Binary Stars & Stellar Structure

Binary Stars & Stellar StructureThe Main Sequence

Principles of Stellar StructureHydrostatic Equilibrium

Opacity of the Sun

February 11, 2021

University of Rochester

Binary Stars & Stellar Structure

I Constructing the main sequencewith binary stars

I Observations of stars which weseek to theoretically explain

I Principles of stellar structureI Hydrostatic equilibrium

I Crude stellar interior modelsI Pressure, density, and

temperature in the center (whereluminosity is produced)

I Opacity of the sun: diffusion oflight from center to surface

Reading: Kutner Ch. 9.4, Ryden Sec.14.1, 15.1, & 15.Appendix

February 11, 2021 (UR) Astronomy 142 | Spring 2021 2 / 26

Calculating binary properties

ExampleAn eclipsing binary is observed to have a period of 25.8 years. The two components haveradial velocity amplitudes of 11.0 km/s and 1.04 km/s and sinusoidal variation of radialvelocity with time. The eclipse minima are flat-bottomed and 164 days long. It takes11.7 hr for the ingress to progress from first contact to eclipse minimum.

Answer the following questions:1. What is the orbital inclination?2. What are the orbital radii?3. What are the masses of the stars?4. What are the radii of the stars?


ExampleCloseup of primary minimum:

t2 � t1 = 11.7 hr t3 � t2 = 164 days


ExampleAnswers:

1. Since it eclipses, the orbits must be nearly edge-on. Since the radial velocities aresinusoidal the orbits must be nearly circular.

2. Orbital radii:

rS = vSP

2p= 1.42 ⇥ 1014 cm

= 9.54 AU

rL = vLP

2p= 1.34 ⇥ 1013 cm

= 0.90 AU

Therefore, the semimajor radius of the system is r = r1 + r2 = 10.4 AU.


Example3. Masses: use Kepler’s third law

mL

mS=

vS

vL=

111.04

= 10.6 mS + mL =(r/AU)3

(P/yr)2 = 15.2 M�

= mS + 10.6mS

mS = 1.3 M�, mL = 13.9 M�

4. Stellar radii (note: solar radius = 6.96 ⇥ 1010 cm):

RS =vS + vL

2(t2 � t1)

= (6.02 km s�1)(11.7 hr)

= 7.6 ⇥ 1010 cm = 1.1 R�

RL =vS + vL

2(t3 � t1)

= 369 R�


Data on eclipsing binary stars

The most useful ongoing big compendium of detached main-sequence double-eclipsingbinary data comes from Oleg Malkov and collaborators (Malkov 1993; Malkov, Piskunov,& Shpil’Kina 1997; Malkov et al. 2006, Malkov 2007; Malkov et al. 2012). Much of the datacome from decades of work by Dan Popper.I Those objects for which orbital velocities have been measured comprise the

fundamental reference data on the dependences of stellar parameters — effectivetemperature Te, radius, and luminosity — on stellar mass.

I A strong trend emerges in plots of mass versus any other quantity, involving all starswhich are not outliers in the mass versus radius plot. This trend, in whatever graphit appears, is called the “main sequence,” and its members are called the dwarf stars.


http://adsabs.harvard.edu/abs/1993BICDS..42...27M

http://adsabs.harvard.edu/abs/1997A&A...320...79M

http://adsabs.harvard.edu/abs/1997A&A...320...79M

http://adsabs.harvard.edu/abs/2006A&A...446..785M

http://adsabs.harvard.edu/abs/2007MNRAS.382.1073M

http://adsabs.harvard.edu/abs/2012A&A...546A..69M

https://en.wikipedia.org/wiki/Main_sequence

The H-R diagram

We can not measure the masses of single stars or binary stars in orbits of unknownorientation. Unfortunately, this includes more than 99% of the stars in the sky.I We can usually measure the luminosity from bolometric magnitude and the

temperature from colors, however. The plot of this relation is called theHertzprung-Russell diagram (or H-R diagram).

I The H-R diagram for detached eclipsing binaries is very similar to that of stars ingeneral: it corresponds to the main sequence in other samples of stars.

I This correspondence allows us to infer the masses of single or non-eclipsing

multiple stars.


https://en.wikipedia.org/wiki/Hertzsprung-Russell_diagram

The H-R diagram


Theoretical principles of stellar structure

I Vogt-Russell “Theorem:” The mass and chemical composition of a star uniquely

determine its radius, luminosity, internal structure, and subsequent evolution. Notcompletely right but a very good approximation.

I Stars are spherical to a good approximationI Stable stars are in hydrostatic equilibrium: the weight of each infinitesimal piece of

the star’s interior is balanced by the pressure differential across the piece.I 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 or P =rRTM

I However, in very hot or giant stars the pressure exerted by light — radiation pressure —can dominate!


Theoretical principles of stellar structure

Energy is transported from the inside to the outside, most of the time in the form of light.I The interiors of stars are opaque. Photons are absorbed and re-emitted many times

on their way from the center to the surface in a random walk process called diffusion.

I The opacity depends on the density, temperature, and chemical composition.

I Most stars have regions in their interiors in which the radial variations oftemperature and pressure are such that hot bubbles of gas can “boil” up toward thesurface. This process, called convection, is a very efficient energy transportmechanism and can frequently be more important than light diffusion.


Basic equations of stellar structure

Hydrostatic equilibriumdPdr

= �GMrr

r2

Mass conservationdMrdr

= 4pr2r

Energy generationdLrdr

= 4pr2re

Energy transportdTdr

= � 316s

kr

r2Lr

4pr2

Adiabatic temperature gradient✓dTdr

◆

rad= �

✓1 � 1

g

◆µ

kGMr

r2

Convection occurs ifTP

dPdT

<g

g � 1

Equations of state

Pressure

P(r, T, composition) =rkTµ

+4sT4

3c

in general throughout most normal stars.Opacity

k = k(r, T, composition) in general

Energy generatione = e(r, T, composition) in general

Boundary conditions are

Mr ! 0Lr ! 0

)as r ! 0

T ! 0P ! 0r ! 0

9>=

>;as r ! Rstar

where Mr and Lr are the mass and luminositycontained within radius r.


Basic equations of stellar structure

There are no analytical solutions to the equations of stellar structure; usually,stellar-interior models are generated numerically. Does that mean we cannot do anythingwithout a computer program?I No; progress can be made by assuming a formula for one of the parameters and then

solving for the rest.

I In this manner we can learn the workings of some of these differential equations andestablish scaling relations useful in understanding the shapes of the empirical R(M),Te(M), L(M), and L(Te) results.

I It will be useful to first derive the stellar-structure equation we will use most: theequation of hydrostatic equilibrium.


Hydrostatic equilibrium

This is the principle that the gravitational force on any piece of the star is balanced by thepressure across the piece. This also holds for planetary atmospheres (see Kutner 23.3,Ryden 9.2, 14.1).I Because planetary atmospheres are thin compared to the radii of planets, it is

sufficient to approximate atmospheres as plane parallel slabs with constantgravitational acceleration. Thus,

dPdz

= �rg

a 1D Cartesian differential equation in z which can be solved in various cases.I In stellar interiors, we still get a 1D differential equation (because stars are

spherically symmetric), but we cannot ignore the spherical shape or the radialdependence of gravitational acceleration.



Mr

Dm

r

R

Dr

M

Mr: mass contained in radius r

Consider a spherical shell of radius r andthickness Dr ⌧ r within a star with massdensity r(r). Its weight is

DF = �GMrDmr2

= �GMr

r2 4pr2Drr(r)

= �4pGMrr(r)Dr

Note the minus sign: force points inward.



If the star is in hydrostatic equilibrium, then the weight is balanced by pressuredifferences across the shell:

�DF = P(r)4pr2 � P(r + Dr)4p(r + Dr)2

⇡ P(r)4pr2 �

P(r) +dPdr

Dr�

4pr2

= �dPdr

4pr2Dr

to first order since Dr ⌧ r. Therefore,

dPdr

=DF

4pr2Dr= �4pGMrr(r)Dr

4pr2Dr= �GMrr(r)

r2


Crude stellar models

A surprisingly large amount about stellar interiors can be learned by starting with aformula for the mass density r(r) and solving the equations of hydrostatic equilibriumand the equation of state self-consistently for pressure and temperature.I The most useful is if the formula for density resembles the real thing; if not, P and T

will be way off.

I We will consider problems with density in polynomial or exponential form in thiscourse.

I Using simple expressions for the density, the hydrostatic equilibrium equation can beeasily integrated.


Crudest approximation: the “uniform Sun”

We know the distance to the Sun (from radar measurements), its mass (from Earth’sorbital period), and its radius (from angular size and known distance):

r = 1 AU = 1.495978707 ⇥ 1011 m

M� = 1.988 ⇥ 1030 kg

R� = 6.957 ⇥ 108 m

Therefore, we know the average mass density (mass/volume) is

r� =M�V�

=3M�4pR3

�= 1.41 g/cm3

= 26% of Earth’s density of ⇠ 5.5 g/cm3

What would the pressure and temperature of the Sun be if it had uniform density?


The “uniform Sun”

With r(r) = r� = 3M�/4pR3�,

dPdr

= �GMrr

r2 = �Gr2

✓M�

r3

R3�

◆✓3M�4pR3

�

◆= �3GM2

�4pR6

�r

The surface has P(R�) = 0 if it does not move, so

Z 0

P(0)dP = �3GM2

�4pR6

�

Z R�

0r dr

�P(0) = �3GM2�

4pR6�

R2�

2=) P(0) =

38p

GM2�

R4�

) PC = P(0) = 1.34 ⇥ 1015 dyne/cm2 = 1.3 ⇥ 109 atmospheres


The “uniform Sun”Thus, the pressure elsewhere inside the uniform density Sun is

Z P(r)

PC

dP0 = �3GM2�

4pR6�

Z r

0r0 dr0 = P(r)� PC = �3GM2

�8pR6

�r2

P(r) =3

8p

GM2�

R4�

� 38p

GM2�

R6�

r2

=3

8p

GM2�

R4�

✓1 � r2

R2�

◆

Now we can solve for the temperature using the fact that the Sun is an ideal gas:

PV = NkT =) P = nkT =rkTµ

where N is the number of particles, n = N/V is number density, and µ is average particlemass.


The “uniform Sun”

Solving for temperature gives

T(r) =µP(r)

kr=

µ

k4pR3

�3M�

38p

GM2�

R4�

✓1 � r2

R2�

◆

=12

µGM�kR�

✓1 � r2

R2�

◆

Suppose the average mass of the particles in this ideal gas is the same as the mass of theproton, µ = 1.67 ⇥ 10�27 kg. Then

TC = T(0) = 1.2 ⇥ 107 K

Note that in our approximations we end up with T = 0 at the surface, while in reality it isabout 6000 K. However, that is indeed much less than 12 MK.


Central pressure in a star: scaling relationReturn for a moment to the central pressure in a uniform star:

PC =3

8p

GM2

R4

The only part of the equation which depends on the functional form of the density is thedimensionless coefficient 3/8p. Otherwise the central pressure is proportional to M2R�4.As you will see in the homework, for densities such as

r(r) = rC

⇣1 � r

R

⌘and r(r) = rC

1 �

⇣ rR

⌘2�

you obtain the central pressure

PC =5

4p

GM2

R4 and PC =15

16p

GM2

R4


Central pressure in a star: scaling relationIt turns out this is a simple scaling relation for stars:

PC µGM2

R4

and the proportionality constant gets larger as the central density gets larger. A completecalculation for stars of low to moderate mass yields the expression

PC ⇡ 19GM2

R4

For the Sun, M� = 1.99 ⇥ 1033 g and R� = 6.96 ⇥ 1010 cm, so

PC ⇡ 19GM2

�R4�

= 2.1 ⇥ 1017 dyne cm�2 = 2.1 ⇥ 1016 Pa


Central pressure in a star: scaling relationFor other main sequence stars, we can derive an expression in terms of M� and R�:

PC ⇡ 19GM2

R4 = 19GM2

R4

✓M�M�

◆2 ✓R�R�

◆4

= 19GM2

�R4�

✓M

M�

◆2 ✓R�R

◆4

= 19(6.67 ⇥ 10�8 cm g�1 s�2)(1.99 ⇥ 1033 g)2

(6.96 ⇥ 1010 cm)4

✓M

M�

◆2 ✓R�R

◆4

= 2.1 ⇥ 1017✓

MM�

◆2 ✓R�R

◆4dyne/cm2

PC ⇡ 2 ⇥ 1011✓

MM�

◆2 ✓ RR�

◆�4atm


Central density and temperature of the Sun

The central pressure of the Sun is about 100 times larger than that of the “uniform” Sun.So what is the central density?

Guess: rC is 100 times higher than the average density (equivalent to guessing thatinternal temperature does not vary much with radius).I For the Sun, this is not a bad guess; the central density turns out to be 110 times the

average density: rC = 150 g/cm3. So we can obtain another scaling relation fromrC µ MR�3:

rC = 25MR3 = 150

✓M

M�

◆✓R�R

◆3g/cm3


Central density and temperature of the Sun

As we will see in a couple of weeks, the average gas-particle mass in the center of theSun, considering its composition and the fact that the center is completely ionized, isµC = 1.5 ⇥ 10�24 g. But the material is still an ideal gas, so

TC =PCVNCk

=PC

nCk=

PCµC

rCk= 15.7 ⇥ 106 K

And indeed, T does not vary much with radius. We can make a scaling relation out ofthis as well to use on stars with different mass, radius, and composition:

TC =PCµC

rCkµ

GM2

R4R3

MµC = 15.7 ⇥ 106 K for the Sun

TC = 15.7 ⇥ 106✓

MM�

◆✓R�R

◆✓µ

µ�

◆K


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