Today in Astronomy 328: stellar structurephysics.gmu.edu/~satyapal/astro328fall07/course... · 6 September 2007 Astronomy 328, Fall 2007 1 Today in Astronomy 328: stellar structure

6 September 2007 Astronomy 328, Fall 2007 1

Today in Astronomy 328: stellar structure

Binary stars (left over from last time)Principles of stellar structureHydrostatic equilibrium gives scaling relations for• average pressure, density and

temperature• pressure, density and

temperature in the center (where the luminosity is produced)

Diffusion of light from center to surface

Figure: Chaisson and McMillan, Astronomy Today


Eclipsing binary stellar system

Flux

Time

vr1v

2v

PrimarySecondary Minima

P


Stellar masses determined for binary systems

If orbital major axes (relative to center of mass) or radial velocities known, so is the ratio of masses:

If furthermore the period and sum of major axis lengths known, Kepler’s third law can the used with this relation to solve for the two masses separately.

21 2 2

2 1 1 1

r

r

vm a vm a v v

= = =


Stellar masses determined for binary systems (continued)

If only the radial velocity amplitudes v1 and v2 are known, the sum of masses is (from Kepler’s third law)

You’ll prove this in Homework #2.

If orientation of the orbit with respect to the line of sight isknown, this allows separate determination of the masses; that’s why eclipsing binaries are so important (if the system eclipses, we must be viewing the orbital plane very close to edge on: sin i is very close to 1).

31 2

1 2 .2 sin

P v vm mG iπ

+⎛ ⎞+ = ⎜ ⎟⎝ ⎠


Stellar radii determined for totally-eclipsing binary systems

Duration of eclipses and shape of light curve can be used to determine sizes (radii) of stars:

Relative depth of primary and secondary brightness minima of eclipses can be used to determine the ratio of effective temperatures of the stars.

( )

( )

1 22 1

1 23 1

2

2

sv vR t t

v vR t t

+= −

+= −

TimeFl

ux1t 2t 3t 4t


Example

An eclipsing binary is observed to have a period of 8.6 years. The two components have radial velocity amplitudes of 11.0 and 1.04 km/s and sinusoidal variation of radial velocity with time. The eclipse minima are flat-bottomed and 164 days long. It takes 11.7 hours from first contact to reach the eclipse minimum.

What is the orbital inclination?What are the orbital radii?What are the masses of the stars?What are the radii of the stars?


Example (continued)

8.6 years

1.04km/s

11km/s

vr

Flux

Time

Closeup ofprimary minimum

11.7hours

164days


Example (continued)

AnswersSince it eclipses, the orbits must be observed nearly edge on; since the radial velocities are sinusoidal the orbits must be nearly circular.Orbital radii:

11 10.61.04

s

s

vmm v

= = =

141.42 10 cm2

=9.5 AU

s sPr vπ

= = ×

131.34 10 cm2

=0.90 AU

10.4 AU

Pr v

r

π= = ×

=


Example (continued)

Masses:

Stellar radii (note: solar radius = ):

3

2 15.2

10.61.3 , 13.9

s

s s

s

rm m MP

m mm M m M

+ = =

+ =⇒ = =

(Kepler’s third law)

(previous result)

( )3 12369

sv vR t t

R

+= −

=

( )

( )( )

2 1

-1

10

26.02 km s 11.7 hr

7.6 10 cm 1.1

ss

v vR t t

R

+= −

=

= × =

6 96 1010. × cm


Data on eclipsing binary stars

Latest big compendium of eclipsing binary data is by O. Malkov. See following slides.This, and vast amounts of other data, can be found on line at the NASA Astrophysics Data Center:

http://adc.gsfc.nasa.gov/adc.html

Why do the graphs appear as they do? That’s what we’ll try to figure out, as we study stellar structure during the next few lectures.


Luminosities of eclipsing binary stars (Malkov 1993)

0.1 1 10Mass (solar masses)

10-410-310-210-1100101102103104105106

Lum

inos

ity (s

olar

lum

inos

ities

)


Radii of eclipsing binary stars (Malkov 1993)


0.1

1

10

Radi

us (s

olar

radi

i)


Effective temperatures of eclipsing binary stars (Malkov 1993)


2

3

4567

104

2

3

45

Effe

ctiv

e te

mpe

ratu

re (K

)


Hertzsprung-Russell (H-R) diagram for eclipsing binary stars (Malkov 1993)

40000 30000 20000 10000 0Effective temperature (K)

10-3

10-2

10-1

100

101

102

103

104

105

106

Lum

inos

ity (s

olar

lum

inos

ities

)


H-R diagram for binaries and other nearby stars

Stars within 25 parsecs of the Sun (Gliese and Jahreiss1991)Nearest and Brightest stars (Allen 1973)Pleiades X-ray sources (Stauffer et al. 1994)Binaries with measured temperature and luminosity (Malkov1993)

1 105 1 104 1 1031 10 5

1 10 4

1 10 3

0.01

0.1

1

10

100

1 103

1 104

1 105

1 106

Effective temperature (K)

Lum

inos

ity (s

olar

lum

inos

ities

)


Principles of stellar structure

Vogt-Russell “theorem:” the mass and composition of a star uniquely determine its radius, luminosity, internal structure, and subsequent evolution.Stars are spherical, to good approximation.Stable stars are in hydrostatic equilibrium: the weight of each infinitesimal piece of the star’s interior is balanced by the force from the pressure difference across the piece. • Most of the time the pressure is gas pressure, and is

described well in terms of density and temperature by the ideal gas law, PV = NkT.

• However, in very hot stars or giant stars, the pressure exerted by light -- radiation pressure -- can dominate.


Principles of stellar structure (continued)

Energy is transported from the inside to the outside, most of the time in the form of light.• The interiors of stars are opaque. Photons are

absorbed and reemitted many many times on their way from the center to the surface: a random-walk process called diffusion.

• The opacity depends upon the density, temperature and chemical composition.

• Most stars have regions in their interiors in which the radial variations of temperature and pressure are such that hot bubbles of gas can “boil” up toward the surface. This process, called convection, is a very efficient energy-transport mechanism and can be more important than light diffusion in many cases.


The equations of stellar structure

2

2

2

Hydrostatic equilibrium:

Mass conservation:

4

Energy generation:

4

r

r

r

GMdPdr r

dM rdr

dL rdr

ρ

π ρ

π ρε

= −

=

=

2 2

2rad

Energy transport:3

16 4Adiabatic temperature gradient:

11

Convection occurs if:

1

r

rH

LdTdr r r

GMmdTdr k r

T dPP dT

κρσ π

µγ

γγ

= −

⎛ ⎞⎛ ⎞ = − −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

<−

4

Equations of statePressure:

( , , composition) in general

4 throughout most normal stars3

Opacity:= ( , , composition) in general

Energy generation:= ( , , composition) in general

H

P P T

kT Tm c

T

T

ρ

ρ σµ

κ κ ρ

ε ε ρ

=

= +

Boundary conditions:

as

as star

ML

r

TP r R

r

r

→→

⎫⎬⎭

→

→→→

⎫

⎬⎪

⎭⎪

→

00

0

000ρ

M Lr

r r, : mass or luminosity containedwithin radius .


The equations of stellar structure (continued)

We show these here for illustration only; we will not attempt to solve this network of differential equations. (Take Astronomy 530 to learn more.)In fact, there are no general, analytical solutions to the equations; usually stellar-interior models are generated by computer solution.Instead of solving these equations, we shall employ some crude approximations to develop several scaling relationships that will also be crude, but will correctly illuminate the physical processes important in stars. But the equation of hydrostatic equilibrium is useful by itself, so we will take the time here to derive it.


Hydrostatic equilibrium

Consider a spherical shell of radius r and thickness ∆r rwithin a star with mass density (mass per unit volume) ρ(r). Its weight is

(minus sign: force points inward)

∆r r

M

M(r)

∆m

R

( )

( ) ( )

( ) ( )

2

22 4

4 .

GM r mF

rGM r

r r rrGM r r r

π ρ

π ρ

∆∆ = −

= − ∆

= − ∆


Hydrostatic equilibrium (continued)

If the star is in hydrostatic equilibrium the weight is balancedby the pressure difference across the shell:

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

22

2 2

2

2 2 2

4 4

4 4

4 ,

4so .

4 4

F P r r P r r r rdPP r r P r r rdr

dP r rdr

GM r r r GM r rdP Fdr r r r r r

π π

π π

π

π ρ ρ

π π

−∆ = − + ∆ + ∆

⎡ ⎤≅ − + ∆⎢ ⎥⎣ ⎦

= − ∆

∆∆= = − = −

∆ ∆

( )r r∆

(to first order in pressure)

This differential equation, one of the equations of stellar structure, can be solved easily under certain assumptions.


The Sun’s interior, on the average

Since we know its distance (from radar), mass (from Earth’s orbital period) and radius (from angular size and known distance):

we know the average mass density (mass per unit volume):

13

33

10

1 AU 1.4960 10 cm

1.98843 10 gm

6.9599 10 cm

r

M

R

= = ×

= ×

= ×

-33

31.41 g cm

4= only 26% of Earth's.

M MV R

ρπ

= = =


The Sun’s interior, on the average (continued)

For the average pressure, we use the equation of hydrostatic equilibrium, and assume (very crudely) that the gas pressure varies linearly from some central value to zero at the surface:

centersurface center

centersurface

2 2

P P PdP Pdr r r r R

GMGMr R

ρρ

−∆≈ = = −∆ −

= − = −

( )( )

center

8 33-2

10

15 -2 9

Then 2 2

6.67 10 1.99 10 1.41 dyne cm

2 6.96 101.34 10 dyne cm 1.3 10 atmospheres

GMPPRρ

−

= =

× ×=

× ×

= × = ×


The Sun’s interior, on the average (continued)

The Sun is an ideal gas:

Average particle mass is about the mass of the proton ( ), so

Now for some less-crude estimates….

PV NkT

P nkT kTm

=

= =ρ

(N= number of gas particles, n=number density: number of particles per unit volume, = avg. particle mass) m

( )( )( )( )

24 15

16

6

1.67 10 1.34 10K

1.41 1.38 10

=11.5 10 K .

mPT

kρ

−

−

× ×= =

×

×

241.67 10 gm−×


Central pressure in a star

The star’s center, being its densest and hottest spot, will turn out to be the site of virtually all of the star’s energy generation, so we will make somewhat more careful estimates of the conditions there.From hydrostatic equilibrium equation again:

But the pressure at the surface, P(R), better be zero because the star’s surface doesn’t move and there’s nothing outside the surface to push back, so

( ) ( ) ( ) ( )2 .

R R

r r

GM r rdPP R P r dr drdr r

ρ′ ′′ ′− = = −

′ ′∫ ∫

( ) ( ) ( )2 .

R

r

GM r rP r dr

rρ′ ′

′=′∫


Central pressure in a star (continued)

We can’t do the integral unless we know the density as a function of position, so instead we make a crude approximation:

and the integral becomes

3 M MV R

ρ ≈ ≈(ignoring dimensionless factors like 4π/3, because we’re just trying to get the order of magnitude right),

( ) ( ) 3

3 2 3 2 3

2

6

1R R

r r

R

r

M rGM GM rP r dr M drR r R r R

GM r drR

⎛ ⎞′ ′′ ′≈ ≈ ⎜ ⎟⎜ ⎟′ ′ ⎝ ⎠

′ ′≈

∫ ∫

∫



Central pressure (r = 0):

Lo and behold, a complete calculation for stars of moderate to low mass (Astronomy 553 style) yields

so we have derived a pretty good scaling relation for PC.

P GMR

r dr GMR

R

GMR

CR

≈ ′ ′ =

≈

z26 0

2

6

2

2

4

2

(still ignoring dimensionless factors)

2

419CGMP

R=



For the Sun:

So, for other main sequence stars, to adequate approximation,

33 101.99 10 g 6.96 10 cmM R= × = ×

217 -2

4

11

19 =2.1 10 dyne cm

10 atmospheres

CGM

PR

≅ ×

>

42 217 -2

4 2 419 =2.1 10 dyne cm .CRGM MP

R M R

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟≅ ×⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠


Central density and temperature of the Sun

Central pressure is a little more than a factor of 100 larger than average pressure. Guess: central density 100 times higher than average? (That’s equivalent to guessing that the internal temperature does vary much with radius.)

For the Sun, that’s not bad; the central density turns out to be 110 times the average density, Thus, since

As we will see in a couple of weeks, the average gas-particle mass in the center of the Sun, considering its composition and the fact that the center is completely ionized, is .

-3150 gm cm .Cρ =

241.5 10 gmCm −= ×

3 ,C M Rρ ∝3

-3325 150 gm cmC

RM MM RR

ρ⎛ ⎞⎛ ⎞

= = ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠


Central density and temperature of the Sun (continued)

But the material is still an ideal gas, so

Compare to : T doesn’t vary very much with radius.We can make a scaling relation out of this as well, to use in extrapolating to stars similar to the Sun but having different masses, sizes and composition:

615.7 10 K.C C C CC

C C C

P V P P mTN k n k kρ

= = = = ×

2 36

4

6

15.7 10 K for the Sun;

15.7 10 K .

C CC C

C

C

P m GM RT mk MR

RM mTM R m

ρ= ∝ = ×

⎛ ⎞ ⎛ ⎞⎛ ⎞= × ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

T


Opacity and luminosity in stars

At the high densities and temperatures found on average in stellar interiors, matter is opaque. The mean free path, or average distance a photon can travel before being absorbed, is about = 0.5 cm for the Sun’s average density and temperature (given above).Photons produced in the center have to random-walk their way out. How many steps, or mean free paths, does it take for a photon to random-walk from center to surface? (See PU problem 5.11.)

Suppose photon starts off at the center of the star, and has an equal chance to go right or left after each absorption and re-emission. Average value of position after N steps is

x x x x NN N= + + + =( )/1 2 0


Opacity and luminosity in stars (continued)

However, the average value of the square of the position is not zero. Consider step N+1, assuming the chances of going left or right are equal:

But if this is true for all N, then we can find by starting at zero and adding , N times (i.e. using induction):

x x x

x x x x

x

N N N

N N N N

N

+ = − + +

= − + + + +

= +

12 2 2

2 2 2 2

2 2

12

12

12

2 12

2

b g b g

.

xN2

2

x NN2 2= .



Thus to random-walk a distance , the photonneeds to take on the average

So far, we have discussed only one dimension of a three-dimensional random walk. Three times as many steps need to be taken in this case, so to travel a distance R, the photon on the average needs to take

x LN2 =

N L=

2

2 steps.

N R=

3 2

2 steps.



For the Sun, and for a constant mean free path of 0.5 cm,

Each step takes a time , so the average time it takes for a photon to diffuse from the center of the Sun to the surface is

Note that the same trip only takes for a photon travelling in a straight line.

N =×

= ×3 6 96 10

0 55 81 10

10 2

222

.

..

e ja f steps.

∆t c=

211 43

9.7 10 s 3.1 10 years.R

t N tc

= ∆ = = × = ×

/ 2.3 sR c =

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Today in Astronomy 328: stellar structurephysics.gmu.edu/~satyapal/astro328fall07/course... · 6 September 2007 Astronomy 328, Fall 2007 1 Today in Astronomy 328: stellar structure