PHYS3651 The Physical Universe - Department of …phys3651/notes.pdfis what we mean by a celestial sphere in the followings. Earth Celestial sphere Figure 1.1: Earth inside the in

PHYS3651 The Physical Universe

Dr. S.C.Y. [email protected]

mailto:[email protected]

Contents

Overview 1

1 Spherical Astronomy 4

1.1 Horizontal Coordinate System . . . . . . . . . . . . . . . . . . . . . 4

1.2 Sky and Celestial Sphere . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Longitude and Latitude . . . . . . . . . . . . . . . . . . . . 6

1.2.2 Motion of the Sun . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.3 Special Points . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 Equatorial Coordinate System . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Circumpolar Stars . . . . . . . . . . . . . . . . . . . . . . . 9

1.3.2 Angular Separation . . . . . . . . . . . . . . . . . . . . . . . 10

1.3.3 Great Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4 Other Celestial Coordinate Systems . . . . . . . . . . . . . . . . . . 13

1.5 Limitations of Coordinate Systems . . . . . . . . . . . . . . . . . . 14

1.5.1 Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.5.2 Aberration of Light . . . . . . . . . . . . . . . . . . . . . . . 15

1.5.3 Parallax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Light and Telescopes 18

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CONTENTS ii

2.1 Electromagnetic Wave . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2 Magnitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Spectrum, Spectral Lines, and Atoms . . . . . . . . . . . . . . . . . 22

2.4 Optics and Telescopes . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4.2 Refracting telescopes . . . . . . . . . . . . . . . . . . . . . . 27

2.4.3 Reflecting and catadioptric telescopes . . . . . . . . . . . . . 27

2.4.4 Magnification and resolution . . . . . . . . . . . . . . . . . . 28

2.4.5 Lens speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.5 CCD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3 Celestial Mechanics 34

3.1 Newton’s Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . 34

3.2 Newton’s Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.2.1 Roche Lobe . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.2.2 Critical Density of the Universe . . . . . . . . . . . . . . . . 44

3.2.3 Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.3 Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.3.1 Kepler’s Laws of Planetary Motion . . . . . . . . . . . . . . 47

3.3.2 Orbits in Two-Body Problem . . . . . . . . . . . . . . . . . 47

3.3.3 Proof of Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . 53

3.4 Impact Parameter and Scattering Angle . . . . . . . . . . . . . . . 55

3.5 Restricted Three-Body Problem . . . . . . . . . . . . . . . . . . . . 56

4 Introduction to Radiative Processes 61

CONTENTS iii

4.1 Solid Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.2 Specific Intensity and Flux . . . . . . . . . . . . . . . . . . . . . . . 63

4.3 Emission and Absorption . . . . . . . . . . . . . . . . . . . . . . . . 67

4.4 Scattering Cross Section . . . . . . . . . . . . . . . . . . . . . . . . 68

4.5 Basics of Statistical Mechanics (Optional) . . . . . . . . . . . . . . 72

4.5.1 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 72

4.5.2 Isolated Systems . . . . . . . . . . . . . . . . . . . . . . . . 73

4.5.3 Systems in a Heat Bath . . . . . . . . . . . . . . . . . . . . 74

4.5.4 The Perfect Classical Gas . . . . . . . . . . . . . . . . . . . 75

4.5.5 The Partition Function . . . . . . . . . . . . . . . . . . . . . 76

4.5.6 The Perfect Quantal Gas and Quantum Statistics . . . . . . 79

4.5.7 The Partition Function . . . . . . . . . . . . . . . . . . . . . 80

4.5.8 Derivation of Blackbody Radiation . . . . . . . . . . . . . . 80

4.6 Physics of Blackbody Radiation . . . . . . . . . . . . . . . . . . . . 82

4.6.1 Stefan-Boltzmann law . . . . . . . . . . . . . . . . . . . . . 83

4.6.2 Rayleigh-Jeans Law . . . . . . . . . . . . . . . . . . . . . . . 84

4.6.3 Wien Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.6.4 Wien’s Displacement Law . . . . . . . . . . . . . . . . . . . 85

4.6.5 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . 85

4.6.6 Temperature Definitions . . . . . . . . . . . . . . . . . . . . 86

4.7 Chemical Potential and Saha Equation (Optional) . . . . . . . . . . 87

Overview

• What is this course about?This is an introductory course to astrophysics, likely your first formal course inastrophysics.

• What will I learn in this course?The aim is to provide essential knowledge on mathematical tools and physi-cal concepts that used in astrophysics, in order to help appreciate the physicalprinciples of how the Universe works and to set the stage for more serious coursesin astrophysics. Throughout this course, you may NOT see pretty astronomy pic-tures, but a lot of physics equations. These are what professional astrophysicistssee most of the time, and how we extract scientific results from observations.

• What is astrophysics about?Astrophysics is the study of the physical properties of objects in the Universe andtheir interactions. It has a close connection with many other fields in physics,including mechanics, electromagnetism, statistical mechanics, relativity, and evenquantum physics in some cases. As we believe that the laws of physics are uni-versal, we can apply those laws we developed on Earth to the celestial objects tounderstand how they work. In some cases, we can also do it the other way round:use the Universe as a laboratory to test the laws of physics, in particular, underthe most extreme conditions that can never be reproduced on Earth. Some mayconsider astronomers as those study the physical properties of celestial objectsand astrophysicists as those use the celestial objects to study physics.

• Why observations?Unlike other branches of physics, astrophysics relies heavily on observations. Thisis because due to the distance and scale of the celestial objects, it is very difficult(if not impossible) to collect samples for detailed studies or to reproduce themin laboratories. Fundamentally, there are only four things one can measure withastronomical observations: position, spectrum (flux), time, and polarisation. InChapter 1, we will introduce the coordinate systems, which are the fundamentalsin positional astronomy. They are also like the language of astronomy when wewant to specify an object to another astronomer. Different coordinates and theirlimitations will be discussed.

• Course overview:As mentioned, Chapter 1 will cover coordinate systems. In traditional astronomy,

1

CONTENTS 2

electromagnetic radiation is the main cosmic messenger we rely on. In Chapter2, we will talk about the basics of light and how we detect them using telescopes.In Chapter 4, we will study how radiations are generated and propagated. Thiscan give insights into the physical processes happening inside celestial objects.Together these topics provide an introduction to spectroscopy.

In Chapter 3, we will review Newtonian mechanics and apply it to simple two-bodyproblems. We will see how the orbital motions of celestial bodies are governed bysimple laws of physics.

A note on units: Throughout this course, we will follow the tradition in thisfield to use cgs unit, i.e. cm, g, and s. This is what you will see in researchpapers. Force will be expressed in terms of dyn (=g cm/s2) and energy is in erg(=g cm2/s2). Charge is in electrostatic unit of charge (esu), so that Coulombs lawbecomes F = q1q2/r

2.

Exercise: 1N= dyn and 1 J = erg.What is the cgs unit of pressure?

Syllabus

Ch. 1 Spherical Astronomy:Sky and Celestial Sphere, Equatorial Coordinate System, Other Celestial Coordi-nate Systems, limitations of the coordinate systems.

Ch. 2 Light and Telescope:Electromagnetic Wave, Magnitudes, Spectrum, Spectral Lines, and Atoms, Opticsand Telescopes, CCD.

Ch. 3 Celestial Mechanics:Newton’s Laws of Motion, Kepler’s Laws, Two-body Problems, Scattering, Re-stricted Three-body Problem.

Ch. 4 Introduction to Radiation Processes:Solid Angle, Specific Intensity and Flux, Emission and Absorption, Blackbody Ra-diation.

Learning objectives

Ch. 1 Spherical Astronomy:Understand the definition and limitations of coordinate systems.Manage to apply the equatorial coordinates and to determine the angular separationbetween points.

CONTENTS 3

Ch. 2 Light and Telescope:Be able to convert between magnitude and flux.Understand the formation of spectral lines.Be able to compare the working principles of different types of telescopes and tocalculate the diffraction limit.

Ch. 3 Celestial Mechanics:Manage to calculate orbits of celestial bodies using Newton’s gravitation law.Manage to solve the two-body problem and derive Kepler’s laws.Understand the physical significance of Lagrangian points.

Ch. 4 Introduction to Radiation Processes:Understand the basic terminology of radiative transfer and the transfer equation.Manage to apply the blackbody radiation to astrophysical situations.

Chapter 1

Spherical Astronomy

(Chapters 1 and 3.1 in textbook.)

Astronomy is the science studying objects in the sky. We need to have a coordinatesystem to tell others where those objects are. How do we define such a system? Weneed to introduce the celestial sphere and study the spherical geometry.

1.1 Horizontal Coordinate System

The first natural coordinate system you may come up is the altitude-azimuthcoordinates (or horizontal coordinates). When you take a picture of the sky,you specify the altitude (elevation) and azimuth (angle around the horizon). How-ever, what is the problem with this system? Objects are constantly moving in thesky and this coordinates depend on the observer location! Therefore, we need acoordinate system that moves with the sky, or more precisely, with the celestialsphere.

1.2 Sky and Celestial Sphere

What we see as the sky is, in fact, the mostly empty space outside the Earth’ssurface. During day time, the scattering of Sun light by the atmosphere brightensup the sky and we cannot see much beyond the atmosphere or even the clouds. Atnight, without the interference of the Sun, we can see much further, including thestars, galaxies and many other objects in the universe.

Since the sky or the universe is all around us, our ancestors thought that we wereinside a large sphere, the celestial sphere, and all stars, including the Sun, were

4

CHAPTER 1. SPHERICAL ASTRONOMY 5

moving on the celestial sphere. We now know that celestial sphere is not real, butthe concept that we are inside an imaginary infinite sphere is still very useful inastronomy, Fig. 1.1, especially when we want to consider the positions of stars. Thisis what we mean by a celestial sphere in the followings.

Earth

Celestial sphere

Figure 1.1: Earth inside the infinite large celestial sphere.

A fine point about the celestial sphere is where exactly is its center. When weobserve objects near to us, like the artificial satellites or even the Moon, we have tobe precise because their positions relative to the distant stars depend on the center.Two common positions for the center are the center of the Earth, which defines thegeocentric coordinate system, and the position of the observer on the surface ofthe Earth, which defines the topocentric coordinate system. As seen in Fig. 1.2there is a small difference between the two systems, but since the distance to thestar ≫ the Earth radius R⊕, the difference is negligible.

R d

Figure 1.2: Geocentric vs. topocentric coordinate systems

Because the Earth is rotating from west to east, everything in the sky will movefrom east to west. As a result, the Sun and almost all the stars will rise from theeast and set in the west. If the celestial sphere is fixed with the Earth, the positionsof the stars will change from minute to minute. That won’t be very convenient.

The celestial sphere is, hence, fixed with the stars. (Here we implicitly assume thatstars do not move. In fact, they do move in space. However, they are so far awaythat we can only detect the motions of very few of them.) For observers on theEarth, it will rotate once a day, just like the stars do. However, how far the stars


are away from us is not shown on the celestial sphere. In other words, celestialsphere is just a two dimensional projection of the three dimensional universe. Weare not going to talk much about distance measurement of celestial objects becauseit is very difficult although very important.

1.2.1 Longitude and Latitude

Before we go into the details of the coordinate systems on the celestial sphere, wefirst briefly review the coordinate system on the Earth’s surface.

Earth’s surface is also a sphere. We use longitude and latitude to specify theposition of a city, say, on the Earth. The latitude is defined as the angle sustainedat the center of the Earth from the equator, where equator is the great circlemid-way between the north and south poles.

Longitude is defined as the angular distance east or west from an imaginary line, ameridian, running from the north pole to south pole. This meridian is chosen asthe one through the Greenwich Observatory in England. For example, the latitudeand longitude of Hong Kong are about 22.5N and 114.2E, i.e. 22.5 north ofequator and 114.2 east of Greenwich.

Hong Kong

Greenwich

22.5

114.2

Vernal equinox

North celestial pole

Line of right ascensionSouth celestial pole

0h

2h

Line of declination

Equator

22h

20h

30

60

−30

0

Ecliptic

Figure 1.3: Coordinate system on Earth and the celestial sphere with equatorialcoordinate system.

We now back to celestial sphere, Fig. 1.3. Like the surface of the Earth, it has twopoles: the north and south celestial poles. They lie directly above the Earth’spoles. The celestial equator lies directly above the Earth’s equator.

An observer standing on the Earth’s surface can only see half of the celestial sphereat one time. The other half is blocked by the Earth itself. The visible half isbounded by the observer’s horizon. The point on the celestial sphere directlyabove the observer is called the zenith. Due to the rotation of the Earth, zenith


not only depends on the position of the observer on the Earth, but also is not afixed point on the celestial sphere.

1.2.2 Motion of the Sun

Figure 1.4: Solar day vs. sidereal day.

We use solar time in daily life. A solar day is 24 hours, which is defined as thetime period for the Sun to return to the same position in the sky as observed onEarth. Since the Earth orbits around the Sun, it has to rotate about 361 degreein 24 hours (see Figure 1.4). On the other hand, a sidereal day is the time fora distant star to return to the same position in the sky. The Earth rotates

during this period. Therefore, a day is shorter and stars rise earlier/latereveryday by min.

As the Earth rotates around the Sun, the latter appears to move respect to back-ground stars. Over one year, the Sun appears to move around the celestial sphereonce. The path of this motion is called ecliptic. Since the rotation of the Earth isnot perpendicular to the plane of revolution, the ecliptic does not coincide with thecelestial equator. It makes an angle of 23.5 with the celestial equator, the sameangle that the rotational axis of the Earth tilts from the revolution axis.

Questions: There are 88 constellations in the sky but only 12 are zodiac, why?

How does the ecliptic of the Sun look like in the equatorial coordinate system?

1.2.3 Special Points

The two points that the ecliptic intersects the celestial equator are called theequinoxes. Vernal equinox is the point where the Sun crosses the celestial equa-tor from the southern to the northern half of the celestial sphere, around March 21each year. It is in the constellation Pisces. Vernal equinox also marks the origin ofthe celestial coordinate system, as we will talk about below.


Autumnal equinox is the point where the Sun goes from the northern half to thesouthern half, around September 23 each year. It is in the constellation Virgo. TheSun rises and sets due east and due west respectively at the days of equinoxes.

There are other two special points on the ecliptic. At the summer solstice, theSun reaches the greatest distance from the celestial equator in the northern half ofthe celestial sphere, around June 21. For an observer on Earth, the Sun rises andsets at different directions on different day during the year. On summer solstice, itrises and sets at the northern most points. It is the longest day and shortest nightfor the northern hemisphere. While it is summer for the northern hemisphere, it iswinter for the southern hemisphere.

At the winter solstice, the Sun reaches the greatest distance from the celestialequator in the south, around December 22. The Sun rises and sets at the southernmost points of the year. It is shortest day and winter for the northern hemisphere.The names of the solstices are biased to people in northern hemisphere.

As a note, have you ever wonder why Easter, unlike many other festivals, is not onthe same day every year? This is because Easter is held on the first Sunday afterthe first full moon occurring on or after the vernal equinox.

Discussion: At the summer solstice, on which part of the Earth one will see theSun passes directly overhead? This is called the Tropic of Cancer.

What is the difference of the Sun’s path in a year north and south of this line?

On the same day, on which part of the Earth one will not see Sun rise and Sunset? (Ans: the Antarctic Circle and the Artic Circle, respectively. What are theirlatitudes?)

A counterpart of the Tropic of Cancer in the southern hemisphere is called theTropic of Capricorn. What latitude is it and what is special about this line?

Question: Do you know why are there four seasons?

1.3 Equatorial Coordinate System

The most common coordinate system on the celestial sphere is the equatorial co-ordinate system. As discussed above, this system is like an extension of longitudeand latitude on Earth to the sky. The reference plane is the celestial equator andthe coordinates used are the right ascension and declination. Just like longitude,we need to choose a reference point. (How is 0 longitude defined on Earth?) Thisis chosen using the vernal equinox.


Declination (dec., symbol δ) is the celestial equivalent of latitude on the Earth. Itis measured in degrees, from 0 at the celestial equator to 90 at the poles, positivevalues or with an additional symbol “N” for the northern half and negative valuesor “S” for the southern half of the celestial sphere.

Right ascension (RA, symbol α) is the equivalent of the longitude. The zero lineof right ascension is chosen to pass through the vernal equinox. Right ascension ismeasured eastwards from the vernal equinox in hours, minutes and seconds, from 0to 24 hours.

RA Dec.h m

vernal equinox 0 0 0summer solstice 6 0 23.5autumnal equinox 12 0 0winter solstice 18 0 −23.5

Table 1.1: The coordinates of the four special points on the ecliptic.

In the equatorial coordinate system, vernal equinox is at 0h0m and 0; autumnalequinox at 12h0m and 0; summer solstice at 6h0m and 23.5 and winter solsticeat 18h0m and −23.5, Table 1.1. (It is customary to write the hour and minute ofright ascension as superscripts.) Be careful! One common source of confusion isthat the minute and second in RA are in units of time, but those in declination arearcminutes and arcseconds.

Questions:

1h = , 1m = ′, and 1s = ′′.

Why the zodiac signs start with Aries, but the Vernal equinox is currently in Pisces?

1.3.1 Circumpolar Stars

We now talk about which stars an observer can see and which cannot. As we havementioned before, at any particular time, an observer on Earth can only see halfthe celestial sphere. The other half is blocked by the Earth itself (see Fig. 1.5).If we take the rotation of the Earth into account, can the observer see the wholecelestial sphere? Usually not.

If seen from the poles, stars move in circles parallel to the horizon and never riseor set. If seen from the Earth’s equator, the rotation of the Earth does allow theobserver to see the whole celestial sphere. At intermediate latitudes, say L north,some stars never rise (see homework), and hence the observer cannot see it at all.


At the other extreme, some stars never set. They are called circumpolar stars.For people in northern hemisphere, there is a star, called Polaris, only 1 away fromthe northern celestial pole. It is often called the North Star.

LEarth

Visible

Invisible

North

South

L

(x,y,z)θ

φ

Figure 1.5: Left: an observer on Earth can only see half the celestial sphere at atime. Right: spherical and Cartesian coordinate systems.

1.3.2 Angular Separation

What is the angular separation between two points? We would like to first relatethe equatorial coordinate system with a Cartesian coordinate system. Assume thatthe celestial sphere is the unit sphere with the z-axis passing through the northcelestial pole and the x-axis passing through vernal equinox (see Fig. 1.5).

A point P with Cartesian coordinates r = (x, y, z), x2 + y2 + z2 = 1 has sphericalcoordinates

x = sin θ cosϕy = sin θ sinϕz = cos θ .

(1.1)

For the equatorial coordinate system, α = ϕ and δ = π/2− θ, therefore,

r1 =

xyz

=

. (1.2)

The angular separation ∆Θ between two points with coordinates (α1, δ1) and (α2, δ2)is

cos(∆Θ) = cos δ1 cosα1 cos δ2 cosα2+cos δ1 sinα1 cos δ2 sinα2+sin δ1 sin δ2 . (1.3)

The proof is left as an exercise (hint: use the dot product).

Exercise: Star A is at (17h55m0s,−600′0′′), Star B is at (18h5m0s,−600′0′′).


What is their angular separation in the sky in arcminutes? Is it simply 10m × 15?

For small angular separation ∆Θ ≪ 1, write α1 = α, δ1 = δ, α2 = α + ∆α, andδ2 = δ +∆δ, then Eq. (1.3) becomes

cos(∆Θ) = cos δ cosα cos(δ +∆δ) cos(α+∆α) + cos δ sinα cos(δ +∆δ) sin(α+∆α)

+ sin δ sin(δ +∆δ) .

We use the identities

sin(a+ b) =

cos(a+ b) =

to obtain

cos(∆θ) = cos δ cosα cos(δ +∆δ) cos(α +∆α) + cos δ sinα cos(δ +∆δ) sin(α +∆α)

+ sin δ sin(δ +∆δ) .

For x≪ 1, we can use the Taylor expansion

sinx = x− x3

3!+x5

5!− . . . ≈ x

cosx = 1− x2

2!+x4

4!− . . . ≈ 1− x2

2

so that

1− ∆Θ2

2= cos δ cosα(cos δ cos∆δ − sinα sin∆δ) + cos δ sinα(cos δ cos∆δ − sin δ sin∆δ)

(sinα cos∆α + cosα sin∆α) + sin δ(sin δ cos∆δ + cos δ sin∆δ)

= cos δ cosα

(cos δ −∆δ sin δ − ∆δ2

2cos δ

)(cosα−∆α sinα− ∆α2

2cosα

)+cos δ sinα

(cos δ −∆δ sin δ − ∆δ2

2cos δ

)(sinα +∆α cosα− ∆α2

2sinα

)+sin δ

(sin δ +∆δ cos δ − ∆δ2

2sin δ

)


= cos δ cosα

(cos δ cosα−∆α cos δ sinα−∆δ sin δ cosα− ∆α2

2cos δ cosα

+∆δ∆α sin δ sinα− ∆δ2

2cos δ cosα

)+ cos δ sinα (cos δ sinα

+∆α cos δ cosα−∆δ sin δ sinα− ∆α2

2cos δ sinα−∆δ∆α sin δ cosα

−∆δ2

2cos δ sinα) + sin δ

(sin δ +∆δ cos δ − ∆δ2

2sin δ

)= 1− ∆α2

2cos2 δ − ∆δ2

2cos2 δ − ∆δ2

2sin2 δ

∆Θ2 = (∆α cos δ)2 +∆δ2 . (1.4)

This can be used to calculate the proper motion of a star, dΘ/dt, on the celestialsphere in terms of small changes in R.A. and Dec. (see Fig. 1.6).

Figure 1.6: Proper motion of a star on the celestial sphere.

Exercise: Now use Eq. (1.4) to re-do the previous exercise and compare yourresults.

Question: What is the unit of proper motion?

How does the proper motion of an object relate to its space velocity?

1.3.3 Great Circle

A great circle is a circle which is the intersection of the sphere with a plane passingthrough the origin. Let n0 = (x0, y0, z0) be a unit vector perpendicular to the plane.


Then, all points (x, y, z) on the plane satisfy

xx0 + yy0 + zz0 = 0 . (1.5)

Substituting Eq. (1.1) into the above equation, we have an equation of θ and ϕdefining the great circle.

For example, the ecliptic is horizontal in Fig. 1.3. Hence, the unit vector is verticalin the figure and is (18h, 66.5), for which the spherical coordinates are (θ, ϕ) =(23.5, 270). This point is the ecliptic north pole, see below. The equation ofecliptic is

0 = − cos δ sinα sin 23.5 + sin δ cos 23.5

tan δ = tan 23.5 sinα . (1.6)

1.4 Other Celestial Coordinate Systems

n0

rv

Ecliptic

Celestial equatorβ

λ

P

r’

r

Figure 1.7: The ecliptic coordinate system.

The ecliptic coordinates are given by the ecliptic latitude β and ecliptic longitudeλ. Their definitions are similar to those of declination and right ascension, but nowreferring to the ecliptic instead of celestial equator. In particular, β of a point Pis the angle between the point and the ecliptic, positive for northern hemisphereand negative for southern hemisphere. λ is the angular distance between vernalequinox and the intersection point of the great circle passing through the point andthe north ecliptic pole and the ecliptic, Fig. 1.7.

For the point P , use the dot product,

cos(90 − β) = r · n0

sin β = −y sin 23.5 + z cos 23.5

= − sin 23.5 cos δ sinα + cos 23.5 sin δ . (1.7)


The component of r perpendicular to n0 is r−(r ·n0)n0. Hence, r′ is the unit vector

along this direction,

r′ =(cos δ cosα, cos δ sinα, sin δ)− sin β(0,− sin 23.5, cos 23.5)

the magnitude. (1.8)

The Cartesian coordinates of the vernal equinox are rv = (1, 0, 0), and cosλ = rv ·r′,which is just the x-component of r′,

cosλ =cos δ cosα

|(cos δ cosα, cos δ sinα + sin β sin 23.5, sin δ − sin β cos 23.5)|. (1.9)

The readers could work out the expression in the denominator, which is not veryilluminating.

Another common coordinate system is the Galactic coordinate system, wherethe defining great circle is the Galactic plane, with the north Galactic pole atα = 12h51.4m and δ = 277′ and the zero point of Galactic longitude at the directionof Galactic center (17h45.6m,−2856′). Galactic coordinates are expressed in (l, b),where l is the Galactic longitude b is the Galactic latitude. Finally, there is alsoSupergalactic coordinate system, which is used in the studies of nearby galaxyclusters, including the Virgo Supercluster.

1.5 Limitations of Coordinate Systems

As a final remark, we discuss some limitations of the coordinate systems. It isactually more complicated to specify the position of objects in the sky. First of all,stars are moving in space. But they are very far away, hence, the proper motionsare generally small, typically in the order of millisecond per year. Even if the objectsare not moving, their apparent positions in the sky are changing at different timesof a year due to the motion of the Earth. The two major causes are aberration oflight due to the finite speed of light and parallax for nearby objects.

1.5.1 Precession

The major problem with the equatorial coordinate system is that the vernal equinoxis not fixed with respect to the stars. It is constantly moving due to precessionof the Earth’s rotational axis. This is caused by the fact that the Earth is nota perfect sphere but has a larger diameter at the equator than at the pole. Thegravitational pull of the Sun and the Moon on the near and far sides of the Earthare thus different, causing a torque perpendicular to the rotational axis. (You canfind more details at http://courses.physics.northwestern.edu/Phyx125/Precessionof the Earth.pdf)

http://courses.physics.northwestern.edu/Phyx125/Precession%20of%20the%20Earth.pdf

http://courses.physics.northwestern.edu/Phyx125/Precession%20of%20the%20Earth.pdf


The precession of the Earth has a period of 26,000 years, which means the celestialnorth or south pole, traces out a circle with that period. 13,000 years from now, theEarth spin axis will be 47 away from the Polaris. Note that direction of precessionis opposite to the rotation of the Earth. As a result, the intersection between theecliptic and the celestial equator, i.e. the equinoxes are constantly shifting westwardabout the ecliptic pole, with a period of 26,000 yr, i.e. about 1.38 per century. Thisis also the reason why the zodiac signs start with Aries, since the vernal equinoxwas in Aries when the constellations were introduced in the past, and only movedinto Pisces in 67 B.C.

Because of precession, when we talk about the equatorial coordinates, we also needto specify the date and time used for comparing star coordinates, the epoch. Thestandard epoch commonly used now is the beginning of the year 2000, denotedJ2000.0. In some old books, you could find the epoch J1975.0 or B1950.0. Inprofessional observatories, the exact epoch, i.e. the observation date, is needed inorder to point the telescope to the right direction at very high accuracy.

The changes in equatorial coordinates relative to J2000.0 can be approximated by

∆α = M +N sinα tan δ (1.10)

∆δ = N cosα , (1.11)

where

M = 1.2812323T + 0.0003879T 2 + 0.0000101T 3

N = 0.5567530T − 0.0001185T 2 − 0.0000116T 3 .

M and N are in degrees and T = (t− 2000.0)/100 with t in fractions of a year.

Question: What is the effect of precession on the seasons?

1.5.2 Aberration of Light

Imagine sitting inside a moving vehicle on a rainy day, the rain drops would appearto travel down at an angle, even there is no wind. Same is true for light, due itsfinite speed and the observer’s motion. This effect depends on the position of thesky and time of the year. For example, there is no aberration along the directionof motion, and it is maximum when perpendicular. The observer’s motion can bedue to: 1. the Earth’s rotation, 2. the Earth’s orbital motion around the Sun. 3.the Sun’s motion around the Galaxy.

We can estimate the maximum displacements using classical mechanics since v ≪ c,although strictly speaking, relativity is needed. Aberration caused by the Earth’srotation is called diurnal aberration. It changes every day, but the magnitude is


vEarth

Figure 1.8: Aberration of light.

Sun1AU

θ

d

Figure 1.9: parallax.

relatively small. Aberration caused by the motion of the Earth around the Sun iscalled annual aberration. It has a larger amplitude and a longer period of 1 year.

Exercise: At the equator, what is the rotation velocity of the Earth? Hence, whatis the amplitude of diurnal aberration?

What is the velocity of the Earth around the Sun and the corresponding aberration?

For annual aberration, It is interesting to note that this motion is always perpen-dicular to the Sun, therefore, the Sun always appears to be ′′ off from its trueposition. Finally, motion of the Sun in the Galaxy results in secular aberration,which is of the order of arcminutes. However, the Sun takes 230million years torevolve around the center of the Galaxy. In practice, this aberration never changesand hence it is often ignored.

1.5.3 Parallax

Parallax arises because stars are not at infinite distance. The change of the ob-server’s viewpoint, mostly due to the motion of the Earth around the Sun, causes anearby star appears to move relative to distant objects at the background. We canturn it around to use the parallax to determine distance, which is the most funda-mental parameter we wish to know, but also the most difficult one to measure.


1 parsec is defined as the distance that gives an annual parallax of 1′′. (Rememberthat 1 = 60′ and 1′ = 60′′. How large is an arcsecond? Put a 10c/ coin across theharbour. The diameter as seen from campus is about 1′′!) Hence, from Fig. 1.9with 1 A.U. = 1.50× 1013 cm,

tan 1′′ =

=

d = cm. (1.12)

The nearest star, Proxima Centauri1, has a parallax of 0.7687′′, therefore, its dis-tance is pc. Stars in the Milky Way have distances from a few pc to kpc. OurSun is 8.5 kpc from the Galactic center and the Milky Way has a diameter of 30 kpc.The Andromeda Galaxy is 0.78Mpc away and other galaxies are over Mpc away.The observable Universe has a radius of 15Gpc.

The Hipparcos satellite from ESA was able to measure the position and parallaxof 118,281 stars in the solar neighborhood brighter than magnitude 9, providingvaluable information on their distances. It will be succeeded by the Gaia satellite,which was launched to L22 on 2014 Jan 8 and it will measure the distance to 1 billionobjects (see https://arxiv.org/abs/1609.04303).

1Scientists recently discovered an Earth-like planet orbiting Proxima Centaurihttp://www.nature.com/nature/journal/v536/n7617/full/nature19106.html.

2We will talk about the second Lagrangian point, L2, in Chapter 3.

https://arxiv.org/abs/1609.04303

http://www.nature.com/nature/journal/v536/n7617/full/nature19106.html

Chapter 2

Light and Telescopes

(Chapters 3.2, 3.3, 5 and 6 in textbook.)

Although we can now go to the Moon and bring back samples of soil and rocks, wecan still only study other objects by investigating their radiations; could it be neu-trinos, charged particle cosmic rays, gravitational waves, or electromagnetic waves.Among them, the electromagnetic waves, or EM waves for short, is dominating. Inthis chapter, we will discuss the properties of EM waves and the detection methods.

2.1 Electromagnetic Wave

EM waves are oscillations of the electric and magnetic fields, Fig. 2.1. It can beproduced by the acceleration of charged particles, and in turn, EM waves, or ingeneral electric field and magnetic field will affect the motion of charged particles.They have no effect on neutral particles.

Electric field

direction ofpropagation

λ

Magnetic fielddirection ofpropagation

Horizontalpolarization

Vertical polarization

Figure 2.1: Properties of electromagnetic waves.

Light, radio waves, infrared, ultraviolet, X-rays and gamma rays are EM waves ofdifferent frequencies. Like other kinds of waves, the three fundamental properties of

18

CHAPTER 2. LIGHT AND TELESCOPES 19

EM waves are the speed (the speed of light is usually denoted by c), the frequency,f , and the wavelength, λ. They are related by

c = fλ . (2.1)

For EM waves in vacuum, the speed of light is independent of frequency (no disper-sion), and is equal to 2.99792458× 1010 cm/s. This value is exact in the sense thatwe define the length of one meter by this value and the definition of time (which isdefined by the transition of some atoms). The constant speed of light is the startingpoint of special relativity.

EM waves can have a very wide range of wavelengths, from shorter than an atomor as long as the size of the Universe. Radio waves are about 1mm to 100m, thenfollowed by microwave, and then infrared IR and visible light. The wavelengthsof visible light are from 400 nm to 700 nm. Our atmosphere is transparent to EMwaves in radio and over a few windows in IR and visible light. The most importantwindow to human eyes is the optical window between 300 nm to 1100 nm. Theatmosphere is opaque to all shorter wavelength EM waves. For example, if we wantto carry out X-ray (wavelengths from about 10−7m to about 10−9m) or gammaray (wavelengths less than about 10−10m) astronomy, we have to get above theatmosphere, for example, from a satellite.

Apart from the three basic properties, different EM wave can also carry differentpolarization, which is the direction that the electric field in the EM wave points,Fig. 2.1. The direction of polarization must be perpendicular to the direction ofpropagation. Thus, there are two kinds of polarizations. The EM wave from a sourcewill in general contain a mixture of waves with different directions of propagation,different wavelengths and different polarizations.

When EM waves propagate, energy is transported from one place to another. Theamount of energy radiated by a star per unit time is called the luminosity, whichin units of erg /s. Then at the observer, the amount of energy received per unitarea per unit time is called the intensity or energy flux, which is in units oferg /cm2/s. Notice that flux describes the energy received and it can be understoodas the brightness of an object. If the source is far away, even it radiates enormousamount of energy, the intensity observed could be low. To be more specific, considera sphere with distance d from the source, the total surface area is 4πd2. Hence, thetotal energy passing through the sphere per unit time is flux F times the area, i.e.F × 4πd2, which should be equal to the total energy emitted per unit time, i.e. thesource luminosity L. Mathematically,

F =L

4πd2. (2.2)

See Fig. 2.2 below. (What will be the relation looked like in a 2-D universe?)

So far we have discussed the wave nature of the EM wave, but in some situations,we find that EM waves behave as particles. For example, atoms can only absorb


sphere areaIntensity at

surface of sphere

source strength

Figure 2.2: Inverse square law.

one, two or any integral multiple of certain unit of light. We call the basic unitphoton and said that atoms can only absorb one photon or two photons, etc, butnot half photon. (Of course, atoms can absorb no photon at all.) The wave natureof EM wave is the collective behavior of a lot of photons. The energy, E, of eachphoton is given by

E = hf (2.3)

where h is the Planck’s constant, with value about 6.63 × 10−27 erg s= 4.14 ×10−15 eV s, and f is the frequency of the photon. Higher the frequency, higherthe energy of each photon has. We find that particle nature of light is prominentwhen the frequency is high. Thus, for gamma rays, we will often speak of thephotons, but for radio waves, we often think them as waves.

2.2 Magnitudes

The brightness of objects in the sky varies a lot. For example, Sirius, the brighteststar (apart from the Sun), is about 500 times brighter than the dimmest stars wecan see with naked eyes. Therefore, if we use the values of intensity to describe theirbrightness, we have to write a lot of zeros. We use a log scale instead. Examplesof log scales include the Richter scale for earthquakes and decibel (dB) for soundintensity.

The visual magnitude or apparent magnitude m of a star tells the brightnessthe star as we see it. This system dates back to Hipparchus in ancient Greece. Inmodern terms, stars visible to human eyes are classified into 6 magnitudes from thebrightest (m = 1) to the faintest (m = 6), and an m = 1 star is 100 times brighterthan an m = 6 star. In other words, if star A is 100 times brighter than star B, themagnitude of A is 5 units less than the magnitude of star B. Thus, a bright starhas a smaller or even negative magnitude number.

Exercise: From the definition above, show that each grade of magnitude is about


2.5 times brighter than the next one.

What is the relation between the intensity and magnitude of a star? Similar toabove, one can show that

m = −5

2log10

(I

I0

), (2.4)

where we arbitrarily choose one fixed intensity I0 as reference.

We thought the star Vega had constant brightness and chose it as the reference.Hence it had magnitude zero. However, we later found out that it is in fact avariable star, and we have abandoned it as the reference. Its average magnitudeis about 0.03. We don’t have such “standard” reference star anymore and I0 isjust a number. The apparent magnitude of the Sun is about −26.8, the full Moonabout −12, Sirius about −1.3, the naked eye limit about 6 and the dimmest imagetaken by the Hubble Space Telescope is about 30. The actual situation is morecomplicated: one star may be brighter than another in the blue band, but fainter inred. Therefore, if you look up a star in research papers or star catalogues, you mayfind its magnitude specified in different bands, e.g. visual, blue, red, or IR. Theyare denoted by U, B, V, R, I, Z, etc. The most common one is the V (meaningvisual) band magnitude, which centers on yellow color (551 nm), close to the peakresponse of human eye response.

If a star is moved from distance d to D, then md will change to mD, where

mD = md − 5 log

(d

D

). (2.5)

Note that if D is large, then d/D is small, log(d/D) is negative and mD is large.This only means that an object farther away is dimmer. If we know the distance toa star and have measured its apparent magnitude, we want to compare its intrinsicbrightness (i.e. luminosity), then it is more convenient to convert to a standarddistance of 10 parsec. This is the definition of the absolute magnitude M . Hence,it follows from Eq. (2.6) that the apparent and absolute magnitudes are related by

M = m− 5 log

(d

10 pc

). (2.6)

The difference m−M is called the distance modulus. Type Ia supernovae haveabsolute magnitude of M = −19.3. This has been used to determine the distanceto their host galaxies, leading to the discovery of acceleration in the Universe ex-pansion. This was awarded the Nobel Prize in 2011. Finally, we note that inscientific measurements, the apparent magnitude has to be corrected for the ab-sorption through the atmosphere. Additionally, absorption from the interstellarmedium also need to be accounted for in the calculation of absolute magnitude.


Exercise: The apparent magnitude of the Sun is −26.8, what is its absolute mag-nitude? Which one is more luminous when compared with Sirius (M = 1.4) andVega (M = 0.6)?

2.3 Spectrum, Spectral Lines, and Atoms

It was discovered by Newton that white light is composed of all the colors of therainbow. The image of this range of colors is called the optical spectrum. Theentire electromagnetic spectrum is much wider than the optical spectrum, includingthe infrared, ultraviolet, etc. In astronomy, spectrum also means the graph ofintensities versus the frequencies or wavelengths. A typical spectrum is shown inFig. 2.3. This graph tells us a lot about the nature of the source of the radiations.What should be the unit of the y-axis in the figure? Since the integrated area underthe curve is flux, the y-axis is the flux density (i.e. flux per unit frequency), andit has units of erg/s/cm2/A for optical spectrum. For radio spectrum, it is usuallyflux per Hz, and per keV for X-ray spectrum.

Frequency

Intensity Intensity

Frequencyf2f1

Figure 2.3: A continuum spectrum (left) and a spectrum with spectral lines (right).

Showing on the left of Fig. 2.3 is a continuum spectrum, since the variationsof the intensity is smooth. Usually, a spectrum contains some abrupt changes inintensities. This is a spectrum with spectral lines, or simply line spectrum asshown in the right of Fig. 2.3. Here we can see two kinds of spectral lines. The lineat frequency f1 is called an absorption line because radiation at this frequency isabsorbed and hence the intensity is lower than the underlying continuum. The lineat frequency f2 is an emission line.

A German physicist Gustav Kirchhoff studied the formation of spectral lines andsummarized into Kirchhoff’s three laws of spectroscopy:


1. A hot, dense gas or hot solid object produces a continuous spectrum with nodark spectral lines.

2. A hot, diffuse gas produces bright spectral lines (emission lines).

3. A cool, diffuse gas in front of a source of a continuous spectrum produces darkspectral lines (absorption lines) in the continuous spectrum.

Figure 2.4: A classical pic-ture of atom with the nucleusat the center and several elec-trons orbiting around it.

To derive these laws, we have to know some prop-erties of atoms. In conditions common to us, every-thing is made up of atoms. The classical picture ofan atom is shown in Fig. 2.4. The nucleus containsprotons, which is positively charged, and neutrons,which is electrically neutral. Almost all mass of anatom concentrates at the nucleus. There are usuallyseveral electrons going around the nucleus. Elec-trons are negatively charged. For a neutral atom,the number of electrons is equal to the number ofprotons in the nucleus. There are about 110 dif-ferent kinds of atoms, the simplest being hydrogen,with only one proton in the nucleus. Hence, neutralhydrogen has one electron.

Quantum mechanics, which is the theory for atomic or smaller systems, tells usthat electrons can only be in certain configurations relative to the nucleus. Theseallowed configurations are called states. Each state corresponds to some definiteenergy. After solving the Schrodinger equation, it can be shown that the possibleenergies of a hydrogen atom are

En = −mee4

2h21

n2in cgs (2.7)

= −mee4

8h2ϵ20

1

n2in MKS, (2.8)

where h ≡ h/2π, me is the electron mass, e = 4.8 × 10−10 esu is its charge in cgsunit (or e = 1.6 × 10−19C in MKS), ϵ0 is the vacuum permittivity and n is anypositive integer. Different n labels different state. Numerically,

En = −13.6

n2eV, (2.9)

where eV is an unit for energy with 1eV = 1.6×10−12 erg. With the leading negativesign, higher states (larger n) have higher energies (less negative). The least energystate (n = 1) is called the ground state and others are called the excited states.

Since electron can only be in those states, when it jumps from one state to another,it can only emit or absorb energy equal to the difference between the energies of


the two states. For example, if the electron of a hydrogen atom in the excited staten = 5 jumps to a lower state n = 3, it will emit energy

E =

= eV. (2.10)

By Eq. (2.3), the photon emitted is of frequency 2.3 × 1014 Hz, which is in theinfrared. If it jumps to a higher state, it has to absorb energy. In general, theenergy differences between states of hydrogen atom is given by

En − Em = −13.6 eV

(1

n2− 1

m2

)(2.11)

for various positive integers n andm. We can calculate the wavelength of the photonemitted by E = hf = hc/λ, such that

1

λ=

2π2mee4

h3c

(1

m2− 1

n2

)≡ R∞

(1

m2− 1

n2

), (2.12)

where R∞ = 1.097×105 cm−1 is called the Rydberg constant. It corresponds to thetransition from m = 1 to n = ∞.

The transition lines of hydrogen is particularly important, since H is the mostabundant element in the Universe. The transitions from n ≥ 2 to n = 1 is calledthe Lyman series, with n = 2 → 1 called the Ly-α, n = 3 → 1 called the Ly-β,n = 4 → 1 called the Ly-γ, and so on. The Balmer series is transitions from n ≥ 3to n = 2, i.e. n = 3 → 2 called the Balmer-α, n = 4 → 2 called the Balmer-β,etc. These are also called the Hα, Hβ, Hγ, etc. For completeness, the transitionsto n = 3 is called the Paschen series, to n = 4 is called the Brackett series, and ton = 5 is called the Pfund series.

Exercise: What are the wavelengths of Hα and Hβ? Which part of the EMspectrum (or color) do they correspond to? Hence, why are they so important?

Now, it is easy to understand Kirchhoff’s laws. The continuous spectrum comesfrom blackbody radiation emitted by any objects with temperature above abso-lute zero. We will discuss more on the blackbody radiation in Chapter 4 later inthis course. Emission lines are produced by electrons making downward transition(“falling”) from a higher orbit to a lower orbit. When radiations with continuumspectrum pass through gas of atoms in low pressure, those atoms will absorb pho-tons with energy equal to the differences of energies of their states. The atomswill be excited. When they fall back down, it will produce emission lines, but thephotons emitted will travel in all directions. As a result, depending on the viewpoint, the observer will see emission or absorption lines, Fig. 2.5.

Note that different atoms have different set of states and hence, spectral lines. Thespectral lines of hydrogen correspond to energies given by Eq. (2.11). We can tell


low density gas

continuum

absorptionline spectrum

emission line spectrum

Figure 2.5: Radiations with continuum spectrum passing through low density gas.It depends on viewing angle whether the observer sees an emission or an absorptionspectrum.

from the spectrum of a star which elements are present on the outer atmosphere ofthe star.

Not just atoms, molecules also have their own sets of spectral lines. We can findall those lines in laboratories on the Earth. However, the spectral lines observed ofsome star are often shifted to some other wavelengths, due to the relative motionof the star and the Earth. This is the Doppler effect. Unlike the classical case,EM wave does not require a medium to popagate, Doppler effect is a result oftime dilation according to special relativiity. The change in wavelength, called theredshift is given by

z =∆λ

λ0=

√1 + v/c

1− v/c− 1, (2.13)

where ∆λ is the observed wavelength minus the original wavelength, λ0, v is therelative speed between the source and the observer, and c is the speed of light. themotion of the star affects all of its spectral lines with the same factor β. Note thatfor v ≪ c, z ≈ v/c.

This is an extremely useful technique in astronomy for velocity measurements, lead-ing to all kinds of important discoveries, including extrasolar planets, binary starsystems, rotation of galaxies (hence dark matter), and the expansion of the Universe(hence big bang and dark energy).


2.4 Optics and Telescopes

2.4.1 Basics

Astronomical telescopes have only one main purpose: to collect more light, not tomagnify or to focus. Therefore, larger is better. Not all telescopes can focus light.Since high energy radiation can penetrate into materials, it is very difficult to deflectthe X-ray photons (need grazing at shallow incidence angles) and impossible to focusgamma rays. Therefore, all gamma-ray telescopes and some X-ray telescopes arenon-focusing. (What is the advantage of focusing telescopes?)

In this section, we mainly discuss optical telescopes, since they are most commontype and have the longest history. However, the general principles apply to all kindsof telescopes. A large collecting surface enables us to detect dimmer objects. Inthe dark adapted condition, the pupil of a human eye can relax to about 7mm indiameter and stars of magnitude 6 can be seen. If a telescope of diameter 20 cmis used, the intensity of stars is amplified by a factor of (200/7)2 = 816. Thus, byEq. (2.4), through this telescope, stars of magnitude 6 + 5

2log10(816) = 13 can be

seen. In general, if the diameter of the telescope is Dmm, the dimmest star can bedetected is of magnitude

6 +5

2log10

(D

7

)2

= 6 + 5 log10

(D

7

). (2.14)

This is a rough estimate. Many other factors affect what we can see. Also, if we useother detectors, like CCD, instead of our eyes, we usually can detect much dimmerobjects (why?).

One common misconcept is to ask “how far I can see with this telescope?” This isnot a well-defined question. Provided that the object is bright enough, no matterhow far, we can still see that object. If it is very dim, we cannot see it even if it isnear us. Another common mistake is paying too much attention on the “maximummagnification” of a telescope.

focal lengthfocal length

Figure 2.6: Principles of refracting (left) and reflecting (right) telescopes.


2.4.2 Refracting telescopes

There are two ways to bend lights: refraction with lenses or reflection with mirrors.Therefore, there are three kinds of telescopes. When EM wave enters a medium,its speed will change. (If it enters the medium from vacuum, its speed must slowdown. Nothing can travel faster than the speed of light in vacuum.) If the light rayenters the medium at an angle, the ray will be bent. This is refraction. It can bedescribed by Snell’s Law

sin θ1sin θ2

=n2

n1

, (2.15)

where n1 and n2 are the indices of refraction of the two medium, and θ1 and θ2are the incident and refraction angles, respectively, measured from the normal. Forexample, air has n = 1.0003 and water has n = 1.33 relative to vacuum. Snell’s lawcan be derived using Fermat’s principle (or the principles of least time).

The main component of a refracting telescope or a refractor is a lens, which focusesthe light rays by refraction (Fig. 2.6 left). The distance between the focus andthe lens is called the focal length. The refractors have two main disadvantages.First, there is dispersion: in medium other than vacuum, EM waves with differentfrequencies travel in different speeds. Dispersion leads to chromatic aberrationin refractors. It means that light rays of different colors (different wavelengths)focus at different points. For example, if we pick the best focus for green color,the resulting image will have a red halo around it. This problem can be mitigatedwith a long focal length or using a lens system of two or more lenses, but then theoptics is complicated and the cost will increase. An achromatic lens uses two lenselements made of different materials. It can focus two colors to the same point. Anapochromatic (“APO”) lens can bring three colors to the same focus.

A much bigger problem is that refractor requires a large piece of perfect lens. It isextremely difficult to manufacture a large piece of glass without bubbles in it. Evenif that can be done, the lens is usually too heavy that it would deform differentlyunder self gravity when the telescope points at different direction. As a result, allmodern telescopes for astronomical research are reflectors. The largest refractorstill in use today has a diameter of 102 cm.

2.4.3 Reflecting and catadioptric telescopes

If a thin layer of metal is deposited onto a polished glass surface, the reflectancewill be increased, and we have a mirror. A reflector to converge the light raysto the focus, Fig. 2.6 right. The surface of the mirror can only deviate from thedesired shape by one quarter of the wavelength. This is about 100 nm for visiblelights (2 cm for radio waves). Since we only use one surface of the glass, defectsin the bulk are irrelevant. Also, the angle of reflection is the same for all colors,there is no chromatic aberration. One additional advantage is that we can support


the mirror from “behind,” not just the edge as in refractors. Thus, we can builtvery large reflectors. The largest is 10m in diameter. To focus the light rays frominfinity (i.e. parallel incident rays) into a point, the shape of the mirror has to bea parabola. (do you know how to prove that mathematically?) This kind of mirroris much more difficult to polish than a spherical one. Using a spherical mirror willresult in spherical aberration.

The reflectors are not without shortcomings. The focus in Fig. 2.6 is in front ofthe mirror. We have to divert the light rays to a position convenient for viewing.Usually, a secondary mirror will be introduced. Two possible configurations areshown in Fig. 2.7. Fig. 2.7(a) is the Newtonian design, while in Fig. 2.7(b), a hole isopened at the center of the main mirror and is called the Cassegrain design. Thesedesigns introduce some obstructions to the light path. The obstruction will scatterlight, and hence the image produced by a perfect reflector is not as sharp as theimage by a perfect refractor with a lens of same diameter.

(a) (b)

Figure 2.7: Two possible focus arrangements for reflectors.

The third type of telescope is called catadioptric telescope. It is a hybrid de-sign using both lenses and mirrors. One design very popular among the amateurastronomers is Schmidt-Cassegrain, because of its compact size. Some designs usea spherical primary mirror, which is easy to manufacture, with a corrector plate infront.

2.4.4 Magnification and resolution

Another important component of an optical telescope is the eyepiece. This is notnecessary if CCD or photographic film is used to record the image, but is critical forvisual observations. The simplest design of an eyepiece is just a lens. It is usuallyput at a position that the distance between the eyepiece and the objective (wouldit be the main mirror in reflectors or the main lens in refractors) is equal to thesum of their focal lengths. It is shown in Fig. 2.8 that object with angular sizeθ1 will have an image of angular size θ2. Thus, the angular magnification, or just


Objective Eyepiece

θ1 θ 2

f f1 2

Figure 2.8: The magnification of a telescope is determined by the focal lengths ofthe objective and the eyepiece.

magnification, of the telescope is

magnification =θ2θ1

=f1f2

. (2.16)

Exercise: Using the small angle approximation tan θ ≈ θ for small θ, prove theequation above.

Changing the magnification is accomplished by simply changing the eyepiece witha different focal length. Even for the largest telescopes, the magnification used isseldom over 500, usually between 100 and 200. It is because a large image givenby the high magnification is usually very fuzzy. As we will see below, this is due tothe Earth’s atmosphere most of the time.

Even without considering atmospheric effects, theoretically is there a physical limiton the angular resolution? Due to diffraction of light, a point source, like a star,will not be focused to an infinitely small point by a telescope. The best image isactually a blurred disk, called the Airy disk. If two stars are very close to eachother, their Airy disks could overlap and the observer cannot tell if it is one staror two. We say that the two stars cannot be resolved. For a circular aperture, theblurring produced by diffraction limits the angular resolution to an amount givenby the Rayleigh criterion:

θdiffraction limit =1.22λ

D(2.17)

where D is the diameter of the objective of the telescope and λ is the wavelength ofthe light.1 (The value of θdiffraction limit is in radian.) For example, for yellow light,

1To derive the constant 1.22 requires Fraunhofer diffraction. The diffraction pattern can be


λ = 600 nm, if D = 20 cm, the angular resolution is 0.75′′, which means we wouldsee as one star if in fact the two stars are separated by less than 0.75′′, even in idealconditions.

(b)(a)

Figure 2.9: In (a), the Airy disks produced by a large telescope are small. The twostars can be resolved. In (b), the Airy disks of the same pair of stars by a smalltelescope are larger. The two stars cannot be resolved.

Exercise: Sirius is the brightest star in the night sky it is at a distance of 2.64 pcand a diameter of 3.4 solar radius (i.e. 2.4×106 km). How large a telescope is neededto resolve its image in visible light (λ = 555 nm)?

So practically we can treat stars as unresolved point sources.

If we use a short focal length eyepiece to obtain high magnification, the image ofthe Airy disk will be magnified. This will blur the whole image and degrade thequality. In some department stores, they advertise their telescopes by claiming ahigh magnification, say 600 or more. This is an unsound claim and their telescopescan only be treated as toys.

By Eq. (2.17), it seems that we can obtain high resolution by increasing the size ofour telescope. This is true up to a point. For ground based observers, star light hasto pass through the atmosphere and it acts as a large refractive medium. The air inthe atmosphere is constantly moving and the image of a star will dance around, likethe bottom of a swimming pool when viewed above the water. The effect is called theseeing. The seeing limit is usually about 1′′. Thus, even for a small telescope likethe 20 cm we talked above, its resolution is limited by the seeing, not the diffractionof its optics. This is one more reason why a high magnification is not useful.However, we can go up above the atmosphere to avoid the bad seeing, for examplethe Hubble Space Telescope. Many ground-based telescopes, including the biggest

found by 2D Fourier transform of a circular aperture. The solution is I(θ) ∝ [j1(x)/x]2, where j1

is the Bessel function of the first kind of order one and x = πD sin θ/λ. The first minimum is setby the first zero of J1(x), which occurs at x = 3.832, hence, θ ≈ sin θ = 1.22λ/D.


ones, are seeing-limited and their resolution is nowhere near the diffraction limits.This is why some astronomers advocate to build more medium-size telescopes, whichare cheaper, instead of extremely large ones.

For radio observations, since the wavelength of radio waves is about 105 timesthat of visible light, we have to build an enormous telescope to obtain the sameresolution. One remedy is to use computer to recombine the signals from severalradio telescopes. The several radio telescopes will function as parts of the dish ofan imaginary radio telescope, with effective diameter equal to the distance betweenthe actual radio telescopes. (We call the objective of a radio telescope a dish.) Thisis called the radio interferometry. This technique has been applied to optical aswell, but it is slightly different due to the much higher frequency of visible light.

2.4.5 Lens speed

For unresolved objects, such as a distant stars or quasar, all incident light rays arefocused into a point, hence, the brightness depends on the lens diameter. However,for extended objects, e.g., the moon, planets, nebulae, or nearby galaxies, the lightcoming out from the telescope spreads over some area. Therefore, the surfacebrightness depends on the magnification. One important parameter to consideris the f-number (or f-ratio) of a telescope or a lens, which is defined as

f-number ≡ f

D, (2.18)

where f is the focal length and D is the diameter of the lens. Note that a largerf-number means a smaller diameter. For photographic lenses, the f-number can beadjusted in steps of

√2, in the series of 1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16,... Every

successive step reduces the lens’ effective diameter by√2, such that the amount of

light passing through is reduced by half.

Exercise: For a resolved object with angular size α (in radians), show that thephysical size of its real image by a telescope of focal length f is αf .

The amount of light collected is proportional to D2, from the exercise above, weknow that the light collected is spread out over an area ∝ (αf)2. As a result, thesurface brightness I of the image depends on

I ∝(D

f

)2

=

(1

f-number

)2

. (2.19)


We see that a smaller f-number gives a brighter image (i.e. with larger surfacebrightness), since the lens aperture is larger. Telescopes or lenses with smallerf-numbers are therefore referred to as having a “higher speed” or “faster”.

Exercise: Comparing between the lens on a cell phone with f = 4.1mm and f/2.2,and the primary mirror of the Hubble Space Telescope, which has D = 2.4m andf/24. Which one gives a brighter image? Which lens/mirror is “faster”?

Finally, we should mention that beside the optics, the mount of a telescope is alsovery important. Not only must it support the optics, it must also track the starsacross the sky. As we have discussed in previous chapter, due to the rotation ofthe Earth, stars and every object move on the celestial sphere. There must bemechanism to turn the telescope such that the image of the stars is fixed for us orthe detectors.

2.5 CCD

CCD is Charge Coupled Device. It is a semiconductor device with the appearancesimilar to an ordinary computer chip. On the top of the chip, there is a window,allowing light to go in. After applying the power, each element of the device willconvert photons to electrons. The number of electrons released is proportional tothe number of photons hit the device. Hence, by reading the amount of charge, wecan tell the intensity of the light source.

A typical CCD consists of an array of light detecting elements, pixels, usually in therange of 768×512 to 2048×2048. Thus, we could form a picture of such resolutions.The size of one pixel depends on the model, with a typical value of 9µm × 9µm.For a 1024 × 1024 CCD, the light detecting area is about 1 cm by 1 cm, which isless than the size of photographic films. This is one of the disadvantages of CCDas compared with films. The other disadvantages are low resolution and that onlyblack and white pictures can be taken, because CCD only detects the intensity,not the color of the light source. There are two methods to obtain a color photo.One is called the tricolor photo. The observer takes three photos with three red,green and blue filters. Then, combines the three photos into one with the help ofcomputer software. The second method is to use a color CCD. In which three pixelsform a group. Each pixel in a group is covered by either red, green or blue filters,and the electronics of the CCD would combine the data to output a color photo.

There seems to be many disadvantages of CCD, but there is one overwhelmingly


advantage. The quantum efficiency of a professional grade CCD could go upto 80%, which means that it can detects most of the photons, as compared withonly 2% to 4% of photographic films, and 1% of human eyes. In astronomicalapplications, almost all objects in the sky are very dim. A high efficiency devicewill greatly reduce the exposure time. Note that the quantum efficiency is frequencydependent. It is lowest in the blue.

As mentioned, for an object with angular size α (in radians), the physical size ofits real image by a telescope of focal length f is αf . For example, the image ofthe Moon (angular size of about 0.5) of a telescope of focal length 2m is about

. The CCD must have size larger than this to cover the whole Moon. Tomatch the resolving power of the telescope and the CCD, two pixels should coverthe angular resolution. Hence, for example, if the angular resolution is about 1′′

and the pixel size is 9µm, then the matching focal length is

= 3.7m . (2.20)

Astronomical CCDs are usually cooled to low temperature to avoid thermal noise.

Chapter 3

Celestial Mechanics

(Chapter 2 in textbook.)

Since we believe that the laws of physics we developed on Earth should hold any-where in the Universe, the motions of celestial bodies should be governed by classicalmechanics, which was mainly developed by Newton. In this chapter, we will reviewthe basics of the theory of gravitation, then apply it on the two-body problem. Whystudy the two-body problem? This is the simplest case of celestial motion that canbe solved analytically. Also, it is very useful for describing the motions of objectsin a binary system or planetary system.

3.1 Newton’s Laws of Motion

We will briefly review Newton’s laws in this section. The readers are assumed toknow the material well. This section only serves as a reminder.

The first law of mechanics describes the resistance of matter to change in its stateof motion: A body in motion will remain in motion, unless it is acted upon by someexternal force.

Newton’s formulation of the second law is the familiar

F = ma = m v (3.1)

where F is the force vector, m is the mass of an object and a is the accelerationvector. The mass in this equation is the inertial mass, which relates the responseof the body to external force. The acceleration is the rate of change of the velocity.Velocity describes both the speed and the direction of the motion. Thus, sometimethe acceleration is non-zero even if the speed of the body remains constant.

34

CHAPTER 3. CELESTIAL MECHANICS 35

Eq. (3.1) is valid only in an inertial frame, which is any frame at rest or in constantvelocity with respect to the fixed stars. If we are careful enough, we can find thatthe “rest” frame on the Earth is not an inertial frame by experiments. This conceptof inertial frame becomes very important when we discuss special relativity.

The third law states that whenever there is an action, there will be an equal inmagnitude but opposite in direction reaction. For example, we feel the gravitationalattraction of the Earth pulling us down, at the same time, there is a force of thesame strength pulling the Earth “up.” (Do you know how to prove Newton’s laws?)

Momentum, or linear momentum, of a particle is defined as the product

p = mv . (3.2)

We found that in the absence of any external force, by Eq. (3.1), the total momen-tum of a system remains constant. This is the conservation of momentum. (Do youknow how to prove?)

The addition of velocities in classical mechanics is very simple. For example, if atrain is moving with velocity vt relative to the station and a ball is moving withvelocity vb relative to the train, then relative to the station, the ball is moving withvelocity vt + vb.

The kinetic energy of a particle is given by

K. E. =1

2mv2 =

p2

2m(3.3)

where v and p are the magnitude of velocity and momentum respectively. If wewant to change the velocity of the particle, a force must act on it. The change ofkinetic energy is equal to the work done W , which, for constant force, is the dotproduct of the force vector F and the displacement vector d of the particle

W = F · d . (3.4)

Apart from the kinetic energy, another important form of energy is the poten-tial energy. This is the energy associated with the configurations of the system.The most important example is the gravitational potential energy. For a particlewith mass m at height h above some reference point on the Earth’s surface, thegravitational potential energy is

U = mgh (3.5)

where g is the free fall acceleration constant on Earth’s surface, g ≈ 9.8m/s2.

There are other kinds of energies, like the chemical energy or nuclear energy. If wesum up all kinds of energies in an isolated system, the total energy also remainsconstant. This is the principle of conservation of energy. Energy (strictly speakingshould be mass-energy) cannot be created nor destroyed. It can be converted from


one form into another, and the total energy is always conserved. (Do you know howto prove?)

Other than the linear motion, rotation of a body is another important subject inmechanics. We will mainly talk about the rotation of a particle on a plane aroundsome point, the center. The angular position of the particle is the angle made by theline joining the center and the particle and some fixed reference line. The angularvelocity is the rate of change of the angular position, usually denoted by ω and inthe unit of radian per second. As the name implied, angular velocity also describesthe direction of the rotation. Similarly, we define the angular acceleration.

R

h

To Sun

R

θ

Figure 3.1: Can we see sunset twice a day?

Exercise: A simple investigation on the rotation of the Earth will tell us that wecan indeed see sunset twice, or many more times, a day, Fig. 3.1. Everyone knowsthat if we are at a high mountain, the time of sunset will be later, because it takesmore time for the Earth to rotate the Sun out of our sight. If we are of height habove the Earth surface, how much later will the sunset be? Referring to the figure,the angle θ is given by

θ = (3.6)

The time for the Earth to rotate this amount is

∆t = (3.7)

Hence, if we substitute to R the radius of the Earth and take h as 1.7m, the heightof a typical adult, then ∆t = s. In order to see sunset twice, all you have to dois to sit down to see the sunset. After it just sets, stand up immediately. You cansee the Sun sets again. The Sun has an angular diameter of 0.5, just after sunset,how high you have to climb (assuming it takes no time) to make the Sun totally goabove the horizon?

Corresponding to the mass in linear or translational motion, we have moment ofinertia I in rotational motion, defined as

I = mr2 (3.8)


for a particle with mass m and a distance r from the center. The angular mo-mentum L is defined as

L = Iω = mr2ω = r × p . (3.9)

Obviously this depends on the choice of origin. Angular momentum is a veryimportant quantity in many astrophysical situations. For example, when materialcollapses due to gravity, it always forms a disk first, because angular momentumis conserved and very hard to get rid of. This results in accretion disks, and it isalso the reason why disk structure is seen everywhere in the Universe, from solarsystems to galaxies. In spacecrafts, angular momentum gradually builds up throughpointing. The angular momentum is stored in reaction wheels and eventually needto be cancelled through thrusters or other means.

A force acting on a particle does not necessarily change its angular velocity; thetangential component of the force must be non-zero to do so. We define the torqueτ as the product of the tangential component of the force and the distance of theapplication point from the center

τ = r × F . (3.10)

Newton’s second law applying to rotation becomes

τ =dL

dt, (3.11)

that is the rate of change of the angular momentum equals to the torque. If thenet torque is zero, the angular momentum remains constant. This is the principleof conservation of angular momentum. (Do you know how to prove?) The mostimportant example is the systems of central force. In such a system, a particleis moving under the influence of a force which always points to or points awayfrom a fixed point. Since the force is always radial, its tangential component andhence the torque are always zero. The angular momentum is conserved. Note thatthis depends on the choice of origin: angular momentum can be conserved aboutone point but not another, depending on the net torque (indeed it is ususally notconserved about other points). The kinetic energy of rotation is given by

K. E. =1

2Iω2 . (3.12)

Before closing this section, we will give an application of mechanics on the spacecrafttrajectory Almost always, a spacecraft will visit a few major planets several timesbefore it goes to its destination, which could be Saturn, for example. The majorreason for such a visit is to accelerate the spacecraft, called gravity assist. Thepoint is that if we can appropriately choose the trajectory of the spacecraft, thespeed of the spacecraft relative to the Sun will be increased. A large amount of fuelcan be saved. We now see how it could be in details.


v1 v2

v’1

v’2

in center of mass frame

u

M M mm M

m

Figure 3.2: The figure at the left shows the velocity of the planet and the spacecraft.The middle shows the point of view in the center of mass frame. In this frame, theSun is moving to the left. After the encounter with the planet, the right figureshows that the spacecraft gains speed by gravity assist.

Suppose a spacecraft of mass m is approaching a planet of mass M ≫ m. Theirvelocities are respectively v1 and v2 relative to the Sun, Fig. 3.2. To simplify theproblem, we consider the center-of-mass frame, in which the total momentum iszero. We first determine the center-of-mass velocity u relative to the Sun,

(M +m)u =Mv1 −mv2 . (3.13)

We have

u = (3.14)

Hence, the velocities of M and m in the center-of-mass frame are

v′1 = v1 − u = (3.15)

v′2 = v2 + u = (3.16)

respectively. We can see that the total momentum in this frame is Mv′1 −mv′2 = 0.Assume for simplicity that after they gravitationally interact, the directions of theirvelocities are perpendicular to the original velocities in the center of mass frame.(Do you know how to calculate their speeds if this assumption is not true? Hint:It depends on the angle of scattering.) The conservation of momentum and energyrequires that their speeds do not change. (how to prove?) Transforming back tothe frame in which the Sun is at rest, the velocity of the spacecraft is given by thevector sum of the velocity of the spacecraft in the center of mass frame and thevelocity of the center of mass frame. We found that its speed increases√

v′22 + u2 =

√(v2 + u)2 + u2 > v2 . (3.17)


Discussion: Assuming that the high temperature is not a problem, can a spacecraftuse the Sun for gravity assist acceleration?

3.2 Newton’s Gravitation

Over 300 years ago, Newton proposed a theory for gravitation, which essentiallysays that everything attracts everything. This theory not only explains the fallingof an apple, but also the motions of planets around the Sun and even the motionof distant binary stars.

The Newton’s law of gravitation states that every particle attracts any other particlewith a force

F = −Gm1m2

r2r (3.18)

where m1 and m2 are the masses of the two particles and r is the distance betweenthem. Negative means attraction. G is the gravitational constant, whose valueis

G = 6.67× 10−8 cm3 g−1 s−2 . (3.19)

This law can be derived from the general relativity in the small potential and lowspeed limit. Note that this law only holds between two particles, for extended ob-jects we need to take the sum of every particles (see example below). The directionof the force on one particle is toward the other particle. Thus, the gravitational forcetends to pull them together. This simple statement has great implications. Sincewe have not found any “anti-gravity,” the gravitational force cannot be canceledand is accumulative. A greater mass will create a greater force. In astronomicalscale, gravitational force is the dominant force.

A fine point about the masses in Eq. (3.18) is in order. To be precise, the mass inEq. (3.18) is the gravitational mass, comparing with the inertial mass in Eq. (3.1).The gravitational mass is a property of the particle which describes the magnitudeof its influence on other objects gravitationally. The inertial mass describes itsresponse to force. These two kinds of masses need not to be the same. The factthat they are exactly the same is the called the equivalence principal and it isthe starting point of general relativity.


A brief review on gravitational force, potential, and energy.

Force ? Potential Energy

F = GMmr2

−−−−−−−−−→ E =yy per

? Potential

= GMr2

−−−−−−−−−→ V =

Example: The gravitational potential due to a point mass M is easily deducedto be V (r) = −GM/r at a point of distance r from it. We now find out bydirect integration the gravitational potential due to a thin uniform spherical shellof material.

z 0

x

y

z

θ

φ

P

Figure 3.3: The geometry of a uniform spherical shell of material.

Let the surface mass density of the shell be ρ, the total mass of the shell be M =4πR2ρ, and its radius be R. We choose the coordinate system, Fig. 3.3, such thatthe point of interest P is on the z-axis with coordinates (0, 0, z0), where z0 couldbe greater than or less than R (outside or inside the shell).

The surface element shown has area sin θ dθdϕ. Its distance from P is√R2 sin2 θ + (R cos θ − z0)2. Hence, its contribution to the gravitational potential


is

dV = − GR2(ρ sin θ dθdϕ)√R2 sin2 θ + (R cos θ − z0)2

. (3.20)

The gravitational potential is then given by

V = −∫ ∫

GR2ρ sin θ dθdϕ√R2 sin2 θ + (R cos θ − z0)2

= −2πGR2ρ

∫ π

0

sin θ dθ√R2 − 2Rz0 cos θ + z20

= 2πGR2ρ

∫ π

0

d(cos θ)√R2 + z20 − 2Rz0 cos θ

= −2πGR2ρ

∫ 1

−1

dx√R2 + z20 − 2Rz0x

= −2πGRρ

(−1

z0

√R2 + z20 − 2Rz0x

)∣∣∣∣1−1

=GM

2Rz0

√R2 + z20 − 2Rz0x

∣∣∣∣1−1

. (3.21)

Be very careful on how we take the square root. From the very definition of poten-tial, Eq. (3.20), we have to take the positive roots. If z0 < R,

V =

=

=

(3.22)

which is independent of z0. The potential is constant, the force is zero. If z0 > R,

V =

=

=

(3.23)

As a result, outside the shell, gravitationally, it acts as a point mass.

Exercise: Using the result above, show that for a solid sphere with sphericallysymmetric but arbitrarily radial mass distribution, the potential outside the sphereis still given by the same formula.

Hence, show that the gravitational force between two spheres of masses M1 andM2 separated by R is given by GM1M2/R

2, assuming their density profiles are


spherically symmetric.

The gravitational potential energy in Eq. (3.5) is valid only near the Earth’s surface.For object above the Earth’s surface, the gravitational potential energy is given by

U = −GM⊕m

r(3.24)

where M⊕ is the mass of the Earth, m is the mass of the particle and r is thedistance of the particle from the center of the Earth. By convention, the potentialenergy at infinity is zero. If the particle is near the Earth’s surface, r = R⊕ + hwhere R⊕ is the radius of the Earth and h is small, we can use the Taylor series(1 + x)−1 = 1− x+ x2 − . . . to obtain

1

r=

1

R⊕ + h= ≈ . (3.25)

Eq. (3.24) becomes

U = −GM⊕m

r≈ (3.26)

Up to an irrelevant constant term, the potential is in the form U = mgh. We cannumerically check that g is equal to GM⊕/R

2⊕.

Imagine we throw a rock to the sky, the rock will fall back to the ground. However,if we throw it fast enough, it can escape the gravitational pull of the Earth andnot return. The critical speed is called the escape velocity. By conservation ofenergy, it is easy to calculate the escape velocity. At the Earth’s surface, K.E. ofthe rock is 1

2mv2 and potential energy (P.E.) is −GM⊕m/R⊕. At infinity, both

K.E. and P.E. are zero;

1

2mv2 −G

M⊕m

R⊕= 0

v = ≈ km s−1 (3.27)

Exercise: What are the escape velocities of the solar system and the Milky Way?

As a warm-up for the next section, we recall the uniform circular motion, which isa particle revolving around a center with constant speed v and constant distance rfrom the center. Hence the angular speed of a particle ω = v/r is constant. Theperiod T is given by

T =2π

ω. (3.28)

The acceleration is given by

a =v2

r= ω2r . (3.29)


The particle needs a centripetal force to keep it in uniform circular motion. Ifthe force is provided by the gravitational force of an object with mass M at thecenter, then

GMm

r2= ma = mω2r , (3.30)

which impliesT 2 = (3.31)

This is a special case of Kepler’s third law.

Exercise: assuming no air resistance, how fast a bullet has to travel on the surfaceof the Earth so that it can keep going around in a circular motion? Compared thevalue with the escape velocity.

3.2.1 Roche Lobe

We now study the gravitational potential around a binary star system. Most starsare in binary systems. Mass exchange can occur between the two stars. Rochelobe, is the equipotential surface which just encloses the two stars. The Rochelobe hugs the larger star tighter. The intercepting point is called the Lagrangianpoint. Note that this is not at the same location as the center of mass. It is closerto the lighter star than to the heavier star. If matter flows out from one star of thebinary, for example, if one of them goes to the red giant phase, it will first fill upthe Roche lobe then channel to the companion star via the Lagrangian point.

Roche lobe

l

d

Figure 3.4: The Roche lobe of a pair of stars. Which star is more massive?

Suppose the masses of the two stars are m1 at position with coordinates (0, 0, 0)and m2 at (d, 0, 0). By symmetry, the Lagrangian point must lie on the x-axis. Let


its coordinates be (l, 0, 0). At the Lagrangian point, the gravitational forces due tothe two stars are equal, we have

Gm1

l2= (3.32)

Solve for l,l = (3.33)

Notice that the position of the Lagrangian is between the two stars and it is nearerto the lighter one. This is, in fact, not the exactly correct. See Section 3.5 for moredetailed analysis of Lagrangian points.

3.2.2 Critical Density of the Universe

We can estimate the critical density of the universe by simple Newtonian gravity,although we need general relativity to rigorously derive it. Let the average massdensity of the universe be ρ. Suppose a galaxy be a distance r from us. Then, thetotal mass inside the sphere of radius r is

M =4

3πr3ρ . (3.34)

If the mass of the galaxy is m, the potential energy of the galaxy due to the massin the sphere is given by

U = −GMm

r= (3.35)

Assume that velocity of the galaxy is radial, and the speed is given by the Hubble’slaw: the speed of a galaxy is proportional to its distance from us, i.e. v = H0r,where H0 is the Hubble constant. The kinetic energy T of the galaxy is

T =1

2mv2 = (3.36)

and the total energy isE = T + U = (3.37)

If E < 0, the galaxy is bounded, which means that the galaxy is not energeticenough to escape from the gravitational pull of other mass. If E > 0, the galaxyis unbounded and will fly away. Therefore, the critical density ρc for which is thegalaxy is just bounded is

E = 0

=

ρc = (3.38)

If we take the value of the Hubble constant as H0 = 70 km s−1Mpc−1, ρc =g/cm3. This is about the mass of five hydrogen atoms per cubic me-

tre, but the average density observed from stars, gas, etc (excluding dark matterand dark energy) is found to be only 0.2 hydrogen atoms per cubic metre.


3.2.3 Virial Theorem

This theorem states that for a gravitationally bound system in equilibrium, thetime averaged total energy is one-half of the time averaged potential energy

⟨E⟩ = 1

2⟨U⟩ . (3.39)

The proof goes as follow. For a system of particles, let pi and ri be the linearmomentum and position of the i-th particle at sometime t. Consider the quantity

Q ≡∑i

pi · ri . (3.40)

We note that

dQ

dt=

d

dt

∑i

mdridt

· ri =1

2

∑i

d2

dt2(mr2i ) =

1

2

d2I

dt2. (3.41)

In equilibrium, we expect that I should stay roughly the same. Hence, we assumethat the time average of its derivative is zero⟨

dQ

dt

⟩= 0 . (3.42)

The time derivative of Q is given by

dQ

dt=∑i

(dpi

dt· ri + pi ·

dridt

). (3.43)

We recognize the second term as twice the kinetic energy∑i

pi ·dridt

=∑i

1

mi

p2i = 2T . (3.44)

By Newton’s second law,

dpi

dt=∑j =i

Gmimj

|rj − ri|3(rj − ri) ≡

∑j =i

Fij . (3.45)


Here, we sum only over j. Notice that Fij = −Fji, we have

∑i

dpi

dt· ri =

∑i

(∑j =i

Fij · ri

)

=1

2

(∑i

∑j =i

Fij · ri +∑j

∑i=j

Fji · rj

)

=1

2

(∑i

∑j =i

Fij · ri +∑i

∑j =i

Fji · rj

)

=1

2

∑i

∑j =i

(Fij · ri + Fji · rj)

=1

2

∑i

∑j =i

(Fij · ri − Fij · rj)

=1

2

∑i

∑j =i

Fij · (ri − rj)

= −1

2

∑i

∑j =i

Gmimj

|rj − ri|

= U . (3.46)

We need the factor 1/2 to compensate for the double counting of the number ofpairs of particles. In summary, U + 2T = dQ/dt and after taking time average,⟨U⟩+ 2 ⟨T ⟩ = 0. Total energy is E = T + U and we have Eq. (3.39).

Stars were formed in nebulae. At the beginning, the density of gas and dust in anebula is very low, and they are moving slowly. Hence, the total energy is roughlyzero. On the other hand, a star is gravitationally bounded. Its potential energy isnon-zero and negative. By virial theorem, its total energy is also negative. Thisimplies that energy must be transferred out for a nebula to form a star. Physicallythis is via thermal radiation during star formation.

Another interesting consequence of Virial theorem is that as a star loses energy, thetotal energy becomes more negative. Since ⟨T ⟩ = −⟨U⟩ /2, the KE of the particlesactually increases! The net result is that the star gets hotter while loses energy,somewhat like having a “negative heat capacity”.

3.3 Two-Body Problem

In this section, we talk about the motion of two bodies under their mutual gravi-tational attraction. Why study this? It can tell us different orbits of planets andcomets around the Sun, e.g. circle, ellipse, parabola, and hyperbola. After a brief


review on Kepler’s laws, we will derive the general solution to the two-body problemusing Newtonian mechanics, and apply it to prove Kepler’s laws.

3.3.1 Kepler’s Laws of Planetary Motion

• First Law: A planet orbits the Sun in an ellipse, with the Sun at one focus ofthe ellipse.

• Second Law: A line connecting a planet to the sun sweeps out equal areas inequal time intervals.

• Third Law: The orbital period P of a planet and the semi-major axis of itsorbit a are related by P 2 ∝ a3.

3.3.2 Orbits in Two-Body Problem

Our goal here is to derive the motion of objects in two-body problem using Newton’sgravitational force equation. We will employ transformations (r1, r2 → r andm1,m2 → µ) to solve Newton’s 2nd law.

Let first assume that the force between the two bodies depends only on their relativeposition, the force F acting on the first body by the second is a function of r1−r2,where r1 and r2 are the position of the two bodies. Then, if the masses of the twoare m1 and m2, their equations of motion are

m1r1 = F (r1 − r2) (3.47)

m2r2 = −F (r1 − r2) (3.48)

Let R be the position of the center of mass and r be the relative position,

R =m1r1 +m2r2m1 +m2

(3.49)

r = r1 − r2 (3.50)

Exercise: From Eq. (3.49), we have

R = .

Compare this with the sum of Eqs. (3.47) and (3.48),

R = ,

which just means that the center of mass of the system will move in a straight linewith constant velocity.


If we calculate the difference between Eqs. (3.47) and (3.48), we have

m1m2r1 −m1m2r2 =

m1m2

m1 +m2

r = . (3.51)

If we define µ ≡ m1m2/(m1+m2) then Eq. (3.51) is reduced to a one-body problemF (r) = µr. Therefore, µ is called the reduced mass. If, for example, m2 ≫ m1,r is the position of the first body relative to the second and the reduced mass ism1. This is the case for planets orbiting around the Sun in our solar system.

r

θ

rθ ^^

Figure 3.5: A particle moving on a plane with polar coordinates.

We now assume that F is a central force, which means that the direction of F isequal to r and its magnitude is a function of the distance only, F = F (r)r. Hence,the torque is zero and angular momentum is conserved. If the angular momentumis denoted by L, it is a constant vector. Since r · L = 0, r is always on the planeperpendicular to L. Hence, the bodies are moving on a plane.

On the plane, we usually label the position of a point by its Cartesian coordinates,(x, y). Here, we also need the polar coordinate system (r, θ), Fig. 3.5. Their relationsare

x = r cos θy = r sin θ

r =

√x2 + y2

θ = tan−1(y/x). (3.52)

The unit vectors r and θ are illustrated in Fig. 3.5. In terms of the unit vectors xand y along the x- and y-axes, they are

r = x+ y

θ = x+ y. (3.53)

For Newton’s gravitation, from Eq. (3.51),

µr = −Gm1m2

r2r ≡ −GMµ

r2r , (3.54)


where M ≡ m1 +m2 is the total mass. We would like to calculate r in terms of rand θ. Since x and y are constant vectors, they do not change with time (but rand θ do change). We have

˙r = = θ θ˙θ = = −θ r

. (3.55)

We can now calculate r,

r = r r

r =

= (3.56)

r =

=

= (r − rθ2)r + (2rθ + rθ)θ . (3.57)

Comparing with Eq. (3.54), separating into azimuthual and radial parts, we haver − rθ2 = −GM

r2

2rθ + rθ = 0. (3.58)

These are the equations of motion. We are just interested in the equation of orbit(not interested in time), that is the dependence of r in terms of θ. Notice that theazimuthual part

d

dt(r2θ) = = = 0 (3.59)

by the second equation of Eq. (3.58). Thus, L ≡ µr2θ is a constant of motion. It isin fact the angular momentum. We can rewrite the radial part of Eq. (3.58) as

r = −GMr2

+L2

µ2r3,

or µr = −GMµ

r2+

L2

µr3(3.60)

This is the modified force law. It has a similar form as the gravitation law, butslightly modified. This effective force has an extra term representing the cen-tripetal force. This can be used to derive the effective potential energy of thesystem

Ueff = −GMµ

r+

L2

2µr2. (3.61)

Although this is just a mathematical form instead of real potential energy, it canhelp us easily visualize the orbits of a particle with certain energy (Fig. 3.6).

As L = µr2θ, we can write r in terms of L by

r =dr

dt=

dr

dθ

dθ

dt=

L

µr2dr

dθ. (3.62)


Therefore, the radial part, i.e. the first equation of Eq. (3.58) becomes

L

µr2d

dθ

(L

µr2dr

dθ

)− r

(L

µr2

)2

= −GMr2

d

dθ

(1

r2dr

dθ

)=

1

r− GMµ2

L2. (3.63)

Let u = 1/r, dr/dθ = −1/u2du/dθ. The above equation gives

d2u

dθ2+ u =

GMµ2

L2, (3.64)

for which the general solution is

1

r= u =

GMµ2

L2[1 + e cos(θ − θ′)] , (3.65)

where e and θ′ are two constants of integration. θ′ just tells us how the orbit orientsrelative to our coordinate system. We can set it to 0 or π such that e ≥ 0. e is animportant parameter of the orbit, called the eccentricity.

Exercise: Verfiy that Eq. (3.65) is a solution of Eq. (3.64).

If e = 0, The particle is in constant distance from the center. This is a circle. Moregenerally, for 0 < e < 1, the orbit is called an ellipse. It is closed and bounded.

Figure 3.6: Effective potential.


The particle will revolve around the center of mass periodically, where it is calledthe focus of orbit.

If e = 1, the orbit is called a parabola. If e > 1, the orbit is a hyperbola. Bothparabola and hyperbola are open orbits. The particle will come near the focus onceand then go away. This is the case for some comets, they will enter the inner solarsystem, fly by the Sun only once, then never come back. Particle with parabolicorbit is that its kinetic energy just balances the potential energy, i.e. the total energyis zero. When it gets farther away from the focus, its speed will decrease and tendsto zero, while particle with hyperbolic orbit has a non-zero speed at infinity.

focus

ellipse

parabola

hyperbola

Figure 3.7: Left: three kinds of orbits around the focus. Right: conic sections.

All these shapes are called the conic sections, Fig. 3.7. The point on the or-bit nearest to the focus is called the perihelion (the farthest point is called theaphelion). The distance between perihelion and the focus is given by

rp =L2

GMµ2

1

1 + e. (3.66)

This occurs when the right hand side of Eq. (3.65) is the largest, that is when θ = θ′.

Exercise: Transform Eq. (3.65) to Cartesian coordinates and show that: (a) e = 0is a circle, (b) 0 < e < 1 is an ellipse, (c) e = 1 is a parabola, and (d) e > 1 is ahyperbola.


Exercise: Show that the total energy of a two-body system is: (a) minimum for acircular orbit, (b) < 0 for a elliptical orbit, (c) = 0 for a parabolic orbit, and (d)> 0 for a hyperbolic orbit.

For closed orbits, we can calculate the period. We can put θ′ = 0 in Eq. (3.65)because the period does not depend on it. Then, the equation of orbit is

1

r= u =

GMµ2

L2(1 + e cos θ) . (3.67)

We would like to calculate the area of the orbit. First transform to the Cartesiancoordinate system, we have r cos θ = x and r + ex = L2/GMµ2. Hence,

r =L2

GMµ2− ex

x2 + y2 =

(L2

GMµ2

)2

− 2eL2

GMµ2x+ e2x2

y2 + (1− e2)x2 +2eL2

GMµ2x =

(L2

GMµ2

)2

y2 + (1− e2)

[x+

eL2

GMµ2(1− e2)

]2=

1

1− e2

(L2

GMµ2

)2

. (3.68)

This is the equation of an ellipse with a shifted center. Its area equal to the areaof the following ellipse

y2 + (1− e2)x2 =1

1− e2

(L2

GMµ2

)2

. (3.69)


Let B2 = 11−e2

(L2

GMµ2

)2and w = x

√1− e2/B. The area is

A = 2

∫ B/√1−e2

−B/√1−e2

√B2 − (1− e2)x2 dx

=2B2

√1− e2

∫ 1

−1

√1− w2 dw

=πB2

√1− e2

=π

(1− e2)3/2

(L2

GMµ2

)2

. (3.70)

3.3.3 Proof of Kepler’s Laws

First law: As discussed above, in a close orbit, a planet moves around the centerof mass of the system, which is the focus of the orbit. If the star is much moremassive than the planet, it will essentially sit at the focus.

Exercise: How much does the Sun move due to due to Earth? Due to Jupiter?

Second law: In polar coordinates, the area of a circular sector is given by

A =1

2r2θ . (3.71)

θ

rr

θ21A= r 2

Figure 3.8: Area of a sector.

The rate of change in area with time is

dA

dt=

1

2r2dθ

dt=

1

2r2θ =

1

2

L

µ. (3.72)

For a very small angle dθ, the same formula alsoholds for ellipse. This is a constant because the an-gular momentum L is a constant. To calculate theperiod P , it follows from Eqs. (3.70) and (3.72) that

P =2µA

L=

L3

G2M2µ3

2π

(1− e2)3/2. (3.73)

Third law: The semi-major axis a is the averagedistance of the perihelion and the aphelion. FromEq. (3.66) and a similar expression for the aphelion,

a =L2

2GMµ2

(1

1 + e+

1

1− e

)=

L2

GMµ2

1

1− e2. (3.74)


Figure 3.9: Solid line: observed rotation curve of a galaxy. Dotted line: predictionfrom Kepler’s third law.

Comparing Eqs. (3.73) and (3.74), we have

P 2 =4π2

GMa3 . (3.75)

This is the enhanced version of Kepler’s third law. We have derived the proportionalconstant from Newton’s law. This provides a very powerful tool to estimate the massof celestial objects, from moons to planets to stars to galaxies.

Exercise: From Kepler’s first and second laws, derive Newton’s law of gravitation.

Exercise: Assuming a circular orbit, from Eq. (3.75), derive the angular velocity,and hence the linear velocity, as a function of distance from the center.

In a galaxy, stars are concentrated at the center, therefore, the rotation curve is ex-pected to follow Kepler’s third law except at the very center. However, the observedrotation curves is much flatter (see Figure 3.9). The discrepancy is attributed todark matter. For the Milky Way, it is estimated that 90% of the mass is darkmatter, while ordinary matter is only 10%.


3.4 Impact Parameter and Scattering Angle

We now study the two body problem in another point of view. Consider the casethat a light body approaches a very heavy body from far away, for example, asatellite approaching a planet. (Does it sound familiar?) Results of last sectiontell us that the reduced mass is very closed to the mass of the light body and thetrajectory is a hyperbola.

Let the mass of the heavy body beM , the mass of the light body bem. We considerM ≫ m such that µ ≈ m. When the light body is far away from the heavy body,let its incident velocity be v and the impact parameter, b, be the perpendiculardistance between the heavy body and the incident velocity, Fig. 3.10. We wouldlike to determine the scattering angle, Θ.

rpvp

M

b

v

Θ

m

θ r

Figure 3.10: Impact parameter of a two body system.

When the light body is far away, it is obvious that the angular momentum L isgiven by L = mbv, which is a constant of motion. At perihelion, let its velocitybe vp, it is perpendicular to the line joining the two bodies. Hence, vp = rθ atperihelion. (At other point on the trajectory, the velocity has radial component,see Eq. 3.56.) As L = mr2θ, we have

mbv = L = mr2θ = mrp(rpθ) = mrpvp (3.76)

and by Eq. (3.66),

vp =bv

rp= . (3.77)

By conservation of energy,

1

2mv2 =

v2 =

e = . (3.78)


We see from this equation that e > 1, and hence we have a mathematical proofthat the trajectory is a hyperbola. For simplicity, let’s define our coordinate systemsuch that θ′ = 0 in Eq. (3.65) when r = rp. Then from Eq .(3.66), the light bodywill go infinitely far away at the angles

1 + e cos θ = 0

cos θ = −1

e. (3.79)

Negative cos θ means that θ > π/2. We can also write

θ = π − cos−1 1

e. (3.80)

Finally, from the geometry in Fig. 3.10, the scattering angle Θ is given by θ+ (θ−Θ) = π. Hence,

Θ = 2θ − π

= π − 2 cos−1 1

e

= . (3.81)

Let us check that if the impact parameter is very large, the light body should notbe affected much by the heavy body. We expect the scattering angle is small. If bis large, e is large, cos−1 1/e is closed to π/2, and Θ is small.

This calculation also sheds some light on the problem of gravity assist. Since inthis section we have assumed that heavy body does not move, we are essentially inthe center-of-mass frame. We have mentioned in Section 3.1 that the final speed ofthe spacecraft depends on the scattering angle. We now know that we can controlthe scattering angle by adjusting the impact parameter and the incident velocity.

3.5 Restricted Three-Body Problem

In general, three-body problem is very difficult. It is sometimes chaotic and doesnot have analytical solution. In this section, we will discuss a special case and somemore about Lagrangian points.

We assume that two point masses m1 and m2 revolve around their center of massin circular motion, as shown in Fig. 3.11. The origin of the coordinates is chosento be at the center of mass, and at time t = 0, m1 is at the positive x-axis. If a andb are the distances of the masses from the origin, then a/b = .


m1

m2

R1

R2

X

Y

a

b

(X,Y)

Figure 3.11: The positions of the three bodies.

Exercise: Using Eq. (3.75), find the angular speed, n, of the point masses.

Let c ≡ cosnt and s ≡ sinnt. Then, the positions of the m1 and m2 are respectively(cb, sb) and (−ca,−sa).

A third body is assumed to always lie on the orbital plane defined by the twomasses. We also assume that it is very light (i.e. a test particle) and will notinfluence the motion of the other two. This is the case, for example, if the twomasses are the Sun and Jupiter, while the third body is an asteroid.

If the position of the third body is (X, Y ), then R1 and R2 defined by the figureare

R1 = (X,Y )− (cb, sb) = (X − cb, Y − sb) (3.82)

R2 = (X,Y )− (−ca,−sa) = (X + ca, Y + sa) . (3.83)

The equations of motion are

X = −Gm1

R31

(X − cb)− Gm2

R32

(X + ca) (3.84)

Y = −Gm1

R31

(Y − sb)− Gm2

R32

(Y + sa) (3.85)

where Ri are the magnitude of Ri.

We now transform to a moving coordinate system which rotates with the two masses.This is called the co-rotating frame. It is not an inertial reference frame, meaningthat there will be extra terms in Newton’s 2nd law. Physically, it could mean that


we see the Sun and the asteroid from Jupiter. Mathematically, we define (x, y) by

X ≡ cx− sy (3.86)

Y ≡ sx+ cy . (3.87)

Note that even if x and y are time independent, X and Y still depend on timethrough c and s. In terms of x and y, R2

1 = (x− b)2 + y2 and R22 = (x + a)2 + y2.

Since

X = −nsx+ cx− ncy − sy (3.88)

X = −n2cx− 2nsx+ cx+ n2sy − 2ncy − sy (3.89)

Y = ncx+ sx− nsy + cy (3.90)

Y = −n2sx+ 2ncx+ sx− n2cy − 2nsy + cy , (3.91)

the equations of motion are

−n2cx− 2nsx+ cx+ n2sy − 2ncy − sy

= −Gm1

R31

(cx− cb− sy)− Gm2

R32

(cx+ ca− sy) (3.92)

−n2sx+ 2ncx+ sx− n2cy − 2nsy + cy

= −Gm1

R31

(sx− sb+ cy)− Gm2

R32

(sx+ sa+ cy) . (3.93)

If we calculate the sum of c times Eq. (3.92) and s times Eq. (3.93), we have

− n2x+ x− 2ny = −Gm1

R31

(x− b)− Gm2

R32

(x+ a) . (3.94)

Subtract s times Eq. (3.92) from c times Eq. (3.93), we have

2nx− n2y + y = −Gm1

R31

y − Gm2

R32

y . (3.95)

These two equations are the main equations. We will not study the general solutions.Instead, we will only find out the equilibrium points, which means the solutions withfixed x and y values.

If x and y do not depend on time, x = x = y = y = 0. Substitute into Eq. (3.94)and Eq. (3.95), we have

n2x =Gm1

R31

(x− b) +Gm2

R32

(x+ a) (3.96)

n2y =Gm1

R31

y +Gm2

R32

y . (3.97)

If y = 0, Eq. (3.96) gives us three solutions. They are the Lagrangian points L1,


x

y

m2

L1L2 L3

m1

L5

L4

a b

Figure 3.12: Left: The positions of the five Lagrangian points relative to the twomasses. Right: Gravitational potential of the system in the co-rotating frame.

L2 and L3. For m1 ≫ m2, Eq. (3.96) can be solved to obtain

L1 :

(−a+ (a+ b)

(b

3a

)1/3

, 0

)

L2 :

(−a− (a+ b)

(b

3a

)1/3

, 0

). (3.98)

L3 :

(a+ b+

5

12b, 0

)

L1 is in between the two masses, and it is in fact the Lagrangian point that wefound in Eq. (3.33), but now we have the correction (the left hand side of Eq. (3.96))coming from the rotation of the two masses. All these three Lagrangian points lieon the line joining the two masses, Fig. 3.12.

If y = 0, by Eq. (3.97), we have

n2 =Gm1

R31

+Gm2

R32

. (3.99)

Substitute this into Eq. (3.96), we have(Gm1

R31

+Gm2

R32

)x =

Gm1

R31

(x− b) +Gm2

R32

(x+ a)

0 = −Gm1

R31

b+Gm2

R32

a

R31 = R3

2 (3.100)

because m1b = m2a. Hence, x = (b − a)/2. From the facts that n2 = G(m1 +


m2)/(a+ b)3, R31 = R3

2 and Eq. (3.99), we have

R32 = (a+ b)3(

a+ b

2

)2

+ y2 = (a+ b)2

y = ±√3

2(a+ b) . (3.101)

Thus, the coordinates of the fourth and fifth Lagrangian points L4, L5 are

L4 :

(1

2(b− a),

√3

2(a+ b)

)

L5 :

(1

2(b− a),−

√3

2(a+ b)

). (3.102)

It can be shown that L1, L2 and L3 are unstable,1 in the sense that material atthese points would fly away if slightly perturbed. We saw that material will betransferred from one star to another through L1. On the other hand, L4 and L5

are stable if m1/m2 > 25. These are the cases for the Sun-Earth and Earth-Moonsystems.

The Lagrangian points have been used as parking lots for satellites. For the Sun-Earth system, the solar observatory SOHO is in L1, such that it can observe thesun continuously. L2 is used by the microwave probes WMAP and Planck, whichcan observe the sky without any interference from the Sun and the Earth. It willalso be used by the James Webb Space Telescope in future. L3 is too far to beuseful, but it is a popular place for the hypothetical counter-Earth. The STEREOsatellites were able to observe L3 and have ruled out the existence of any largeobjects there. L4 and L5 are the homes to many asteroids. This class of objects arecalled Trojans asteroids. The STEREO satellites have visited L4 and L5 to detectthe Trojans asteroids. We have also discovered many asteroids at L4 and L5 of theSun-Jupiter system.

Question: In Fig. 3.12 right, why do the Lagrangian points seem to lie on top ofthe potential? And why does the potential fall off at large distance?

1See http://wmap.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf.

http://wmap.gsfc.nasa.gov/media/ContentMedia/lagrange.pdf

Chapter 4

Introduction to RadiativeProcesses

(Chapters 3.4–3.6, 9.1–9.4 in textbook.)

In astrophysics, one studies distant objects through their emission. It is very im-portant to understand how the radiation is generated and transmitted. We willbriefly introduce the radiative transfer and blackbody radiation.

4.1 Solid Angle

Recall that one way to define an angle is that the angle sustained by an arc isthe ratio of the length of the arc to the radius of arc, the left diagram of Fig. 4.1,θ = l/r. The angle of a whole circle is, of course, 2π.

Ω

r

Al

rθ

Figure 4.1: Angle is the ratio of arc length to radius. The solid angle is the ratio ofthe area to the square of the distance.

Suppose there is a sphere from a distance from us. We could talk about the angularsize of it, with units in degrees or radians. However, if we like to talk about howmuch of the sky is blocked by the sphere, we are talking about the solid angle,Fig. 4.1. Its definition is the ratio of the area to the square of the distance, Ω = A/r2,and has a unit of sr. The solid angle of the whole sky is 4π sr.

To find out the infinitesimal solid angle, consider the infinitesimal area in Fig. 4.2.

61

CHAPTER 4. INTRODUCTION TO RADIATIVE PROCESSES 62

The two sides are rdθ and r sin θdϕ, hence the area is dA = r2 sin θdθdϕ anddΩ = dA/r2. This gives

dΩ = sin θ dθ dϕ . (4.1)

dΩ = sin d d

φ dφ

θd

y

z

x

θ

θθ φ

Figure 4.2: Differential solid angle.

Example: If you are at the center of a cube, what will be the solid angle sustainedby one face of the cube?

By symmetry, there are six faces and the solid angle of whole sky is 4π, hence thesolid angle of one face should be 2π/3. Let’s do it the hard way.

2a

2a

2a

x

φ

φ

α

θ

z

y

Figure 4.3: Solid angle sustained by a face of a cube.

Let the length of one side of the cube be 2a. We consider the shaded area shownin Fig. 4.3. We will use Eq. (4.1) to integrate the differential solid angle for everypoint in the area, then multiple the result by eight times to get the answer. Weneed to figure out the ranges of ϕ and θ. It is easy to see that 0 ≤ ϕ ≤ π/4. For


θ, it runs from 0 to a maximum value α which depends on ϕ. To find out α, weconsider the red line in the figure. Its length can be expressed in terms of cosϕ ortanα. We therefore have

a tan θ = . (4.2)

Hence, the range of θ is 0 ≤ θ ≤ α, where α = .

Finally, we integrate to obtain the required solid angle

Ω = 8

∫ π/4

ϕ=0

∫ α

θ=0

sin θ dθ dϕ

=

= . (4.3)

To evaluate the integral, we need

cos2 α =1

1 + tan2 α=

1

1 + 1/ cos2 ϕ=

cos2 ϕ

1 + cos2 ϕ. (4.4)

If we put u = sinϕ, then

Ω = 8

∫ π/4

0

1− cosϕ√1 + cos2 ϕ

dϕ

= 2π − 8

∫ √2/2

0

1√2− u2

du

= 2π − 8 sin−1

(u√2

)∣∣∣∣√2/2

0

= 2π − 8(π/6)

=2π

3. (4.5)

4.2 Specific Intensity and Flux

In this section, we will discuss various flux, intensity and energy density. These arescientific terms to describe a radiation field and light rays.

Imagine inside a region filled with radiation, or photons, with all kinds of frequenciesand going in all directions. Consider a photon (or a light ray) goes in a specificdirection n, the specific intensity, Iν , is defined as the energy carried by photonspass through a small area dA perpendicular to n, pointing to similar directionswithin a small solid angle dΩ, between frequency ν and ν + dν, and time dt (seeFig. 4.4)

Iν ≡ dEν

dA dt dΩdν. (4.6)


n

dΩdA

Figure 4.4: For the definition of specific intensity.

Physically, the specific intensity could be understood as the brightness. It isgenerally a function of Ω, position, ν and time, and has a dimension of energy perunit area per unit time per unit solid angle per unit frequency. In c.g.s. the unitsare erg/s/cm2/sr/Hz. Note that Iν is independent of distance! It is like surfacebrightness (magnitude per square arcsec). Stars do not fade with distance, but justget smaller, i.e. smaller solid angle. If we take two pictures of the Sun, one fromEarth and one from Venus, using the same camera setting, the surface brightnessof the sun will look identical in the two pictures.

We can also express the specific intensity in terms of per unit wavelength insteadof frequency. The physical requirement is that∫ ∞

0

Iν dν =

∫ 0

∞Iλ dλ , (4.7)

orIν dν = −Iλ dλ (4.8)

in the differential form. Using the relation between wavelength and frequency,c = νλ, we have

Iλ =ν

λIν =

c

λ2Iν . (4.9)

What is the unit of Iλ?

dΩdA

θ

Figure 4.5: For the definition of energy flux.

Next, we consider the case when the light ray direction is not perpendicular to dA.As shown in Figure 4.5, it is obvious that if θ = 90 the flux passing through thearea dA will be zero, and the flux is maximum when θ = 0. For a general θ, the


effective area is reduced by a factor of cos θ, therefore, the specific energy fluxpassing through a small area dA with direction dΩ is

dFν = Iν cos θ dΩ . (4.10)

To obtain the net flux (actually flux density, as it is per unit frequency), we haveto integrate all directions

Fν =

∫Iν cos θ dΩ . (4.11)

It is in units of erg s−1 cm−2Hz−1.

Exercise: Show that if the radiation is isotropic, i.e., Iν does not depend on Ω,then Fν = 0.

Hint: You will see this type of integral∫

dΩ a lot in astrophysics. In the actualcalculation, one needs to express dΩ in terms of θ and ϕ first. Also, it is sometimesuseful to make a substitution µ = cos θ, which gives dµ = − sin θ dθ, such that∫ π

0

f(θ) sin θ dθ =

∫ 1

−1

f(µ) dµ . (4.12)

Example: Constancy of specific intensity along rays in free space.

Consider any ray L and any two points along the ray. As shown in Fig. 4.6, constructareas dA1 and dA2 normal to the ray at these points. By energy conservation, energycarried by the set of rays passing through both dA1 and dA2 can be expressed intwo ways:

dE1 = Iν1dA1dtdΩ1dν1 = dE2 = Iν2dA2dtdΩ2dν2 .

Here dΩ1 is the solid angle subtended by dA2 at dA1 and so forth. Since dΩ1 =dA2/R2,

dΩ2 =dA1/R2, and dν1=dν2, we have Iν1 = Iν2 , i.e. Iν =constant along a ray.

dA1dA2

R

Figure 4.6: Constancy of intensity along rays.


We now prove the inverse square law. For a distant star, one can take cos θ = 1.Since Iν is independent of distance,

Fν =

∫Iν cos θ dΩ = IνΩ . (4.13)

Ω = πR2/d2, where R is the stellar radius and d is the distance. Hence, Fν ∝ d−2.Physically, the star does not get fainter with distance, but the solid angle getssmaller, therefore, the total flux is smaller.

For a photon with energy E, it carries a momentum of E/c. The momentumflux pν is the momentum per unit time per unit area perpendicular to dA, whichalso equals to the pressure. Imagine a photon strikes a wall, only the perpendicularcomponent of the momentum will change and exert pressure. Therefore, only theperpendicular component matters and this introduces an extra factor of cos θ in themomentum flux

pν =1

c

∫Iν cos

2 θ dΩ . (4.14)

Ωd

dA

ds= dtc

Figure 4.7: For the definition of energy flux.

We define the specific energy density, uν , as the energy per unit volume per unitfrequency. It has units of erg/cm3/Hz. Consider a cylinder in Figure 4.7 of crosssection dA and length ds=cdt, the total energy in the volume is uν dAcdt dν. Onthe other hand, after time dt, all photons inside will come out, passing through thearea dA to all solid angle. From the definition of Iν , the total energy passing canalso be expressed as (

∫Iν dΩ) dA dt dν. Equating these two,(∫Iν dΩ

)dA dt dν = uν dAc dt dν

uν =1

c

∫Iν dΩ . (4.15)

Finally, we are going to prove that for isotropic radiation, the radiation pressurep is equal to 1/3 of the energy density. Consider a system consists of a containerwith isotropic radiation field inside. Since the photons have to turn around on theboundary, the radiation pressure on the boundary is twice the momentum flux; butwe integrate only over 2π solid angle,

p = 2

∫pν dν =

2

c

∫ π/2

θ=0

Iν cos2 θ dΩdν . (4.16)


By isotropy, Iν does not depend on Ω, therefore,

p = . (4.17)

On the other hand, the energy density is

u =

= 3p . (4.18)

This relation is the equation of state for radiation. In cosmology, it can be used todescribe the radiation-dominated era in the early Universe.

Similarly, the energy flux flowing out of the boundary is

F ≡∫Fν dν =

∫Iν dν

∫ π/2

θ=0

cos θ dΩ = π

∫Iν dν . (4.19)

It is related to the energy density by F = cu/4.

4.3 Emission and Absorption

The monochromatic emission coefficient jν is defined as the energy emittedper unit time per unit solid angle per unit volume:

jν ≡ dE

dV dΩdt dν(4.20)

and it has units of erg/cm3/s/sr/Hz. When a beam of cross section dA travelsthough a emission region for a distance ds, the volume it covers is dV =dAds. Theenergy it gains is

dIν = jν ds . (4.21)

For absorption, we first note that the amount of absorption depends on the intensityof the incident beam, e.g., no absorption could occur if the beam contains no energy.Therefore, the absorption coefficient αν is defined as the change in the beamintensity after traveling for a distance ds:

dIν = −ανIν ds . (4.22)

αν is in units of cm−1. Putting Eqs. (4.21) and (4.22) together, the general form ofthe radiative transfer equation is

dIνds

= −ανIν + jν . (4.23)

For pure emission, αν = 0, the solution is

Iν(s) = Iν(s0) +

∫ s

s0

jν(s′) ds′ . (4.24)


For pure absorption, jν = 0, the solution is

Iν(s) = Iν(s0)e−

∫ ss0

αν(s′) ds′ . (4.25)

The integral in the exponent is called the optical depth

τν ≡∫ s

s0

αν(s′) ds′ . (4.26)

This is an important parameter to describe the emission properties of plasmas inastrophysics. A medium is optically thick or opaque when τν ≫ 1, meaning thatphotons cannot transmit for a long distance without being absorbed. A mediumwith τν ≪ 1 is optically thin or transparent, such that photons can propagatemore or less freely. For example, the early Universe had a high temperature, mostmaterials were ionized, absorbing all electromagnetic radiation. Therefore, it wasopaque to light (i.e. optically thick). It was until the last scattering surface whenmatter decoupled from photons, resulting in the cosmic microwave background wesee today. Also, the surface i.e. photosphere of a star (e.g. the Sun) is defined asthe point where τ = 2/3, such that photons can escape freely into space.

We can rewrite equation 4.23 as

1

αν

dIνds

= −Iν +jναν

. (4.27)

From the definition of τν , we have dτν = αν ds. Define the source functionSν ≡ jν/αν , then

dIνdτν

= −Iν + Sν . (4.28)

It is easy to see that if Iν > Sν , then dIν/dτν < 0. Physically, it means that Iνtends to decrease along the ray until it is the same as Sν . Conversely, Iν increasesif Iν < Sν . In other words, Iν always tries to approach Sν . The general solution toEquation 4.28 is

Iν = Iν(0)e−τν +

∫ τν

0

e−(τν−τ ′ν)Sν(τ′ν) dτ

′ν . (4.29)

For the simple case of a constant source function Sν , this reduces to

Iν = Iν(0)e−τν + Sν(1− e−τν ) . (4.30)

Again, τν → ∞ implies Iν → Sν , i.e., given a large optical depth (e.g., travel forsufficient distance in a optically thick medium), the observed intensity will approachto the source function.

4.4 Scattering Cross Section

How strong do two particles interact with each other? How can we describe it?Experimentally, we usually send a uniform beam of particles, with same mass and


energy, to certain target particle. Some incident particles will be absorbed, somewill be scattered, some will be even chemically changed to others. The number ofincident particles affected is proportional to the number of incident particles. Ifthis ratio is larger, the interaction between the incident particles and the target isconsidered to be stronger.

(a) (b)

Figure 4.8: The total cross section depends on the orientation.

The intensity of the beam is defined to be the number of incident particles crossingunit area normal to the beam in unit time. The total cross section is defined tobe

σT =number of particles affected per unit time

incident intensity. (4.31)

Note that the dimension of the total cross section is same as an area, becausethe numerator has dimension of pure number per time, while the denominator hasdimension of pure number per time per area.

To illustrate the idea, let say there is a board with area A which absorbs everyparticles incident on it. If the board is normal to the incident particles, Fig. 4.8a,it is immediately that the total cross section is

σT = A . (4.32)

A larger total cross section does mean a stronger interaction. If the board is parallelto the direction of motion of the incident particles (and the board is very thin),Fig. 4.8b, the total cross section is zero. Hence, the cross section depends on theorientation in general.

For simplicity, from now on, we assume that the incident particles will just bescattered. To have a finer description on scattering, we define the differentialcross section, σ(Ω), as

σ(Ω) dΩ =number of particles scattered into solid angle dΩ per unit time

incident intensity.

(4.33)In this formula, we have assumed that the place where we actually do the measure-ment is far away from the scattering center. If an incident particle is scattered, itmust be scattered to some direction. We have

σT =

∫σ(Ω) dΩ (4.34)


dΩ

Figure 4.9: The differential cross section.

where the integration is over the 4π solid angle.

θ+ θd

θb

Figure 4.10: The geometry for the calculation of differential cross section.

We are going to calculate the differential cross section of the gravitational inter-action. We know the relation between impact parameter and the scattering anglefrom Eq. (3.81). Notice that there is a cylindrical symmetry. We only have tocalculate the θ dependency of the differential cross section, σ(Ω) = σ(θ).

Referring to Fig. 4.10, all incident particles passing through the annulus at the left,with radii b and b + db, will be scattered to the annulus at the right, with anglesbetween θ and θ + dθ. If the incident intensity is I particles per unit normal areaper unit time, then the number of particles passing through between b and b + dbis I 2πb db per unit time. This number should be equal to 2πIσ(θ) sin θ dθ by thedefinition of differential cross section, where the 2π comes from the integration ofthe ϕ dependency. We have

σ(θ) =b

sin θ

∣∣∣∣dbdθ∣∣∣∣ . (4.35)

We take the absolute value because b and θ usually vary in opposite directions.


Then, by Eq. (3.81), we have

cos

(π − θ

2

)=

(1 +

b2v4

G2M2

)−1/2

1

sin2 θ/2= 1 +

b2v4

G2M2

b2v4

G2M2= cot2

θ

2bv2

GM= cot

θ

2v2

GM

db

dθ= − 1

2 sin2 θ/2

db

dθ= −GM

2v21

sin2 θ/2. (4.36)

The differential cross section is

σ(θ) =b

sin θ

GM

v2 sin2 θ/2

=G2M2

2v4cot θ/2

sin θ sin2 θ/2

=G2M2

4v4 sin4 θ/2

=(GMm)2

16E2 sin4 θ/2, (4.37)

where E = mv2/2 is the energy of the incident particles. M ≫ m is assumed suchthat µ ≈ m. In principle, we could get the total cross section by integration

σT =

∫σ(θ) sin θdϕ dθ . (4.38)

If we actually do the integration, we find that σT diverges to infinity:

σT =

∫(GMm)2

16E2 sin4 θ2

sin θ dϕ dθ (4.39)

= −2π(GMm)2

16E2

∫ π

0

2 sin θ2cos θ

2dθ

sin4 θ2

= −2π(GMm)2

4E2

∫ π

0

d(sin θ2)

sin3 θ2

=2π(GMm)2

8E2

1

sin2 θ2

∣∣∣∣∣π

0

= ∞ . (4.40)

Physically, this is because the total cross section describes how far away the targetcan affect the incident particles. But gravitational interaction is infinitely longrange, all incident particles will be affected, although the effect could be tiny.


4.5 Basics of Statistical Mechanics (Optional)(Chapter 8.1 in textbook. Chapters 2, 7, 9, 10, 11 in F. Mandl: StatisticalPhysics, John Wiley, 1988, 2nd ed.)

4.5.1 Thermodynamics

We very briefly review the basics of thermodynamics. The zeroth law of thermo-dynamics states that if two systems A and B are in thermal equilibrium with eachother and B and C are in thermal equilibrium, then A and C are in thermal equi-librium. We say that they all have same temperature. There are many kinds oftemperature scales, for example, the length of a rod at different temperatures.

We can verify by experiments that same amount of work, no matter which kind,produce same temperature rise. We call the form of energy transfer heat. The firstlaw states that change in energy of a system is equal to net heat input plus network done on the system. Note that energy of a system is a function of state, whileheat and work done on it are not. They depend on its history.

For an isolated system in equilibrium, usually energy E, volume V and numberof molecules N are fixed. We call it the macrostate: (E, V,N); or (E, V,N, α)where α are other macroscopic variables, for example, the dependence of densityon location. On the other hand, a microstate specifies the positions, velocities,internal states of each particles. It is almost impossible to fully describe.

We need to count the number of microstates corresponding to the same macrostates.The states are discrete for quantum system and continuous for classical system. Wewill adopt the quantum system viewpoint. We denote the number of states forenergy between E and E + δE by Ω(E, V,N, α).

We give an example. Consider spins in magnetic field. For single paramagneticatom, energy E = −µ ·B where µ is the magnetic moment, B is the magnetic field.Let assume that the spin can only take two values ±h/2. Thus, energy can only be±µ ·B.

For N such atoms, if n of them align with the field and (N −n) anti-align with thefield, total energy and the number of states are

E = n(−µB) + (N − n)(µB) = (N − 2n)µB , (4.41)

Ω =N !

n!(N − n)!. (4.42)

We assume that each microstate compatible with the constraints has equal a prioriprobabilities. Then, (one form of) the second law states that value of α will evolve insuch a way that Ω(E, V,N, α) is always non-decreasing and equilibrium correspondsto value of α for which Ω(E, V,N, α) attains its maximum.


We define entropy by

S(E, V,N, α) = k lnΩ(E, V,N, α) (4.43)

First law is about energy conservation. Second law is about direction. All ex-periments show that isolated systems tend to equilibrium, not the opposite. Realprocesses are non-reversible. Hence, entropy is always non-decreasing for isolatedsystems. We also see that the more disordered the system is, the larger its entropy.Equivalently, larger entropy, less information.

4.5.2 Isolated Systems

If an isolated system is partitioned into two subsystems and they are nearly inde-pendent, then

E = E1 + E2 (4.44)

V = V1 + V2 (4.45)

N = N1 +N2 (4.46)

Ω(E, V,N,E1, V1, N1) = Ω1(E1, V1, N1)Ω2(E2, V2, N2) . (4.47)

Hence,S(E, V,N,E1, V1, N1) = S1(E1, V1, N1) + S2(E2, V2, N2) (4.48)

entropy is an extensive quantity (proportional to the size of the system). (Comparedto intensive quantity, e.g. temperature.)

To define absolute temperature scale, first consider diathermal wall (not permeableto everything, except heat). At equilibrium, entropy is maximum,

0 =

(∂S

∂E1

)E,V,N,V1,N1

=

(∂S1

∂E1

)V1,N1

+

(∂S2

∂E2

)V2,N2

dE2

dE1

(4.49)

Since dE2/dE1 = −1, we have(∂S1

∂E1

)V1,N1

=

(∂S2

∂E2

)V2,N2

(4.50)

We see that (∂Si/∂Ei)Vi,Niis a measure of temperature, and define the absolute

temperature T by (∂Si

∂Ei

)Vi,Ni

=1

T. (4.51)

Defined as this, the perfect gas temperature scale is equal to the absolute temper-ature scale. By second law,

0 <dS

dt=

(1

T1− 1

T2

)dE1

dt. (4.52)

Heat flow from high temperature to low temperature.


4.5.3 Systems in a Heat Bath

To consider systems in constant temperature, we study an isolated system suchthat our system of interest is subsystem 1 and the subsystem 2 is a heat bath,which means it can absorb or provide as much energy as subsystem 1 needs withoutchanging its temperature. The probability that subsystem 1 is in a definite state ris proportional to the number of states of the heat bath compatible with it

pr = const.Ω2(E0 − Er) = const. exp[S2(E0 − Er)/k] (4.53)

where E0 is the total energy of our system and the heat bath. By the definition ofheat bath, Er ≪ E0,

S2(E0 − Er)

k=S2(E0)

k− Er

k

∂S2(E0)

∂E0

+1

2

E2r

k

∂2S2(E0)

∂E20

+ · · · (4.54)

By Eq. (4.51), the second term is Er/kT . The third is the change of temperatureof the heat bath, which, by definition, negligible. Eq. (4.53) is then

pr =1

Ze−βEr (4.55)

where Z =∑

r exp(−βEr) and β = 1/kT . This is Boltzmann distribution,which gives the probability of a microstate of a system at some fixed temperature.Z is called the partition function.

If there are degeneracies g(Er), the probability of the system at particular energyis

p(Er) =1

Zg(Er)e

−βEr . (4.56)

The mean energy of the system is

E =∑r

prEr = −∂ lnZ∂β

. (4.57)

Scientists usually will employ a conceptual construction. The energy, for example,of a particular system will have some particular time dependent value. There willbe fluctuations. We could consider a large number of identical systems at somefixed temperature, called a canonical ensemble. The average of the energy ofeach system will be given by Eq. (4.57), without any fluctuation.

Example: From the Boltzmann equation above, the ratio of probability that asystem will be in state a and in state b is given by

P (Eb)

P (Ea)=gbgae−(Eb−Ea)/kT . (4.58)

At what temperature a gas of neutral hydrogen will have equal number of atoms inthe ground and first excited states?


For hydrogen atoms, ground state is n = 1 and first excited state is n = 2. Thedegeneracy is gn = 2n2. Therefore,

1 =2(22)

2(12)e−[(−13.6 eV/22)−(−13.6 eV/12)]/kT

10.2 eV

kT= ln 4

T = 8.54× 104 K .

This is even higher than the surface of the Sun. However, we know from observationsthat some hydrogen atoms are ionized in the Sun. How can that be possible? (Theanswer lies in the Saha Equations, which will be discussed later in this chapter.)

4.5.4 The Perfect Classical Gas

Gas consists of molecules moving about fairly freely in space. Perfect gas rep-resents an idealization in which the potential energy of interaction between themolecules is negligible compared to their kinetic energy of motion. If the energystates of one single molecule are εr, the partition function of a single molecule is

Z(T, V, 1) =∑r

exp(−βεr) . (4.59)

The partition function of many identical molecules is not [∑

r exp(−βεr)]N (oth-

erwise this will lead to the Gibbs paradox). Since the molecules are identical, wecannot distinguish the cases: the first molecule is at state r and the second is atstate s; or the first molecule is at state s and the second is at state r. We can onlysay that one molecule is at state r and one is at state s. The partition function fortwo molecules is then

Z(T, V, 2) =∑r

exp(−2βεr) +1

2!

∑r,s

r =s

exp[−β(εr + εs)] . (4.60)

The partition function for N molecules is then

Z(T, V,N)

=∑r

exp(−Nβεr) + . . .

+1

N !

∑r1,...,rN

all ri different

exp[−β(εr1 + . . .+ εrN )] . (4.61)

We define classical regime, in which the probability that any single-particle stateis occupied by more than one molecule is very small. If we define the occupationnumber for a state as the number of particles in that state, then classical regimeis in which the occupation number for any state is much less then one.


For classical perfect gas, only the last term in Eq. (4.61) is important. Consider thefunction

1

N !

∑r1,...,rN

exp[−β(εr1 + . . .+ εrN )] (4.62)

The difference between this and the last term of Eq. (4.61) is not significant forclassical perfect gas. So, our final result for this section is

Z(T, V,N) =1

N !

∑r1,...,rN

exp[−β(εr1 + . . .+ εrN )]

=1

N !

[∑r

exp(−βεr)

]N(4.63)

for classical (occupation number small) perfect (interaction negligible) gas (particlesmoving freely).

4.5.5 The Partition Function

To calculate the partition function of the gas, it is reduced to calculate the partitionfunction for a single molecule. The energy of a molecule can be written as

εr = εtrs + εintα (4.64)

where εtrs is the energy for the translational motion and εintα is the energy of theinternal excitations. Hence,

Z(T, V, 1) =

[∑s

exp(−βεtrs )

][∑α

exp(−βεintα )

]≡ Ztr

1 Zint . (4.65)

The internal energy εintα depends on the internal details of the molecules, for exam-ple, type, excited states, etc. It does not depend on the volume. We now evaluateZtr

1 , which can be applied to any perfect gas. The translational energy is

εtr =p2

2m. (4.66)

We would like to find the number of states f(p)dp with momentum of magnitudebetween p and p + dp. f(p) is called the density of states. Consider a cube ofsides with length L. From quantum mechanics, the allowed wavefunctions are

ψ = const. sin(nxπ

Lx)sin(nyπ

Ly)sin(nzπ

Lz)

(4.67)

with nx, ny, nz = 1, 2, . . .. The magnitude of the wave vector k is defined by

k2 =π2

L2(n2

x + n2y + n2

z) . (4.68)


Hence, the volume per allowed point in k-space is (π/L)3. Since the n’s can onlybe positive, we have to count only the positive octant, and the volume of the regionlying between the radii k and k+dk in the positive octant is 1

84πk2dk. The number

of states in this region is (1

84πk2dk

)1

(π/L)3=V k2dk

2π2. (4.69)

The relation between the wave vector and momentum is k = 2πp/h, the final resultof the density of states is

f(p) dp =V 4πp2dp

h3. (4.70)

Note that we have only considered the translational motion. For example, if theparticle has non-zero spin, we have to multiply the above formula by the numberof internal degrees of freedom.

We come back to calculate the partition function Ztr1 .

Ztr1 =

∑s

exp(−βεtrs )

=

∫ ∞

0

exp(−βp2/2m)f(p)dp

=4πV

h3

∫ ∞

0

exp(−βp2/2m)p2dp

=4πV

h3

(2m

β

)3/2 ∫ ∞

0

exp(−x2)x2dx

=4πV

h3

(2m

β

)3/2 √π

4

= V

(2πmkT

h2

)3/2

. (4.71)

The full partition function is

Z(T, V,N) =1

N !V N

(2πmkT

h2

)3N/2

ZNint . (4.72)

If a particle is in state r and the energy of this state Er depends on the volume,then dEr/dV ≡ −Pr is by definition the negative of pressure (contributed by thisparticle), because the derivative is the work done per unit change in volume. The


Gravity

Pressure

II

I

Figure 4.11: The outward pointing pressure balances the inward pointing gravita-tional force.

average pressure is just

P =∑r

prPr

=∑r

pr(−dEr

dV)

=∑r

1

Ze−βEr(−dEr

dV)

=1

Zβ

∑r

(∂e−βEr

∂V

)β

=1

Zβ

(∂Z

∂V

)β

=1

β

(∂ lnZ

∂V

)β

. (4.73)

Substitute Eq. (4.72) into the above equation and notice that Zint is independentof the volume, we have

P =1

β

N

V=NkT

V, (4.74)

or PV = NkT , which holds irrespective of the internal molecular structure.

To illustrate one simple application of ideal gas law in astronomy, we could roughlyestimate the temperature at the core of a star. We hypothetically divide the starinto to halves, the regions I and II in Fig. 4.11. The mass of each half is M/2 if thetotal mass of the star is M . They are separated by a distance R/2 if the radius of

the star is R. Thus, the gravitational attraction of the two halves is G(M/2)2

(R/2)2. The


area between them is 4π(R/2)2. The average pressure is in the order of

⟨P ⟩ =G(M/2)2

(R/2)21

4π(R/2)2

=4

3

GMρ

R(4.75)

where ρ is the average density of the star. For ideal gas with N particles, ideal gaslaw gives us the temperature

T =PV

Nk

=Pmp

(Nmp/V )k

=Pmp

ρk

=4GMmp

3kR(4.76)

where k is the Boltzmann’s constant, mp is the mass of proton. We have thesecond equality because in a main sequence star, most of it are protons. The fourthequality is given by Eq. (4.75). If we substitute the data of our Sun, the result isT = 3×107K. A more detailed calculation of the core temperature gives 1.5×107K.

4.5.6 The Perfect Quantal Gas and Quantum Statistics

One basic assumption of classical gas of identical particles is that the mean occupa-tion number for single-particle states is much less than one. For quantal gas, themean occupation number could be near or even greater than one. The main quan-tum effect is the quantum statistics: how particles occupy single-particle states.Eq. (4.63) is no longer correct.

Let nr be the occupation number of the single-particle state with energy εr. Whatare the possible values of nr? There are two mutually exclusive classes.

Bose-Einstein statistics (BE): there is no restriction on nr, i.e. nr = 0, 1, 2, . . ..They are called bosons, with integral spin 0, h, 2h, . . ..

Fermi-Dirac statistics (FD): the nr can only be 0 or 1. They are called fermions,with half-integral spin 1

2h, 3

2h, . . .. Another way to say about fermions is that they

satisfy the Pauli exclusion principle: no two fermions can be in the same single-particle state.

The particles in consideration could be fundamental or composite, as long as theyare identical. A composite particle consisting of odd number of fermions is a fermion.A composite particle consisting of even number of fermions is a boson.


4.5.7 The Partition Function

We will give the general form of the partition function in this subsection. Sup-pose the energy of the single-particle states are ε1, ε2, ε3, . . ., and the correspondingoccupation numbers are n1, n2, n3, . . .. For gas of N particles,∑

r

nr = N . (4.77)

Also

nr =

0, 1 fermions0, 1, 2, 3, . . . bosons

. (4.78)

Any set of nr that satisfies these two conditions defines a state of the gas. Thepartition function is then

Z(T, V,N) =∑

n1,n2,...

exp(−β∑r

nrεr) . (4.79)

where the first sum is over all sets of nr. The mean occupation number is

ni =

∑n1,n2,...

ni exp(−β∑r

nrεr)∑n1,n2,...

exp(−β∑r

nrεr)

= − 1

β

(∂ lnZ

∂εi

)T,εr(r =i)

. (4.80)

4.5.8 Derivation of Blackbody Radiation

In this section, we discuss the thermal gas of photons, using the method developedin the last section. An ideal black body is an object that absorbs all radiations fallon it. A black body is also a perfect radiator and its radiation is a thermal gas ofphotons. Most hot objects, including stars and a piece of hot metal, behave roughlylike black bodies.

Photons are of spin 1, they are bosons and obey BE statistics. They also do notinteract with each other (Maxwell’s equations are linear). Hence, photon gas isperfect gas.

Since photons can be emitted or absorbed, photon number is not a constant.Eq. (4.77) does not apply. The occupation number for each state r can take any


values, nr = 0, 1, 2, . . . . Partition function is

Z(T, V ) =∑

n1,n2,...

exp(−β∑r

nrεr)

=

(∑n1

e−βn1ε1

)(∑n2

e−βn2ε2

)· · ·

=1

1− exp(−βε1)1

1− exp(−βε2)· · ·

=∞∏r=1

1

1− exp(−βεr)(4.81)

because ∑n

e−βnε = 1 + e−βε + (e−βε)2 + · · ·

=1

1− exp(−βε). (4.82)

Thus, we have lnZ(T, V ) = −∑∞

r=1 ln(1− e−βεr) and the mean occupation numberof state r is

nr = − 1

β

∂ lnZ

∂εr

=e−βεr

1− e−βεr

=1

eβεr − 1. (4.83)

We now derive the Planck’s law of black-body radiation. Energy and momentumof a photon of frequency ν are ε = hν and p = hν/c. The density of states is givenby Eq. (4.70). However, there is one more complication for photons. There are twopolarizations for each translational degree of freedom, two perpendicular directionsof the linear polarization, for example. The number of internal degrees of freedom istwo. The label r specifies the translational motion, the frequency, and polarization.In terms of frequency ν, we have

f(ν)dν = 2V 4π(hν/c)2hdν/c

h3

=8πV ν2dν

c3. (4.84)

Combining the mean occupation number (Eq. 4.83) and the density of state above,we have the number of photons in the frequency range ν and ν + dν as

dNν =8πV

c3ν2dν

exp(βhν)− 1. (4.85)


The total energy of photons in volume V in this frequency range is

dEν = hνdNν =8πV h

c3ν3dν

exp(βhν)− 1. (4.86)

The energy density per unit frequency is defined as

uν = Eν/V =8πhν3

c3(ehν/kT − 1). (4.87)

The energy density per unit wavelength uλ is defined as uλ |dλ| = uν |dν| whereλ = c/ν is the wavelength. We have |dν/dλ| = c/λ2, and

uλ =c

λ28πh(c/λ)3

c3(ehc/λkT − 1)

=8πhc

λ51

exp( hcλkT

)− 1. (4.88)

4.6 Physics of Blackbody Radiation

Astrophysical emissions can be classified as thermal and non-thermal origins. Ther-mal radiation is emitted by the thermal motion of charged particles, such as black-body radiation and thermal Bremsstrahlung. Non-thermal radiation is generatedby other processes, and the particles do not follow a thermal distribution. Examplesare synchrotron radiation and inverse Compton scattering.

In this course, we will only discuss the blackbody radiation. It is one of the mostcommon and important radiation mechanisms. You will find it in the Sun and otherstars and even the cosmic microwave background of the Universe. When we say theUniverse has a temperature of 3K, how do we measure it?

Figure 4.12: A blackbody.

A blackbody is an idealized object that ab-sorbs all radiation in any frequencies. It canbe approximated by cavity with a small holein it, such that any incident light can nevercome out (Figure 4.12). The photons inside arethen in thermal equilibrium with the surround-ing. Their distribution follows the Bose-Einsteinstatistics. This gives the blackbody radiation.

Every object has a finite temperature emitsblackbody radiation. As we will see below, thepeak frequency only depends on temperature.This can be used to explain why objects glowat different color at different temperature, e.g.from red to blue as it heats up.


The specific intensity of blackbody radiation is given by Planck’s law, which canbe derived from Eqs. (4.15) and (4.87):

IBBν ≡ Bν(T ) =

2hν3/c2

exp(hν/kBT )− 1, (4.89)

where h is the Planck constant and kB is the Boltzmann constant. Examples ofblackbody spectrum are shown in Figure 4.13. Note that thermal radiation is NOT

Figure 4.13: Blackbody radiation spectrum.

the same as blackbody radiation. For the former, the source function Sν is equal tothe blackbody intensity Bν

Sν = Bν(T ) , (4.90)

so that the emission and absorption coefficients are related by

jν = ανBν(T ) . (4.91)

This is the Kirchhoff’s law. For blackbody radiation, Iν = Bν . It is clear thatfor thermal radiation in optically thick media, it becomes blackbody. We describebelow some properties of blackbody radiation.

4.6.1 Stefan-Boltzmann law

The flux from blackbody radiation is

F = σSBT4 , (4.92)

where σSB is the Stefan-Boltzmann constant

σSB =2π5k4B15c2h3

= 5.67× 10−5 erg cm−2 K−4 s−1 . (4.93)


It can be proved by integrating the Planck spectrum directly

F =

∫ ∞

0

Fν dν =

∫ ∞

0

∫Bν cos θ dΩdν =

∫ ∞

0

∫ π/2

0

∫ 2π

0

Bν cos θ sin θ dϕ dθ dν .

(4.94)We consider only radiation going out from a surface, therefore, θ goes from 0 toπ/2 only.

F = 2π

∫ ∞

0

∫ π/2

0

Bν cos θ sin θ dθ dν = 2π

∫ ∞

0

Bν dν

∫ 1

0

µ dµ = π

∫ ∞

0

Bν dν ,

(4.95)where we have substituted µ = cos θ. Finally,

F = π

∫ ∞

0

2hν3/c2

exp(hν/kBT )− 1dν =

2πh

c2

(kBT

h

)4 ∫ ∞

0

u3

eu − 1du , (4.96)

where u = hν/kBT . The integral is nontrivial but it can be shown that the answeris π4/15. This gives Stefan-Boltzmann law above.

This law tells us the total energy emitted per unit area per unit time. As anexample, we know that the Sun has a surface temperature T = 5800K, to estimatethe solar luminosity L⊙, we just need to integrate over its surface area 4πR2

⊙,

L⊙ = 4πR2⊙σSBT

4 . (4.97)

4.6.2 Rayleigh-Jeans Law

At the low energy limit, hν ≪ kBT , we can expand

exp

(hν

kBT

)− 1 =

hν

kBT+ . . . (4.98)

This is the Rayleigh-Jeans limit of the Planck’s law

IRJν (T ) = (4.99)

Physically, it is the classical limit and it leads to the ultraviolet catastrophe.

4.6.3 Wien Law

At the high energy limit, hν ≫ kBT , we can expand

exp

(hν

kBT

)− 1 ≈ exp

(hν

kBT

)(4.100)

This is the Wien limit of the Planck’s law

IWν (T ) = (4.101)

Unlike Rayleigh-Jeans law, Wien law is a quantum effect, as you can guess fromthe fact that it contains the constant h.


4.6.4 Wien’s Displacement Law

At what frequency the blackbody radiation peaks at? We can find the answer bysetting

∂Bν

∂ν

∣∣∣∣ν=νmax

= 0 . (4.102)

This results inx = 3(1− e−x) , (4.103)

where x ≡ hνmax/kBT . This can only be solved numerically

hνmax = 2.82kBT , (4.104)

orνmax = 5.88× 1010 T Hz , (4.105)

where T is in K. This is Wien’s displacement law. Similarly, we can also derivethe law in wavelength:

∂Bλ

∂λ

∣∣∣∣λ=λmax

= 0 . (4.106)

Solving y = 5(1− e−y), we obtain

λmax = 0.290T−1 cm , (4.107)

T is in K. However, note that λmaxνmax = c. (why?) For example, the sun has atemperature of 5800K, which corresponds to λmax=500 nm. This is the wavelengthof green light, coincident with the peak sensitivity of human vision.

4.6.5 Monotonicity

One important property of blackbody radiation is that the curves of different tem-peratures in Figure 4.13 never cross, i.e. a curve of higher temperature is entirelyabove the lower temperature one. This can be proved by

∂Bν

∂T=

2h2ν4

c2kBT 2

exp(hν/kBT )

[exp(hν/kBT )− 1]2(4.108)

always > 0. This has two consequences. First, although the blackbody peak shiftstoward shorter wavelength (blue color) as temperature increases, the intensity atlong wavelength (red color) always increases with temperature. Second, given ν,there is one-to-one correspondence between Iν and T . As we will show below, thiscan be used to define temperature.


4.6.6 Temperature Definitions

Brightness temperature

At a given frequency ν, one can equate the brightness of an object Iν to the black-body intensity. The temperature obtained this way is called the brightness tem-perature Tb, i.e.

Iν = Bν(Tb) . (4.109)

We can then measure the brightness in units of K. This is often used in radio as-tronomy, where we express the surface brightness of a nebula in terms of brightnesstemperature, and also the system noise temperature of antennas. In radio frequen-cies, the Rayleigh-Jeans law is usually applicable, so that

Tb = Iνc2

2kBν2. (4.110)

This is essentially how infrared thermometers (pyrometers) work, which provides auseful way to measure temperature when conventional methods are not practical,e.g., fast moving objects, objects far away, or having too high temperature to con-tact. The brightness temperature of giant pulses from some pulsars can go beyond5× 1039K, the highest known brightness temperature in the Universe.

Color temperature

If the measured spectrum of an object have a shape more or less like a blackbody,we can perform a fit to obtain the temperature. Even simpler, one could just fitthe peak of the emission then apply Wien’s displacement law. This gives the colortemperature. This is the same as the color temperature you set in digital photog-raphy or TV screens. Why a lower color temperature gives a “warmer” tone, whilea high temperature gives a “cool” tone? Some advanced infrared thermometers maymeasure two or more frequency bands to estimate the color temperature from theintensity ratio. This can give more accurate measurements.

Effective temperature

Recall that the Stefan-Boltzmann law relates the total flux to the temperature. Thiscan also be used to define the temperature. The effective temperature Teff is thetemperature of a blackbody that emits the same amount of flux as the observedobject:

Fobs = σSBT4eff . (4.111)

The Sun is not a perfect blackbody, but we can derive the effective temperature of5800K from the total flux, same for other stars.


4.7 Chemical Potential and Saha Equation (Op-

tional)

We would like to study the relative abundance of the reactants and products of areaction in equilibrium. We will only derive the simplest case in the first half ofthis section and state the main result, Saha equation, in the second.

The total energy of a particle moving with momentum p is given by E2 = m2c4 +p2c2. If it isn’t moving near speed of light, we have

E =√m2c4 + p2c2 ≈ mc2 +

p2

2m. (4.112)

Note that the rest mass energymc2 includes the internal energy, for example, bound-ing energy. Consider a simple reaction

A B , (4.113)

where A and B could be two excited states of a single particle. Let the total numberof particles be N . We have the constraint N = NA +NB in obvious notations.

The total energy of the microstate that there are NA particles of type A withmomenta pAi , i = 1, . . . , NA and NB particles of type B with momenta pBj , j =1, . . . , NB is ∑

i

(mAc

2 +(pAi )

2

2mA

)+∑j

(mBc

2 +(pBj )

2

2mB

). (4.114)

There are N !/(NA!NB!) microstates with the same energy. Hence, the probabilitythat the system is in a state with NA particles of type A with momenta pAi andNB particles of type B with momenta pBj is, according to Eq. (4.55), proportionalto

N !

NA!NB!exp

−β

[∑i

(mAc

2 +(pAi )

2

2mA

)+∑j

(mBc

2 +(pBj )

2

2mB

)]. (4.115)

We do not care about the momenta of the particles, and sum (integrate) up allmicrostates of different momenta. We have already done this in Eq. (4.71). Theprobability that the system is in a state withNA particles of type A andNB particlesof type B (with any momenta) is proportional to

N !

NA!NB!e−βNAmAc2 V NA

(2πmAkT

h2

)3NA/2

e−βNBmBc2 V NB

(2πmBkT

h2

)3NB/2

.

(4.116)


By Stirling formula for large s, ln s! = s ln s − s, we can rewrite the factor in theabove formula as

1

NA!e−βNAmAc2 V NA = e−βNAmAc2+NA−NA lnNA+NA lnV

= e−βNA[mAc2+kT ln(NA/V )−kT ] . (4.117)

For classical ideal gas of particles with mass m at temperature T , the chemicalpotential is defined by

µ = mc2 − kT ln(gsnQ

n

)(4.118)

where n is the number density of particles in the gas, nQ is called the quantumconcentration

nQ =

(2πmkT

h2

)3/2

(4.119)

and gs is the internal degree of freedom of the particle. It depends on the spin of theparticle and, for example, the excited states of the hydrogen atom. For electron,proton or neutron, gs = 2. We have implicitly assumed that for our particles A andB, gA = gB = 1.

In terms of these, Eq. (4.116) is

N ! e−βNA(mAc2+kT ln(NA/V )−kT )+NA lnnQA e−βNB(mBc2+kT ln(NB/V )−kT )+NB lnnQB

= N ! eN e−βNA(mAc2−ln(nQA/nA)) e−βNB(mBc2−ln(nQB/nB))

= N ! eN e−βNAµA e−βNBµB

= N ! eN e−βNµB exp[−βNA(µA − µB)] . (4.120)

Since µB depends only very weakly on NA (through NB = N−NA), the probabilityis greatest when µA = µB. This is the main result: in equilibrium, the chemicalpotentials of the reactant and product equal.

Let ∆E = mAc2 −mBc

2. The equality of chemical potentials implies

mAc2 − kT ln

(nQA

nA

)= mBc

2 − kT ln

(nQB

nB

)∆E = −kT ln

(nAnQB

nBnQA

). (4.121)

If ∆E is much smaller than mAc2, then nQA ≈ nQB and

nA = nBe−∆E/kT . (4.122)

Chemical potential can be interpreted as how much energy is needed to put onemore particle into the system. It depends on the physical parameters of the system.For example, to keep the temperature unchanged, we have to speed up the particle


before injecting it to the system. Since it costs arbitrarily low energy to create aphoton with very long wavelength, we claim that the chemical potential of photonis zero.

For reactions with more reactants,

A+B C +D , (4.123)

at thermodynamical equilibrium, the energy needed to create particles A and Bmust equal the energy needed to create particles C and D. Hence, we claim that

µ(A) + µ(B) = µ(C) + µ(D) . (4.124)

Substitute the expressions of the chemical potential in Eq. (4.118) in this, theresulting equation is the Saha equation.

Let us consider a very important example,

γ +Hn e− + p , (4.125)

the ionization of hydrogen atom at the n-th excited state by absorbing a photonγ. To satisfy the assumption that they are ideal gases, their density must be muchless than the quantum concentration, n≪ nQ. As mentioned, µ(γ) = 0 and

µ(e) = mec2 − kT ln

(genQe

ne

), (4.126)

µ(p) = mpc2 − kT ln

(gpnQp

np

), (4.127)

µ(Hn) = m(Hn)c2 − kT ln

[g(Hn)nQH

n(Hn)

]. (4.128)

The energy of the n-th excited hydrogen atom is En, Eq. (2.7), hence we have

m(Hn)c2 = mec

2 +mpc2 + En . (4.129)

Substitute these into the chemical potential equation, we have

− En = kT ln

[n(Hn)

g(Hn)nQH

genQe

ne

gpnQp

np

]. (4.130)

Since the mass of hydrogen is approximately equal to the mass of a proton, nQH =nQp. Also, let εn = |En|. The Saha equation becomes

exp(−εn/kT )g(Hn)

gegpnQe

nenp

n(Hn). (4.131)

For free electrons, ge = 2

np

n(Hn)=

2gpneg(Hn)

(2πmekT

h2

)3/2

exp(−εn/kT ) . (4.132)


In most cases, there is no net charge, ne = np and all the gs are of order of unity.We finally have

n2e

n(Hn)∼(2πmekT

h2

)3/2

exp(−εn/kT ) . (4.133)

We see that there is significant change in the percentage of ionization when tem-perature is around ε1/k, which is about 160,000K.

Example: In Sun’s photosphere, ne = 1.88 × 1013 cm−3, T = 5777K, what is thenumber ratio between Ca II (singly-ionized calcium) and Ca I (neutral calcium)?Given that gII = 2.30 and gI = 1.32 and ionization energy χI = 6.11 eV for Ca I.

Using the Saha equation,

NII

NI

=2gIIgIne

(2πmekT )3/2

h3e−χI/kT

=2× 2.30

1.32× 1.88× 1013(2πmekT )

3/2

h3e−6.11 eV/kT

≈ 927 .

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