Binary Stars & Stellar Structure
Transcript of 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
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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?
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ExampleCloseup of primary minimum:
t2 � t1 = 11.7 hr t3 � t2 = 164 days
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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.
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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�
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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.
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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.
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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!
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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.
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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.
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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.
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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.
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Hydrostatic equilibrium
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.
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Hydrostatic equilibrium
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
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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.
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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?
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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
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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.
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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.
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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
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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
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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
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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
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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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