binary stars & stellar structure - university of...

Binary Stars & StellarStructure

Direct measurements of stellar mass, radius,and temperature in eclipsing binaries

Principles of Stellar StructureHydrostatic Equilibrium

Opacity of the Sun

January 28, 2020

University of Rochester

Binary Stars & Stellar Structure

I Direct measurements of stellarmass, radius, and temperature

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

January 28, 2020 (UR) Astronomy 142 | Spring 2020 2 / 1

Measurement of binary star massesArmed with the definition of the location of the center of mass (m1a1 = m2a2), if orbitalmajor axes (relative to the center of mass) or radial velocities are known, so is the ratio ofthe masses: m1

m2=

a2

a1=

v2r

v1r=

v2

v1

Furthermore, from Kepler’s third law,

m1 + m2 =P

2πG

(v1r + v2r

sin i

)3

For unresolved binaries (the vast majority) we can only measure P and the two velocityamplitudes. This gives two equations and three unknowns: m1, m2, and sin i.

This is why eclipsing binaries are so important. If the system eclipses, we must beviewing the orbital plane close to edge-on. Thus, sin i ≈ 1, leaving two equations withtwo unknowns, m1 and m2.


Measurement of stellar radiusEclipses do not just turn on and off instantaneously; the transitions are gradual. Theshape of the system’s light curve is sensitive to the sizes of the stars.

I If one star can completely block thelight of the other, the “bottom” of theeclipse light curve will be flat.

I The extent of the flat part of the lightcurve is determined by the radius of thelarger companion.

I The slope of the transitions isdetermined by the radius of the smallercompanion.

I If the light curve never flattens out thenthe eclipses are not “total,” i.e., neitherstar completely blocks the other.


Measurement of stellar radius

I Much can be made of the details of theshapes of the light curve at the onset(ingress) and end (egress) of the eclipse.

I The “corners” are usually labeled asindicated at right.

I During an eclipse, the stars move at speedsv1 and v2 in opposite directionsperpendicular to the line of sight.

I v1 and v2 are measured from orbital periodsand radial velocity amplitudes. Thus: time

flu

x

t1 t2 t3 t4

Ingress Egress

2RS = (v1 + v2)(t2 − t1) Diameter of “small” star2RL = (v1 + v2)(t3 − t1) Diameter of “large” star


Measurement of stellar Te

I A useful measurement of the ratioof the effective temperatures Tecomes from the relative depths ofthe primary (deeper) andsecondary eclipses.

I When the stars are not eclipsed,their total flux is

f = πR2SI0S + πR2

LI0L

I What configuration defines theprimary and the secondary eclipse?

Sf

Pf

Flux

Time


Measurement of stellar TeIf the smaller star is hotter, the primary eclipse is when it is behind the larger star:

fP = πR2LI0L

Then the secondary eclipse is when the small star is in front:

fS = π(R2L − R2

S)I0L + πR2SI0S

We have three equations and two unknowns, but usually these are used to construct oneratio to remove uncertainties in the distance r:

f − fPf − fS

=πR2

S I0S + πR2L I0L − πR2

L I0L

πR2S I0S + πR2

L I0L − π(R2

L − R2S)

I0L − πR2S I0S

=πR2

S I0S

πR2L I0L − π

(R2

L − R2S)

I0L

=πR2

S I0S

πR2L I0L

=I0S

I0L


Measurement of stellar Te

Note that the way we defined the I0s is

f = πR2 I0 =L

4πr2 =4πR2σT4

e4πr2 =⇒ I0 =

σT4e

πr2

Thus,f − fPf − fS

=I0S

I0L=

T4eS

T4eL

If r is known, the luminosity and Te =(L/4πr2σ

)1/4 of the brighter star is determinedaccurately from the primary minimum flux, and thus Te of the fainter star from this ratio.


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

2π= 1.42× 1014 cm

= 9.54 AU

rL = vLP

2π= 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%26A...320...79M

http://adsabs.harvard.edu/abs/1997A%26A...320...79M

http://adsabs.harvard.edu/abs/2006A%26A...446..785M

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

http://adsabs.harvard.edu/abs/2012A%26A...546A..69M

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

Radii of eclipsing binary stars

Dwarfs (main sequence)

Giants


Luminosity of eclipsing binary stars


Effective temperature of eclipsing binary stars


The H-R diagram

We cannot 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-eclipsingmultiple stars.


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

The H-R diagram


H-R diagram for eclipsing binary starsConvention: the T axis is plotted backwards in H-R diagrams.

Dwarfs (main sequence)

Giants


Eclipsing binaries vs. nearby starsStars within 25 pc (Gliese & Jahreiss 1995)


Eclipsing binaries vs. young stars in open clustersX-ray selected Pleiades (Stauffer et al. 1994)


Eclipsing binaries vs. bright or very nearby starsBright/nearby from Allen’s Astrophysical Quantities


Theoretical principles of stellar structure

I Vogt-Russell “Theorem:” The mass and chemical composition of a star uniquelydetermine 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 difference 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 =ρRTM

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 of

temperature 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

I Hydrostatic equilibriumdPdr

= −GMrρ

r2

I Mass conservationdMr

dr= 4πr2ρ

I Energy generationdLr

dr= 4πr2ρε

I Energy transportdTdr

= − 316σ

κρ

r2Lr

4πr2

I Adiabatic temperature gradient(dTdr

)rad

= −(

1− 1γ

)µ

kGMr

r2

I Convection occurs ifTP

dPdT

<γ

γ− 1

Equations of stateI Pressure

P(ρ, T, composition) =ρkTµ

+4σT4

3cin general throughout most normal stars.

I Opacityκ = κ(ρ, T, composition) in general

I Energy generationε = ε(ρ, T, composition) in general

I Boundary conditions

Mr → 0

Lr → 0

}as r→ 0

T → 0

P → 0

ρ → 0

as r→ Rstar

where Mr and Lr are the mass and luminosity containedwithin 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 and

establish 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

= −ρg

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

∆m

r

R

∆r

M

Mr: mass contained in radius r

Consider a spherical shell of radius r andthickness ∆r� r within a star with massdensity ρ(r). Its weight is

∆F = −GMr∆mr2

= −GMr

r2 4πr2∆rρ(r)

= −4πGMrρ(r)∆r

Note the minus sign: force points inward.



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

−∆F = P(r)4πr2 − P(r + ∆r)4π(r + ∆r)2

≈ P(r)4πr2 −[

P(r) +dPdr

∆r]

4πr2

= −dPdr

4πr2∆r

to first order since ∆r� r. Therefore,

dPdr

=∆F

4πr2∆r= −4πGMrρ(r)∆r

4πr2∆r= −GMrρ(r)

r2


Crude stellar models

A surprisingly large amount about stellar interiors can be learned by starting with aformula for the mass density ρ(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 this

course.I Using simple expressions for the density, the hydrostatic equilibrium equation can be

easily 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

ρ� =M�V�

=3M�4πR3

�= 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 ρ(r) = ρ� = 3M�/4πR3�,

dPdr

= −GMrρ

r2 = −Gr2

(M�

r3

R3�

)(3M�4πR3

�

)= −3GM2

�4πR6

�r

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

P(0)dP = −3GM2

�4πR6

�

∫ R�

0r dr

−P(0) = −3GM2�

4πR6�

R2�

2=⇒ P(0) =

38π

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

∫ P(r)

PC

dP′ = −3GM2�

4πR6�

∫ r

0r′ dr′ = P(r)− PC = −3GM2

�8πR6

�r2

P(r) =3

8π

GM2�

R4�− 3

8π

GM2�

R6�

r2

=3

8π

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 =ρkTµ

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)

kρ=

µ

k4πR3

�3M�

38π

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’sabout 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

8π

GM2

R4

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

ρ(r) = ρC

(1− r

R

)and ρ(r) = ρC

[1−

( rR

)2]

you obtain the central pressure

PC =5

4π

GM2

R4 and PC =15

16π

GM2

R4


Central pressure in a star: scaling relation

It turns out this is a simple scaling relation for stars:

PC ∝GM2

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 relation

For 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

)4

dyne/cm2

PC ≈ 2× 1011(

MM�

)2 ( RR�

)−4

atm


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: ρC 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: ρC = 150 g/cm3. So we can obtain another scaling relation fromρC ∝ MR−3:

ρC = 25MR3 = 150

(M

M�

)(R�R

)3

g/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

ρCk= 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

ρCk∝

GM2

R4R3

MµC = 15.7× 106 K for the Sun

TC = 15.7× 106(

MM�

)(R�R

)(µ

µ�

)K


Opacity and luminosity in starsAt the high densities and temperatures found on average in stellar interiors, matter isopaque. The mean free path, or average distance a photon can travel before beingabsorbed, is about

` = 0.5 cm

for the Sun’s average density and temperature.

Photons produced in the center of the Sun have to randomly walk their way out, aprocess called diffusion.


Opacity and luminosity in starsHow many steps (in number of mean free paths) does it take for a photon to get from thecenter of the sun to the surface?

Let us work in 1D. Suppose a photon stars off at the center of the star and has an equalchance to go right or left after each absorption and re-emission step. The average positionafter N steps is

〈xN〉 = (x1 + x2 + · · ·+ xN)/N = 0

However, the average value of the square of position is nonzero. Consider step N + 1,assuming an equal chance of going left or right:

〈x2N+1〉 =

12〈(xN − `)2〉+ 1

2〈(xN + `)2〉

=12〈x2

N − 2xN`+ `2〉+ 12〈x2

N + 2xN`+ `2〉

= 〈x2N〉+ `2


Opacity and luminosity in stars

If this expression for 〈x2N+1〉 is true for all N, then we can find 〈x2

N〉 by starting at zero andadding `2 N times (using induction):

〈x2N〉 = N`2

Thus, to randomly walk a distance√〈x2

N〉 = L the photon needs to take, on average,

N = L2/`2 steps

In 3D, the photon needs to take 3 times as many steps, so to travel a distance R it needs

N =3R2

`2 steps


Opacity and luminosity in stars

For the Sun, assuming a constant mean free path ` = 0.5 cm and usingR = 6.96× 1010 cm,

N =3(6.96× 1010)2

(0.5)2 = 5.81× 1022 steps

This is very opaque!

Each step should take a time ∆t = `/c, so the average time required for a photon todiffuse from the center of the Sun to the surface is

t = N∆t =3R2�

`c= 9.7× 1011 s ≈ 31, 000 yr

Note that the same trip takes only t = R�/c = 2.3 s for a photon traveling in a straightline!


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