets are yet another

source of information about the solar system abundances.

vi) Outside the solar system, spectra of other stars and interstellar clouds are

important sources of information about the cosmic abundances.

Cosmic abundances of various elements have been determined using a variety of

methods including those discussed above. Refer to Fig. 10.1 which depicts the

variation of cosmic abundances with mass number of elements.

Fig.10.1: Abundances of various elements in the universe as a function of mass number

The same data is shown in greater detail in Fig. 10.2. The abundances have been

expressed in terms of a unit in which the abundance of silicon (Si) is exactly 106. This

is because the abundance of Si is very close to this number.

Fig.10.2: A detailed version of Fig. 10.1 for elements up to mass number 80

The salient features of Figs. 10.1 and 10.2 are as follows:

1. Hydrogen (H1) and helium (He

4) are the most abundant elements in the universe.

About 90% of the particles in the universe are hydrogen atoms. Helium is the next

most abundant element, accounting for about 10% of all the particles.

2. Heavier elements constitute less than 1% of the total matter in the universe.

Mass number

Ab

un

dan

ce (

Lo

g s

cale

)

0 50 100 150 200 -2

2

6

10

Mass number

Ab

un

dan

ce (

Lo

g s

cale

)

0 20 40 60 80 -2

2

6

10

26

From Stars to Our

Galaxy 3. If we leave out H

1 and He

4, we observe that abundances generally increase with

mass number up to the mass number around 60. This is in the neighbourhood of

iron (Fe56

). Around this mass number, there is a broad peak.

4. Beyond the mass number 60, the abundances decrease. At first, the decrease is

faster and then it gradually tapers off.

5. There are peaks of abundances corresponding to elements with mass numbers 12,

16, …. and so on (multiples of 4). Moreover, elements with mass numbers, 14, 18,

… and so on (multiples of 2) are more abundant as compared to those with odd

mass numbers.

On the basis of these features of cosmic abundance data, we can conclude that:

a) The origins of hydrogen and helium are perhaps different from the origin of

heavier elements in the universe.

b) Peaks of abundances at mass numbers that are multiples of four could involve a

particle such as the α-particle or the helium nucleus, which has mass equal to 4

atomic mass unit (amu).

SAQ 1

On the basis of relative number of atoms of hydrogen and helium in the universe,

calculate the fractional mass of the matter in the universe contributed by hydrogen and

helium.

Having learnt about the cosmic abundances, a logical question that may come to your

mind is: Where and how are these elements formed? Astronomical studies tell us

that all the elements, except hydrogen and helium, have been synthesised in the stars

during their evolution. This is also reinforced by the observation that the older stars in

our Galaxy contain much less heavier elements than the younger stars. Thus, we can

visualise the following roadmap for creation of elements and how the process is

related with the evolution of stars:

a) Elements are formed inside the stars. Since the birth and death of stars is a

continuous process, the formation of elements is also an on-going process.

b) The oldest stars in the Galaxy, called Population II stars, were formed from the

original matter of the Galaxy which was mostly hydrogen. These stars had to

manufacture their own heavy elements. Therefore, they are relatively poor in

heavier elements.

c) At the end of their life, some of these stars explode and return the heavier

elements formed by them to the interstellar medium.

d) From this enriched interstellar material, new stars are formed. These relatively

younger stars, also called Population I stars, are rich in heavier elements. In

addition, they also manufacture elements in their cores which constitute the raw

material for the subsequent generations of stars.

The hypothesis that elements are made inside the stars gets support from the detection

of elements like technetium in the spectra of some stars. This element is not found in

the solar system. Where could this element have been formed except in the stars in

which it is observed?

In the language of astronomy,

any element heavier than He4

is called a heavy element.

Spend

5 min.

27

Nucleosynthesis and

Stellar Evolution Now, the question is: What is the origin of the major constituents, namely

hydrogen and helium, of interstellar medium? An acceptable theory in astronomy

tells us that hydrogen and helium were formed in a different process (see Unit 15).

You could question this theory since you have learnt earlier that He4 is formed from

H1 in the core of the Sun and the other main-sequence stars. The fact is that if we take

account of all the helium that could have been formed in the stars in all the galaxies,

it falls much short of the total helium estimated to be present in the universe (about

30% by mass). To appreciate this statement, solve the following SAQ.

SAQ 2

The atomic weights of hydrogen and helium are 1.0079 and 4.0026, respectively. In

the fusion reaction converting hydrogen into helium, one gram of hydrogen produces

about one gram of helium and approximately 6 × 1018

ergs energy is released. Given

that the luminosity of the Sun is 4 × 1033

erg s−1

and its estimated age is 5 × 109 years,

show that only about 5% of its mass has been converted into helium. Take the solar

mass as 2 × 1030

kg.

Having solved SAQ 2, you might conclude that only a small fraction of the total

helium present in the universe has been manufactured in the stars. It is, therefore,

reasonable to believe that light elements like hydrogen, helium, deuterium (D2), He

3,

and Li7 did not form inside the stars. You may ask: Do we have any clue about the

origin of these elements? According to one theory about the origin and evolution of

the universe, these elements were formed in the first minute after the birth of the

universe. At that time, the universe was hot and dense and the conditions were

suitable for the formation of light elements. (This issue is discussed in detail in unit 15

of this course.) This theory is supported by the fact that the abundances of light

elements predicted by it in the early universe agree very well with the observed

abundances. The coincidence is considered a very strong evidence supporting the

idea that the early universe was very hot and dense and that it was born in a

violent event called the Big-Bang.

A clue to support the hypothesis that heavier elements are manufactured inside the

stars was provided by Bethe (in US) and Weizsacker (in Germany) in 1938.They

showed the possibility of converting hydrogen into helium through nuclear reactions

which would take place at high temperatures and high densities. Such conditions are

readily available in the interior of stars such as the Sun, which also has plenty of

hydrogen. The work of Bethe and Weizsacker gave birth to the field of nuclear

astrophysics and subsequently, scientists were able to show that other heavier

elements could also have been formed in stars through the process of nucleosynthesis

involving a variety of nuclear reactions. Would you not like to know about

nucleosynthesis? We discuss this in the next section.

10.3 STELLAR NUCLEOSYNTHESIS

You may recall from your school physics that nuclear fusion is the process of coming

together of two or more light nuclei to form a new nucleus. Since a fusion reaction

involves coming together of charged particles, it requires that they have sufficient

energy to overcome the Coulomb repulsion. Such energies are readily available in the

form of thermal energy in stellar interiors. The nuclear reactions taking place in such

environments are called thermonuclear reactions. The process of creation of new

elements in such reactions is called nucleosynthesis.

A variety of nuclear reactions can take place depending upon the elements present,

temperature and density in the interior of stars. The information about these reactions

helps us understand stellar evolution better because on the basis of this information,

Spend

10 min.

28

From Stars to Our

Galaxy we can determine the age of the stars as well as their future. Let us now discuss the

major processes by which elements are synthesised in the stellar core.

1) Hydrogen Burning

The first stage of nucleosynthesis is the fusion of hydrogen nuclei and consequent

formation of helium. You have already learnt in Unit 5 about the chain of

reactions, called the pp-chain, which causes fusion of four hydrogen nuclei

(protons) to form helium. This is the process by which the Sun and other similar

stars generate their energy:

eeHHH 211 ++→+ +

HeHH 312 +→+

1433 H2HeHeHe +→+

Another chain reaction has helium as its end product and it produces energy in the

main sequence stars. It starts with carbon and is called the carbon-nitrogen cycle

(CN-cycle). Obviously, for this reaction, it is necessary that the stars have some

carbon to begin with. Such stars are called the second generation stars. The stars

which start life with only hydrogen and helium are called first generation stars.

The temperature required for CN-cycle is higher than the temperature required for

the pp-chain. The nuclear chain reactions involved in the CN-cycle are given

below:

C12

+ H1

→ N13

+ γ

N13

→ C13

+ e+ + νe

C13

+ H1 → N

14 + γ

N14

+ H1 → O

15 + γ

O15

→ N15

+ e+ + νe

N15

+ H1 → C

12 + He

4

You may note that the end result of the CN-cycle is to combine four hydrogen

nuclei to form helium; carbon merely acts as a catalyst for the reaction and is not

consumed in the process. You may ask: If we know that CN-cycle is active in a

star, what information can we obtain about that star? Recall from Unit 7 that,

as we go up in the H-R diagram, the luminosity increases. Since luminosity of a

star is proportional to some power (generally 3.5) of its mass, the mass also

increases upwards. Further, the internal temperature of a star is generally

proportional to its mass. Therefore, the stars in which energy is generated by the

CN-cycle are generally found in the upper region of the main sequence. They

have high internal temperature and were formed from material enriched in heavy

elements.

SAQ 3

On the main sequence in the H-R diagram, where would you find stars which have

internal temperatures lower than that of the Sun?

2) Helium Burning

When all the hydrogen in a stellar core has been used up, pp-reactions and

CN-cycle are no longer possible. As a result, the energy generation stops and the

Spend

2 min.

The temperature inside the stars

is very high. Therefore, all the

atoms are ionised. The nuclear

reactions take place between

nuclei and the products are also

nuclei. In this context, therefore,

whenever we talk of atoms or

elements, we really mean nuclei.

29

Nucleosynthesis and

Stellar Evolution pressure in the core decreases. It is no longer able to match the inward

gravitational pull of the particles and the core contracts. In some stars, the

contraction continues till the temperature has risen to about 2 × 108

K. When the

core temperature of a star attains this value, it is possible for helium nuclei to fuse

together and produce carbon.

If you look at the periodic table carefully, you will find that at mass number eight,

there is a gap. This means that the element at this position is unstable. The

question is: How do reactions between helium nuclei overcome this gap? The

two He4 nuclei (or α-particles) combine to form Be

8. Since Be

8 is unstable, it

disintegrates into two α-particles in a very short time. In the presence of to and fro

reactions of this kind, the stellar core becomes a sea of He4 nuclei with a few Be

8

nuclei floating. These floating Be8 nuclei combine with α-particles to form C

12

nuclei. Despite very low population of Be8 nuclei, C

12 is formed because the

reaction between α-particle and Be8 has a very high probability of occurrence. So,

carbon is formed when three α-particles combine according to the following

reactions:

He4 + He

4 → Be

8 + γ

Be8

+ He4 → C

12 + γ

The above nuclear reaction is also called the triple-αααα reaction. Once carbon is

formed in the stellar core, formation of heavier nuclei becomes possible. You may

recall that carbon and some other heavier elements are absolutely essential for the

origin of life. Thus, it can be argued that we are here and discussing

nucleosynthesis today because triple-αααα reactions took place in some stellar cores

in the distant past!

SAQ 4

a) Why are three α-particles needed to initiate helium reactions?

b) Explain the importance of triple-α reaction in the formation of heavy nuclei. In

what way is triple-α reaction related to the origin of life on the Earth?

3) Burning of Carbon and Heavier Nuclei

Once carbon is formed, it can combine with an α-particle and form oxygen (O16

).

Afterwards, O16

can react with an α-particle to form Ne20

. When all the helium

has been converted into carbon in the core of the star, helium reactions stop and

energy generation is terminated once again. This leads to further gravitational

contraction of the core and temperature of the interior of the star increases. At the

enhanced temperature, it becomes possible for two carbon nuclei to combine and

form Mg24

.

By now, you must have noted that every time a particular type of nuclear fuel is

used up completely, core contraction due to gravitation takes place. As a

consequence, there is a rise in the temperature of the core. When the temperature

has risen to the required level, yet another nuclear reaction becomes possible. This

cycle continues till all the nuclei in the core have become iron nuclei.

You may ask: Why does the series end at iron? To answer this question, refer to

Fig. 10.3 which depicts the binding energy curve for nuclei. Note that the average

binding energy per nucleon increases with mass number till we reach the mass

number of iron. This means that with increasing mass number, the nuclei are more

tightly bound and are more stable. It also means that iron is the most stable

element. It cannot combine with other nuclei to produce still heavier nuclei. The

binding energy curve also gives us a clue for understanding why energy can be

Spend

5 min.

30

From Stars to Our

Galaxy derived by fusing light elements as well as by splitting (fission) the nuclei of

heavy elements.

Fig.10.3: Average binding energy per nucleon as a function of mass number

It is, therefore, understandable why elements heavier than iron cannot form as a

result of thermonuclear fusion reaction. However, the fact of the matter is that

elements heavier than iron do exist in nature. The question, therefore, is: How did

they form? It is believed that such heavier elements formed in some special types

of nuclear reactions called s- and r-processes.

4) s- and r- processes

The elements heavier than iron are probably synthesized by the absorption of one

neutron at a time. For example, let us consider the nucleus of atomic number Z

and mass number A. It is written as (Z, A). When it absorbs one neutron, its mass

number increases by one and it becomes (Z, A+1). If the new nucleus absorbs yet

another neutron, it becomes (Z, A+2). If this nucleus emits a β-particle before it

can absorb another neutron, it becomes (Z+1, A+2). The last one, nucleus (Z+1,

A+2), may absorb another neutron to become (Z+1, A+3). In this way, all the

elements up to Bi209

are formed. This process of absorption of one neutron at a

time is a slow process and is named as s-process. The neutrons required for this

process are produced as a by product of reactions of the following types:

He4

+ C13

→ O16

+ n

O16

+ O16

→ S31

+ n

Well, you can further ask: Why does the s- process stop at Bi209

? It is because the

elements heavier than Bi209

are unstable and emit β-particles before they can absorb a

neutron. However, if neutrons become available in large numbers, then these nuclei

can absorb neutrons rapidly. Due to this rapid process, or the r-process, elements

right up to uranium are synthesized in stars.

So far, you have studied about cosmic abundances and stellar nucleosynthesis. You

now know that all the heavy elements are formed in stars due to thermonuclear

reactions. The relative proportion of heavier elements in stars tells us whether it is a

first or a second generation star. The various types of nuclear reactions are associated

with different stages in the life of a star and also with the location of the star on the

HR-diagram. You will now learn about the evolution of stars.

0 50 100 150 200 250 5

6

7

8

9

Mass number

Bin

din

g e

ner

gy p

er n

ucl

eon

(M

eV)

31

Nucleosynthesis and

Stellar Evolution 10.4 EVOLUTION OF STARS

You may recall from Unit 9 that gravitational collapse of a gas cloud gives birth to

stars. A stable star is formed and finds its place on the main sequence only when it

attains hydrostatic equilibrium, that is, when the pressure inside the star balances the

gravitational force acting inwards. In the following, we shall discuss the life of stars

on the main sequence and afterwards.

10.4.1 Evolution on the Main Sequence

For most of their lives, stars live on the main sequence on the H-R diagram. We

would, therefore, like to know: i) How long does a star live on the main sequence?

and ii) What happens in the core of a star while it is on the main sequence and

when it departs from the main sequence?

When a star arrives on the main sequence, it is said to have been born. That is why the

main sequence is called zero-age main sequence (ZAMS). While the star is on the

main sequence, it burns hydrogen either through pp-chain or through CN-cycle. There

is little change in its luminosity. The best example of the main sequence star is the

Sun. It is known that the luminosity of the Sun has not changed much during its

lifetime of 5 billion years. You may ask: How do we estimate the life of a star on

the main sequence? Fortunately, for the Sun, we have enough data to make an

intelligent guess. It is estimated that the core of the Sun has about 10% of its total

mass and from SAQ 2, we know that so far it has burnt only about half of this

(hydrogen) mass to make helium. Therefore, it is estimated that the Sun would stay on

the main sequence for another 5 billion years.

Let us now ask ourselves a more general question: How long does a star live on the

main sequence? To address this question, we need to look critically at the mass-

luminosity relation of stars. We know that the ultimate source of energy for a star is its

mass. When it is burning hydrogen to form helium, it is actually converting its mass

into energy. The relation between luminosity (L) and mass (M) of a main sequence

star is given by:

L ∝ M 3.5

(10.1)

Since luminosity is the energy radiated by a star per second and M is its total mass

which can be converted into energy, the time for which it will stay on the main

sequence can be written as:

τ ≈ M / L. (10.2)

From Eqs. (10.1) and (10.2), we can write:

τ ∝ M −2.5

(10.3)

The constant of proportionality is determined by the expected lifetime of the Sun.

Eq. (10.3) shows that the more massive a star is, the shorter is its life on the main

sequence. To get a feel of the numerical values of the age of stars on the main

sequence, you should solve the following SAQ.

SAQ 5

The estimated lifetime of the Sun on the main sequence is ~ 1010

years. Calculate the

main sequence lifetime of a star of mass (i) 10 MΘ and (ii) 0.5 MΘ.

Spend

5 min.

32

From Stars to Our

Galaxy

Fig.10.4: Lifetime and luminosity of stars on the main sequence as function of mass

Refer to Fig. 10.4 which shows the lifetime of stars on the main sequence as a

function of the stellar mass. You may note that a star of mass 10 MΘ will stay on the

main sequence for about 3 × 107 years and a star of mass 0.5 MΘ will stay on the main

sequence for about 6 × 1010

years. You may ask: How come a more massive star

which has more nuclear fuel has shorter lifetime on the main sequence? Recall

that massive stars must burn their fuel at a faster rate to emit more energy from their

surfaces per second. Thus, they run out of fuel sooner. By the same token, lower mass

stars burn fuel at a slower rate. In this context, it is interesting to note that the

estimated age of the universe is ~ 1 to 2 × 1010

years, a time period much shorter than

the estimated lifetime of low mass stars on the main sequence!

Having got a fairly good idea about lifetimes of stars of different masses on the main

sequence, you may like to know: How does the core of a main sequence star

evolve?

While a star is on the main sequence, its core becomes progressively richer in helium.

When the core consists only of helium, no nuclear reaction takes place because the

temperature is not high enough for the next stage of nuclear reactions. As a result, the

pressure in the core decreases and the core must contract under its own weight. The

gravitational energy released due to contraction raises the temperature of hydrogen in

a thin shell around the core so much that it starts burning (Fig. 10.5). The helium

made in the shell adds to the mass of the core whose contraction is accelerated. The

energy produced in the shell and the gravitational energy due to the contracting core

push out the envelope of the star due to radiation pressure. As a result, the star

expands in size. Its surface becomes cooler but its luminosity increases enormously

because of increased surface area. It becomes a giant star.

SAQ 6

It is estimated that after its life on the main sequence, the Sun will swell to 200 times

its present radius. If, at that time, its surface temperature is half of its present

temperature, calculate the luminosity of the Sun in terms of its present luminosity.

Let us now look at the events that take place in the life of a star when it leaves the

main sequence.

Spend

3 min.

5 10 15 20 10

4

106

108

1010 10

4

102

100

10-2

Lifetime

Lif

etim

e (y

r)

Lu

min

osi

ty (

sola

r u

nit

) Luminosity

Mass (solar unit)

0

Fig.10.5: When hydrogen in

the core of a star is

exhausted, the core

contains pure

helium and

hydrogen burns in

a thin shell around

the core

Helium in

the core Expanding

outer layers

Hydrogen in

the shell fusing

to helium

33

Nucleosynthesis and

Stellar Evolution 10.4.2 Evolution beyond the Main Sequence

When a star leaves the main sequence, it becomes a giant star because of its increased

size. Since the surface temperature of these stars is low, they appear red and are also

known as red giant stars. The more massive (> 10 MΘ) stars become so huge because

of expansion that they are called supergiant stars. The time taken by a star to travel

from the main sequence to the giant or supergiant branch is much shorter than its stay

on the main sequence. The Sun will also become a giant star after about 5 billion

years.

Meanwhile, the cores of these stars continue to contract and their temperatures rise

further. When the temperature of the core is about 2 × 108 K, triple α - reactions

become possible. The energy released in these reactions builds up the pressure in the

core and further contraction of the core is halted. Further details of evolution depend

on the mass of the star. Let us discuss some illustrative cases now.

Mass of the star is ~ 1 MΘΘΘΘ

If the mass of the star is approximately 1 MΘ, the helium burning is rather abrupt and

a large amount of energy is suddenly released. This phenomenon is called a helium

flash. Several circumstances, some not yet completely understood, combine at this

stage to force the star to throw away its outer envelope. The ejected matter surrounds

the star. This object is called a planetary nebula (Fig. 10.6). In a small telescope, it

appears like a planet, hence the name planetary nebula.

Within a short time, the gas surrounding the star vanishes due to interaction with the

interstellar medium. Simultaneously, the core again begins to contract and becomes

very dense and the matter in the core becomes a degenerate gas.

Degenerate gas is a particular configuration of a gas composed of fermions (electrons,

neutrons, etc.) whose behaviour is regulated by a set of quantum mechanical laws, in

particular, Pauli’s exclusion principle (explained in the physics electives PHE-11

entitled Modern Physics). This configuration of a gas is usually reached at high

densities. Recall that according to Pauli’s exclusion principle, no more than two

fermions (of opposite spin) can occupy the same quantum state of a system. As

the density of electrons in the core of a star increases in a fixed volume, these particles

progressively fill the lower energy states. The additional electrons are forced to

occupy states of higher and higher energy.

This process of gradually filling in the higher-energy states increases the pressure of

the electron gas. The pressure of the degenerate gas depends only on the density of the

gas and is independent of its temperature. Since degeneracy occurs when the density

is high, the pressure of the degenerate gas is high. The equation of state of such a gas

is independent of temperature unlike the normal gas.

In stars of mass ~ 1 MΘ , only the electron component of the matter becomes

degenerate. The pressure of degenerate electrons is sufficient to halt the contraction of

the star. A state of equilibrium is established. The star is then called a white dwarf

star. The maximum mass of a white dwarf star is ~ 1.4 MΘ. This limit on the mass of a

white dwarf star was predicted theoretically by S. Chandrasekhar, an Indian

astrophysicist. It is, therefore, called the Chandrasekhar limit.

In the H-R diagram, the white dwarf stars occupy the left bottom corner, much below

the main sequence as shown in Fig. 10.7. The numbers above the lines in Fig. 10.7

indicate the mass of white dwarf stars in terms of solar mass.

Fig.10.6: A planetary nebula

34

From Stars to Our

Galaxy

Fig.10.7: Location of white dwarf stars on the H-R diagram

There is no source of nuclear energy inside a white dwarf star. It is a dead star. It just

utilises the thermal energy of its particles to radiate from its surface. It can live like

this for several billion years and, afterwards, it becomes a cold object. The path of

evolution of such stars on the H-R diagram is shown in Fig. 10.8a. Note that, after the

star leaves the main sequence, it swells and heads towards the giant region.

Subsequently, depending upon its mass, it may become a white dwarf star.

(a) (b)

Fig. 10.8: a) Evolutionary track of a general mass star; and b) evolutionary track of a star of a few

solar mass

Mass of the star is ~ 5 MΘ

After becoming a giant star, a star having mass approximately equal to 5 MΘ traces its

path back on the H-R diagram towards the main sequence along a horizontal line,

called the horizontal branch (see Fig. 10.8b). This happens because of the

commencement of the helium fusion (also called triple alpha process). As helium is

exhausted and the next set of nuclear reactions involving carbon starts, the star

4.5 4.3 4.1 3.9 3.7 3.5

-4.5

-3.0

-1.5

0

Main sequence Sun

0.89 0.51 0.22

log T

log

(L

/L Θ

)

White dwarfs

Formation of planetary

nebula

Giant region

Main

sequence

White

dwarfs

105

103

10

10-1

10-3

10-5

Surface temperature of star (K)

80000 40000 20000 6000 2000

Lu

min

osi

ty (

sola

r u

nit

s)

4.1 4 3.9 3.8 3.7 3.6

0.1

1

10

100

1000

10000

Main sequence

1 solar

radius

Horizontal branch Red

giant

Surface Temperature (K; Log scale)

Lu

min

osi

ty (

sola

r u

nit

)

10 solar radii

35

Nucleosynthesis and

Stellar Evolution changes its evolutionary course backwards towards the giant branch. It wriggles

several times between the giant and the horizontal branch. During this time, it also

sheds a lot of mass. Eventually, it also becomes white dwarf after passing through the

planetary nebula phase.

Mass of the star is > 10 MΘ

If the initial mass of the star is several times the solar mass, the helium reactions start

gradually. When helium in the core is exhausted, the core contracts once again. The

helium reactions are now ignited in a thin shell around the core. The temperature of

the contracting core rises and becomes high enough for carbon-carbon reactions to

occur. The cycle of core contraction and burning of the next heavy element continues

till the core consists of iron. No more fusion reactions are possible now.

SAQ 7

Why do fusion reactions stop at iron?

Let us pause for a moment and think of what we have learnt about the evolution of

stars. We have traced the evolution of stars and simultaneous formation of elements.

Depending upon its mass, a star can take different evolutionary courses. Well, you can

further ask: What happens after all the nuclear reactions have stopped and the

stellar core consists of iron only? We have seen from the binding energy curve (Fig.

10.3) that iron has the highest binding energy per nucleon. It cannot, therefore, burn to

give off energy. Thus, the core is forced to contract. The gravitational energy heats

the core resulting in the disintegration of iron nuclei into nuclei of helium. The break

up of iron is an endothermic process, that is, the reaction absorbs energy rather than

giving it out. To feed this process, the only source of energy is the gravitational

collapse. As a result, the collapse of the core is accelerated. Faster release of

gravitational energy due to collapse gives rise to further break up of iron nuclei into

helium nuclei. This, in turn, accelerates the collapse. As the density of the core

continues to increase, even the helium nuclei cannot remain intact as nuclei. They

break up into protons and neutrons.

Coming back to the evolution of the collapsing star, let us first consider its envelope.

Due to transfer of energy from the core to the envelope, very high temperatures are

produced in the envelope and nuclear reactions in various layers of the envelope are

ignited. The energy released due to these reactions heats the envelope so much that it

becomes prone to explosive disintegration in a matter of seconds. In this short time

before explosion, nuclear reactions produce heavy nuclei all the way up to the iron

group. In addition, the large numbers of neutrons released in the nuclear reactions

participate in the r-process and heavy nuclei beyond Bi209

are produced. Finally, the

envelope explodes. The explosion is called a supernova. The elements built inside

the star over hundreds of millions of years of its life are thrown into the interstellar

medium. The new stars born from this enriched medium contain a small proportion of

heavy elements and these stars are called the second generation stars. It has been

suggested that the interstellar material from which the solar system was formed

contained heavy elements released in a supernova that took place nearby.

Well, the envelope explodes but what happens to the core? The core keeps

collapsing. According to our present knowledge, if the initial mass of the star is up to

about 12 MΘ, then the core is left with a mass of about 2 MΘ to 3 MΘ. The matter in

such stars is mostly neutrons. Like electrons, neutrons also obey Pauli’s exclusion

principle. At the extremely high density which exists in the core, neutrons become

degenerate. The pressure exerted by the degenerate neutrons is sufficiently high to halt

the collapse of the core. The core stabilises in the form of a neutron star.

Spend

3 min.

In the nuclear reaction taking

place in the core of stars, a

large number of neutrinos are

also produced. You may be

aware that the neutrino has no

charge and very small mass, if

at all. These particles react

with matter extremely weakly.

So, their mean free path is very

large and these were

considered to be the carrier of

energy from the core to the

envelope. It has, however,

been found now that at the

extremely high densities

developed in the core, the

mean free path of neutrinos is

shorter than the size of the

core. So, they cannot transfer

energy to the envelope. A

different process is now

proposed for the transfer of

energy from the core. When

the density of matter

approaches the nuclear density

(~ 1015

gcm−3

), the short range

nuclear forces come into play.

The particles repel one

another strongly and rebound.

A bounce is said to travel from

the core and into the envelope.

It is the bounce that is thought

to transfer energy to the

envelope.

36

From Stars to Our

Galaxy If the initial mass of the exploding star is close to 15 MΘ or more, the core that is left

behind has a mass more than 3 MΘ. A core of this mass cannot attain equilibrium of

any kind. It keeps contracting. Eventually, its gravitational field becomes so strong

that even light cannot escape it. It becomes a black hole.

10.5 SUPERNOVAE

Supernovae are extremely violent and bright stellar explosions. The energy released in

one such explosion is equivalent to the conversion of 1 MΘ into energy, or about

1047

J! The luminosity of a supernova is of the order of the luminosity of a whole

galaxy. In all supernova explosions, the brightness reaches a peak within a few days

of the explosion. During the short time of its maximum brightness, it shines as a very

bright object. Thereafter, its brightness decreases, first rapidly and then slowly. The

variation of the brightness of an object with time is called the light curve of the

object.

Light curves of supernovae have been studied extensively because of their importance

in many areas of astronomy. Based on the type of light curves, supernovae have been

classified as type I and type II. Fig. 10.9 shows the light curves of type I and type II

supernovae. You may note that the light curve of type I supernova reaches higher

brightness but show a rapid decline. On the other hand, the light curve of type II

supernova shows a lower maximum brightness but slower decline. The difference in

the light curves of the two types of supernovae indicates differences in the stars which

explode.

Fig.10.9: Schematic light curves of supernovae of type I and type II

You may argue: Since supernovae are so bright, it must be easy to observe them.

Surprisingly, it is not so; observations of only four supernovae in our Galaxy have

been recorded in the last two thousand years! The most famous supernova occurred in

the year 1054 A.D. and was observed by the Chinese astronomers in the constellation

Taurus.

Type I supernovae are further subdivided into two classes: Ia and Ib. The light curve

for a supernova of type Ia is shown in Fig.10.10. Again, note that the brightness of

the supernova is maximum at the time of explosion and it drops drastically afterwards.

The light curves of supernovae of type Ia are almost identical. These supernovae are

believed to be caused due to the explosion of white dwarf stars. You may ask: How

can a white dwarf give rise to a supernova explosion because its mass cannot be

greater than 1.4 MΘΘΘΘ? This would be possible if we imagine a white dwarf star in a

binary system with a main sequence or a giant star as its companion. Because of its

strong gravitational field, the white dwarf star sucks matter from the companion. As

0

Bri

gh

tnes

s

Time (days)

Type I

Type II

37

Nucleosynthesis and

Stellar Evolution its mass increases beyond 1.4 MΘ, it explodes as a supernova. Since all stars have the

same mass at the time of explosion, it is possible that supernovae Ia reach the same

brightness or absolute magnitude. Further, the observed apparent magnitude of a

supernova can be used to determine its distance using its distance modulus (defined

as m − M) because all type Ia supernovae have the same absolute magnitude at the

time of the maximum brightness. In recent years, type Ia supernovae have been used

successfully to find distances of distant galaxies. This has been very useful in

understanding the nature and the ultimate fate of the universe.

Fig.10.10: Schematic diagram of the light curve of type Ia supernova

Type Ib supernovae are thought to be due to carbon burning in the degenerate core of

a star. Carbon burning in this situation is abrupt and very rapid. It is called a carbon

flash. It generates so much energy that the star explodes.

The stars which explode to become supernovae of type II have usually masses higher

than 10 MΘ. Their light curves are all distinct. These supernovae leave behind neutron

stars or black holes, which are detected several years after the explosion.

You may ask: How do we know all that we have said above about supernovae? At

the time of explosion, a star throws a cloud of gas which travels into the interstellar

medium with speeds of the order of 10,000 kms−1

. The cloud expands and merges

gradually into the interplanetary medium. It seeds the interplanetary medium with

heavy elements. The explosion also generates ripples in the surrounding medium

which create conditions favourable for the formation of new stars. An example of the

what remains behind a supernova explosion is the Crab Nebula. It is the remnant of

the supernova of 1054 A.D. A neutron star is located at the centre of the nebula which

was formed as a result of the explosion.

Now, let us summarise what you have learnt in this unit.

10.6 SUMMARY

• Chemical composition of the universe refers to the presence of different types of

elements and their proportions in the universe. The chemical composition of the

universe is the same as that of the solar system.

• Relative proportions of elements in the universe are called cosmic abundances.

Observed abundances of elements show that (i) hydrogen and helium are the most

abundant elements in the universe, (ii) there is a broad peak of abundances near

Time (days)

Bri

gh

tnes

s

Supernova explosion

0

38

From Stars to Our

Galaxy A = 56 (iron group of elements), and (iii) there are peaks at mass numbers which

are multiples of 4.

• All elements except hydrogen and helium are formed inside the stars due to

thermonuclear reaction and creation of new elements in such reactions is called

nucleosynthesis.

• Depending upon the elements present, density and temperature in the interior of

stars, nucleosynthesis takes place through one of the three major processes,

namely hydrogen burning, helium burning and burning of carbon and

heavier nuclei. As long as a star stays on the main sequence, it burns hydrogen.

• The life span of a star on the main sequence depends on its mass: 5.2−∝τ M .

Also, the evolution of a star away from the main sequence depends on its mass.

Low mass stars become white dwarfs at the end of their lives.

• Massive stars (> 10 MΘ) explode as supernovae at the end of their lives leaving

behind neutron stars or black holes.

10.7 TERMINAL QUESTIONS Spend 30 min.

1. List the major processes of formation of elements inside stars. Why can elements

beyond iron not be formed by fusion?

2. Suppose that a supernova explosion takes place at the distance of Proxima

Centauri (~ 3 pc). If its luminosity equals the luminosity of our galaxy (~1012

LΘ),

show that it would appear about as bright as the Sun. Take the absolute magnitude

of the Sun as 5 and its apparent magnitude as − 27.

3. For stars having more mass than 10 MΘ, the luminosity is directly proportional to

their masses. Show that their lifetime on the main sequence is independent of their

masses.

4. Aldebaran (α-Tauri, Rohini in India) is a red giant star of spectral class K5. Its

radius is 22 times the radius of the Sun. Its surface temperature is 3800 K.

Calculate its luminosity (in units of solar luminosity) and its absolute magnitude.

The absolute magnitude of the Sun is 5.

10.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. 90% hydrogen and 10% helium by number implies that the universe comprises 90

units of mass of hydrogen and 40 units of mass of helium because the mass of a

helium atom would be 4 units if mass of a hydrogen atom is taken as 1 unit. So, out of 130 units of mass, 90 units is hydrogen and 40 units is helium.

Thus,

Percentage of hydrogen = 70~100130

90×

Percentage of helium = 30~100130

40×

That is, approximately 70 percent of the mass of the matter in the universe is

contributed by hydrogen and remaining by helium.

39

Nucleosynthesis and

Stellar Evolution 2. Total energy radiated by the Sun during its lifetime

= Luminosity of the Sun × Age (in s) of the Sun

= (4×1033

erg s−1

) × (5×109×3×10

7s) [ 1 yr = 3×10

7 s]

= 6×1050

erg

Now, this energy was generated in the nuclear reactions within the Sun converting

hydrogen into helium. So, to produce this much energy, mass of hydrogen

consumed = g10gerg106

erg106 32

118

50

=×

×

−

Therefore, percentage of the total mass of the Sun which has been converted into

helium till date = 100g102

g10

33

32

××

= 5%

3. The internal temperature of a star is roughly proportional to its mass. Since

luminosity of a star is proportional ~ M 3.5

, the stars with lower internal

temperature than the Sun have lower luminosities than the Sun. So, these stars

will be found in the H-R diagram to the right of the Sun and lower than the Sun.

4.a) Since the nucleus with mass number 8, which nucleosynthesis of two α-particles

will produce, is unstable, three α-particles are needed to produce stable nuclei

beyond mass number 8.

b) If triple - α reaction could not take place, carbon and heavier nuclei would not

have been synthesized. Further, since carbon is essential for the origin of life,

without triple - α reaction, life on the Earth would not have been possible.

5. The lifetime of a star on the main sequence is given by (Eq. (10.3)):

τ ∝ 5.2−M

Thus, for the Sun, we have:

Θτ ∝ 5.2−

ΘM

On the basis of above expressions for τ , we can write:

5.2

−

Θ

τ=τ

M

M

i) For a star whose mass is 10MΘ, we can write its lifetime on the main sequence

as:

yr1010=τ

5.2

10−

Θ

M

M

yr1010 5.210 −×=

= yr1010yr10 75.7 =

≅ 3 × 107 yr

40

From Stars to Our

Galaxy ii) For a star whose mass is 0.5MΘ, we can write:

2

1

=M

M

Thus,

yr2

110

2.5

10

−

=

≅ 6 × 1010

yr

6. We know that the luminosity of a star is given as:

L = 424 TR

So, we can write for the Sun:

42 4 TRL π=

After the Sun has expanded, we have, as per the problem:

R = 200 RΘ

and

T = 2

ΘT

Thus, the luminosity of the expanded Sun is:

4

2

2)200(4

σπ= Θ

TRL

= 16

10444

2

4

TR××

4

2

4

2

4

416

1044

TR

TR

L

L

×

××=

Θ

L = Θ× L3105.2

7. Fusion reaction stop at iron because iron is the most stable atom/nucleus. It cannot

combine with any other nucleus to produce heavier nuclei.

Terminal Questions

1. See text.

2. As per the problem, luminosity of the supernova, L = 1012

LΘ .

If the absolute magnitude of supernova is M and that of the Sun is MΘ, we can

write:

)(4.0

10 Θ−−

Θ

= MM

L

L

1012

= )(4.0

10 Θ−− MM

41

Nucleosynthesis and

Stellar Evolution 12 = − 0.4 (M − MΘ)

M = − 30 + 5

= − 25

If m is the apparent magnitude of the supernova, we can write:

5log5 −+= rMm

))3((log530 ×+−= (r = 3pc)

= −30 + 5 × 0.48

5.27−≈

The value (−27.5) of apparent magnitude of the supernova is approximately same

as that of the Sun. So, the supernova will be as bright as the Sun.

3. As per the problem,

L ∝ M

Now, the lifetime of a star can be approximated as:

L

M~

or τ ∝ M

M

= Constant.

So, the life time of such stars is independent of their masses.

4. You have seen in SAQ 2 of Unit 5 that, using Stephan - Boltzmann law, we can

express the luminosity of a star as

42 4 TRL =

where R is the radius of the star and T is its temperature.

We can, therefore, write the luminosity of a star like Aldebaran as:

424 TRLAld σπ=

= 42 K)3800()22(4 R×

And, the luminosity of the Sun can be written as:

=ΘL42

)K(60004 R×

Thus, we have:

4

4

6

)8.3(484×=

L

LAld

95=

42

From Stars to Our

Galaxy Further, the ratio of luminosity of Aldebaran and the Sun can be expressed in

terms of their absolute magnitude MAld and MΘ, respectively, as:

)(4.0

10 Θ−−= MMAld Ald

L

L

Thus, we get:

)50.4(

1095−−= AldM

because MΘ is 5. Taking logarithm on both sides, we get:

log 95 = − 0.4 (MAld − 5)

1.98 = − 0.4 (MAld − 5)

or, M = − 4.95 + 5

= 0.05

43

Compact Stars

UNIT 11 COMPACT STARS

Structure

11.1 Introduction

Objectives

11.2 Basic Familiarity with Compact Stars

Equation of State and Degenerate Gas of Fermions

11.3 Theory of White Dwarf

Chandrasekhar Limit

11.4 Neutron Star

Gravitational Red-shift of Neutron Star

Detection of Neutron Stars: Pulsars

11.5 Black Hole

11.6 Summary

11.7 Terminal Questions

11.8 Solutions and Answers

11.1 INTRODUCTION

In Unit 10, you have learnt about the evolution of stars, that is, how they live their

lives after being formed and take their respective positions on the H-R diagram. You

know that the evolution of a star is governed by two competing forces, namely, the

gravitational contraction and the radiation pressure due to energy generating

thermonuclear fusion reactions. In spite of the huge mass of a star, the amount of its

nuclear fuel is finite and it has to end ultimately. When it happens, a star can no longer

be prevented from contraction due to gravity. The density of the star increases

manifold and it turns into a compact object. In the present Unit, you will study about

such compact objects, also called compact stars.

There are many interesting questions pertaining to compact stars, such as: Does the

gravitational collapse of a star take place uninterrupted? If not, what is the mechanism

which balances the force of gravity? Do all stars end their life by becoming compact

stars of similar type? If not, what determine(s) the nature of the compact stars? You

will discover the answers to these and other related questions as you study this Unit.

A compact star can become a white dwarf, a neutron star, or a black hole depending

upon its initial mass. Further, the gravitational collapse of compact stars like white

dwarf and neutron stars is halted by the degeneracy pressure of fermions − a quantum

mechanical phenomenon. You will learn about the degeneracy pressure in Sec. 11.2.

The theoretical analysis of the relation between the nature of a compact star and its

mass was done by S. Chandrasekhar. This led him to predict a limiting mass for white

dwarf stars. You will learn about the theory of white dwarfs in Sec. 11.3. In Sec. 11.4,

you will study various characteristics of neutron stars and also understand why it is

difficult to detect them optically. All the theoretical predictions about neutron stars

could only be put to test when they were observed in the form of pulsars. In Sec. 11.5,

you will learn about one of the most interesting objects, called the black hole, which

physics has ever predicted. You will discover that the black hole signify the ultimate

victory of gravity in the evolution of stars.

44

From Stars to Our

Galaxy

Objectives

After studying this unit, you should be able to:

• understand the role of mass in deciding whether a compact star becomes a white

dwarf, a neutron star or a black hole;

• explain the concept of degeneracy pressure of fermions and its role in compact

stars;

• discuss the concept of Chandrasekhar limit and obtain an expression for it;

• understand the formation of neutron stars and its internal structure;

• derive an expression for the gravitational red shift of the neutron stars;

• describe the detection of neutron stars in the form of pulsars and discuss its

properties;

• explain the concept of Schwarzschild radius for black holes; and

• describe the geometry of space-time around the non-rotating black holes.

11.2 BASIC FAMILIARITY WITH COMPACT STARS

You may recall from Unit 10 that the gravitational force is balanced by the outward

force due to pressure gradient inside a star. The pressure inside is generated due to

thermonuclear reactions. With the passage of time, a star exhausts all its nuclear fuel

by ‘burning’, that is, due to nuclear reactions. The star cannot support itself against

gravity and begins to collapse. Eventually, the density of the star increases

tremendously and the star turns into a very compact object. At this stage, you would

like to know:

a) In the absence of radiation pressure due to nuclear reactions, what enables

compact stars to withstand gravitational squeeze?

b) Are all the stars similar to one another after their collapse?

To address the first issue, recall from the course entitled Physics of Solids (PHE-13)

that when atoms are very close to one another, their quantum energy states overlap

and the electrons in those orbitals behave as if they are free from their parent atoms. A

similar situation is obtained in compact stars where, due to very high pressure, all the

electrons are separated from their parent atoms. This phenomenon is called pressure

ionisation. A compact star is, therefore, a collection of nuclei and free electrons. You

have studied in Unit 10 how the pressure of the degenerate gas of free electrons

restrains the seemingly unstoppable gravitational contraction. This is because a

quantum state cannot accommodate more than one fermion (e.g., electron, proton,

neutron, etc.). This implies that, when the density of electrons is high, they are forced

to occupy quantum states with higher energies because lower states are full. In such a

situation, the pressure of the gas depends only on the density and is independent of the

temperature. You have learnt that gas of free electrons in such a state is called

degenerate electron gas and the pressure exerted by it is called degeneracy

pressure.

So, the gravitational collapse of compact stars is balanced by the degeneracy pressure

of electrons. It is, however, important to note that Pauli’s exclusion principle holds for

all fermions. You will learn later in this Unit that degeneracy pressure due to neutrons

plays an important role in stabilizing some compact stars.

The answer to question (b) raised above is: No. The nature of the remains of a star

after death depends on its mass. You may recall from Unit 10 that the mass of a star

plays a crucial role in its evolution and determines its luminosity. Similarly,

depending upon its mass, a dying star can turn into any one of the three kinds of

compact stars, namely a white dwarf, neutron star or black hole. You will study about

these compact stars later in the Unit. Here, it should suffice to say that it was genius of

45

Compact Stars the Indian astrophysicist, S. Chandrasekhar, who first showed that a degenerate star

cannot have mass larger than a certain maximum mass. He suggested, on the basis of

theoretical calculations, that the degeneracy pressure of electrons will be able to stop

further collapse of a star and its mass is less than a certain mass called Chandrasekhar

limit. The resulting star is called a white dwarf. If the mass of a collapsing star is

more than the Chandrasekhar limit, but less than 3 MΘ then the degeneracy pressure

of neutrons can halt the collapse. These stars are known as neutron stars. Further, if

the mass of the collapsing star is even higher, there is no way that the collapse can be

halted and the collapsing star becomes a black hole.

Thus, compact stars are simply the end products of ordinary stars and are

characterised by smaller sizes and higher densities. To compare and contrast their

sizes with that of the Sun, recall that the radius of the Sun is 7 × 1010

cm and its mass

is ~ 2 × 1033

g. A white dwarf of the same mass as that of the Sun would have a radius

about 100 times smaller, that is, around 109

cm. A neutron star of similar mass may

have radius of about 10 km only. And a mass equal to that of the Sun is too small for a

black hole! Generally, it is believed that the minimum mass of a black hole is three

times the mass of the Sun (3MΘ). You may ask: What is the radius of a black hole?

Well, that is a somewhat difficult concept. We will talk about it in Sec. 11.5 of this

unit.

Fig.11.1: Typical range of densities of celestial objects as a function of their typical radii

To get an idea about the average densities of compact stars vis-à-vis their radii, refer

to Fig. 11.1. Note that the main sequence stars, such as the Sun, have

density ~ 1 g cm−3

. On the other hand, the density of a white dwarf could be ~ 107 to

108 g cm

−3 and that of a neutron star ~ 10

15g cm

−3. A black hole, however, has no

defined average density, as you will learn later in this Unit. Also, the radius of a black

hole is a theoretical construct which means that if you come within this distance from

the centre (where all the mass is concentrated), you cannot escape from the

tremendous attraction of its gravitational force. Why only you? Even photons, the

fastest particles, cannot escape.

You know from the course

entitled Thermodynamics and

Statistical Mechanics (PHE-06)

that the behaviour of a gaseous

system can be understood on the

basis of its equation of state.

101 106 1011 1016 10-4

10-2

100

102

104

106

108

1010

1012

1014

1016

Sun Earth

White

dwarfs

Neutron

stars

Main sequence

stars

Radius (cm)

Den

sity

(g

cm

-3)

46

From Stars to Our

Galaxy

11.2.1 Equation of State and Degenerate Gas of Fermions

In the preceding paragraphs, you have learnt that the densities of white dwarfs and

neutron stars are very high compared to the densities of objects we come across in

everyday life, or even the densities of ordinary stars like the Sun. Therefore, to know

more about the compact stars, you should have some idea about how matter behaves

when the density is extraordinarily high. You may ask: Why should matter behave

differently? It is because the equation of state of matter at high densities is different

from the equation of state at ordinary densities. When the molecules of a gaseous

system are few and far between, as in our room, or in the atmosphere, collision is the

only way the molecules can interact with one another. Such a gas of molecules is

called an ideal gas and its equation of state is given by:

PV = NRT, (11.1)

where P is the pressure, V is the volume, N is the total number of particles, R is the

gas constant, and T is the temperature. It is evident from Eq. (11.1) that an equation of

state relates different thermodynamic quantities (such as pressure, volume and

temperature) of the gas. Such a relation is very useful in investigating the behaviour of

gaseous systems.

When the density of the gaseous system becomes as high as in compact stars, the

distance between two atoms/molecules becomes comparable to the size of the

atoms/molecules or even the size of the nucleons (i.e., protons, neutrons)! At such

densities, other forces such as the Coulomb force and nuclear forces begin to

influence the dynamical behaviour of the atoms/molecules. As a result, the equation of

state of an ideal gas cannot describe the behaviour of high density gases.

Now, to get an idea about the equation of state which can describe matter at high

densities such as that in compact stars, we note that compact stars consist mainly of

degenerate gas of electrons and neutrons. Further, you may recall from the course

Thermodynamics and Statistical Mechanics (PHE-06) that we need to use statistics to

describe the collective behaviour of a large number of particles (atoms or molecules)

of a system. Ordinary gases obey the so-called Maxwell-Boltzmann (M-B) statistics.

But, the fermions, which make up the compact stars, obey the so-called Fermi-Dirac

(F-D) statistics. So, in the context of the equation of state, our problem reduces to

know how F-D statistics influences the physical parameters such as pressure inside a

compact star.

You may further recall from this course that the number density

=

xdpd

dN

33of

particles in phase space can be written as:

fh

g

xdpd

dN

333= (11.2)

where p and x are the momentum and position variables of the particle, respectively,

h is the Planck’s constant and f is the distribution function. In Eq. (11.2), xdpd33 is

the volume element in the phase space, with d 3p containing the momentum elements

and d 3x containing the position elements; g is called the statistical weight, i.e.,

number of states that a single particle can have for a given value of the momentum p.

If S denotes the spin of particle, then g = 2S + 1. The distribution function f denotes

the average occupation number of a cell in the phase space.

The velocity distribution of

the particles of an ideal gas is

given by the well known

Maxwell’s velocity

distribution law which you

know already from the

kinetic theory of gases.

When the particle density is

low and temperature is high,

F-D as well as B-E (Bose-

Einstein) distributions become

identical to:

f (E) = e−(E−µ)/kT.

which is known as the

Maxwell-Boltzmann

distribution.

47

Compact Stars You may also recall that, for an ideal gas in equilibrium at temperature T, the F-D

statistics gives the distribution function f as:

[ ] 1/)(exp

1)(

+µ−=

TkEEf

B

(11.3)

where kB is the Boltzmann constant, µ is the chemical potential and E is the energy.

The energy distribution function given by Eq. (11.3) is valid for fermions such as

electron, proton and neutron which have half-integral spins (i.e., 1/2, 3/2 ..). From

Eq. (11.3), it is obvious that for T → 0, f (E) becomes a step function (Fig. 11.2), that

is,

f (E) = 1 when E < µ (11.4a)

and

f (E) = 0 when E > µ. (11.4b)

Fig.11.2: Fermi distribution function at T = 0 and at temperature, T > 0

When the distribution of fermions is similar to the T = 0 case shown in Fig. 11.2, the

gas of fermions is called completely degenerate. Physically, this means that all energy

states below a certain energy are fully occupied and the occupancy above this energy

is zero. The energy up to which all states are occupied is called the Fermi energy, EF

and for fermions at T = 0, EF = µ.

SAQ 1

Explain how Eqs. (11.4a) and (11.4b) are satisfied.

So, now you know the nature of f for a degenerate gas. The question is: How can we

use Eq. (11.2) to arrive at the equation of state for compact stars? Suppose you

want to know how many particles are present in a unit volume having all permissible

values of momentum. You have to simply integrate the number density in phase space

(left hand side of Eq. (11.2)) over all possible momenta, i.e.,

= pdxpdd

dNn

3

33 (11.5)

Spend

5 min.

0.5

1.0

0.0

0.0 1.0

T = 0

T > 0

Fra

ctio

n o

f st

ate

s o

ccu

pie

d

Energy (E/FF)

48

From Stars to Our

Galaxy

Now, if you want to know the energy density ε of particles, you have to integrate over

the energies of all the particles:

= ,3

33 pd

xdpd

dNE (11.6a)

where E is the energy of the particle given by:

( )2

1242

pcmE += (11.6b)

m being the mass of each particle and c being the velocity of light. Similarly, the

pressure of the gas defined as the rate of momentum exchanged across an ideal

surface of unit area, can be written as:

= pdxdpd

dNpvP

3

333

1 (11.7)

Actually, Eq. (11.7) shows that pressure can be expressed as the momentum flux of a

gas. Since we are discussing only the isotropic pressure here, the said flux in any one

direction (out of three) is one-third of the net momentum flux. That is why a factor of

1/3 has been put before the integral. Assuming that the motion of fermions in compact

stars is non-relativistic, it can be shown (we have avoided giving the mathematical

details) that Eq. (11.7) reduces to:

3

5

eKP ρ= (11.8)

where K is a constant and ρe is the density of electrons. Eq. (11.8) is the equation of

state of a degenerate gas of electrons at high density. Note that, unlike the ideal gas,

the pressure of a degenerate gas is independent of temperature. This implies that even

when temperature of a degenerate gas of fermions is very very low, it can exert

tremendous pressure.

Let us now discuss white dwarf stars in which only the electrons are degenerate.

11.3 THEORY OF WHITE DWARF

After a star has exhausted its nuclear fuel, it begins to collapse due to gravity. If the

mass of the star is less than about 8 MΘ, the gravitational contraction is accompanied

by expulsion of matter from its outer envelope. The discarded matter forms a ring-like

structure around the collapsing star and is called planetary nebula (see Fig. 11.3).

Fig. 11.3: A planetary nebula

The planetary nebula, so

called because of its planet-

like appearance, is visually

one of the most attractive

astronomical objects.

49

Compact Stars The core of the star continues to contract till its density attains a value in the range

105 − 10

8 g cm

−3. At this density, a new equilibrium sets in and these stars are called

white dwarfs. The evolution of a medium mass star into a white dwarf is shown

schematically in Fig. 11.4.

Fig.11.4: Evoluation of medium mass star leading to the formation of white dwarf

One of the interesting features of white dwarfs is that they have mass almost equal to

the mass of the Sun, but their radii are only about five to ten thousand kilometres. The

characteristic features of white dwarfs were explained theoretically by

S. Chandrasekhar.

11.3.1 Chandrasekhar Limit

S Chandrasekhar investigated the effect of the gravitational field on a degenerate gas

of electrons to estimate the mass-radius relation of white dwarfs. One of the

characteristics of a degenerate gas is that its equation of state does not involve

temperature. When particles are non-relativistic, the pressure exerted by such a gas is

given by Eq. (11.8). Now, inside a star in hydrostatic equilibrium, the pressure must

balance the gravitational pull towards the centre. For a star in hydrostatic equilibrium,

(recall from Unit 8 that) the pressure gradient can be expressed as:

2

1

R

GM

dr

dPe

e

=ρ

(11.9)

where RM and are the mass and the radius of the star respectively. If we assume that

the density of the star is uniform, we can write:

eRM ρπ= 3

3

4 (11.10)

Formation of planetary nebula Giant

region

Main

sequence

White

dwarfs

105

103

10

10-1

10-3

10-5

Surface temperature of star (K)

80000 40000 20000 6000 2000

Lu

min

osi

ty (

sola

r u

nit

s)

50

From Stars to Our

Galaxy

Substituting for e

ρ from Eq. (11.10) in Eq. (11.9) and integrating from the centre to

the surface, we get:

4

2

R

MPe ∝ (11.11)

But, from Eq. (11.8) and Eq. (11.10), we find that:

5

3

5

R

MPe ∝ (11.12)

Comparing Eqs. (11.11) and (11.12), we obtain:

3/1

1

MR ∝ (11.13a)

We can also write Eq. (11.13a) as

2

M∝ρ (11.13b)

Eq. (11.13) shows that, as the mass of the white dwarf increases, its radius decreases.

The mass-radius relation is plotted in Fig. 11.5. The shrinking of radius is

understandable because increase in mass would mean increase in the force of

gravitational contraction.

Fig.11.5: Mass-radius plot for white dwarfs

A logical question at this point is: What will happen if we go on adding mass to a

white dwarf? Eq. (11.13a) indicates that if enough mass is added to a white dwarf, its

radius would ultimately shrink to zero! What is the value of this mass?

Chandrasekhar showed that if mass of the white dwarf is about 1.4 times the solar

mass, its radius will shrink to zero. This is called the Chandrasekhar limit. If the

mass of the white dwarf is more than this limiting mass, its gravitational contraction

cannot be balanced by the degeneracy pressure of the electrons. Thus, a star of mass

greater than 1.4 solar mass cannot become a stable white dwarf unless it ejects mass in

some way. No white dwarf has been discovered which has a mass higher than

Chandrasekhar limit.

In the absence of energy generating nuclear reactions, there is actually no source of

energy left in a white dwarf. Therefore, these stars would go on shining by radiating

S. Chandrasekhar was

awarded Nobel Prize in

1983 for his extensive

theoretical work on white

dwarfs.

0 0.5 1.0 1.5 2.0 2.5 0

0.01

0.02

Chandrasekhar limit

Mass (solar unit)

Ra

diu

s (s

ola

r u

nit

)

51

Compact Stars their thermal energy and in the process, their temperature would decrease. Ultimately,

the entire thermal energy will be lost. The star becomes a cold object.

Would you not like to know how long a white dwarf star takes to cool down? The

internal temperature (say, Twd), of a white dwarf is almost constant. Its total thermal

energy can, therefore, be written as:

wdBe TkNU2

3=

wdB

p

Tkm

M

µ=

2

3 (11.14)

where M is the mass of the white dwarf. Eq. (11.14) has been written by taking the

energy of an electron at temperature Twd as (3/2)kBTwd, (1/2)kBTwd for each degree of

freedom. In Eq. (11.14), pmµ is the mean molecular weight of nuclei inside the star.

For stars which consist entirely of heavy elements, 2=µ . If we take M = 1MΘ, and

Twd = 107K, the total thermal energy of a white dwarf is ~ 10

48 ergs! This is a very

significant amount of energy. The question is: How long will this amount of thermal

energy enable a white dwarf to shine? To know this, solve the following SAQ.

SAQ 2

Suppose the luminosity of a white dwarf star of mass 1MΘ is 10−3

LΘ. If the luminosity

of the Sun, LΘ is 4 × 1026

Js−1

, calculate the time for which the white dwarf will keep

shining with its present luminosity.

Having solved SAQ 2, you know that the thermal energy of a white dwarf can keep it

shining for billions of years! Sirius B, the companion of the ‘Dog Star’ Sirius A, is the

earliest white dwarf star to be observed. Fig. 11.6 shows a white dwarf star.

Fig.11.6: A white dwarf star shown by arrow

Fig. 11.7 shows the distribution of the observed white dwarfs on the H-R diagram.

Different lines such as 0.89, 0.51 and 0.22 in the Figure indicate the masses of stars in

terms of the solar mass. White dwarfs are dimmer compared to the main sequence

stars of the same surface temperature because of their smaller size. The difference in

absolute magnitudes of white dwarfs compared to main sequence stars is in the range

5 – 10.

Spend

5 min.

52

From Stars to Our

Galaxy

Fig. 11.7: Location of white dwarfs in the H-R diagram

As we have mentioned earlier, white dwarfs are compact stars resulting from the death

of medium mass stars like the Sun. You may ask: What are the remnant of stars

which are much more massive than the Sun? Such stars end their lives as neutron

stars. You will learn about it now.

11.4 NEUTRON STAR

To understand the formation of neutron stars, you may recall from Unit 10 that when

the nuclear reactions stop, the core of a massive star (~ 10 MΘ to 100 MΘ) collapses

and a supernova explosion takes place. The supernova explosion blows away the outer

shell of the star. The question is: What is the remnant of a supernova explosion? In

1934, Walter Baade and Fritz Zwicky suggested that the core of a supernova

explosion could be a small, high density neutron star.

When the core of a star begins to collapse in the absence of nuclear reactions, its

density increases and attains a value comparable to the density of white dwarfs. The

degeneracy pressure due to electrons can balance the gravitational contraction only

when the mass of the object is smaller than the Chandrasekhar limit. If the mass is

more than this limiting value, the gravitational collapse continues and the density of

the star increases further. When the density is in the range ~ 1014

− 1015

g cm−3

, the

following two things happen:

a) protons and electrons combine to produce neutrons and neutrinos according to the

following inverse β-decay reaction:

e + p → n + v

b) atomic nuclei begin to disintegrate due to pressure ionisation.

Since neutrons are fermions, their degeneracy pressure halts the gravitational

contraction and a stable neutron star is formed. It has been calculated that the neutron

stars have radii of about 10 km and densities of their core is about 1014

to 1015

g cm−3

.

Note that neutron stars are much denser than the white dwarfs.

You may ask: Is there any limiting mass for neutron stars similar to the

Chandrasekhar limit for white dwarfs? Yes; but the precise value of the limiting

mass for the stars is difficult to determine because the behaviour of matter at such

high densities is not well understood yet. The estimated limiting mass is in the range

2 – 3 MΘ.

4.5 4.3 4.1 3.9 3.7 3.5

-4.5

-3.0

-1.5

0

Main sequence Sun

0.89 0.51 0.22

log T

log

(L

/L Θ

)

White dwarfs

53

Compact Stars

11.4.1 Gravitational Red-shift of Neutron Stars

You have just learnt that the neutron stars are very compact. Their gravitational pull

must be very strong indeed. How strong is its gravitational pull? To get an idea, we

may estimate the value of gravitational acceleration, g, on a neutron star, say, of mass

Mn ~ 1.5MΘ and radius Rn ~ 10 km. The expression for gravitational acceleration on a

neutron star can be written as:

2n

nn

R

GMg =

Substituting the values of Mn and Rn, we get, gn ~ 2 × 1014

cm s−2

. The value of gn is

~ 2 × 1011

times higher than the gravitational acceleration on the Earth! That is, the

gravitational force on a neutron star is 1011

times stronger than that due to the Earth.

Well, to get a feel for the perceptible effect of such a strong gravitational field, let us

do a thought experiment. Suppose you are standing on the surface of the star which is

emitting a yellow light at λ = 5800

A . The light reaches your eyes approximately one

metre above the ground. The interesting question is: Will you see the light as yellow?

If you think you will, you are wrong! Let us find out the reason.

A photon leaving the surface of the neutron star has energy hν and equivalent mass

hν/c2. The neutron star will attract this photon and, as a result, the photon has to work

‘hard’ against the star’s gravity to reach your eyes (just as a stone thrown upward

from the earth slows down as it goes higher). The work done by the photon to reach

your eye is, Hgc

hW n2

ν= , where H ~ 100 cm, is the height of your eyes from your

feet. Further, ν can be calculated using the relation λ = c / ν, i.e.,

ν ~ 5.172413 × 1014

Hz. Now, to arrive at your eyes, the photon works against gravity

and hence its energy, that is, frequency will decrease and wavelength will increase. To

calculate the change in wavelength or frequency, we can use the energy conservation

principle. Let ν′ be the frequency of the light (photons) that you observe. Thus, from

the energy conservation principle, we can write:

Whh −ν=ν′ (11.15)

Hgc

hh n2

ν−ν=

−ν=

21

c

Hgh

n

In terms of the wavelength of photons, Eq. (11.15) can be written as:

2

~c

Hgd n

λ

λ (11.16)

where dλ = λ′ − λ and λ′ = c / ν′. Substituting the values of gn, c and H in Eq.

(11.16), we get:

5102.2~ −×λ

λd (11.17)

Very often, we do thought

experiments: the experiments

which cannot be done for

practical reasons, but they

could be done, in principle.

54

From Stars to Our

Galaxy

Thus, yellow light of wavelength λ = 5800

A leaving your feet (the surface of the

neutron star) will arrive at your eyes as light of

wavelength

A13.5800~2

c

Hgn

λ+λ=λ′ . Such differences in the wavelengths can

be measured easily these days.

The increase in the wavelength of a photon when it comes out of a strong gravitational

field is called gravitational red-shift. The significance of this phenomenon lies in

the fact that scientists must make necessary corrections in the observed frequency of

photon presumably coming from objects such as neutron stars in order to understand

physical processes on their surface.

It is difficult to observe neutron stars optically because, being very small in size, these

are very faint objects. However, with radio telescopes, neutron stars have been

detected in the form of pulsars: the stars which emit regular pulses of radiation, very

often several times a second.

11.4.2 Detection of Neutron Stars: Pulsars

In 1967, a remarkable discovery was made in the history of astronomy by a student

named Jocelyn Bell who was doing her doctoral work under Prof. Anthony Hewish in

Cambridge, England. She observed certain periodic pulses of radio waves coming

from a certain direction in the sky which were repeated precisely every 1.337s

(Fig. 11.8). Very soon, she discovered a few more such objects with different

periodicities. These objects are known as pulsars. A year later, it was already clear

that these pulses must have been emitted by rotating neutron stars. Periodicities of

the pulsars have been found to be between 10−3

s and 4s.

(a) (b)

Fig. 11.8: a) Detection of pulses of radio waves, an evidence of the existence of neutron stars; and b)

pulsar at the centre of a nebula (Image credit: Chandra X-ray Observatory)

You may ask: How can we be sure that a strong pulse, whose periodicity is about

a second, is emitted by a neutron star and not by a white dwarf? When a star

rotates, each of its layers experiences centrifugal force directed outward as well as the

pointing gravitational force directed inward. If Ω be the angular velocity and R be the

radius of the rotating star, the two forces balance in a Keplerian orbit and we can

write:

2

2

R

GMR =Ω (11.18)

Time

Nu

mb

er o

f p

ho

ton

s

55

Compact Stars For a white dwarf, we can write .gcm10~~ 37

3

−ρR

M So the typical value of Ω is:

Ω ~ 2.58 s−1

. (11.19)

Therefore, the time period can be written as:

T = Ω

π2

~ 2.433s. (11.20)

From Eq. (11.18), it is clear that for objects with lower densities, the corresponding Ω

will also be lower, and hence the time period would be higher! Therefore, white

dwarfs and other bigger celestial objects are too big, their densities are too low and

they cannot emit pulses with periods as short as those observed in pulsars. In addition,

it has been observed that there are pulsars having periods of only a few milliseconds!

It is, therefore, almost impossible for white dwarfs to show such periodicities. So, the

belief that pulsars must be neutron stars became even stronger. Now, before

proceeding further, you should solve an SAQ.

SAQ 3

Calculate the gravitational red shift for the yellow light )A5800(

=λ on the surface of

Sirius B when the photon travels a distance of 1m. Take the mass of Sirius B as

Θ= MM SiB 1 , and its radius as 16000=SiBR km.

Emission mechanism and observed profiles

Regarding pulsars, a logical question could be: What causes emission of pulses from

a neutron star? It is suggested that a neutron star is like a gigantic light house. In a

light house, a powerful light-source rotates and ships see it periodically. In exactly the

same way, emission of radiation takes place continuously, but due to rotation of the

neutron star the emission is detected only periodically by an observer on the Earth

(Fig. 11.9).

Fig. 11.9: Light-house model of pulsars

Spend

5 min.

Magnetic axis

Radiation

Neutron star

Rotation axis

Radiation

56

From Stars to Our

Galaxy

You may argue: If emission takes place from the surface of a neutron star, it

should have been detected all the time and not periodically! Well, to understand

the periodic emission, you need to know that it is linked to the rapid rotation and

strong magnetic fields of neutron stars. Suppose a molecular cloud of radius, Rmc ~ 1

pc having a typical magnetic field, Bmc ~ 10−6

G collapses into a neutron star of size,

Rn ~ 10 km. From Unit 8, you know that if the matter is highly conducting, the

magnetic flux linked to it is conserved:

B r 2 ~ constant, (11.21)

Thus, we can write:

22mcmcnn RBRB = (11.22)

Substituting the typical values of Rn, Bmc and Rmc, we get:

Bn ~ 9.5 × 1018

G.

Thus, we find that the value of magnetic field associated with a neutron star is very

high compared to the Earth’s magnetic field which is only a fraction of a Gauss! The

above estimate has been arrived at under the assumption that there is no loss of

magnetic field and thus the value indicates the upper limit. In reality, after some

dissipation in the process of the formation of a neutron star, the field is smaller, close

to 1011

to 1014

Gauss or even less. Generally, this field is bipolar and the magnetic axis

is not aligned with the spin axis of the star (Fig. 11.9).

As the neutron star spins at great speeds, its magnetic field induces a very strong

electric field. As a result, electrons present in the atmosphere of the neutron star are

accelerated and attain very high energies. These electrons then gyrate round the

magnetic lines of force and emit radiation called synchrotron radiation which is

directed along the lines of force. Every time the neutron star rotates, each of the two

radiating magnetic poles may point towards us (Fig. 11.9) and we may see two pulses

of radiation per rotation of the star. In many objects, you can see these two peaks

very distinctly (Fig. 11.8). Thus, the lighthouse model tells us that pulsars are not

mysterious objects at all; they are rotating neutron stars.

You may further argue: If the neutron star is emitting radiation continuously, it

would lose energy; in the absence of nuclear energy, what energy is it losing −−−− the

gravitational energy, the rotational energy or the magnetic energy? If it loses

gravitational energy, the star would collapse further and the spin period may go down.

If it loses magnetic energy, the intensity of radiation will go down with time.

However, if it loses rotational energy, the pulsar would slow down and its time period

would increase. Observations support the argument that the most dominant component

of the energy loss is the loss of rotational energy.

So far, you have learnt that stars having mass up to 1.4 MΘ stabilise as white dwarfs

and those having mass up to 3 MΘ stabilise as neutron stars. Now, suppose that a star

has mass greater than 3 MΘ and all thermonuclear reactions have ceased in it. The

question is: What is the fate of such a star? What force will oppose gravity and

prevent the complete gravitational collapse of such stars? Obviously, the degeneracy

pressures due to electrons and neutrons are insufficient to halt the collapse because the

stellar mass is greater than 3 MΘ. In fact, there is no force which can halt the complete

gravitational crunch of such a star. This theoretical collapse of a star into a singular

point of zero volume and infinite density is called a black hole. You will learn about

black holes now. But, before that, how about solving an SAQ?

57

Compact Stars SAQ 4

Suppose the Sun shrinks to the size of a neutron star of radius 106 cm. Calculate the

magnetic field strength at the surface of the neutron star. Take the radius of the Sun to

be 1011

cm and the magnetic field at its surface equal to 1 gauss.

11.5 BLACK HOLE

When the mass of a star is more than the limiting mass (~3 MΘ) for neutron stars and

all nuclear reactions in its core have stopped due to the lack of nuclear fuel, there is no

force which can stop its continuous gravitational collapse. Due to such an unhindered

collapse, the object collapses to zero radius and infinite density! It is very difficult to

visualise such an object physically. However, such an object has an interesting

property. Its gravity is so strong that not even light can escape from it. Therefore, such

a body is called a black hole.

The physics of black hole involves very sophisticated mathematics which is beyond

the scope of this course. Nevertheless, we can apply simple principles of physics to

conclude that black holes do exist. Let us ask ourselves: For a given mass, what

should be the size of a body so that even light (that is, photons - the fastest

moving thing) cannot escape from it? The answer to this question has great bearing

on our understanding of black holes. The first theoretical attempt to address this

question was made by Schwarzschild who used the principles of General Theory of

Relativity given by Einstein. The main conclusions of Schwarzschild can also be

derived using Newtonian mechanics and the concept of escape velocity. You may

recall from school physics that, for a body of mass M and radius R, the escape velocity

is given by:

R

GMvesc

2= (11.23)

Since we are interested in a black hole and trapping of light by it, we put vesc = c, the

velocity of light. Then, Eq. (11.23) reduces to:

2

2

c

GMRg = (11.24)

Rg in Eq. (11.24) is called the Schwarzschild radius. Eq. (11.24) signifies that the

size of an object of mass M must shrink to Schwarzschild radius to become a black

hole. For example, if the object has mass equal to 1MΘ, its radius must be 3 km if it

has to behave like a black hole. In other words, the Sun must shrink to a radius of 3

km to be able to trap light and become a black hole!

SAQ 5

Calculate the Schwarzschild radius for the Earth.

You may ask: If black holes are point like objects, what is the physical

significance of Schwarzschild radius? Schwarzschild radius essentially means that

any object (including a photon) which comes within Rg (= 2GMbh /c2) of a black hole

is trapped. This limiting value Rg is also called the event horizon, since no event that

takes place within Rg from the centre of the black hole can be viewed by any observer

at R > Rg. Why is it so? This is because the gravity of the point like object is so strong

that even radiation (i.e. photons) cannot escape from the region inside the

Schwarzschild radius.

Spend

3 min.

Spend

5 min.

58

From Stars to Our

Galaxy

Although the concept of Schwarzschild radius is helpful in visualising a black hole,

you have to be careful in not stretching the meaning of escape velocity too far. This is

because of Einstein’s General Theory of Relativity (GTR) which treats gravity

entirely differently from Newtonian gravity. According to GTR, we cannot talk of

escape velocity at all. Einstein proposed that near a gravitating object, the photons

move in a curved trajectory as seen by an observer at a large distance. The black hole

being compact and massive, its gravitational force is very strong which bends the

photon path so much that a photon trying to escape would immediately return back.

Bending of light close to a black hole

As you know from the well known Einstein mass-energy equation (E = mc2), every

form of energy E has an equivalent mass m. So, the photons with energy E = hν will

also have a mass, mphoton = hν/c2. This is not the true mass since photons are really

massless. Nevertheless, this mass can be attracted by any other massive body such as

the Sun. This will bend the path of a photon. In fact, the bending of light was observed

by the famous British astronomer Arthur Eddington in the year 1919. He observed a

star during the total eclipse whose location in the sky was near the edge of the Sun and

found that the apparent location of the star has been changed by a small angle. This

convinced him that the light from that star must have been bent by the Sun. Thus,

strong bending of light can be taken as an evidence that black holes exist.

Types and location of black holes

The exact process of the formation of a black hole is not known but there are several

possibilities. Unlike the other types of compact stars, black holes do not have any

narrow range of masses. There seem to be two populations of black holes: one

population has a mass typically 6 to 14 times the mass of the Sun. It is formed due to

gravitational collapse after a supernova explosion. These so-called stellar mass black

holes could be detected all around the galaxy. Black holes of the second population

are very massive, with masses ranging from a few times 106MΘ to a few times 10

9MΘ.

These seem to be located at the centres of galaxies.

Detection of a black hole

Since radiation cannot escape from a black hole, it cannot be observed / detected

directly. However, it can be detected indirectly through its gravitational field. Any

matter close-by is strongly attracted by it. Suppose that the black hole is a member of

a binary. Then, it sucks matter from its companion. The matter flowing into the black

hole gets heated to a very high temperature and emits X-rays. Thus, to search for

black holes, we should look for X-ray binaries.

Now, let us summarise what you have learnt in this unit.

11.6 SUMMARY

• When the nuclear fuel of a star is exhausted completely, it contracts due to self-

gravity and becomes a compact star having density much higher than a main

sequence star.

• In the absence of radiation pressure due to nuclear reactions, the gravitational

force in the compact stars is counter-balanced by the degeneracy pressure of

fermions like electrons and neutrons.

• The nature of the remnants of a star after death depends on its mass; the dying

star turns into any one of the three kinds of compact stars, namely, white dwarf,

neutron star or black hole.

59

Compact Stars

• If the mass of a dying star is less than the Chandrasekhar limit (~ 1.4MΘ), it

turns into a white dwarf in which the gravitational force is balanced by the

degeneracy pressure of electron gas.

• If the mass of the dying star is more than the Chandrasekhar limit but less than

another limiting value (~ 3MΘ), it turns into a neutron star in which the

gravitational force is balanced by the degeneracy pressure of neutrons. If the mass

of the collapsing star is even higher, the gravitational collapse cannot be halted

and the collapsing star becomes a black hole.

• The degeneracy pressure of a degenerate gas of electrons is given by:

3

5

eKP ρ=

• Chandrasekhar showed that, for white dwarfs, the mass-radius relation is:

3

1

1

M

R ∝

and the density-mass relation for such stars is:

2M∝ρ

• Pulsars are stellar objects which emit periodic radio frequency pulses. Pulsars are

rotating neutron stars, because white dwarfs cannot emit pulses of this

periodicity.

• Emission of pulses from a rotating neutron star is explained on the basis of light-

house model.

• Black holes are objects of zero radius and infinite density. Such an object is

difficult to visualise physically.

• The Schwarzschild radius is given as:

2

2

c

GMRg =

and it signifies that the size of an object of mass M must shrink to Rg to become

a black hole.

• The Schwarzschild radius is also called the event horizon because an event that

takes place within Rg from the centre of the black hole cannot be observed.

• The physics of black hole cannot be understood on the basis of Newtonian

mechanics; we need to invoke Einstein’s General Theory of Relativity (GTR)

for this purpose.

11.7 TERMINAL QUESTIONS Spend 30 min.

1. For a completely degenerate electron gas, the number density of particles with

momenta in the range p and p + dp is given by:

60

From Stars to Our

Galaxy

dpph

dppn2

3

8)(

π=

Fpp ≤

0= F

pp >

where F

p is the Fermi momentum (corresponding to Fermi energy). Show that

F

p increases with n as 3

1

n . Also show that the average momentum < p > varies

as 3

1

n .

2. The masses and radii of a typical neutron star (NS), a typical white dwarf (WD)

and a typical main sequence star (MS) are given below:

Mass Radius

NS ΘM1 10 km

WD ΘM1 104 km

MS ΘM1 106 km

Calculate the rotational time periods in all these cases and show that only neutron

stars satisfy the pulse time periods observed for pulsars.

3. Assume that the peak wavelength of X-rays coming from an X-ray binary is

A1 .

Calculate the temperature that the falling matter onto a black hole must attain to

emit this radiation.

11.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. At T = 0, the exponent TkE B/)( µ− in the denominator of Eq. (11.3) goes to

infinity. Thus, for µ>E , the denominator of f (E) becomes infinite and we have

f (E) = 0.

For µ<E , the exponent TkE B/)( µ− in the denominator of Eq. (11.3) becomes

∞− at T = 0. Thus, the denominator of f (E) becomes 1 and f (E) = 1.

2. On the basis of Eq. (11.14), we find that the thermal energy of a star of mass

4810is1 ΘM erg. As per the problem, luminosity of the white dwarf star is:

Θ−= LL

310

)Js104(10 1263 −− ××=

130 s.erg104 −×=

Thus, the time for which the white dwarf will keep shining can be written as:

130

48

s.erg104

erg10

−×=t

61

Compact Stars

4

1018

=7103

1

×yr

×= s103yr1 7

≅ 1010

yr.

3. On the basis of Eq. (11.16), we can write the expression for the change in

wavelength on the surface of white dwarf (Sirius B) as

22

2.

R

GM

c

H

c

Hgd

wd

wd

×λ

=

λ=λ

29

332138

2220 cm)106.1(

g)102()sgcm1067.6(

)scm109(

cm)100()A5800(

×

××××

×

×=

−−−

−

8106.16.19

267.68.5 −×××

××=

A

≅ 8103.3 −×

A (Negligible)

4. From Eq. (11.21), we can write the magnetic field linked to the Sun and to the

neutron star (to which the Sun converts) as:

NSNS RBRB22 =ΘΘ = Constant

where BNS and RNS , respectively, are the magnetic field and radius of the neutron

star. Thus,

NS

NSR

RBB

2

2Θ

Θ=

= 1 gauss 26

211

cm)10(

cm)10(×

1010= gauss

5. From Eq. (11.24), we can write the expression for Schwarzschild radius for the

Earth as:

2

2

c

GMR E

Eg =

2220

272138

scm109

g)106()sgcm1067.6(2

−

−−−

×

××××=

= 0.9 cm

62

From Stars to Our

Galaxy

Terminal Questions

1. As per the problem,

dpph

dppn2

3

8)(

π=

π

=F

p

dpph

n

0

2

3

8

3

83

3

Fp

h

π=

Thus, we can write:

3

1

npF

∝

Now, the expression for average momentum can be written as:

< p > =F

p

dppnpn

0

)(1

×π

=F

p

dppnh

0

3

3

18

4

184

3

Fp

nh××

π=

4

18 34

3

n

nh××

π=

Thus, we can write:

< p > 3

1

n∝

2. From Eq. (11.18), we can write the expression for the angular velocity of a

rotating star as:

3

R

GM=Ω

So, the time period is:

Ω

π=

2T

GM

R3

2π=

63

Compact Stars Thus, the time period for the neutron star is:

g)102()sgcm1067.6(

cm102

332138

318

×××π=

−−−NST

267.6

10102 4

××π= − s

4104.5 −×= s

Time period for white dwarf is:

g)102()sgcm1067.6(

cm102

332138

327

×××π=

−−−WDT

≈ 17 s

And the time period for a main sequence star is:

g)102()sgcm1067.6(

cm102

332138

333

×××π=

−−MST

≈ 31017× s

Thus, we find that the time period of the pulses emitted by a neutron star is of the

same order as of the pulses from pulsars.

3. We know that,

Kcm3.0=λ Tm

As per the problem:

810A1 −==λ

m cm

Thus,

K103

K10

3.0

7

8

×=

=−

T

64

From Stars to Our

Galaxy

UNIT 12 THE MILKY WAY

Structure

12.1 Introduction

Objectives 12.2 Basic Structure and Properties of the Milky Way

12.3 Nature of Rotation of the Milky Way

Differential Rotation of the Galaxy and Oort Constant

Rotation Curve of the Galaxy and the Dark Matter

Nature of the Spiral Arms

12.4 Stars and Star Clusters of the Milky Way

12.5 Properties of and Around the Galactic Nucleus

12.6 Summary

12.7 Terminal Questions

12.8 Solutions and Answers

12.1 INTRODUCTION

In Unit 11, we discussed the death of stars and its consequences and thereby

completed our discussion about stars. You now know how stars are formed, how they

live their lives and how and why they die. If you wish to know more about the

Universe, you may ask questions like: What is the structure of the Universe? Is it a

single entity comprising unrelated and independent stars or does it consist of

substructures in which stars are arranged according to a scheme? If stars constitute

communities, what is the nature of such communities and what mechanism brings

them into existence? It is now believed that stars arrange themselves into billions

of self-contained systems called galaxies. Our star, the Sun, is one of about 200

billion stars in the galaxy called the Milky Way – our home galaxy. In the present

Unit, you will learn about the Milky Way galaxy also called the Galaxy.

On a clear night sky, you can see a broad white patch running across the sky. It is seen

during winter months in the northern hemisphere. The patch is actually made up of

hundreds of billions of stars, so close to each other that they cannot be seen

individually. This is a part of our home galaxy, the Milky Way. Just as Copernicus

discovered that the Earth and the planets revolved around the Sun, astronomers in the

early twentieth century discovered that the Milky Way system is indeed a galaxy in

which our solar system is situated.

In Sec. 12.2, you will learn the basic structure and properties of the Milky Way. It is

found that the Sun revolves once in 200 million years around the centre of this galaxy

in a nearly circular orbit. In addition, the Galaxy itself is rotating. You will learn the

consequences of the rotation of the Galaxy in Sec. 12.3. You will also discover that

the Galaxy is a spiral galaxy with several spiral arms. In Sec. 12.4, you will learn

about the various constituents of the galaxy, namely the stars, globular clusters,

compact stars and the interstellar medium. At the present time, we know a lot about

our Galaxy, its shape and its size. Much of the contemporary focus has been to

understand the central region of the galaxy and the activities surrounding it. Today, it

is believed that the central region contains a massive black hole of mass M ~ 2.6 ×

106MΘ. In Sec. 12.5, you will learn about the nature and characteristics of the central

region of the Milky Way.

65

The Milky Way Objectives

After studying this unit, you should be able to:

• describe the shape and size of the Milky Way galaxy;

• explain the consequence of rotation of the Galaxy;

• describe the major building blocks of the Galaxy including the stars and star

clusters;

• explain the nature and persistence of the spiral arms of the Galaxy; and

• describe the activities taking place near the centre of the Galaxy particularly the

nature of the central compact object.

12.2 BASIC STRUCTURE AND PROPERTIES OF THE

MILKY WAY

The Milky Way is a highly flattened, disk shaped galaxy comprising about 200 billion

(2 × 1011

) stars and other objects like molecular clouds, globular clusters etc. It is so

huge that, to travel from one edge of the Galaxy to the other, light takes about

100,000 years! Its radius is about 15,000 pc. The solar system is located roughly at a

distance of about 8.5 kpc from the centre of the Galaxy. The total mass of Milky Way

has been estimated to be about 2 × 1011

MΘ. Refer to Fig. 12.1 which shows a

schematic diagram of the Milky Way Galaxy. You may note that, broadly speaking,

the Galaxy can be divided into three distinct parts: a central bulge (B), the flattened

galactic disk (A) and a halo (H) which surrounds the Galaxy.

Fig. 12.1: a) A schematic diagram of the Milky Way Galaxy viewed edge on, b) view of the Galaxy

from the top; and c) part of the Milky Way visible in the sky

The Milky Way belongs to the Local Group, a group of 3 big and 30 odd small

galaxies. After the nearby Andromeda galaxy (M31) shown in Fig. 12.2, ours is the

30,000 pc

Sun 8500 pc

A B

H

(a) (b)

(c)

66

From Stars to Our

Galaxy

largest galaxy in the group. It has a few dwarf galaxies as satellites or companions.

Prominent among them are Large and Small Magellanic clouds.

Fig. 12.2: a) The Andromeda galaxy; b) the Large Magellanic Cloud; and c) the Small Magellanic

Cloud

Let us now discuss the nature of the components of Milky Way.

The Central Bulge

The central bulge (Fig. 12.1a) is a more or less spherical cloud of stars. Being located

in the disk region of the Galaxy, we cannot see this region in optical wavelengths. It is

so because the disk region consists of gas and dust which absorbs optical wavelengths

and obstructs our view. The total mass of the bulge is estimated to be about 1010

MΘ.

Apart from stars, this region consists of gas in the form of molecular clouds and

ionised hydrogen. The motion of the stars and the gas near the centre of the bulge

suggests that there could be a massive black hole at the centre.

The Disk Component

The flattened disk component has a radius of about 15,000 pc. But its thickness is very

small. Most of the stars are located along the central plane of the disk and as we move

away from this plane, the density of stars decreases.

(a)

(b) (c)

67

The Milky Way The most significant feature of the disk component is the existence of spiral arms.

Condensation of stars has been observed along the spiral arms. These arms have very

young stars called Population I stars, star-forming nebulae, and star clusters. The arms

are named after the constellations in the direction of which a large portion of the arm

is situated. Our solar system is located on a Local or Orion arm.

The Halo Component

The bulge and the disk components are surrounded by another, not so well defined,

and not so well understood spherical component called the halo component. This is

mainly made up of gas and older population of stars. These stars exist in very dense

clusters; each cluster having 105 to 10

6 stars. These are called globular clusters. Stars

in these clusters are so densely packed that they cannot be resolved, and clusters

appear like a circular patch of light (Fig. 12.3).

Fig. 12.3: A globular cluster (NGC 1916) containing millions of stars

The nature of galactic rotation (about which you will learn later in the Unit) suggests

that there is a large amount of matter which is governing the motion of the stars in the

disk. This matter is not visible in any wavelengths and scientists call it the dark

matter. It is believed that the halo contains at least an equal amount of matter as the

disk itself, if not more, in the form of dark matter. What could be the nature of this

matter? We do not know yet.

12.3 NATURE OF THE ROTATION OF THE MILKY WAY

In the early nineteenth century, Jan Oort and Bertil Lindblad studied the motion of a

large number of stars located near the Sun. Their study indicated that stars in the

Galaxy constitute a gravitational system. Like many other gravitating systems, our

galaxy also rotates, though very slowly. The rotational velocity enhances the stability

of the Galaxy since the outward centrifugal force can counterbalance the inward pull

due to gravity.

The assumption that the Galaxy is a gravitating system is at the core of the

traditional hypothesis about how the Galaxy came into being. Since the centrifugal

force is only along the equatorial plane, there is no obstacle for matter (other than the

pressure) to fall along the vertical direction. As a result, the initially spherical

distribution of matter has become, after over 10 billion years, the highly flattened

galaxy of today.

68

From Stars to Our

Galaxy

Refer to Fig. 12.4 which shows the initially spherical distribution of matter (Fig.

12.4a) gradually evolving into the present day flattened Galaxy (Fig. 12.4d).

(a) (b) (c) d)

Fig. 12.4: Evolution of the Galaxy (from a to d) starting from a spherical cloud of gas and settling

into the flattened disk shape of today

The rotational velocity of stars in the Galaxy is very slow compared to what we are

familiar with in our solar system. For instance, at a distance of roughly 8 kpc away

from the centre, the Sun takes about 240 million years to rotate once around the

galactic centre. The question is: How do we know about this rotation? How do we

measure it? What are the consequences of this rotation on the structure and

further evolution of the Galaxy? Let us now learn about these aspects of the Galaxy.

12.3.1 Differential Rotation of the Galaxy and Oort Constants

To appreciate the rotation of stars and other objects in the Galaxy, you need to

understand the concept of differential rotation. Let us consider the motion of a star in

the Galaxy. Suppose, for the sake of argument, that the whole mass of the galaxy is

concentrated at its centre and the stars move like planets round the Sun on orbits

called the Keplerian orbits. The angular velocity of a star at a distance r from the

centre can, therefore, be written as (Eq. (5.1), Unit 5):

.)(

2/1

3

=ω

r

GMr Gal

Kep (12.1)

where MGal is the mass of the Galaxy. We see from Eq. (12.1) that the angular

velocity is not a constant. Further, using Eq. (12.1) and the relation v = ωr, we can

write the rotational velocity of the star as:

vφ (r) ∝ r−1/2

. (12.2)

Now, if stars in the Galaxy are embedded as particles in a rigid body, then the angular

velocity, ω of stars would have been constant, independent of its distance from the

centre and the rotational velocity vφ (r) of a star in such a system is given by:

vφ (r) ∝ r. (12.3)

Comparison of Eqs. (12.2) and (12.3) clearly shows that the nature of the dependence

of rotational velocity on the distance from the centre is different in the two cases – the

Keplerian motion and rigid body rotation.

When different components of a system rotate independently, the rotation is known as

differential rotation. On the basis of observations, we can say that the stars indeed

have differential rotation. We do see the signature of this differential rotation in

astronomical observations.

69

The Milky Way

Fig.12.5: Geometry of the differential galactic rotation for stars closer to the Sun

Let us now obtain expression for the velocity of an object in the Galaxy. To do so,

refer to Fig. 12.5, which depicts the velocity vectors of the Sun (S) and a star (Q) with

respect to the galactic centre (C). Let us assume that the Sun and the star are at a

distance of 0R and R , respectively from C and let D be the distance between the Sun

and the star. Let V0 be the Sun’s rotational speed and V be the star’s rotational speed;

both assumed here to be on circular orbits for simplicity. Let us also assume that l is

the angle between the direction of the galactic centre and the direction of the star from

the Sun.

This is the so-called galactic longitude of the star. (Galactic longitude of the centre is l

= 0 by this definition). The Sun-star direction makes an angle α with the velocity

vector of the star. If ω0 and ω be the angular velocities of the Sun and the star,

respectively, we can write:

000 / RV=ω (12.4a)

and

RV /=ω (12.4b)

We are interested in finding the radial speed of the star with respect to the Sun. The

radial velocity is the velocity along the line joining the Sun and the star. We can,

therefore, write the star’s radial velocity with respect to the Sun, Vr, as:

Vr = V cosα − V0 sin l . (12.5)

where the first term on the right hand side is the component of the star’s space

velocity along the Sun-star direction and the second term is the component of the

Sun’s own velocity along the Sun-star direction. Thus, radial velocity is essentially the

difference between the projected velocities along the Sun-star direction. Substituting

Eq. (12.4) in Eq. (12.5), we get:

Vr = ω R cos α − ω0R0 sin l (12.6)

In triangle QSC in Fig. 12.5, the angle SQC is 90 + α.

S V0

l

90o +

V

Q

R

R0

C

D

70

From Stars to Our

Galaxy

Thus, using the law of sines, we can write:

or,

lR

R

R

l

R

sincos

sin)90(sin

0

0

=α

=α+

Substituting this value of cos α in Eq. (12.6), we get:

Vr = (ω − ω0) R0 sin l (12.7)

Similarly, the tangential component of the velocity of star with respect to the Sun can

be written as:

Vt = V sin α − V0 cos l = (ω − ω0) R0 cos l − ωD. (12.8)

In the solar neighbourhood, D << R0, and we can approximate the angular velocity of

the star, ω in terms of ω0 by using the Taylor series expansion. Thus, we can write:

)( 00

0

RRdR

d

R

−

ω+ω=ω (12.9)

If we define a constant A as,

−=

ω−=

000

00

2

1

2RR

dR

dV

R

V

dR

dRA (12.10)

then, the radial component of the velocity of a nearby star can be written as:

Vr = − 2A (R − R0) sin l = AD sin (2 l ) (12.11)

where, we have made use of the fact that for D << R0, (R0 − R) ~ D cos l (Fig. 12.6).

Further, using the same procedure and approximations, you can show that the

tangential velocity of the star is given by:

Vt = D [A cos (2 l ) + B] (12.12)

where B (= A − ω0 ) is another constant given as:

00

02

ω−

ω−=

RdR

dRB (12.13)

Since V0 = ω0 R0, Eq. (12.13) can be written as:

+−=

00

0

2

1

RdR

dV

R

VB (12.14)

The constants A and B are called Oort constants. Fig. 12.7 depicts the observed

variation of a star’s radial velocity as a function of galactic longitude. Note that the

Fig.12.6

C

Q

D l

R R0

R0 - R = D cosl

S

71

The Milky Way variation of radial velocity is periodic in longitude with a period of 180o and this

observation is consistent with Eq. (12.11). The value of R0 is 8.5 kpc. The estimated

values of Oort constants are:

A = 15 kms−1

kpc−1

and

B = −10 kms−1

kpc−1

Fig. 12.7: Variation of the radial velocity of stars as seen from the Earth

SAQ 1

a) Derive Eq. (12.12).

b) Show that

−=

00

0

2

1

RdR

dV

R

VA

and

+−=

00

0

2

1

RdR

dV

R

VB

On the basis of the above discussion, you know that the galactic disk in our

neighbourhood is indeed differentially rotating. You may ask: Do the stars rotate as

if the entire mass is concentrated at the galactic centre? To answer this question,

we must know the rotation velocity of galactic objects as a function of their distances

from the centre of the Galaxy. This is known as the rotation curve of a galaxy which

contains useful information about the mass distribution in a galaxy. Let us learn about

it now.

12.3.2 Rotation Curve of the Galaxy and the Dark Matter

The rotation curve of a galaxy is obtained by plotting the rotational velocities of

galactic components against their distances from the galactic centre. Refer to Fig. 12.8

which depicts the rotation curve of our galaxy. Note that it is not a smooth curve;

rather, it has various ups and downs. This nature of the curve is not understandable in

-40

-20

0

+20

+40

0 90 90 180 270 360

Galactic longitude (degrees)

Ra

dia

l s

pee

d (

km

s-1)

Spend

10 min.

72

From Stars to Our

Galaxy

terms of Keplerian motion of stars in the vicinity of the Sun which assumes that the

galactic mass is concentrated at the centre and the rotation velocity must be

proportional to R −1/2

(Eq. (12.2)).

Fig. 12.8: Rotation curve of the Milky Way

Rotation curves, such as the one shown in Fig. 12.8, indicate that the matter is spread

out all across the galaxy. This inference is, however, contrary to the concentration of

stellar mass around the nucleus of the galaxy. You may ask: What is the way out of

this paradoxical situation? If the rotation curve flattens out instead of falling

monotonically, we can infer that much of the matter in the disk cannot be in the form

of stars. How do we infer this? Let us try to understand it using the fact that the

centrifugal force RV2

φ must balance gravity and we can write:

2

2)(

R

RGM

R

V=

φ (12.15)

where, M(R) is the mass of the Galaxy up to radius R and Vφ is the rotational velocity

of a stellar object at a distance R from the galactic centre. If Vφ = V0, a constant, we

have from Eq. (12.15):

.)(2

0

G

RVRM = (12.16)

Eq. (12.16) indicates that the mass within radius R must increase linearly with R, and

not be concentrated entirely at the centre. But we do not observe this spread out mass.

This means that sufficient amount of matter in the Galaxy exists in a form which is

not detectable. What is this matter made up of ? No one knows. In the field of

Astronomy and Astrophysics, this is known as the missing mass problem. The

matter perhaps exists in a form that does not emit radiation and hence is also called

dark matter.

12.3.3 Nature of the Spiral Arms

In Fig. 12.1, we showed a sketch of our galaxy as it would be seen by someone

situated very high above and outside our galaxy. Note that it consists of spiral arms.

The prominent arms are (a) Norma Arm, (b) Scutum-Crux Arm, (c) Sagittarius Arm,

(d) Orion Arm, (e) Perseus Arm and (f) Cygnus Arm. The Sun’s location is on Orion

Arm (see Fig. 12.9). The spiral arms do not form a single entity that was originally

0 4 8 12 16 150

200

250

300

Sun

Rotation curve

Keplerian motion

Radius (kpc)

Ro

tati

on

al

vel

oci

ty (

km

s-1)

73

The Milky Way present. These are the result of dynamical interaction of the Galaxy with other

galaxies and the matter present in the inter-galactic space.

Fig. 12.9: Schematic diagram depicting some of the spiral arm structures of the Galaxy in which

dots denote stars

SAQ 2 How many times would the Sun have revolved around the centre of the Galaxy if it is rotating with a velocity of 250 kms

−1 at a distance of 8.5 kpc from the galactic centre?

Assume the age of the Sun to be 4.6 × 109 years. This age is different from the age of

the Galaxy, since the Sun is relatively younger.

To have a qualitative understanding of the spiral arms of the Galaxy, we need to ask

ourselves: How do we determine that the disk of the Galaxy comprises spiral

arms? In view of the differential rotation of the galactic disk, how do the spiral

structures persist for so long?

The first evidence of the spiral nature of the Galaxy came from the observations that

the distribution of O and B stars is not uniform in the Galaxy. To appreciate the

significance of this observation, you should recall from Unit 10 that O and B stars are

very massive, bright and short-lived. These massive stars, after their death, return

some of the stellar materials back to the interstellar medium. It is observed that these

stars occur around the locations where star formation can take place; that is, the

interstellar medium containing gas and dust. Therefore, it was suggested that the spiral

arms of the Galaxy contain gas and dust in the form of molecular clouds, and young

stars like O and B stars. But the problem was how to detect gas and dust at such

distances? This could be made possible by radio astronomy, particularly after the

discovery of 21 cm radiation. This enables astronomers to map the Galaxy and probe

its spiral nature. Astronomers use the intensity of the neutral hydrogen and carbon

monoxide emission lines to trace the spiral arms. Further, the young stars emit

ultraviolet light and ionise the gas which surrounds them. These ionised gaseous

regions are known as the H II (H II stands for ionised hydrogen) regions. They are

very luminous. Simultaneously, due to the recombination process taking place in these

H II regions, neutral hydrogen (H I) regions are formed which are detected by their

21-cm radio emission.

Spend

7 min.

Perseus arm

Orion-Cygnus arm

Sagittarius arm

To centre

1 kpc

Sun

74

From Stars to Our

Galaxy

Next logical question could be: How do these spiral structures persist because the

differential rotation of the galaxy should have destroyed them? To answer this

question, C.C. Lin and Frank Shu proposed the so called density wave model.

According to this model, spiral arms are not a simple fixed array of stars; rather, spiral

arms are the areas where the density of gas is greater than in other places. As such, the

arms and the space between the arms contain roughly the same number of stars per

unit volume. However, the arms contain larger number of brighter (O and B) stars.

According to Lin and Shu, the high density waves move through the galactic disk

which gives rise to the formation of stars. As interstellar clouds approach the density

wave, it is compressed, the collapse of interstellar gas is triggered and new stars are

formed. Thus, density waves are capable of generating all the constituents of spiral

arms. You must note that according to the density wave model, a spiral arm is not a

static collection of slow moving gas and stars; rather it is a dynamic entity which

always contains the same type of objects.

This brings us to the question: How does a density wave come into existence? The

density wave model does not provide a satisfactory answer to this question.

Astronomers believe that the death of massive stars causing supernovae explosions

may produce density waves. It has been found by computer simulation (experiments

on the computer) that the supernovae explosions combined with the Keplarian motion

can lead to the formation of spiral arms.

So far, you have learnt how the Milky Way looks like from the edge as well as from

above and what are its large scale basic components. Now we discuss the nature of the

building blocks, namely, the stars and the star-clusters of these components.

12.4 STARS AND STAR CLUSTERS OF THE MILKY WAY

You have learnt in the previous section that the components of the Galaxy consist of

stars and star clusters. The question is: Are the stars in all the regions of the Galaxy

similar? A detailed knowledge about the types of stars in each region can provide

valuable information about various aspects of the Galaxy itself.

Types of Stars in our Galaxy

Stars can be classified into the so-called population (I and II) depending on the

abundance of the metals (any element heavier than hydrogen and helium is called a

metal in astronomy) in them. To quantify, let us represent the fraction of hydrogen per

unit mass by X, of Helium by Y and of metals (all metals combined) by Z. The stars

which are very much metal deficient are called the population II stars. For such stars,

Z < 0.001. Population II stars are observed far from the galactic plane and are

primarily located in the halo. On the other hand, the metallicity of the stars increases

(Z → 0.01) as we approach the disk closer to the plane. These stars constitute the disk

population. Stars in the plane of the disk and in spiral arms are relatively young,

bright and blue in colour. They are called population I stars and they are metal rich.

You may ask: Why do the different regions of the Galaxy have different

metallicity? Actually, when the galaxy was originally formed, there was very little

metal in it, since no metal had been produced by that time. Initially, the stars formed

from material which had no metal (Z ~ 0). But as the density of the initial cloud, from

which the Galaxy was formed, increased towards the plane of the disk, massive stars

were forming closer to this plane and they evolved more rapidly than the halo stars.

They underwent supernova explosions and spewed out metals formed in them (refer to

Unit 10) in the interstellar gas. The interstellar gas was thus enriched with metals as it

approached the galactic plane. The second generation stars formed from this gas are

naturally more metal rich. The population I stars, therefore have higher metallicity

(Z ≥ 0.01).

75

The Milky Way Further, the globular clusters (consisting of roughly 104 to 10

5 population II stars

each) which formed a spherical halo around the Galaxy are found throughout the

Milky Way and other galaxies. Like the stars, the globular clusters also display a high

degree of gradient in their metal content depending upon their location in the Galaxy.

They are found in almost every component of the Galaxy: from galactic centre to very

far away in the halo. There are a few hundreds of these objects in our galaxy. Fig. 12.3

shows one such globular cluster. The typical size of a globular cluster is about 5 pc. It

has been found that the majority of stars in the globular clusters are the main sequence

stars while other stars in them belong to the red-giant branch, sub-giant branch and the

horizontal branch.

Since our galactic centre is about hundred times closer to us than the nearest well-

known galaxy Andromeda, we may assume that the astronomers would be most

excited to know the nature of the galactic centre. However, as we said earlier, we

reside on the plane of the Galaxy, and millions of stars and dust obscure our direct

view of the centre in optical, ultraviolet and soft X-rays. In infrared and radio waves,

the observation becomes easier. Given that the galactic centres are usually bright and

contain many stars, you may naturally be curious to know as to what may be going on

in and around the galactic centre of the Galaxy. In the next section, we shall describe

the motion of matter and gas within about 100pc of the galactic centre including the

properties of the possible black hole at the centre of the Galaxy and how it can be

detected.

12.5 PROPERTIES OF AND AROUND THE GALACTIC

NUCLEUS

Astronomers believe that at the very centre of the Galaxy, there exists a black hole of

mass 2.6 × 106MΘ. Modern radio telescopes using very large baseline arrays have

been able to resolve features of the galactic centre in great detail. It has been estimated

that about 10% of the entire mass of the galactic interstellar matter (that is, about

108MΘ) resides within a 100pc of the centre of the Galaxy. The number density of

hydrogen is about 102 cm

−3 which is about a hundred times greater than the average

density in the entire Galaxy. This interstellar gas is mostly concentrated in giant

molecular clouds (GMCs), of few parsec size, which are very hot (40-200K)

compared to the average gas in the galaxy.

Since the observed X-rays and γ-rays have energies ≥ keV, the gas must be very hot

(~ 107K or more). At least a dozen, very compact and energetic sources have been

detected by satellites such as EINSTEIN and ROSAT (sensitive to X-rays/γ-rays)

within the central region. Many of these compact objects have been identified to be

black holes and neutron stars. At the centre, Sgr A* itself is not very strong in X-rays,

and it is believed that this is due to the fact that not much matter is falling on it.

As we go closer to the centre, the activity increases and there is evidence of very high

rates of star formation. But since the disk does not contain enough mass supply to

continuously carry on star formation, the matter has to come from outside the region.

It is believed that interstellar gas accumulates at a rate of about 10−2

MΘ for some time

(say 107 years) and the accumulated mass then collapses and a fresh star formation

event is triggered due to density waves. This is probably what is happening from time

to time close to the centre of our galaxy.

One can easily compute the mass of the central compact object by using the well

known virial theorem:

2U + Ω = 0, (12.17)

where U is the total kinetic energy and Ω is the total potential energy. That is,

76

From Stars to Our

Galaxy

M < v2

> + Ω = 0, (12.18)

where < v2 > is the mean square velocity and

−=ΩR

RdMR

RMG

0

)()(

(12.19)

Using Eqs. (12.18) and (12.19), we can determine the enclosed mass at a given radius

if we know the stellar velocities data.

Fig. 12.10: Graph showing variation of the enclosed mass as a function of the radial distance close

to the galactic centre

In Fig. 12.10, we show the enclosed mass as a function of distance from the centre of

the Galaxy. It is clear that even though there are no measurements below R < 0.015pc,

the enclosed mass seems to have converged to a fixed number. The mass can be read

out from the plot itself: it is 2.6 ×106MΘ. The only acceptable solution seems to be that

a massive black hole of this mass resides at the centre. For a spiral galaxy this is not

very small, but probably in the lower end. Generally, in the centres of spiral galaxies,

black holes of mass M ~ 107MΘ are quite common.

SAQ 3

What is the mass density of the region around the galactic centre in units of (MΘ pc−3

)

below r = 0.015 pc?

Now, let us summarise what you have learnt in this unit.

12.6 SUMMARY

• The Universe comprises billions of self contained systems called galaxies. The

Sun – only star in our solar system – is one of about 200 billion stars in the galaxy

called the Milky Way, our home galaxy.

• The Milky Way is a highly flattened, disk shaped galaxy. Its radius is 15,000 pc

and its total mass is estimated to be about 2×1011

MΘ.

Spend

5 min.

0.001 0.01 0.1 1.0 10.0 105

106

107

Radius (pc)

En

clo

sed

ma

ss (

sola

r u

nit

)

77

The Milky Way • Broadly, the Milky Way can be divided into three distinct portions: a central

bulge, the flattened galactic disk, and a halo.

• The central bulge is a spherical cloud of stars and most of the mass (~1010

MΘ) of

the Galaxy is contained in it.

• The disk component consists of spiral arms; most of the stars are located along

the central plane of the disk.

• The halo is made up of gas and older population stars; these stars exist in very

dense clusters called globular clusters. The halo contains an equal amount of

matter as the disk itself in the form of dark matter.

• Like many gravitational systems, our galaxy also rotates. The rotational velocity

of stars in the Galaxy is very slow compared to what we are familiar with in our

solar system. The stars in the Galaxy undergo differential rotation.

• Rotation curve of a galaxy contains useful information about the distribution of

mass in it and it helps us understand whether or not most of the mass of the galaxy

is concentrated at the galactic centre.

• The spiral arms of the Galaxy do not form a single entity that was originally

present; rather, they result from the dynamical interaction of the Galaxy with

other galaxies and the matter present in the inter-galactic space.

• The persistence of spiral arms, despite differential rotation, is explained on the

basis of density wave model. According to this model, spiral arms are the areas

where density of gas is greater than other places. The arms and the space between

them contain roughly the same number of stars per unit volume. However, the

arms contain larger number of brighter (O and B) stars.

• Investigations indicate that a massive black hole, having mass 2.6×106MΘ,

resides at the centre of the Milky Way.

12.7 TERMINAL QUESTIONS Spend 30 min.

1. Suppose the contribution of the stellar disk to the mass of our galaxy is negligible

and the mass of the entire Galaxy is concentrated at the centre located at a

distance of 8.5kpc. What would be the mass you need to put at the galactic centre

in order that the estimated Oort constants agree roughly with those observed?

2. Describe the spiral arms of the Galaxy. Explain how they can persist for a long

time.

3. Distinguish between stars of population I and II. Where are they found in the

Galaxy?

4. It is believed that, in its early phase, the universe was very hot. The average

energy of particles at that time was ~ 15 GeV. What was the temperature in

degree Kelvin?

12.8 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. a) From Eq. (12.8), we have the expression for the tangential component of the

velocity of star as:

DlRVt ω−ω−ω= cos)( 00

78

From Stars to Our

Galaxy

And, from Eq. (12.9) we have

)( 00

0

RRdR

d

R

−

ω=ω−ω

0

cosRdR

dlD

ω−=

because lDRR cos~)( 0 − . Thus, we can write:

DdR

dRlDV

Rt 0

02

02

cos2 ω−

ω−=

because we can approximate, 0ω≈ω . Thus, we get

Vt

ω−

ω−= 0

02

02

cos2RdR

dRlD

ω−

ω−+= 0

0

02

)2cos1(RdR

dRlD

ω−

ω−

ω−= 0

00

002

2cos2 RR dR

dRl

dR

dRD

ω−

ω−

ω−= 0

00

002

2cos2 RR dR

dRl

dR

dRD

[ ]BlAD += 2cos

b) We can write:

00 RR R

V

dR

d

dR

d

=

ω

2

0

0

0 0

1

R

V

dR

dV

R R

−

=

where ω and V, respectively, are the angular velocity and radial

velocity of a star. Thus, we can write:

−=

00

0

2

1

RdR

dV

R

VA

and

+−=

00

0

2

1

RdR

dV

R

VB

79

The Milky Way 2. We can write the time period of the Sun’s rotation as:

0

02

V

RT

π=

Since the age of the Sun is 4.6×109 yrs, the number of times (n) it would have

revolved around the galactic centre can be written as:

n T

79 103106.4 ×××= s ( 1 yr = 3 × 10

7)

m)101.3105.8(2

)ms10250(s)1036.4(

163

1316

××××π

××××=

−

21

1.35.82

25036.4

≈

××π

××=

3. Assuming that the mass of the galactic centre is ~ 2.6 Θ× M610 , we can write:

Mass density 3

6

pc)015.0(3

4

106.2

×π

×= ΘM

if we take the mass enclosed within pc015.0≈r to be Θ× M6106.2 . So, we

have:

mass density 33

6

10pc)15.0(3

4

106.2

−

Θ

××π

×=

M

39

3pc10

)15.0(4

6.23 −Θ×

×π

×= M

311 pc108.1 −Θ×≈ M

Terminal Questions

1. From Eqs. (12.10) and (12.14), we have:

BAR

V−=

0

0

= (15+10) kms−1

. kpc−1

= 25 kms−1

.kpc−1

substituting the values of Oort’s constants A and B. Thus,

kpc)(8.5kpc.kms25 110 ×= −−

V

5.212= kms−1

80

From Stars to Our

Galaxy

So, for the given values of Oort’s constant A and B, the value of the star’s (the

Sun) rotational speed (V0) is 212.5 kms− 1

. Now, to determine the mass, MG at the

galactic centre whose gravitational force needs to be counter balanced by the

centrifugal force experienced by the Sun rotating with speed V0, we can write:

0

20

20

R

V

R

MG G =

G

RVM G

02

0=

21311

163621

skgm107.6

m)101.3105.8(10)ms5.212(

−−−

−

×

×××××=

( 1 pc = 3.1×1016

m)

Θ− ×××

×××××= M

3011

16362

102107.6

101.3105.810)5.212(

Θ− ×××

××××××= M

3011

163642

102107.6

101.3105.81010)125.2(

30102( ×=ΘM kg)

Θ

Θ

≈

×

××=

M

M

11

210

10

27.6

1.35.8)125.2(.10

2. See text.

3. See text.

4. You know from the kinetic theory of gases that, the average energy of the

particles at temperature T is given by:

TkE B2

3=

As per the problem, average energy

GeV15=E

J106.11015 199 ×××=

Thus, we have:

TkB2

3J106.11015 199 =××× −

or,

)JK1038.1(3

J)2106.15.1(

123

9

−−

−

××

×××=T

K1014≈

5

Galaxies UNIT 13 GALAXIES

Structure

13.1 Introduction

Objectives

13.2 Galaxy Morphology

Hubble’s Classification of Galaxies

13.3 Elliptical Galaxies

The Intrinsic Shapes of Ellipticals

de Vaucouleurs Law

Stars and Gas

13.4 Spiral and Lenticular Galaxies

Bulges

Disks

Galactic Halo

The Milky Way Galaxy

13.5 Gas and Dust in the Galaxy

13.6 Spiral Arms

13.7 Active Galaxies

13.8 Summary

13.9 Terminal Questions

13.10 Solutions and Answers

13.1 INTRODUCTION

On a clear, dark night, you can see the diffuse faint, narrow band of the Milky Way

stretching across the sky. It becomes broadened towards the constellation Sagittarius,

and is seen to be covered here and there with dark areas. The Italian astronomer and

physicist Galileo Galilei was the first to observe the Milky Way through a small

telescope. He saw in the faint band an array of stars and clusters of stars, interspersed

with dark patches.

Following Galileo’s pioneering observations in the 17th century, the Milky Way has

been studied extensively with a variety of telescopes of increasing sensitivity and

sophistication. We now know that the Milky Way is composed of more than a

hundred billion stars, spread in a large, thin disk with a bloated centre, which is

known as the bulge. The disk has a diameter of about 30,000 pc, but it is only about

1000 pc thick. There are large spiral features, known as spiral arms in the disk. An

object with a structure like that of the Milky Way is called a spiral galaxy. The Sun is

an inconspicuous star, situated in the disk of the Milky Way, at a distance of about

8,500 pc from the centre.

Our Galaxy is not the only one in existence. Within the observing limits of even a

moderately large telescope there are about 10 billion galaxies, covering a wide range

of sizes and shapes. While about half the galaxies have shapes like the Milky Way, a

large fraction of the rest have the appearance of ellipses. Many galaxies have irregular

shapes, while some are mere dwarfs compared to the larger systems. The spiral galaxy

NGC 4622, shown in Fig. 13.3, is similar to the Milky Way. If we could move out of

our Galaxy, and observe it from a great distance along a line of sight which is

perpendicular to the disk, it would show a more or less similar appearance.

Some galaxies occur as single objects, while others occur in groups of a small number

of galaxies or in large clusters containing thousands of galaxies. It is not unusual to

find two galaxies in collision with each other, or interacting with each other from a

6

Galaxies and the Universe

distance. The centres of a small fraction of galaxies contain what is known as an

active galactic nucleus. This is a tiny object compared to the whole galaxy, but emits

energy which can exceed by far the entire energy output of the rest of the galaxy.

Using very large telescopes on the Earth, or the Hubble Space Telescope (HST) which

is in orbit around the Earth, it is possible to obtain images of galaxies which are

located extremely far away. Light from these galaxies takes a long time to reach us.

Light from the galaxies that we detect now, began its journey towards the Earth so

long ago that the galaxies, and the Universe itself, were significantly younger at that

time. The galaxies at these early epochs are found to be significantly smaller and less

well-formed than the galaxies closer to us, which we observe at a much later time in

the history of the Universe. From these observations of distant galaxies, some idea is

now emerging about the formation of galaxies, and their subsequent evolution to their

present state. We shall consider some of these matters in some detail in the following

sections.

Objectives

After studying this unit, you should be able to:

• explain Hubble’s classification of galaxies;

• describe the properties of elliptical galaxies;

• state de Vaucouleurs law;

• describe the properties of lenticular and spiral galaxies; and

• distinguish between normal and active galaxies.

13.2 GALAXY MORPHOLOGY

A galaxy is a system of a very large number of stars, which are bound together by

their mutual gravitational attraction. A typical galaxy, like our own Milky Way,

contains ~ 1011

stars, spread over a region of size ~ 30 kpc, and has luminosity

~ 1011

LΘ, where LΘ is the luminosity of the Sun (Fig. 13.1).

Fig.13.1: The Milky Way as seen in the sky

7

Galaxies

Fig. 13.2: An artist’s sketch of the Milky Way Galaxy

In spite of their very great energy output, such galaxies appear to be very faint when

observed from the Earth, because of their vast distances from us. A few galaxies are

visible to the naked eye as faint patches of light on dark nights, and many more are

visible when a small telescope is used.

Faint, diffuse objects observed in the night sky are called nebulae. The nature of these

nebulae was the subject of intense debate in the 1920s. While some of the nebulae are

clearly objects in our own Galaxy, like remains of supernovae, it was not clear

whether some of the objects were inside our Galaxy, or at great distances outside it.

The matter was finally settled by Edwin Hubble who determined that the distance to

the Andromeda nebula is about 700 kpc by observing variable stars in it. The modern

value of this distance places the nebula far outside the confines of our own galaxy.

Hubble’s observation established that the nebula was an independent spiral galaxy; he

went on to study and classify many other galaxies. Fig.13.3: The spiral galaxy NGC 4622. The galaxy is located at a distance of about 70××××106 pc from us

8

Galaxies and the Universe

13.2.1 Hubble’s Classification of Galaxies

The observed images of galaxies show that they come in a wide variety of brightness,

shape, size and structure. Each observed galaxy is different in detail from other

galaxies, and when a large number of galaxies are examined, it becomes obvious that

there are some basic types into which galaxies can be classified. The first detailed

classification system was introduced by Edwin Hubble in 1936. This pioneering work

has been followed by other more sophisticated classification systems which take into

account the observed properties of galaxies in greater detail, but Hubble’s scheme is

the most widely used even at the present time, because of its simplicity and the insight

that it provides from observational details which can be easily obtained even with a

modest sized telescope.

Hubble’s scheme can be illustrated by his tuning fork diagram shown in Fig. 13.4.

At the left of the diagram, along the base of the tuning fork, are the elliptical galaxies,

which have simple elliptical shapes and appear to be smooth and without any

additional structures. Starting with the almost spherical galaxies of type E0, as one

moves towards the right, the images of galaxies become increasingly elliptical. The

sequence of elliptical galaxies terminates at the point where the two arms of the tuning

fork begin. Along the upper arm are the so called normal galaxies, which come in two

types: lenticular or S0 galaxies and spiral galaxies. As we move to the right along the

upper arm, from spiral type Sa to Sb to Sc, the bulges in the spiral galaxies become

less prominent, and the spiral arms appear to be more open. We will describe the

various galaxy types in greater detail in the following sections.

Fig.13.4: Hubble’s tuning fork diagram

The lower arm of the fork again has lenticular and spiral galaxies, but with a linear

central feature called a bar. These barred galaxies constitute about half of all

lenticular and spiral galaxies. The distinction between normal and barred galaxies is

not absolute, in the sense that most galaxies have some faint bar like features, but a

galaxy is called barred only when the bar is very prominent. Every galaxy type along

the upper arm of the fork has a barred counterpart along the lower arm.

The classification of galaxies as suggested by Hubble is based on very luminous

galaxies, with absolute magnitude MB ≤ − 20, which he called giant galaxies.

However, when galaxies in our neighbourhood are observed, it is found that the most

numerous galaxies are significantly less luminous and more compact than the giants.

These dwarf galaxies are designated as dE.

Sa

SBc SBb SBa

Sb Sc

S0 E7 E4 E0

Normal spiral galaxies

Barred spiral galaxies

Elliptical galaxies

Lenticular galaxy

9

Galaxies Many galaxies have highly irregular shapes, and prominent features like jets, tails and

rings. It is believed that these features are often produced because of interactions

between galaxies, which can lead to large scale disturbances in the distribution of stars

and gas in the galaxies. Many examples of galaxies in on-going close interaction can

be seen. Large galaxies can also swallow significantly smaller companions, and this

process of cannibalisation can lead to significant changes in the structure of the large

galaxy. It is difficult to classify galaxies with highly irregular structures because of

their complexity.

SAQ 1

A galaxy of absolute magnitude M = − 20 is at a distance of 700 kpc. Would it be

visible to the naked eye?

13.3 ELLIPTICAL GALAXIES

When elliptical galaxies were first photographed, they were observed to have elliptical

shapes and a smooth distribution of light (see Fig. 13.5). They were lacking in features

which are very prominent in spiral galaxies, like spiral arms and dark patches and

lanes of dust. Later observations with highly sensitive detectors have shown that

elliptical galaxies often do have faint features produced by dust and other factors.

These faint features turn out to be important indicators of the origin and evolution of

elliptical galaxies to their present form.

Fig.13.5: The elliptical galaxy M87

Elliptical galaxies have an enormous range of optical luminosities. The so called giant ellipticals have luminosities L ≥ L*, where L* ≈ 2 × 10

10LΘ is a characteristic galaxy

luminosity. The number density of galaxies declines sharply for luminosity L > L*. A

galaxy with luminosity L ≈ L* has an absolute magnitude of M ≈ − 20. The most

luminous elliptical galaxies can have L ≥ 100L*.

Elliptical galaxies with L ≤ 3 × 109 LΘ, i.e, M ≥ − 18, are called dwarf ellipticals.

Spend

5 min.

10

Galaxies and the Universe

13.3.1 The Intrinsic Shapes of Ellipticals

When the distribution of light in an elliptical galaxy is studied, it is found that the

isophotes, or curves of equal light intensity, are elliptical in shape; this in fact gives

this type of galaxy its name. In the simplest elliptical galaxies, all the ellipses have the

same centre, their major axes are oriented in almost the same direction, and the

ellipticities are nearly constant.

It was believed at one time that elliptical galaxies acquired their shape due to rotation.

The rotation would cause the galaxy to bulge in directions normal to the axis of

rotation, because of the centrifugal force. However, observations of luminous

elliptical galaxies show that they do not rotate fast enough for the observed flattening

to be due to the rotation. It is now known that the shape comes about because of the

way the stars in the galaxy move.

13.3.2 de Vaucouleurs Law

Most galaxies are so far away that the stars in them cannot be seen individually, and

we can only observe the integrated light from stars in different regions of the galaxy.

The appearance of an external galaxy is therefore like that of a diffuse object. It is

therefore appropriate that we measure the surface brightness of the galaxy, which is

the amount of light received per unit angular area of the galaxy, say one square arc

second.

The surface brightness of light along the isophotes of an elliptical galaxy decreases as

we move away from the centre. It was discovered by G. de Vaucouleurs that the

surface brightness is a very simple function of the length of the semi-major axis of the

isophote. If r is this length, then the surface brightness I(r) is given by

4/1)/(33.3

10)0()( errIrI

−= (13.1)

where re is called the effective radius and I(0) is the surface brightness at r = 0. The

total light emitted by the elliptical galaxy is given by

∞

− π×=π=0

23 )0(1037.32)( IrdrrrIL eE (13.2)

Half the total light of the galaxy is emitted from inside re:

E

r

LdrrrIe

=π0 2

12)( (13.3)

de Vaucouleurs’ law takes on a particularly simple form if the intensity is expressed in

magnitudes.

It can then be written as:

+µ=µ

4/1

1/4

325.8)0()( r

rr

e

(13.4)

Here µ(r) is the surface brightness of the galaxy expressed in magnitudes per square

arc second, and µ(0) is the corresponding magnitude at the centre of the galaxy. It

follows from this equation that a plot of the surface brightness against r1/ 4

should be a

straight line. The surface brightness becomes fainter by 8.325 magnitudes in going

from r = 0 to r = re.

11

Galaxies The light distribution in most elliptical galaxies does follow de Vaucouleurs law fairly

closely. NGC 661 is an excellent example of this; we have shown, in Fig. 13.6, a plot

of the surface brightness in magnitude against r1/4

. The plot is seen to be a straight

line, except in the central bright region where there is significant flattening of the

curve. Much of the deviation seen here is due to the effect of the Earth’s atmosphere.

Fig.13.6: The surface brightness distribution of the galaxy NGC661 (centred around 5500 Å

wavelength). On the x axis is shown r1/4, where r is the major axis distance of the elliptical isophotes from the centre. On the y axis is shown the surface brightness in magnitude, with the origin shifted for convenience

13.3.3 Stars and Gas

A study of the colour and spectrum of elliptical galaxies shows that they lack blue

stars. These stars are highly luminous, and are much more massive than the Sun. The

total lifetime τN of a star, during which the usable nuclear fuel in it is exhausted,

depends on its mass,

yr10~

5.310

−

Θ

τ

M

MN (13.5)

The blue stars being more massive than the Sun have a lifetime significantly less than

1010

yr, with the most massive stars having a lifetime as short as ~ 107 yr. The lack of

such stars means that there has been no star forming activity in ellipticals in relatively

recent times. This points to the absence of substantial amounts of cool gas mixed with

dust from which stars can be formed.

Giant ellipticals contain ≤ 108 − 10

9 MΘ of cool gas. Observations from X-ray

satellites have, however, shown that ellipticals can be highly luminous at X-ray

wavelengths, which indicates the presence of significant amounts of hot gas, at

temperatures of a few times 107 K. In very bright ellipticals, the mass of the gas can

be as high as ~ 1011

MΘ, which constitutes ~ 10 − 20% of the visible mass of the

galaxy.

SAQ 2

Explain in your own words why we expect the gas in elliptical galaxies to be hot.

Spend

5 min.

1 2 3 (Distance from centre)1/4

Surf

ace

brig

htne

ss in

mag

nitu

des

−−−− 10 −−−− 9 −−−− 8 −−−− 7 −−−− 6 −−−− 5 −−−− 4 −−−− 3

NGC 661

12

Galaxies and the Universe 13.4 SPIRAL AND LENTICULAR GALAXIES

Spiral galaxies are identified by a disk-like structure in which are present the spiral

arms. An image of the famous spiral galaxy M31, which is also known as the

Andromeda galaxy, is shown in Fig. 13.7.

A characteristic of the disk is that it is much extended but rather thin. When a spiral

galaxy is viewed face-on, i.e., when the normal to the disk is along the line of sight,

the disk appears to be circular, as in the case of NGC 4622 (see Fig.13.3). When there

is a non-zero angle between the normal and the line of sight, the disk then appears to

be elliptical, as in the case of the Andromeda galaxy. When the disk is viewed edge-

on, the disk appears to be rather thin, as in Fig. 13.8.

Apart from the disk, spiral galaxies contain a central bulge, which is quite obvious in

all the spiral galaxies. They also have a very large but faint halo, whose existence

becomes apparent from a detailed study of the distribution and motion of stars.

Lenticular galaxies, like the spirals, have a bulge and a disk, but the disk does not

contain spiral arms. The bulge and the disk here are of approximately equal

prominence. The bulge has properties very similar to elliptical galaxies, except that it

contains more gas and dust.

Fig.13.7: The Andromeda galaxy, which is spiral galaxy of type Sb, at a distance of 2.2 million light

years from us. Two dwarf galaxies, which are satellites of the Andromeda galaxy are seen in the image

13.4.1 Bulges

The bulge of spiral galaxies is a dense system of stars in the inner region of a galaxy,

more or less spherical in shape. The bulge of our own galaxy, the Milky Way, can be

seen, towards the constellation Sagittarius, and a bulge is clearly visible in the spiral

galaxies M31 and NGC891 (see Figs. 13.7 and 13.8). The properties of the bulges are

rather similar to the properties of elliptical galaxies.

In addition to the systematic motion, the stars in the bulges also have significant

random motion, which supports them against gravity. The number density of stars in

the bulge is about 104 higher than the number density in the neighbourhood of the

13

Galaxies Sun. Bulges have very little gas in them, except near the centre, and so there is not

much ongoing star formation. This means that the bulge does not have the short lived,

high mass blue stars which are present in the disks, because of the continuous star

formation taking place there.

The scale size of the bulges is typically a few kiloparsec (kpc), while the radius of the

disk of a spiral galaxy like ours is about 15 kpc. The prominence of the bulge relative

to the disk decreases along the Hubble sequence (see Fig. 13.4), with the bulges being

most conspicuous in spiral galaxies of type Sa and being almost absent in the galaxies

at the end of the Hubble sequence and beyond.

Fig.13.8: The spiral galaxy NGC 891, which is seen edge on. The thin disk and the bulge are clearly

seen. Along the plane of the disk is seen a dark band, which is a layer of dust present in the galaxy

13.4.2 Disks

The disks in the more luminous (MB ≤ − 20) spiral galaxies extend to ~ 15 kpc or

more, while their thickness is only a few hundred parsec, which makes them rather

thin. The disks are generally taken to be circular in shape (even though they may

appear to be elliptical, due to the projection effect mentioned above). The surface

brightness of the disk follows a simple exponential law. At a distance r from the

centre, measured along the mid-plane of the disk, it is given by

dr

r

dd eIrI

−

= )0()( (13.6)

where rd is the disk scale length. This is a few kpc for the typical spiral galaxy.

If we move normal to the disk keeping r constant, the surface brightness decreases,

and at a distance z from the mid-plane, it is given by

Id (r, z) = Id (r) exp −z/h (13.7)

where Id (r) is as in Eq. (13.6). The scale length h is a few hundred parsec. A galaxy,

whether spiral, elliptical or of any other type, does not have a sharp boundary. The

surface brightness, and therefore the number density of stars which produce the light,

decreases as one moves away from the centre, and gradually reduces to zero. In such a

14

Galaxies and the Universe

circumstance, it is best to characterize disk size by scale lengths like rd and h, and the

sizes of bulges and ellipticals with the effective radii.

The total light emitted by the disk is obtained by integrating Eq. (13.6) over the

surface of the disk:

∞

π=π=0

2 )0(2)(2 dddd IrdrrIrL (13.8)

If we assume that the distribution of light in the bulge is of the de Vaucouleurs type,

then the total light Lb emitted by it is given by Eq. (13.2). The ratio of the emission

from the disk to the emission from the bulge is then

22

)0(

)0(1094.5

×=

e

d

b

d

b

d

r

r

I

I

L

L (13.9)

For a galaxy without a significant disk component, like an elliptical, this ratio is zero.

For lenticular galaxies it is ~ 1, and it increases along the Hubble sequence towards

the late type spirals.

As star formation is an ongoing process in the disk, massive blue stars can be

observed in it. Since such stars have a short lifetime, they cannot be found in

environments where there is no star formation taking place like bulges of spiral

galaxies or elliptical galaxies. The presence of massive stars makes the disk bluer on

the whole than the bulge. Stars in the disk are in differential rotation around the centre

of the galaxy. You have already read about the differential rotation of stars in our own

galaxy, the Milky Way in Block 3.

About half of the disks of spiral galaxies, as well as lenticulars, have a linear structure

which is known as a bar. We have seen in Section 13.2.1 that the lower arm of

Hubble’s tuning fork diagram contains these objects.

SAQ 3

Explain in your own words why older galaxies should be redder.

13.4.3 Galactic Halo

A halo of stars and globular clusters is found to be surrounding the bulge and disk of

our Galaxy. The halo extends to substantial distances beyond the disk, and has the

shape of a flattened spheroid (see Fig. 13.9). Such halos are believed to exist in all

galaxies to a greater or lesser extent.

Globular clusters are gravitationally bound systems of 105 − 10

6 stars. These clusters

are found in the disk plane and close to it, as well as far from the plane. There are

about 150 globular clusters associated within our Galaxy, and these are found to be

distributed approximately in a sphere around the centre of the galaxy. A characteristic

of the globular clusters is that they are very old stars. The clusters far from the plane

are very metal poor, the abundance of the heavy elements in them being as small as

1/300 of the solar value. These clusters are estimated to be at least 11 − 12 Gyr old,

which makes them the oldest structures in our Galaxy.

We expect that other galaxies too would have halos like our own galaxy. One way of

tracing a halo in external galaxies is through the system of globular clusters, which

can be observed to great distances. Globular clusters have been observed around many

galaxies, and elliptical galaxies are found to be particularly rich in these objects.

Spend

5 min.

15

Galaxies SAQ 4

Explain why metal poor stars are very old.

13.4.4 The Milky Way Galaxy

Our Galaxy, called the Milky Way, is a spiral galaxy of type Sbc, which is

intermediate to Hubble types Sb and Sc. The bulge of the galaxy is clearly visible to

the naked eye in the direction of the constellation Sagittarius. The disk of stars is

visible as a diffuse band going across the night sky. A sketch of the structure of the

galaxy is shown in Fig. 13.9.

Fig.13.9: A sketch of our Galaxy, showing the bulge, disk and the halo

The visible bulge is about a few kiloparsec in radius, while the extent of the radius of

the disk is ~ 15 kpc. The Sun is located about 8 kpc or 28000 lys from the galactic

centre, some distance away from the mid-plane of the disk. The luminosity of the disk,

i.e., the total amount of energy per second emitted by all the stars in the disk,

is ~ 2 × 1011

LΘ, where LΘ = 4 × 1033

erg s−1

is the luminosity of the Sun. The total

mass of stars in the disk is ~ 6 × 1011

MΘ, where MΘ = 2 × 1033

g is the mass of the

Sun.

13.5 GAS AND DUST IN THE GALAXY

It was discovered through observations in the early decades of the 20th century that

inter-stellar space, i.e., the space between the stars in our Galaxy contains matter in

the form of gas and dust. The density of the gas is very low and it is difficult to detect

it. The dust usually occurs well mixed with the gas, and constitutes only about one

percent of the total material. The dust nevertheless is able to significantly obscure the

light coming to us from distant stars, and therefore can be easily detected. The gas and

dust together are called the inter-stellar medium (ISM) of the Galaxy.

Inter-stellar dust often occurs in the form of clouds, as is evident from the many dark

nebulae, like the Horsehead nebula (Fig. 13.10a), which are observed in the Galaxy.

The Eagle nebula and the details of the great clouds of dust observed by the Hubble

Space telescope are shown in Fig. 13.10b and c. The dust is mostly confined to the

disk of the Galaxy, and can be clearly seen in the Milky Way, even with the naked

eye, as great dark patches covering the galaxy here and there. It is specially so towards

the Galactic centre.

Spend

5 min.

16

Galaxies and the Universe

Fig.13.10: a) The Horsehead nebula; b) great columns of dust and gas in the Eagle nebula with c) its details as observed by the Hubble Space Telescope

The distribution of dust in the disk is particularly evident when a spiral galaxy is

viewed close to the edge of the disk, as in the case of the galaxy NGC891, which is

shown in Fig. 13.8. Dust affects light passing through it by scattering as well as by

absorption. Both absorption and scattering by dust remove a fraction of light coming

to an observer from a star, and the effect of the two together is termed as extinction.

The fraction of light lost due to extinction depends upon the wavelength of light, the

size of the dust grains, and the quantity of dust in the path of light.

For the dust found in the ISM of our Galaxy, the extinction is approximately

proportional to the reciprocal of the wavelength. This means that blue light suffers the

extinction most and near-infrared light the least. Because dust is concentrated in the

Galactic plane, a star in the plane observed by us is significantly dimmed and

reddened.

Gas and dust have been observed in other spiral galaxies as well. It was believed for

long that elliptical galaxies are free of gas and dust. But X-ray observations have

shown that these galaxies contain hot gas. Careful observation in recent years has

shown that a surprising number of elliptical galaxies also contain small quantities of

dust. This can be distributed in the form of a disk in the central region of the galaxy

and sometimes larger disks are also present. Some elliptical galaxies contain a large

(b)

(a)

(c)

17

Galaxies quantity of dust, which can be distributed over a region larger than the visible extent

of the galaxy. A good example of a giant elliptical galaxy containing dust very

prominently is the radio galaxy Centaurus A shown in Fig. 13.11.

Fig.13.11: The elliptical galaxy Centaurus A which is a highly luminous radio source. A prominent

dust lane is seen to be running across the face of the galaxy

SAQ 5

Explain why stars towards the centre of our galaxy appear fainter and redder.

13.6 SPIRAL ARMS

The disks of spiral galaxies contain spiral arms, which make these galaxies very

photogenic. The typical spiral galaxy has two arms, but many galaxies have three or

four arms. Spiral arms are found to contain many massive, blue stars. Since such stars

can only live for ≤ 10 Myr, the arms must be sites of continuing star formation. The

young blue stars are hot, and they emit radiation which can ionize any gas which may

be present around them. The gas produces emission lines which can be detected. Such

regions ionized by stars are known as HII regions. HII regions are abundant in spiral

arms, indicating the presence of large quantities of gas, as well as populations of

young massive stars. The arms also contain significant quantities of dust.

The disk in a galaxy rotates differentially, that is, stars which are closer to the centre

generally rotate with higher angular speeds than those which are further away. This

differential motion should lead to tighter winding of spiral arms. It can be shown that

for our Galaxy, the arms should have tightened significantly in less than 109 yr. Since

the Galaxy was formed more than 10 billion years ago, the arms should have been

much more wound up than what is observed. The fact that we do not observe such

winding up in our Galaxy and other similar galaxies is explained by the density wave theory of spiral arms. According to this theory, the stars and gas present in the arms

are not fixed there. As stars and gas move along their orbits in the disk, at some point

they pass through the arms and slow down. This leads to crowding of stars and gas in

the arms, and therefore to star formation. The arms can therefore be looked upon as

density waves.

Spend

5 min.

18

Galaxies and the Universe

Fig.13.12: Spiral arms in galaxies

13.7 ACTIVE GALAXIES

A very small fraction of all galaxies are found to be active, in that they emit very

significant quantities of energy which is not produced by the stars. The energy is

produced in a very compact region at the centre of the galaxy, and in many active

galaxies exceeds by far the energy produced by the stars. The compact region from

which the energy is produced is known as the active galactic nucleus (AGN).

A “normal” galaxy like the Milky Way contains ≥ 1011

stars, each of which emits

1033

erg s−1

. The total emission from all the stars is therefore ~ 1044

erg s−1

. This

energy is produced from a region which extends over a few tens of kiloparsec. An

AGN, on the other hand, produces energy at the rate of 1044

− 1047

erg s−1

, from a

region which is a fraction of parsec in size. The radiation produced by stars is thermal

in character, i.e., the spectrum of the radiation is similar to the spectrum produced by a

hot gas. The spectrum of radiation produced by an AGN is quite different, and in

simple cases has a power-law form. This means that the intensity varies as some

power of the frequency:

γν ν∝I (13.10)

where is a constant.

The spectrum of active galaxies is very broad, extending from the radio region to the

infra-red, optical, ultraviolet, X-ray and gamma-ray regions. The shape of the

spectrum indicates that the physical processes that produce the radiation are different

from the processes which produce the spectrum of a hot gas. These processes are

called non-thermal. The spectrum from stars has a number of absorption lines in it,

which are produced by the absorption of thermal radiation by atoms of specific

chemical elements in the cooler region of the stars. While the spectra of AGN do have

absorption lines, the spectra also exhibit very prominent emission lines, which are

produced by non-thermal processes.

There are different kinds of active galaxies. Seyfert galaxies are generally spiral

galaxies with an AGN which is moderately luminous. Radio galaxies are always

19

Galaxies elliptical, and are associated with highly luminous radio emission. BL-Lacs are active

galaxies which are ellipticals. Their characteristic is that, unlike as in the other AGN,

emission lines are either completely absent from their spectra, or are weak. The most

luminous AGN are the quasars. The luminosity of the AGN here is so high that it

outshines the galaxy. The appearance of the object is therefore that of a point source,

like a star. Observations by very large telescopes are needed to discern the galaxy

associated with a quasar. You will study more about active galaxies in Unit 14.

The source of the immense quantity of energy produced by an AGN is believed to be

a black hole, resident at the centre of the galaxy. The black hole is thought to be super

massive, i.e., it has mass in the range of ~ 106 − 10

9 MΘ. Matter from the region

surrounding the black hole falls onto it, and gravitational energy is released in the

process. This energy is taken up by electrons, which in turn emit the energy through

different processes, producing the observed spectrum. The different kinds of AGN

observed can be explained in terms of variations of this general theme. While the

above picture is more or less accepted, it is not completely proven yet.

It has been established that AGN were far more common early in the life of the

Universe than at present. Many galaxies which are normal now must therefore have

been active in the past. They must therefore contain super-massive black holes which

are dormant. Recent observations have shown that a number of normal galaxies may

indeed host such black holes, and it is even thought possible that all galaxies have

super-massive black holes at their centres. The role of black holes in the formation of

galaxies, and whether the black hole or the galaxy came first, is not understood yet

and is an area of active investigation.

SAQ 6

Explain the difference between thermal and non-thermal radiation.

In this unit you have learnt about galaxies. We now summarise its contents.

13.8 SUMMARY

• A galaxy is a system of stars, gas and dust, held together by the mutual

gravitational pull of these components. There are billions of galaxies in the

universe.

• Hubble classified these galaxies as elliptical, lenticular and spirals, both normal and barred spirals.

• The surface brightness of ellipticals varies according to the de Vaucouleur’s law:

+µ=µ

4/1

1/4

325.8)0()( r

rr

e

• Spiral galaxies are characterised by spiral arms and bulges. The spiral arms

contain clouds gas and dust and new stars are continuously being formed there.

Most galaxies have halos around them which contain extremely old stars.

• In the nuclei of many galaxies unusual phenomena take place involving release of

huge amounts of energy. These nuclei are called active galactic nuclei. It is

believed that this activity is caused by massive black holes sitting in the nuclei.

The Milky Way galaxy also has a 106 MΘ black hole at its centre.

Spend

5 min.

20

Galaxies and the Universe 13.9 TERMINAL QUESTIONS Spend 30 min.

1. Explain Hubble’s scheme of galaxy classification. Why has this scheme proved

enduring? What class has been assigned to the Milky Way Galaxy?

2. State de Vaucouleurs law and define the effective radius of an elliptical galaxy. Is

this law obeyed by elliptical galaxies?

3. Explain how the surface brightness in the disc of a spiral galaxy varies with the

distance from the centre of the galaxy. Show that the total light emitted by the disc

is )0(2 2dd

Irπ , where rd is the scale length of the disc and Id (0) is the surface

brightness at r = 0. What is the meaning of rd?

4. Define an active galaxy. What is the source of its activity?

13.10 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. M = m − 5 log r + 5 , where r is in parsec, here 7 × 105 pc.

− 20 = m − 5 (5.8451) + 5 m = − 20 + 29.2255 − 5 ≈ 4.2

Since human eye can see up to magnitude 6, this object will be visible to the

naked eye.

2. See Text.

3. In older galaxies, there are fewer younger (blue) stars but a large number of red

stars as you have studied in Unit 10.

4. See Text.

5. See Text.

6. See Text.

Terminal Questions

1. See Text.

2. See Text.

3. For the first part, see Text. For the second part,

dreIrL drrdd

/

0

)0(2−

∞

π= put r/rd = x

dxxerIx

dd−

∞

π=0

2)0(2

1 integral thesince ,)0(2 2 =π=dd rI

For the third part,

drrdd eIrI

/)0()(

−=

At r = rd, Id becomes 1/e of Id (0), i.e., intensity reduces to 1/e of its initial value.

4. See Text.

21

Active Galaxies

UNIT 14 ACTIVE GALAXIES

Structure

14.1 Introduction Objectives 14.2 ‘Activities’ of Active Galaxies

How ‘Active’ are the Active Galaxies? Classification of the Active Galaxies Some Emission Mechanisms Related to the Study of Active Galaxies

14.3 Behaviour of Active Galaxies Quasars and Radio Galaxies Seyferts BL Lac Objects and Optically Violent Variables

14.4 The Nature of the Central Engine Unified Model of the Various Active Galaxies

14.5 Summary 14.6 Terminal Questions 14.7 Solutions and Answers

14.1 INTRODUCTION

In Unit 13, you have become familiar with various types of galaxies. However, there are galaxies which do not fit into the scheme we described in the last unit. They require more detailed understanding. In this Unit, we shall study the nature of these galaxies. These galaxies are ‘active’, namely, there are signs of activity in them. For instance, their intensities may change significantly in a matter of hours to days. The activities may be manifested in the entire bandwidth of the electromagnetic waves, ranging from radio waves to γ-rays. One important point should be clear to you by now and that is: nothing in this Universe is truly inactive. So why should we learn about ‘active’ galaxies in a separate Unit, when all the galaxies are ‘active’? Truly, every galaxy is evolving and therefore changing with time. Only the degree of activity varies. Roughly speaking, we call those galaxies to be active which show properties such as very strong intensity variation in a relatively short period, and show ejection of matter at very high speeds, etc. Since they are compact, the nuclei of these galaxies show more ‘activity’. So, naturally, we are interested in studying these compact regions, which are known as Active Galactic Nuclei or simply AGNs. Several classes of objects may be called ‘AGNs’: These are: Radio Galaxies, Quasars, BL Lac Objects, Optically Violent

Variables, Seyfert Galaxies, Star-Burst Galaxies, etc. In this unit, we will try to understand how some of these objects behave and why they behave that way. Objectives

After studying this unit, you should be able to:

• describe the nature of activities near the active galactic nuclei;

• explain how these objects are classified;

• explain the importance of synchrotron radiation for the study of AGN; and

• describe the structure of matter very close to the active galactic nuclei.

22

Galaxies and the

Universe 14.2 ‘ACTIVITIES’ OF ACTIVE GALAXIES

You have learnt in the last Unit that there is something unusual going on in the nuclei of many galaxies. The ‘unusual’ may be the tremendous amount of energy released, or may be the emission of extremely high energy particles in the form of narrow jets. It is obvious that all galaxies are not equally active. Therefore, there is a need to discuss further how active an active galaxy really is.

14.2.1 How ‘Active’ are the Active Galaxies?

Active galaxies exhibit evidences of activity in a variety of ways. Their luminosities may be around 1044 erg s−1 (compare this with the luminosity of the Sun, which is 4 × 1033 erg s−1) and can go up to 1048−49 erg s−1. This tremendous energy release is possible only if these objects swallow a few stars per year and convert them into energy. Sometimes the luminosity may vary by tens of percents in a matter of hours! This gives you an idea of the compactness of the size of the nuclei of these galaxies. Recall that light travels at a speed of 3 × 1010cm/s and at this rate in, say, three hours time (~ 104s), it will travel 3 × 1014cm. This is roughly the size of the region from where most of the energy is released. You already know from Unit 11 that a black hole of mass M has a size of 3 × 105 (M/MΘ) cm. Thus, if the activity of the Active Galaxies is governed by a black hole at the centre, it must harbour a black hole of mass of around 109 MΘ or so. Another signature of the activity is that these galaxies emit non-thermal emissions in all the wavelengths, i.e., radio to γ-rays. Compared to a normal galaxy, where the nuclear brightness never masks the spiral arms or the rest of the galaxy, in active galaxies, the nucleus is so bright that the rest of the galactic structure is not seen in most cases. Active galaxies also produce (generally) jet-like outflows of very high energy particles on both sides, normal to the galactic plane. These jets are highly collimated and are continuously ejected. Normal galaxies do not produce these jets. Apart from these, sometimes the emission lines are broad and the state of polarization of the emitted light changes with time in a matter of minutes! Since a tremendous amount of energy is released continuously (rather than a one-time event as in a super-novae), it is natural to assume that the main reason of this is the gravitational energy released by the accretion of matter. What is the source of this matter? What is the nature of the accretion disks? How would the radiation be emitted and at what frequency? Why do the jets form? How are the jets collimated? All these important questions still puzzle astrophysicists today and only some of these questions have been answered satisfactorily. 14.2.2 Classification of the Active Galaxies

Active galaxies are classified according to some, often very vague, criteria. Broadly speaking they may be classified in the following way: a) Radio Galaxies: In these galaxies, the emission in radio waves is very strong,

larger than 1040 ergs/s. They could be further classified into extended radio galaxies and compact radio galaxies, depending on the size of the emission region.

b) Quasars: These objects were called Quasi-Stellar-Objects or Quasars because

they ‘looked’ like stars. Basically, in these objects, the nucleus emits most of the radiation. The behaviour in radio is often similar to that of the radio galaxies. In some, radio emission is not prominent and the quasars are called radio quiet quasars.

23

Active Galaxies c) Seyfert Galaxies: These are usually spiral galaxies with unusual nuclear brightness. Some, say, 5 to 10% of the Seyfert galaxies may be ellipticals as well. They have very broad emission lines in their spectra which distinguish them from other classes.

d) BL Lacertae Objects: These are commonly known as BL Lac objects. They are

identified by very rapid variability in radio, infrared and optical emissions. The light is strongly polarized and the polarization varies rapidly. They have no prominent emission lines and the spectrum is dominated by a continuum.

e) Optically Violent Variables: These are similar to the BL Lac Objects, but they

have weak, broad emission lines. f) Star-Burst Galaxies: In these galaxies the star formation rate is much higher

compared to the rates in normal galaxies. The radiation is emitted in the infra-red region. Star formation may be triggered by merger or collision of galaxies.

Fig.14.1: a) Radio galaxy 3C433 (image courtesy of NRAO/AUI sourced from www.nrao.edu/.../3c433_

3.5cm_opt_6in_med.jpg; b) Quasar 3C273; c) central region of the Seyfert galaxy NGC 1068; d) Starburst galaxy

M82

(a) (b)

(c) (d)

24

Galaxies and the

Universe Let us now learn about some of these objects which show major activities. 14.2.3 Some Emission Mechanisms Related to the Study of Active

Galaxies

One of the most important processes that you need to know while studying active Galaxies is the synchrotron radiation. You may also know this radiation as cyclotron radiation, a terminology used for radiation emitted by low energy electrons in the presence of a magnetic field. To emit this radiation two ingredients are required, namely, large number of high energy electrons and a strong magnetic field. Since the magnetic field is present everywhere in AGN, synchrotron radiation is a very common phenomenon. A relativistic electron of rest mass m0, charge e, velocity v (and Lorentz factor γ = (1 − v2/c2)−1/2) in a magnetic field B feels the force,

)()( 0 Bvv ×=γc

em

dt

d (14.1)

The electron moves in a helical path in the magnetic field (Fig. 14.2) with an angular frequency

15

11

0s rad

10

10108.1 −

γ

×=

γ=ω

G

B

cm

eBg (14.2)

Fig.14.2: Motion of an electron in a magnetic field and the emission of synchrotron radiation

(image credit Gemini Observatory)

The emission is in the shape of a broad spectrum with a peak at critical frequency ωc ~ γ2 ωg. Because of this continuous radiation, the electron loses energy and starts cooling with time. The cooling time scale is clearly the time in which the whole energy is radiated away: Tcool ~ γm0c

2/P s. (14.3)

where P is the energy emitted by an electron in one second.

Electrons in a helical path

Magnetic field

25

Active Galaxies To obtain the net emission from the entire gas you need to consider the velocity distribution of the electrons, which may or may not be thermal. In AGN the

distribution is non-thermal. For the non-thermal electrons, the emission spectrum looks like a power-law, i.e., the intensity of radiation is expressed as: F() = F0

−α (14.4) where α is a constant known as the spectral index. When α ≤ 0.4, we call the spectrum a flat spectrum, and when α is larger, we call the spectrum a steep

spectrum (Fig. 14.3). Synchrotron radiation is largely linearly polarized. Crudely speaking, an observer viewing the gyrating electron sideways will see an oscillating electron which emits a linearly polarized radiation. From above and below, oppositely directed circular polarization will be seen. These circular polarizations will cancel each other when overall integration is made but the linear polarization will survive and it can be detected easily. Another form of radiation which is very important is the emission line

component. Emission lines come from atomic transitions from one bound state to another (Fig. 14.4). Fig.14.4: Schematic diagram of an atomic transition from an excited state to a lower state and

photon emission

Similarly, when a photon of the ‘right’ frequency ν = (Ef − Ei)/h is incident on an atom, the photon may be absorbed. This absorption line will be very thin if the emitting atom is at rest, but if the atoms are moving around, the line may become broad due to Doppler effect: The line width d will then correspond to the velocity v of the emitting matter: d/ ~ v/c, where is the frequency of the radiation emitted in the rest frame and c is the velocity of light. From the broadening of the lines from Seyfert galaxies, sometimes the velocity inferred is anywhere close to 7 × 106 − 2 × 107 m/s. In the next few sections, we shall discuss some of the important classes of the Active Galaxies in more detail. After that, you will learn about the detailed nature of the activities close to the centre, i.e., the nature of the ‘central engines’ which drive these powerful sources in space. But before that, you may like to try an SAQ! SAQ 1

Explain why synchrotron radiation is largely linearly polarized.

Flat

Steep

ν

F (

ν)

Fig.14.3: Schematic diagram

of flat and steep

spectra

Level 1

Level 2

Level 3

Level 4

etc…

Energy

Spend 5 min.

26

Galaxies and the

Universe 14.3 BEHAVIOUR OF ACTIVE GALAXIES

In this section, you will learn about the properties and behaviour of quasars, radio galaxies, Seyferts, BL Lac objects and optically violent variables.

14.3.1 Quasars and Radio Galaxies A quasar (contraction of QUASi-stellAR radio sources) is an astronomical source of electromagnetic energy, including light that dwarfs the energy output of the brightest stars. A quasar may readily release energy in levels equal to the output of dozens of average galaxies combined. The best explanation for quasars is that they are powered by supermassive black holes.Quasars can be observed in many parts of the electromagnetic spectrum including radio, infrared, optical, ultraviolet, X-ray and even gamma rays while most quasars are found to emit in the infrared. The spectra of quasars exhibit characteristic strong emission lines rising above a broad continuum, which are red-shifted due to the high recession velocity of the quasar relative to us. For example, the Lyman line formed by transitions between the n = 1 and n = 2 energy levels in neutral hydrogen normally produces spectral lines with a wavelength of 121.6 nm or 1216 Å which is in the ultraviolet part of the spectrum. In quasar spectra, these are seen at wavelengths in the visible part of the spectrum (see Fig. 14.5). Fig.14.5: Spectrum of a quasar showing relative intensity on the y axis as a function of wavelength

(from www.eso.org) The nuclei of quasars emit UV radiation so strongly that the rest of the galaxy is masked. The radiation that is observed from radio wave to X-ray is totally non-thermal. Synchrotron radiation is the most likely source of this emission. Some quasars could be intense in radio emission while other quasars may have no

27

Active Galaxies significant radio emission. The power emitted in radio waves is usually smaller than that emitted in other wavelengths. In the infra-red region the emitted power is significant.

Fig. 14.6: Spectrum of a quasar in the X-ray region showing relative intensity on the y axis

(Source: www.estec.esa.nl)

Quasars are found to vary rapidly in a matter of few minutes to weeks, months and even years especially in the optical and radio regions. Since nothing can travel faster than light, the shortest time-scale of appreciable variation gives an idea of the upper limit of the size of the object as well. For instance, if a quasar intensity doubles in a day (tvar ~ 105s), the size of the region emitting most of the radiation must be smaller than the light-crossing time, i.e., Lsize ~ ctvar ≈ 3 × 1015cm. So the central region must be even smaller. This indirectly gives an estimate of the mass of the black hole:

M = ,MLsize

Θ×

1015

10103

where Lsize is in units of cm. Emission lines from the quasars are very broad and show characteristic red shifts because these objects are located at very large distances. The red shift indirectly tells you how far the object is situated. The width of the emission lines are determined by large scale motions of the emitting matter. The emitted frequency ν0 is related to the observed frequency ν by, ν = ν0 (1 + v/c). If the emitting atoms follow the Maxwell’s velocity distribution at temperature T, then we can obtain a simple expression for the intensity distribution I(ν) dν.

28

Galaxies and the

Universe We can write

( )

ν

ν−ν−=ν

20

20

2

0 2exp)(

Tk

cMII

B

pd (14.5)

Here I0 is the maximum intensity which occurs at ν = ν0, and Mp is the mass of the atom of the gas. As mentioned earlier, the width directly gives the information about the velocity, and therefore the temperature of the emitter. If we define half-width of the line as the width at half of maximum intensity, then we can show that

half width = 2ln2

2)(2200

cM

Tk

p

Bν=ν−ν (14.6)

SAQ 2

What is the half width d of a line of wavelength λ = 6300

A when the temperature of the gas is 105K? Assume O atoms to be the emitters.

The quasars which are located far away, i.e., quasars with large red shifts also exhibit a strong degree of absorption features. The red shifts of the absorption lines are almost always found to be less than or equal to the red shift of the emission line itself. This indicates that the absorbing gas is located in between the quasar, or is a part of the quasar itself. Radio Galaxies Those Active Galaxies which emit very large amount of energy in radio waves, in excess of, say, 1040 erg/s, are known as radio galaxies. Some of these could be compact radio galaxies, i.e., the emission is concentrated near the nuclear region. The others could be extended radio galaxies, i.e., the emission is seen in the form of two very large blobs which are well separated and usually co-aligned with the central, optically bright nucleus. Fig. 14.7 shows the nature of the radio emitting regions in a radio galaxy.

Fig.14.7: Radio emitting regions in a radio galaxy: jets from the radio galaxy 3C296. The infrared

regions are shown in red

Spend

10 min.

29

Active Galaxies A typical radio lobe in a bright radio galaxy has a luminosity of about 1043 or 10

44

ergs/s. But as we discussed in the last section, this is due to synchrotron radiation and it cools rapidly. In order that the blobs are produced and propagated for millions of years, they must be supplied with energy constantly. Indeed, higher resolution radio images point to an important finding: These radio emitting blobs are supplied with

energy by very narrow jets of rapidly moving matter. If one observes these jets at higher and higher resolution, the matter will be found to propagate in the same direction even close to the centre. This indicates that the prime cause of the jet production lies in the activity at the centre. We shall discuss this in Section 14.5.

14.3.2 Seyferts

Seyfert galaxies are named after Carl K. Seyfert who, in 1943, described their central regions as having peculiar spectra with notable emission lines. Seyfert galaxies appear to be normal spiral galaxies, but their cores fluctuate in brightness. It is believed that these fluctuations are caused by powerful eruptions in the core of the galaxy. In some Seyferts the nucleus outshines all the stars in the host galaxy. These active galaxies are distinguished from the others because of the presence of very ‘broad’ and ‘narrow’ (but broader compared to those from the normal galaxies) emission lines. These are known as the Type-I and Type-II Seyferts, respectively. Often in a given galaxy, features of both the types are seen. Fig. 14.8 shows the emission sectrum of a Type-I Seyfert galaxy.

Fig.14.8: Emission spectrum of the Type I Seyfert NGC 5548

In the presence of rapidly moving emitters, a line emitted at a fixed frequency ν in the rest frame of the atom, will be observed at a different frequency ′ and as we have discussed before, the shift in frequency is related to the velocity of the atom by

./~ cvd

ν

ν−ν′=

ν

ν (14.7)

Since the sign of v could be both negative and positive, the shift occurs on both the sides of the emitted frequency. Close to the centre of the Seyferts, there are emitting clouds randomly rotating around the nucleus and are responsible for the broad emission lines. These lines show

Rel

ati

ve

flu

x

Wavelength

30

Galaxies and the

Universe variability in a matter of weeks to months. From this, one can estimate the distance from the centre where the emitting clouds must be located. It turns out that they could be as close as 1016cm. The corresponding velocities could be as high as 7×106 − 2×107 m s−1. SAQ 3 Estimate the distance that an emission cloud must have from the centre of a Seyfert galaxy (having a central compact object of mass 107

MΘ) in order to produce a velocity of 106 m s−1. Use Kepler’s law.

Lines emitted from the clouds which are located very far away are narrow and they correspond to velocities of the order of only 200 − 500 km/s. Assuming that the rotation velocities of these clouds are Keplerian, a factor of twenty lower in velocity corresponds to a factor of four hundred larger in distance (for a given central object). Thus, narrow line regions (NLRs) are located very far away.

14.3.3 BL Lac Objects and Optically Violent Variables These types of active galaxies are characterized by a high degree of variability in radio, infra-red and optical wavelengths. For instance there are cases in which the intensity varies by a factor of 10 or more in a matter of weeks. From one night to another, variation up to 10-20% is very common. These objects have no emission lines and the polarization is large and rapidly varying. The emitted radiation is most concentrated at the centre and there is a weak sign, if any, of jets or radio lobes. In optically violent variables, there is a weak broad emission line component. Together these are known as Blazars. Fig. 14.9 shows a typical spectrum from a BL Lac object.

Fig.14.9: Spectrum of the BL Lac object Markarian 421 (Source: www.astr.ua.edu/ keel/agn)

Spend 5 min.

31

Active Galaxies 14.4 THE NATURE OF THE CENTRAL ENGINE

So far you have learnt about various classes of active galaxies. But what is really happening close to the centres of these galaxies? We have already mentioned that from the nature of rapid variability, one can infer that the masses of the central objects must be very high and must be concentrated in a compact region. Also, you have learnt that energy is released at a tremendous rate. For instance, a rate of 1047 erg s−1 corresponds to a destruction of matter of about 1026 g s−1 (using the famous formula of Einstein E = mc

2). So in one year, i.e., in 3.15 × 107 s, these galactic centres completely destroy about 3 × 1033 g of matter per year (which is equivalent to 1.5 times the mass of the Sun!). One of the simplest solutions to this energy budget is that the matter can be accreted on to the central compact object and the gravitational energy could be released. This energy is eventually converted into kinetic energy and then to thermal energy and radiation:

νhTkvmr

GMmBp

p~~

2

1~ 2 (14.8)

where, M is the mass of the central compact object, mp is the mass of the particle, T is the temperature of the gas, kB and h are the Boltzmann constant and Planck’s constant, respectively, v is the velocity of matter and is the emission frequency. As matter falls closer and closer to the centre, the velocity is increased. The gas becomes hotter and radiation is emitted at a much higher frequency. Now, what could the nature of the compact object be? In Unit 11, you have seen that only stable compact objects which are also massive are necessarily black holes. So, for the sake of argument, assume that the main cause of the activity is due to matter falling on to a super massive black hole. Does this picture explain everything? Roughly speaking, the answer is ‘yes’! 14.4.1 Unified Model of the Various Active Galaxies

So far, you have learnt how matter is accreted on to a black hole and how the radiation is emitted. What we have described so far is what goes on, say, within the inner 1000 rg where rg = 2GMBH /c

2 is the Schwarzschild radius of a black hole. When the black hole is super-massive, as in the case of our present discussion, the region farther out emits optical and infra-red radiation since there is supposed to be a large dusty torus. There are smaller clouds of gas, some of which are closer than this torus and move very rapidly. They are responsible for broad line emissions (BLR) since high velocity causes larger broadening of lines. Similarly some of the clouds move around very farther away and move slower. These clouds produce narrower emission lines. Thus, depending on the angle at which the observation is made, the dusty torus may block the BLR only selectively. The possibility that different classes of active galaxies may be actually due to such observational effects has been proved in the past most conclusively. This is because after prolonged observation, objects of a given class have been found to change their class altogether. In Fig. 14.10 we draw the unified model of AGN that is widely adopted in the literature, i.e., all AGN are the same. The differences in lines, jets and spectra may be due to different viewing angles. As the plane of accretion can be randomly oriented in the sky, different lines of sight will result in different kinds of observations. So, the same object may be seen differently and classified differently. It all depends on our line of sight!

32

Galaxies and the

Universe

Fig.14.10: The unified scheme for understanding AGN

In this unit you have learnt about the active galactic nuclei. Let us now summarise the contents of this unit.

14.5 SUMMARY

• Active galactic nuclei (AGN) are broadly classified as Radio Galaxies, Quasars,

BL Lac Objects, Optically Violent Variables, Seyfert Galaxies, Star-Burst

Galaxies, etc.

• The spectra of AGN reveal the nature and extent of activities in them, which serves as the basis of their classification.

• The emission spectrum of AGN obeys the power law in the radio region:

F() = F0 v−α

where α is a constant known as the spectral index. When α ≤ 0.4, the spectrum is called a flat spectrum, and when α is larger, it is called a steep spectrum.

• One of the most important processes occurring in AGN is the synchrotron

radiation.

• According to the current understanding, all the AGN may actually be the same object seen at different angles.

non-HBLR

Seyfert 2

Narrow line clouds

Jet

Black hole

Accretion

disk

Dusty

torus Broad line

clouds

Hot

dust Warm

dust

Cool

dust Cool

dust Warm

dust

Hot

dust

HBLR

Seyfert 2

Seyfert 1

33

Active Galaxies

14.6 TERMINAL QUESTIONS Spend 30 min.

1. Electrons having Lorentz factor γ = 100 are gyrating in a magnetic field of 1010G. At what frequency would its synchrotron emission peak?

2. Derive the expression for the half width (Eq. 14.6) caused by Doppler effect. 3. Explain the origin of broad-line and narrow-line regions in a Seyfert galaxy. 4. Describe briefly the nature of the central engine in an AGN.

14.7 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs) 1. See text.

2. Half width = 2ln2

220

cM

Tk

p

Bν

2ln106.116

101038.12

106300

224

516

8 −

−

− ××

×××

×= (in cgs units)

10693.06.116

38.12

63

10102 66××

×

×××=

10693.06.116

38.12

63

2001010 ××

×

×=

10107.2 ×= Hz.

3. 2

22

2

v

GMR

R

GMv

R

GMm

R

mv= = =

16

3378

10

10210107.6 ××××=

−

(in cgs units)

pc 0.043 cm1034.1 17 =×= Terminal Questions

1. 15

11 srad10

10108.1 −

γ×=ω

Bg

100

10

10

10108.11010

5

1011442 ××=ω=ωγ=ω ggc

1srad108.1 -19×=

34

Galaxies and the

Universe 2.

( )

ν

ν−ν−=ν

20

20

2

0 2exp)(

Tk

cMII

B

pd

00 at ν=ν= II max At half the maximum intensity

,20I

I =

Therefore,

( )2

1

2exp

20

20

2

=

ν

ν−ν−

Tk

cM

B

p

( )2

2exp

20

20

2

=

ν

ν−ν

Tk

cM

B

p

2ln)(

2 20

20

2

=ν

ν−ν

Tk

cM

B

p

Thus, Half width 2ln2

2)(2200

cM

Tk

p

Bν=ν−ν=

3. See Text. 4. See Text.

35

Large Scale Structure and

the Expanding Universe UNIT 15 LARGE SCALE STRUCTURE AND

THE EXPANDING UNIVERSE

Structure

15.1 Introduction

Objectives

15.2 Cosmic Distance Ladder

An Example from Terrestrial Physics

Distance Measurement using Cepheid Variables 15.3 Hubble’s Law

Distance-Velocity Relation

15.4 Clusters of Galaxies

The Virial Theorem and Dark Matter

15.5 Friedmann Equation and its Solutions

15.6 Early Universe and Nucleosynthesis

Cosmic Background Radiation

Evolving vs. Steady State Universe

15.7 Summary

15.8 Terminal Questions

15.9 Solutions and Answers

15.1 INTRODUCTION

In previous units you have learnt about stars and galaxies and their properties. You

have learnt that there are billions of galaxies in the universe. It is now time to estimate

the size of the universe and discuss the current ideas about its major components and

its origin. The science which deals with the origin of the universe is called cosmology.

In this unit, we discuss some aspects of cosmology. First we learn how distances of

distant objects can be estimated so that we can get an idea of the size of the universe.

Estimation of distances is a very tricky problem because there is no way of verifying

these distances directly. So, we look for internal consistency in various methods,

which means that the distance estimates given by them agree with one another.

In this unit we discuss the distribution of matter on very large scales also called the

large scale structure of the universe. We also look at the kind of matter that may be

forming the bulk of the universe. This matter is not visible and is called the ‘dark

matter’. It shows itself up only through its gravitational effect.

Objectives

After studying this unit, you should be able to:

• explain how astronomical distances are measured;

• state Hubble’s law and explain how Hubble’s constant indicates the age of the

universe;

• explain the need to postulate the existence of dark matter in the universe;

• derive Friedmann equation and solve it in simple cases; and

• explain why a hot and dense phase in the early universe is needed to explain the

existence of cosmic background radiation and to synthesise light elements.

36

Galaxies and the

Universe 15.2 COSMIC DISTANCE LADDER

An important step in this direction is to estimate the physical size of distant objects

and their distances from us. You know that to specify the position of an object in

three-dimensional space, we need 3 coordinates. Since we are observing from the

Earth (or its vicinity from satellites), it is best to use spherical polar coordinate system

with us as the origin. Two of the coordinates, namely, θ and φ (which are related to

the declination and right ascension, respectively) are easily fixed by pointing a

telescope in the direction of the object. Fixing the distance to an astronomical object is

relatively trickier.

In this section, we give an introduction to the basic principles that are used to measure

distances. The basic principle involved is to use the properties of the nearby objects

and deduce distances of similar objects farther off using these properties. Then we use

the latter objects to deduce the distances to objects still farther, and so on. This series

of steps which takes us from one step (in terms of distances) to the next step (in terms

of distances even farther) is termed the Cosmic Distance Ladder.

All this is best understood through an example of estimating distances on the Earth.

15.2.1 An Example from Terrestrial Physics

Consider the following situation. You are in a house and there are small plants around

the lawn in your house. Outside the house also there are similar plants and in addition

there are also some trees. We do not know the height of these trees. Very far away

there is the sea and on the beach also there are a number of such trees. The problem is

to estimate the distance of the sea from your house. To begin with we do not know

about the height of the trees.

The obvious thing would be to walk down to the sea (keeping in mind that all the

steps should approximately be of the same size) and count the number of steps. The

length of the step multiplied by the number of steps gives us the distance to the beach.

But you cannot use this method if you are not allowed to, or are unable to come out of

the house and go to the beach. The situation is more like this in the context of distance

measurement in astronomy. We can make direct measurement only on the Earth’s

surface. Since it will take about 1,00,000 years, even to reach the other end of the

galaxy by moving at the speed of light, we need to devise a better strategy.

Let us come back to our problem of measuring the distance to the sea beach. We will

put an extra condition that we are not allowed to come out of the gates of the house.

With this constraint we can proceed in the following way.

We first measure the heights h of the plants inside the garden of the house. Next we

measure the angle θ subtended by similar plants outside which are in the vicinity of

the trees. The distance of these plants from us equals .θ

=h

d Since the trees are in the

vicinity of these plants we know that their distance from us is also d. Further, we can

measure the angle θ1 subtended by the heights of these trees. The height h1 of these

trees is now given by h1 = d θ1. In this way we have achieved the first step in the

ladder of finding distances. We next use this information to estimate the distance to

the trees near the seaside. Since the trees near the beach are similar to the ones nearby,

we assume that their height is also h1. By measuring the angle subtended by these

trees we can compute the distance of these trees from us. Because of the physical

association of these trees with the sea side we know the distance of the sea from your

house. This is the second step of the ladder. Like this we can add more steps and go

on to estimate the distances of far off objects.

37

Large Scale Structure and

the Expanding Universe 15.2.2 Distance Measurement using Cepheid Variables

We will now give an example of using this principle to measure distances in

astronomy. In Unit 1 you have learnt the technique of measuring distances by parallax

method. In this section we will discuss how this method can be used to calibrate

another technique for distance measurements using Cepheid variable stars, which in

turn are used to measure even further distances. The brightness of Cepheid variable

stars is a periodic function of time (Fig. 15.1b).

Fig.15.1: a) Cepheid variable stars in NGC 300; b) the brightness of Cepheid stars as a function of time

We have a large number of such stars in our neighbourhood. Their distances can be

measured by the parallax method. From these distances and from their observed

apparent magnitudes, their absolute magnitudes and luminosities can be calculated. It

turns out that their absolute magnitudes are directly proportional to their periods

(Fig. 15.2). This is called the period-luminosity relation for the Cepheids.

Fig.15.2: Period-luminosity relation for Cepheids

0 5 10

Days

Vis

ua

l M

ag

nit

ud

e

3.0

3.5

4.0

4.5

5.0

(a) (b)

0 0.5 1.0 1.5 2.0 2.5

Log period (days)

Ab

solu

te m

ag

nit

ud

e

−−−− 8

−−−− 6

−−−− 4

−−−− 2

0

38

Galaxies and the

Universe This relation is firmly established for the local sample. Now we assume that the

distant sample of these stars also obeys this relation. So, the observed period of a

member of the distant sample is used to find its absolute magnitude. The apparent

magnitude can be observed directly. Using the relation between the absolute

magnitude and the apparent magnitude:

M = m + 5 − 5 log r, (15.1)

we can find the distance of this star. In this way, the Cepheid variable stars have been

used to find distances of nearby galaxies. Used in this manner, Cepheids are called

standard candles.

We can now use Cepheids to define some other objects, such as supernovae, to act as

standard candles to estimate even larger distances.

What is the result of these investigations? We find that distant galaxies are rushing

away from us with velocities which are proportional to their distances. This is called

Hubble’s law.

SAQ 1

Explain the concept of a distance ladder.

15.3 HUBBLE’S LAW

Hubble’s law is probably the single most important step in our attempt to understand

the Universe. This law was discovered by Edwin Hubble and it relates the distances of

galaxies with the velocities with which they are receding away from us (Fig. 15.3).

Fig.15.3: Hubble’s law

The velocity of any object can be split into a component that is along the line-of-sight

and another component that is transverse (perpendicular to) the line-of-sight. The line-

of-sight component of the velocity can be determined very accurately by Doppler shift

of the light that we received from the object. Hence this gives the velocity with which

the object is coming towards us or receding away from us. Let us look at this process.

Spend

5 min.

0 500 1000 1500 2000 2500 3000 3500 4000

Distance (Mpc)

300,000

250,000

200,000

150,000

100,000

50,000

Vel

oci

ty (

km

s−− −−

1)

39

Large Scale Structure and

the Expanding Universe 15.3.1 Distance-Velocity Relation

Using the methods similar to those mentioned above, Hubble estimated the distances

and velocities of a set of galaxies and plotted them. He found that the galaxies in

general seem to be receding away from us. This is popularly known as the expanding

universe. Further, the velocities with which they are receding away, are directly

proportional to their distances from us (Fig. 15.3). This prompted him to propose a

law, now known by his name.

Hubble’s law

v = Hr, (15.2)

where v is the line-of-sight velocity of an object, r its distance from us and the

proportionality constant H is called the Hubble constant.

The importance of this relation is that, once we know the velocity of a galaxy (by red

shift measurement), we can calculate the distance at which it is located. It is important

to point out here that Hubble’s law holds even if we were on some other galaxy. Our

location in the universe does not have any special importance.

Notice that in the above relation H is the slope of the curve shown in Fig. 15.3. Notice

also that 1/H has the dimensions of time. In a very simple picture, we can imagine that

all the galaxies which are today moving away from one another were at some time in

the past together at one point. Some event occurred at that time which triggered the

expansion of the universe. This event is usually called the Big Bang. The quantity 1/H

measures the time since that even, or the age of the universe.

SAQ 2

In astronomy, the velocity is measured in km s−1

. The distance of galaxies is measured

in Mpc or million parsec. Find the dimensions of H.

Unfortunately the measured value of H has lots of errors. But we know today that it is

roughly 70 km s−1

/Mpc−1

. Estimate the age of the universe.

15.4 CLUSTERS OF GALAXIES

Galaxies mostly exist as members of large groups called galaxy clusters

(see Fig. 15.4). A cluster of galaxies contains about a thousand galaxies. A galaxy

cluster consists of a variety of galaxies. It is an observed fact that the gross features of

these clusters are very similar. This fact alone has a far reaching implication. We

expect that roughly the same kind of physical processes are responsible for their

evolution. Further, it is natural to assume that they were created at different times and

began to evolve. This is because, in order to initiate the process of creation and

subsequent evolution of these clusters one needs a certain combination of

astrophysical conditions.

These conditions need not be the same at all points at a given time. Let us illustrate

this situation in the following manner:

Consider a set of systems, A, B, C, … Let the evolution in each of these systems be

governed by the same physical processes. If they started to evolve at different times,

we would expect that at any time, in particular today, they should be in different

stages of evolution. Hence, we would expect they should not show great similarity.

Spend

5 min.

40

Galaxies and the

Universe

Fig. 15.4: Cluster of galaxies

Alternatively, if we find that they look similar today, we may naively think that they

must have got created at the same time, so that they have had the same time for

evolution. For the systems under consideration, namely, the clusters of galaxies, we

know that they look similar but at the same time we have reasons to believe that they

started their evolution at different times.

Our simple-minded reasoning leads to the suggestion that they may have been created

at different times but they have reached some kind of steady state today. Such systems

have a simple, but important relationship for their kinematic parameters such as their

mass and velocity. This relationship is called the virial theorem which we now

describe.

15.4.1 The Virial Theorem and Dark Matter

The virial theorem says that if a system is bounded and is in equilibrium, then its

moment of inertia does not change with time. Such a system will have the following

relation between its total kinetic and potential energies:

2T + V = 0, (15.3)

where T is the total kinetic and V is the total potential energy of the system.

Suppose we have a spherical system consisting of N particles that are interacting

gravitationally. Let the position of the ith

particle be ri and its velocity vi. The total

kinetic energy is then

==

N

i ivmT1

2 .2

1 (15.4)

The total potential energy is obtained by summing over the potential energy of all the

pairs. We have

> −=

jiji

jimGmV

rr (15.5)

41

Large Scale Structure and

the Expanding Universe For a spherically symmetric distribution, V will come out to be proportional to

M (R) R−1

, where R is the radius of the system and M the total mass enclosed in a

sphere of radius R. The total kinetic energy can be estimated in the following manner.

If the velocities are in random directions, some of the particles of the system will

contribute a blue shift and some red shift. This will result in broadening of the spectral

lines. Hence, from the width of the spectral lines we can estimate the root mean

square (rms) velocity. This gives the total kinetic energy. If the system is in

equilibrium, we should have

T = − V/2 (15.6)

or

12 −∝ Rvrms

. (15.7)

When one tries to estimate the mass of the system in the above manner, one finds that

there is much more mass in the system as compared to that suggested by the luminous

mass alone. Hence, it is postulated that there should be a significant fraction of mass

in the form of dark matter. What form this dark matter takes, we do not yet know.

One possibility is that it consists of cold, burnt out stars which emit very little

radiation. Another possibility is that it is in the form of particles which interact very

weakly with normal matter. It is a very active area of research at present.

15.5 FRIEDMANN EQUATION AND ITS SOLUTIONS

Cosmology is the study of the overall features of the universe. In this framework, one

does not bother about local features of the universe like planets or stars. In cosmology,

we set up equations and solve them to obtain the very large scale features of the

universe.

To set up the basic equations governing the evolution of the universe as a whole,

Newtonian theory of gravity is inadequate and is, rigorously speaking, inapplicable.

The correct theory to use is Einstein’s General Theory of Relativity. This, however, is

outside the scope of this course. All is, however, not lost. It so happens that if we go

ahead and apply Newton’s laws (which strictly speaking is not the correct thing to do)

the final equations which we get are the same equations that result from the correct

theory, namely, the General Theory of Relativity.

The aim of this Section is to study the evolution of the Universe using these equations.

Hence, we will not be disturbed by the fact that the derivation of these is not rigorous.

We will happily go ahead and use these equations since we are aware that the final

equations are the correct ones.

We know that over a variety of scales the universe is not homogeneous. We have

planets, stars, galaxies and clusters of galaxies, which indicate that the universe is far

from being homogeneous. However, if we consider the universe as a whole, then at

large enough scales the universe seems to be homogeneous and isotropic. This last

point needs some explanation.

Consider a well maintained lawn. From a distance, the lawn appears uniformly green.

But as we start analysing the lawn on smaller scales, it begins to lose its homogeneity.

Consider a grass hopper sitting in the lawn. It observes the lawn at a scale which is

about the blade of grass. Clearly it will notice that the lawn is not at all homogeneous.

This means that the lawn is homogeneously green over distance scales which are

much bigger than the size of a blade of grass. In a similar way, we say that over scales

of sizes, much bigger than the size of galaxies, the universe is homogeneous.

42

Galaxies and the

Universe The equations that we now derive are the equations that govern the overall evolution

of a homogeneous universe. Consider two points A and O. With O as centre, and AO

as radius, draw a sphere. The force with which a test particle at A is gravitationally

pulled towards O can be calculated by just using the mass enclosed by the sphere. As

we know, the mass outside the sphere will not exert any net force on A.

Let the density of matter be ρ. Notice that density does not depend upon space. This is

because of our condition of homogeneity. However, there is no such restriction on

time dependence. So we will include time dependence for generality. The mass of the

sphere of radius R = OA is

M = ),()(3

4 3tRtρ

π (15.8)

The gravitational potential at A due to this sphere is − GM/R. Further, let Rv = be the

velocity with which A is moving with respect to O. Depending on the magnitudes of

the kinetic energy and the potential energy, the particle at point A may keep moving

away from O or may turn back and fall towards O. This is the well known condition

for escape velocity.

The total energy per unit mass of a test particle at A is

2

2v

R

GME +−= (15.9)

or

R

GMER

222 =− (15.10)

If the energy E is positive, then the distance between A and O will keep increasing and

if E is negative, A will attain a maximum distance from O and then begin to fall

towards O.

This condition on E can alternatively be expressed by replacing −2E by Ek 2. ; k = +1

corresponds to the case E < 0.

An alternative way of writing the last condition is then

R

GMEkR

22.2 =+ (15.11)

Dividing both sides by 2R and expressing the mass M in terms of the density ρ as

ρπ= 3

3

4RM

we get,

ρπ

=+3

82.

22

2G

R

Ek

R

R (15.12)

Defining a new variable ,2 ERa = we can now write the Friedmann equation

governing the evolution of the distance between two particles.

43

Large Scale Structure and

the Expanding Universe

Friedmann equation

3

8

22

2 ρπ=+

G

a

k

a

a (15.13)

We get the same equation from the general theory of relativity. The parameter k then

signifies the curvature of space. In the general theory of relativity, the effect of the

gravitational field is to make the space curved. The curvature of space is denoted by k,

which can take values +1, 0 or −1, depending on the overall density of the universe.

The quantity a is called the scale factor and the nature of a as function of t indicates

the nature of the expansion of the universe.

Fig. 15.5 shows the behaviour of a as a function of t for the three values of k, i.e., +1,

0 and −1. These curves are the solutions of the Friedmann equation. We see that

when k = −1 (which according to general theory of relativity implies that the overall

density of the universe is less than a certain critical density), or k = 0 (the overall

density of the universe equals the critical density), the universe keeps expanding.

When k = +1 (the overall density of the universe is greater than the critical density),

the universe expands up to a point and then starts contracting. Present observations

indicate that the universe will keep expanding, and its expansion will not be followed

by contraction.

In the early phase of the universe, the curvature must have been small, so it is

sufficient to consider the case of k = 0. The solution of the Friedmann equations, of

course, depends on the nature of energy density. If ρ ∝ a−n

, the equation can be solved

for k = 0 and the result comes out to be

nntna

/2/2)2/(∝ (15.14)

Fig.15.5: The variation of a with time

t

a(t

)

k > 0

k < 0

k = 0

44

Galaxies and the

Universe SAQ 3

Verify that Eq. (15.14) is a solution of Eq. (15.13) when ρ ∝ a− n

.

Time-temperature Relationship

We know from the thermodynamics of radiation that it has pressure and energy

density which we denote by prad and ρrad, respectively. They are related to each other

through the relation,

32cp radrad ρ= (15.15)

where c is the speed of light. The distance between points increases as R which, in

turn, is proportional to a. Hence, the volume increases as a3 and the energy contained

in it as ρ a3. Now, using the first law of thermodynamics, dU + pdV = 0, we have

.0][][ 332 =+ρ apdacd (15.16)

Applying it to radiation, we get

4−∝ρ arad

(15.17)

Using this relation in the expression for ρ derived in the last section (with n = 4) we

get,

ta ∝ (15.18)

At the same time we know that the temperature of radiation is related to its energy-

density by

4Trad ∝ρ (15.19)

for isotropic and homogeneous radiation field.

From the last three equations, we get the relationship of temperature with time as

tT /1∝ (15.20)

We see that t = 0 is both interesting as well as disturbing. This is because at that point

of time, the temperature shoots to infinity and so does the energy density. It is this

epoch which is termed as the Big Bang.

15.6 EARLY UNIVERSE AND NUCLEOSYNTHESIS

We saw that the temperature of radiation varies as inverse of the square root of time.

The immediate consequence of this is that the radiation temperature in early times

should have been very high. If matter and radiation were in thermal equilibrium, the

above statements imply that the temperature of matter was also high in early times.

We know from thermodynamics that temperature is a measure of the mean kinetic

energy which in turn implies high energy collisions between particles. The earlier the

epoch, the higher the energy with which these particles collide with each other. Today

we know about the physical phenomena at high energies from experimental

investigations using high energy particle accelerators and colliders. We should expect

Spend

5 min.

45

Large Scale Structure and

the Expanding Universe that the same phenomenon must have taken place in the early universe. In fact, for this

reason, early universe is often called the poor man’s laboratory.

An important class of reactions at high energy is those which lead to the synthesis of

nuclei of elements. Very early on, there were no complex nuclei. The only ones were

the hydrogen nuclei, i.e., protons. At those energies, even if the protons and neutrons

combined to form higher atomic number nuclei (e.g., the Helium nucleus), the kinetic

energies of the particles were so high that the collisions would have immediately

disintegrated them.

As the universe cools, a certain temperature is reached when the energies are low

enough that this backward reaction (namely disintegration) begins to get suppressed.

Hence, stable helium nucleus begins to get formed. As the temperature lowers further,

we expect that higher atomic number nuclei will begin to get synthesized. So the

question is, “Can we proceed in this way and synthesize all the naturally occurring

nuclei”? The answer unfortunately is no!

This line of reasoning works for only the elements with first few atomic numbers. As

the temperature decreases, we can form Lithium and some Boron. The problem comes

up when we need to form Beryllium. In the process of the formation of the stable

Beryllium nucleus, one passes through an intermediate stage where, spontaneous

disintegration is faster than the fusion. So even before there can be fusion the nucleus

which is supposed to participate in the fusion, disintegrates. Hence one cannot form

Beryllium by this procedure. Only after the formation of Beryllium, can the nuclei of

higher atomic numbers be formed. Hence this is called the “Beryllium Bottleneck”.

The universe has to wait for a very long time, namely, till stars form, in order to

synthesize elements of atomic number 5 and higher. (Refer to Unit 10 for details of

nucleo-synthesis inside the stars).

15.6.1 Cosmic Background Radiation

Do we have any signature of the early hot phase of the universe? The answer is yes. In

1965, two scientists at the Bell Telephone Laboratories in America discovered

accidentally a radiation at a very low temperature of only 3 K which seemed to come

from all directions. It was highly isotropic. It was suggested that the radiation fills the

whole universe. Since the wavelength of the peak radiation, ~ 1 mm, falls in the

microwave region, it was called the cosmic microwave background radiation

(CMBR). This is the relic of the era when the universe was very hot and dense. It is

argued that the radiation was once very hot and has been cooled to its present

temperature due to the expansion of the universe over billions of years (recall from

Eqs. (15.20) and (15.19) that T ∝ t1 and ρrad ∝ T 4). Put in another way, the same

energy fills an every increasing volume, so its energy density decreases and so does its

temperature.

The discovery of the CMBR has great significance for cosmology. (The discoverers of

the radiation were awarded the Nobel Prize for their work.) It shows that the universe

was once very hot and dense. This lends great support to theories of the universe

which maintain that the universe is changing with time, that is, it is evolving. Its

existence is a very powerful argument against theories which propose that the

universe is steady, that is, it is unchanging. In the theories of the latter type, it is

extremely hard to produce such a radiation.

CMBR is a topic of intense research today.

15.6.2 Evolving vs. Steady State Universe

You have just seen that CMBR points to a phase of the universe when it was very hot

and dense. This phase is generally known as the ‘hot Big Bang’. The idea is that some

46

Galaxies and the

Universe violent event took place at that time which sent the universe expanding. You have also

seen above that if the early universe had not been hot and dense, it would not have

been possible to synthesise light elements, such as H2, 4

eH and Lithium. In fact, the

prediction of the precise observed abundances of these light elements is a very

powerful argument in favour of the universe that changes with time: an evolutionary

universe.

Yet there is a set of scientists who believe in a steady state universe, a universe

which has no beginning and no end, which appears the same at all points in space and

at all times. Historically, the steady state theory emerged in the1940s and 1950s, when

observational techniques were not much developed, and so the Hubble constant H

could not be measured accurately. Recall that 1/H gives a rough time scale of the age

of the universe. So, the age of the universe inferred from the value of H at that time

turned out to be less than the age of some fossils on the Earth. This was quite

embarrassing. To overcome this age problem, the scientists proposed the steady state

universe. However, CMBR and synthesis of light elements in the early universe are

very powerful arguments against this theory and in favour of the hot Big Bang theory.

With this we come to an end of this unit in which you have studied about the large

scale structure of the universe. We now present its summary.

15.7 SUMMARY

• Distances of far-off objects inform us about the large scale structure of the

universe. To find distances of other galaxies we employ the concept of the

distance ladder. The first step of the ladder is the Cepheid variable stars found in

nearly galaxies. Subsequent steps include objects such as supernovae which can

be detected even in galaxies which are very far off.

• The outcome of the exploration of the universe is that the galaxies are rushing

away from one another and the universe is expanding.

• The expansion of the universe is in accordance with Hubble’s law, v = Hr, so that

1/H gives a rough estimate of the age of the universe.

• The behaviour of the universe is governed by the Friedmann equation:

3

8

22

2 ρπ=+

G

a

k

a

a

• The need for a hot and dense early universe can be explained in connection with

the synthesis of light elements and the existence of the cosmic microwave

background radiation.

• The present understanding is that if the universe was once hot and dense, then it

must be an evolving universe, and not a steady state universe.

15.8 TERMINAL QUESTIONS Spend 30 min.

1. Explain how Cepheid variables have been used to measure astronomical

distances.

2. State Hubble’s law. How can this be used to get an estimate of the age of the

universe?

3. Explain why at one time, the steady state theory appeared necessary. What is its

status now?

47

Large Scale Structure and

the Expanding Universe 15.9 SOLUTIONS AND ANSWERS

Self Assessment Questions (SAQs)

1. See Section 15.2.

2. H = 70 km s−1

Mpc−1

cm10103/s cm1070 61815 ×××= −

s 103

107 1

24

6−

×

×=

∴ Age of the universe = s107

1031

6

24

×

×=

H

76

24

103107

103

×××

×= yr

10104.1 ×= yr = 14 billion years.

3. Eq. (15.13) ρπ

=3

8

2

2G

a

a

Since nCa

−=ρ naC

G

a

a −π=

3

8

2

2

[C is a constant]

∴ 2

3

8 na

GC

a

a −π=

2

1n

aAdt

daa

−==

π=

3

8 GCA

∴ =

−

dtAdaa

n

2

2

tn

an

∝

2

2

( ) nn tna22

.2

∝ .

Hence, proved.

Terminal Questions

1. See Text.

2. See Text.

3. See Text.