Astronomy 142, Spring 2016
26 January 2016
Today in Astronomy 142: stellar structure
q  Observations of stars which we
seek to explain theoretically.
q  Principles of stellar structure
q  Hydrostatic equilibrium
•  Crude stellar interior models
•  Pressure, density and
temperature in the center
(where the luminosity is
produced)
q  Opacity of the Sun: diffusion of
light from center to surface.
Rotational structure of the Sun: the different colors indicate flows slower
(blue) and faster (red) than average. (SOHO/NASA).
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Theoretical principles of stellar structure
q Vogt-Russell “theorem:” the mass and chemical
composition of a star uniquely determine its radius,
luminosity, internal structure, and subsequent evolution
(not completely right, but a very, very good approximation)
q Stars are spherical, to good approximation.
q 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 or P = ρRT/M
•  However, in very hot stars or giant stars, the pressure
exerted by light – radiation pressure – can dominate!
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Principles of stellar structure (continued)
q 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.
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The basic equations of stellar structure
Hydrostatic equilibrium:
Equations of state
Pressure:
P = P( ρ , T , composition) in general
GMr ρ
dP
=−
dr
r2
Mass conservation:
=
dMr
= 4π r 2 ρ
dr
Energy generation:
ρ kT 4σ T 4
+
µ
3c
throughout most normal stars
Opacity:
κ =κ ( ρ , T , composition) in general
dLr
= 4π r 2 ρε
dr
Energy transport:
Energy generation:
ε =ε (ρ ,T , composition) in general
dT
3 κρ Lr
=−
dr
16σ r 2 4π r 2
Adiabatic temperature gradient:
Boundary conditions:
⎛
1 ⎞ µ GMr
⎛ dT ⎞
= −⎜1 − ⎟
⎜ dr ⎟
2
γ
⎝
⎠rad
⎝
⎠k r
Convection occurs if:
T → 0⎫
⎪
P → 0 ⎬ as r → Rstar
ρ→ 0⎪
⎭
T dP
γ
<
P dT γ − 1
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M r → 0⎫
⎬ as r → 0
Lr → 0 ⎭
M r , Lr : mass or luminosity contained
within radius r.
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The equations of stellar structure (continued)
There are no analytical solutions to the
equations; usually stellar-interior models
are generated by computer solution.
Does that mean we get to go home now?
q No. Progress can be made by assuming
a formula for one of the parameters, and
solving for the rest.
q In this manner we can learn the workings of some of
these differential equations, and establish some of the
scaling relations useful in understanding the shapes of
the empirical R ( M ) , Te ( M ) , L ( M ) , and L (Te ) results.
q First it will be useful to derive the stellar-structure
equation we’ll use most, the equation of hydrostatic
equilibrium.
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Hydrostatic equilibrium
In AST 111 we applied hydrostatic equilibrium to the
structure of planetary atmospheres (see FA Sec. 9.2, 14.1).
q Because atmospheres are thin compared to the radii of
planets, it was sufficient there to approximate atmospheres as plane parallel slabs with constant gravitational
acceleration. Thus we got a Cartesian one-dimensional
differential equation, which we solved in various cases:
dP
= −ρ g .
dz
q In stellar interiors we still get a one-D differential equation
(because stars are spherically symmetric), but we can’t
ignore the spherical shape or the radial dependence of
gravitational acceleration.
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Hydrostatic equilibrium (continued)
Consider a spherical shell of
radius r and thickness Δr ≪ r
within a star with mass density
(mass per unit volume) ρ(r). Its
weight is
ΔF = −
=−
GMr Δm
r2
GMr
2
Δm
Mr
Δr
r
4π r 2 Δr ρ ( r )
r
= −4π GMr ρ ( r ) Δr .
M
Mr : mass contained
(minus sign: force points inward)
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R
within radius r
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Hydrostatic equilibrium (continued)
If the star is in hydrostatic equilibrium the weight is balanced
by the pressure difference across the shell:
()
( ) ( )2 ( Δr  r )
⎡
dP ⎤
(to first order
≅ P ( r ) 4π r 2 − ⎢ P ( r ) +
Δr ⎥ 4π r 2
in pressure)
dr
−ΔF = P r 4π r 2 − P r + Δr 4π r + Δr
⎣
⎦
dP
4π r 2 Δr ,
dr
4π GMr ρ r Δr
GMr ρ r
dP
ΔF
=
=−
=−
dr 4π r 2 Δr
4π r 2 Δr
r2
=−
so
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()
Astronomy 142, Spring 2016
()
.
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Crude stellar models
One can learn a surprisingly large amount about stellar
interiors by starting with a formula for the mass density,
ρ(r), and solving the equations of hydrostatic equilibrium
and the equation of state self consistently for pressure and
temperature.
q One learns most, of course, if the formula for density
resembles the real thing.
q In AST 142 we will consider problems in which density of
polynomial or exponential form to be within bounds.
q That is, all forms in which the hydrostatic equilibrium
equation can be integrated straightforwardly.
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Crudest approximation: the “uniform Sun”
Since we know its distance (from radar), mass (from Earth’s
orbital period) and radius (from angular size and known
distance):
r = 1 AU = 1.495978707 × 1011 m
M⊙ = 1.988 × 10 30 kg
R⊙ = 6.957 × 108 m
we know the average mass density (mass per unit volume):
M
3M⊙
ρ⊙ = ⊙ =
= 1.41 g cm -3
3
V⊙
4π R⊙
= only 26% of Earth's (~5.5)
What would the pressure and temperature be, if the Sun had
uniform density?
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The “uniform Sun” (continued)
Let us therefore assume that
integrate away:
⎛
GMr ρ
dP
G
=−
= − ⎜ M
dr
r2
r 2 ⎜⎝
0
∫
()
2
3GM
dP = −
6
4π R
P0
R
∫ r dr
()
6 2
4π R
()
2
⎞ ⎛ 3M ⎞
3GM

⎟⎜
⎟ =−
r
3 ⎟⎜
3 ⎟
6
R
4
π
R
4
π
R
⎠⎝
⎠

r3
Surface has P = 0, if it doesn't move.
0
2
2
3GM
R
−P 0 = −
()
3
ρ r = ρ = 3M 4π R
, and
2
3 GM
⇒ P 0 =
8π R 4

()
.
PC = P 0 = 1.34 × 1015 dyne cm -2 = 1.3 × 109 atmospheres.
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The “uniform Sun” (continued)
Thus, for the pressure elsewhere within the “uniform-density
Sun,”
P( r )
2 r
2
3GM
3GM
d
P
=
−
r
d
r
=
P
r
−
P
=
−
r2
′
′
′
∫
C
6 ∫
6
4π R 0
8π R
P
()
C
2
2
2 ⎛
⎞
3 GM
3 GM 2
3 GM
r2
⎜
⎟
P r =
−
r =
1−
2 ⎟
8π R 4
8π R6
8π R 4 ⎜
R ⎠


 ⎝
()
Now, the Sun is an ideal gas, through and through:
PV = NkT
P = nkT =
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ρ kT
µ
(N= number of gas particles, n= number
density: number of particles per unit
volume, μ = avg. particle mass)
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The “uniform Sun” (continued)
so
()
T r =
=
( ) = µ 4π R3
µP r
kρ
k 3M
2 ⎛
⎞
3 GM
r2
⎜1−
⎟
2 ⎟
8π R 4 ⎜
R ⎠
 ⎝
⎛
⎞
1 µGM
r2
⎜1−
⎟ .
2 ⎟
2 kR ⎜
R
⎝
⎠
Suppose the average mass of the particles in this ideal gas is
the same as the mass of the proton: µ = 1.67 × 10 −27 kg.
Then
TC = T ( 0 ) = 1.2 × 107 K .
q Note that in our approximations we wind up with T = 0
on the surface. We know the temperature is really more
like 6000 K there, but indeed that’s much less than 12 MK.
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Central pressure in a star
Return for a moment to the central pressure in a uniform star:
3 GM 2
.
8π R 4
The only part of the equation which depends upon the
functional form of the density is the dimensionless coefficient,
3 8π . Otherwise, the central pressure is proportional to M2R-4.
You’ll see in the homework, for example, different density
⎡ ⎛ r ⎞2 ⎤
functions:
r⎞
⎛
ρ ( r ) = ρC ⎜ 1 − ⎟ , or = ρC ⎢1 − ⎜ ⎟ ⎥ ,
R⎠
⎢⎣ ⎝ R ⎠ ⎥⎦
⎝
PC =
and obtain
26 January 2016
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5 GM 2
PC =
4π R 4
, or
Astronomy 142, Spring 2016
15 GM 2
=
16π R 4
.
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Central pressure in a star (continued)
Evidently this is a simple scaling relation for stars:
PC ∝
GM 2
R4
,
and the proportionality constant gets larger, the larger the
central density is.
q Lo and behold, a complete calculation for stars of
moderate to low mass yields
PC = 19
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GM 2
R4
.
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Central pressure in a star (continued)
For the Sun:
M = 1.99 × 10 33 g
PC ≅ 19
2
GM⊙
> 10
4
R⊙
11
R = 6.96 × 1010 cm
=2 × 1017 dyne cm -2 = 2 × 1016 Pa
atmospheres
So, for other main sequence stars, to adequate approximation,
PC ≅ 19
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GM 2
R
4
2
=2.1 × 10
17
⎛ M ⎞ ⎛ R ⎞
⎜
⎟ ⎜
⎟
⎝ M ⎠ ⎝ R ⎠
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−4
dyne cm -2
.
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Central pressure in a star (continued)
That is,
2
GM 2 ⎛ M ⎞ ⎛ R ⎞
PC ≅ 19
= 19
⎜
⎟ ⎜
⎟
R4
R 4 ⎝ M ⎠ ⎝ R ⎠
GM 2
2
2 ⎛
GM
M ⎞ ⎛ R ⎞
= 19
4 ⎜M ⎟ ⎜R ⎟
R ⎝  ⎠ ⎝  ⎠
(6.67 × 10
= 19
−8
= 2.1 × 10
−4
)(
cm gm −1 sec −2 1.99 × 10 33 gm
(6.96 × 10
2
17
4
⎛ M ⎞ ⎛ R ⎞
⎜
⎟ ⎜
⎟
⎝ M ⎠ ⎝ R ⎠
10
cm
−4
)
4
) ⎛⎜ M ⎞⎟ ⎛⎜ R ⎞⎟
2
2
−4
⎝ M ⎠ ⎝ R ⎠
dyne cm -2 .
(You’ll need to get used to doing this. )
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Central density and temperature of the Sun
Central pressure is a little more than a factor of 100 larger
than that of the Uniform Sun. Guess: central density 100 times
higher than average? (That’s equivalent to guessing that the
internal temperature doesn’t vary much with radius.)
q For the Sun, that’s not bad; the central density turns out to
be 110 times the average density, ρC = 150 gm cm-3 .
Thus, since ρC ∝ M R3 ,
3
⎛ M ⎞ ⎛ R ⎞
M
-3
ρC = 25
= 150 ⎜
⎟⎜
⎟ gm cm
3
⎝ M ⎠ ⎝ R ⎠
R
q As we will see in a couple of weeks, the average gasparticle mass in the center of the Sun, considering its
composition and the fact that the center is completely
ionized, is µC = 1.5 × 10−24 gm .
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Central density and temperature of the Sun
(continued)
q But the material is still an ideal gas, so
PV
P
P µ
TC = C = C = C C = 15.7 × 106 K.
NC k nC k
ρC k
And, indeed, T doesn’t vary very much with radius.
q 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:
P µ
GM 2 R 3
TC = C C ∝
µC = 15.7 × 106 K for the Sun;
4
ρC k
R M
⎛ M ⎞ ⎛ R ⎞ ⎛ µ ⎞
TC = 15.7 × 106 ⎜
⎟⎜
⎟⎜
⎟ K .
⎝ M ⎠ ⎝ R ⎠ ⎝ µ ⎠
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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 ℓ = 0.5 cm for the Sun’s average density and
temperature (given above; see also Homework #2, prob. 1).
Photons produced in the center have to random-walk their
way out, a process called diffusion. How many steps (=
mean free paths) does it take for a photon to random-walk
from center to surface?
q 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
xN = (x1 + x2 +!+xN )/ N = 0
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Opacity and luminosity in stars (continued)
q 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:
1
1
2
2
x N − ) + ( x N + )
(
2
2
1
1
= x N2 − 2x N  +  2 + x N2 + 2x N  +  2
2
2
2
2
= xN +  .
x N2 +1 =
2
q But if this is true for all N, then we can find x N by
2
starting at zero and adding ℓ , N times (i.e. using
induction):
2
xN
= Nℓ 2 .
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Opacity and luminosity in stars (continued)
q Thus to random-walk a distance
needs to take on the average
N=
L2
ℓ2
2
xN
= L , the photon
steps.
q So far, we have discussed only one dimension of a threedimensional 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
N=
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3R 2
ℓ2
steps.
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Opacity and luminosity in stars (continued)
q For the Sun, and for a constant mean free path of 0.5 cm,
N=
(
3 6.96 ×1010
( 0.5)
2
)
2
= 5.81×10 22 steps.
(Very opaque indeed!)
q Each step takes a time Δt = ℓ c , so the average time it
takes for a photon to diffuse from the center of the Sun to
the surface is
t = NΔt =
2
3R
= 9.7 × 1011 s = 3.1 × 10 4 years.
c
Note that the same trip only takes R /c = 2.3 s for a
photon travelling in a straight line.
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