# Astronomy 201a: Stellar Astrophysics . . . . . . . . . . September 30 ```Astronomy 201a: Stellar Astrophysics . . . . . . . . . . September 30, 2014
Stellar Interiors Modeling
This is where we take everything we’ve covered so far and fold it into models of
spherically symmetric, time-steady stellar structure. (We’ll emphasize insight
over rigorous accuracy.)
Recap the 5 basic interiors equations:
dMr
dP
GMr
= 4πr2 ρ
= − 2 ρ
dr
dr
r
dT
, if conv. stable
=
(dT /dr)ad − ∆∇T , if unstable
dr
dLr
= 4πr2 ρ ǫ
dr
P = P (ρ, T, &micro;)
Let’s assume that &micro; is known, either as an arbitrary input state or on the basis
of pre-computed evolutionary history.
Other knowns:
ǫ(ρ, T, &micro;) ,
κ(ρ, T, &micro;) ,
ionization state
The 5 main unknowns are Mr , ρ, P , T , Lr .
Each of them has a full radial dependence from core to surface.
The whole thing is a two-point boundary value problem.
The values at the boundaries are integration constraints:
Mr
ρ
P
T
Lr
center
0
ρc
Pc
Tc
0
surface (r = R∗ )
M∗
0 ? ρphoto ?
0 ? Pphoto ?
0 ? Teff ?
L∗
For a lot of interiors modeling, ρ, P , T at the surface are not much different
from zero. Computing them in detail is needed for atmospheres modeling.
They can be determined in terms of other unknowns.
But there are 6 boxed unknowns above!
1
With only 5 basic equations, one of the unknowns must just be a free
parameter.
−→ M∗ is usually chosen.
Vogt-Russell “theorem:”
Specifying M∗ and &micro; (i.e., mass &amp; composition) uniquely determines all other
stellar parameters.
Of course, this is only strictly true in the sense that these (time-steady,
spherically symmetric) equations are true. This leaves out magnetic fields,
rotation, pulsation....
..............................................................................
Typically, one must integrate the differential equations numerically. Collins
&sect; 4.7 talks about the Henyey relaxation method that treats the derivatives as
finite differences, &amp; iterates on small changes to the interiors variables to
both
−→ compute a step in an evolutionary calculation.
For insight into the physics, we’ll look at a couple of approximations that let
us reach analytic solutions...
(1) the constant-density model
(2) polytropes
2
The constant density model
Assuming ρ = ρ0 throughout the whole star is a huge approximation, but it
will end up allowing us to do a huge number of useful things.
It will allow us to obtain all of the results in various sections of your textbooks
titled “homology relations” or “homology transformations.” (Usually
those are introduced as something separate, but we get them all “for free.”)
dMr
= 4πr2 ρ0
dr
Z
Easy!
ρ0 =
M∗
dMr = 4πρ0
0
Z
R∗
dr r2
0
M∗
= hρi
(4/3)πR∗3
so if we know any two of {ρ0, M∗ , R∗} this gives the third.
If we stop the integration part-way, we get
Mr (r) =
4
πρ0 r3
3
and this (black curve) often doesn’t look
very much like what one gets numerically
for more centrally condensed stars (red
curve).
Putting aside that bit of incongruity, we can solve the hydrostatic equilibrium
equation for P (r),
dP
GMr
4πGρ20
= − 2 ρ0 = −
r
dr
r
3
Keeping track of limits,
Z
Z P
4πGρ20 r ′ ′
′
dr r
dP = −
3
0
Pc
2
P (r) − Pc = − π Gρ20 r2
3
If we set P = 0 at r = R∗ , we can solve for the central pressure:
3
2π 2 2
3 GM∗2
Gρ0 R∗ =
3
8π R∗4
r2
And also,
P (r) = Pc 1 − 2
R∗
Pc =
Note that these results are independent of the equation of state!
But we do see from the equation of state diagram (T vs. ρ) that stars typically
fall under the curve at which Pgas = Prad .
Thus, let’s assume (for now) that P ≡ Ptotal ≈ Pgas.
So if we know P (r) and ρ0 , we can use the ideal gas equation of state to solve
for T (r),
GM∗
kB T c
r2
ρ0 kB T (r)
, so... T (r) = Tc 1 − 2
=
P (r) =
&micro;mH
R∗
&micro;mH
2R∗
..............................................................................
This looks very fundamental. It is essentially an alternate “proof” of the virial
theorem.
Work out the gravitational potential energy...
Z
GMr
3 GM∗2
EG = − dMr
=;;= −
r
5 R∗
and the integrated thermal energy (assuming ideal gas)...
Z
Z R∗
3
3 GM∗2
EG
dr 4πr2 P (r) = ; ; =
EK =
dV U =
= −
!
2
10
R
2
∗
0
..............................................................................
Moving forward, next is energy conservation. Let’s make use of the power-law
parameterization ǫ = ǫ0 ρT ν .
dLr
= 4πr2 ρ0 ǫ(ρ, T )
dr
4
Let’s not worry too much about getting
the full Lr (r). Energy generation is peaked
near the core, so the star should reach its
full L∗ in the ∼deep interior.
L∗ =
Z
0
L∗
dLr = 4πρ20 ǫ0
Z
0
R∗
dr r2 Tcν
r2
1− 2
R∗
Defining x = r/R∗ and making the integral dimensionless,
Z 1
2
ν 3
dx x2(1 − x2)ν
L∗ = 4πρ0 ǫ0 Tc R∗
{z
}
|0
ν
√
π Γ(ν + 1)
a pure # &middot; &middot; &middot;
4 Γ(ν + 52 )
Now, let’s start looking more at proportionalities (i.e., neglecting the constants
out front) and just keep in mind that we can compute the complete expressions
whenever we want.
(Note, though, that the constant-density model gives ∼okay absolute numbers
for pressure &amp; temperature, but not for luminosity.)
L∗ ∼ ρ20 Tcν R∗3
Thus...
M∗2
∼
R∗6
&micro;M∗
R∗
ν
R∗3
L∗ ∼ &micro;ν M∗2+ν R∗−3−ν
Let’s also look at the case of a star in radiative equilibrium. In this case,
dT
3κρ Lr
= −
dr
4acT 3 4πr2
It’s tricky because our model has T → 0 at the surface.
We can choose an intermediate depth (like r ≈ R∗ /2) where we can be
confident that Lr has reached L∗.
5
Neglecting constants again, let’s solve for Lr ≈ L∗. First note that, at this
intermediate point,
T ≈
Thus,
T 3 r2
L∗ ∼
κρ
3
Tc
4
dT
dr
dT
2 r Tc
Tc
= − 2 ≈ −
dr
R∗
R∗
Tc4 R∗
∼
∼
κ ρ0
&micro;M∗
R∗
4
R∗
κ (M∗/R∗3)
&micro;4 M∗3
Noting that all R∗ ’s cancel out,
L∗ ∼
κ
It’s interesting that we got this result without any reference to nuclear energy
generation.
Over ∼short time-scales (i.e., K-H), a star’s luminosity depends on how
efficiently it transports energy out through the interior (not so much on how
rapidly it generates that energy). Higher κ → radiative interior acts like a
“blanket” keeping the energy in → lower L∗ .
Over longer time-scales, the energy generation does matter. Let’s match the
above scaling with what we got from energy conservation, and we can
eliminate a variable...
&micro;4 M∗3
∼ &micro;ν M∗2+ν R∗−3−ν
κ
We can solve for the main-sequence mass–radius relation, as well as some
other useful things.
What about κ? It depends on ρ &amp; T , so it is likely to depend on radial
distance r (and also on R∗ ).
For now, let’s assume κ is a constant. Absorbing it into the other constants,
we get
(ν−1)/(ν+3)
R∗ ∼ &micro;(ν−4)/(ν+3) M∗
L∗ ∼ &micro;4 M∗3
4/(ν+3)
Tc ∼ &micro;7/(ν+3) M∗
R∗ ∼ &micro;0.36 M∗0.64
L∗ ∼ &micro;4 M∗3
Tc ∼ &micro;0.64 M∗0.36
For PP chain (M∗ &lt;
ν∼4
∼ 1.5 M⊙),
For CNO cycle (M∗ &gt;
∼ 2 M⊙ ), ν ∼ 15–20
6
տ
choose intermediate ν = 8
We can also now determine the position of the main sequence in the
H–R diagram.
x
i.e., we can solve for x in L∗ ∼ Teff
Keeping &micro; fixed and just looking at the M∗ dependence for ν = 8,
4
σTeff
=
L∗
so
4πR∗2
Teff ∼ L∗1/4 R∗−1/2 ∼ M∗3/4 M∗−0.32 ∼ M∗0.43
Thus,
x
L∗ ∼ Teff
corresponds to
M∗3 ∼ (M∗0.43)x so
x ≈ 6.97 .
How does this compare to observations? Taking the whole main sequence into
account (0.1 &lt;
∼ M∗ /M⊙ &lt;
∼ 100), a rough fit gives


 R∗ ∼ M∗0.67 
L∗ ∼ M∗3.37

6.54 
L∗ ∼ Teff
See also the H-R diagrams given out in the 1st lecture, and Pols’ Figure 1.3:
Handouts: ZAMS tabular data for stars &amp; brown dwarfs (online).
Also, see my own mass–radius diagram for stars, planets, asteroids, etc.
7
Mass-radius diagram for stars and planets. ZAMS star/brown-dwarf radii from Baraffe et al. (2003), Girardi et al. (2000),
&amp; Ekstrom et al. (2012). Exoplanet radii from exoplanet.eu. WD/NS model curve from Burrows &amp; Liebert (1993).
QCD strange-matter curve from Franzon et al. (2012). BH curve from Schwarzschild radius. Idealized degenerate model
(yellow dash-dotted curve) from Thomas-Fermi theory → n=1.5 polytrope → n=3 polytrope. Solar system objects from
various JPL web pages. Parallel black lines in lower-left are for constant-density models of metallic H, H2 O ice, silicates,
and iron (top to bottom).
8
Note: Our assumption of constant density may make you think that we
1/3
should have gotten M∗ ∝ volume, i.e., R∗ ∼ M∗ .
However, that would only be true if ρ0 was a universal constant for all stars...
which we never assumed.
(This would also occur for ν = 3, too.)
..............................................................................
Above, we assumed that ∂T /∂r was dominated by the radiative gradient.
∂T
T ∂P
∂T
=
∂r
P ∂r
&micro;mH
∂T
GMr ρ
For an ideal gas,
− 2
∂r
ρkB T
r
Note that both T and ρ cancel, and we can use Mr = (4π/3)ρ0r3 and integrate:
Z T (r)
Z r
4π &micro;mH
′
G ρ0
dr′ r′
dT =
Tc
0
This looks very similar to the pressure integral.
Using the surface boundary condition (T = 0 at r = R∗ ) to solve for Tc , like
before, we get
r2
kB T c
T (r) = Tc 1 − 2
=
and
R∗
&micro;mH
2 R∗
For ∇ad = 2/5, the coefficient is 1/5.
Inconsistent with the earlier coefficient of 1/2, which came purely from
hydrostatic equilibrium &amp; the ideal gas law!
What this tells us is that a constant-density, fully convective star is
internally inconsistent as a concept.
However, it’s only off by about a factor of two... &amp; we’ve never let a little
inconsistency stop us before. Let’s use this model to estimate the convective
velocity vc, as well as ∆∇T , in an unstably stratified region.
9
Assuming efficient convection,
Thus,
vc ≈
L∗
3πρ0 R∗2
1/3
L∗
Lr
≈
4πr2
πR∗2
Fconv ≈ C1 ρ0 vc3 ,
But also,
Fconv =
≈
4 L∗R∗
9 M∗
1/3
≈
(at r ≈ R∗/2).
with C1 ∼ 3
0.04 km/s (for the Sun)
2
kB T
vc
|∆∇T |
where c2s =
≈
.
S≈
cs
&micro;mH
r
3
R∗
3GM∗
, T = Tc , and cs ≈
≈ 270 km/s (for the Sun)
At r =
2
4
8R∗
(If we used the convective Tc , then cs ≈ 170 km/s)
Thus, S ≈ 10−4 ≪ 1 , and the actual temperature gradient is ≈ the adiabatic
gradient in an efficient convective zone.
Lastly, we can also see that convection mixes a star very quickly &amp; efficiently.
ℓ
≈ ∼1 week, for the Sun
vc
≪ τK−H (thermal diffusion timescale ∼ 107 yr)
≪ τnuc
(evolutionary timescale ∼ 109 yr)
τmix ≈
10
Let’s move on from the constant-density model to a slightly more realistic
description... Polytropes
Nowadays, they’re far from being state-of-the-art stellar models, but they’re
still extremely useful at conveying insight about stellar physics.
We’ve explored how adiabatic variations can be expressed using γ exponents...
P ∼ ργ
T ∼ ργ−1
Let’s take from that the idea that the actual equation of state might be
written as
P = K ργ
which is no longer a proportionality for adiabatic changes, but now codified
into the constituitive relation for P (ρ).
(We’ve seen actual examples of this: electron degeneracy, and the
Thomas-Fermi model. Not the ideal gas.)
The traditional thing to do is combine the mass conservation &amp; hydrostatic
equilibrium equations into a single 2nd order ODE:
dMr
= 4πr2ρ
dr
dP
GMr
= − 2 ρ
dr
r
GMr
dρ
= − 2 ρ
dr
r
dρ
GMr
ργ−2 r2
= −
dr
Kγ
Taking the derivative of both sides and using mass conservation,
d
dρ
G
4πr2 ρ
ργ−2 r2
= −
dr
dr
Kγ
1 d
4πG
γ−2 2 dρ
which is a 2nd order ODE for ρ(r).
ρ
r
=
−
r2 ρ dr
dr
Kγ
K γ ργ−1
Solving it needs 2 boundary conditions:
ρ(0) ≡ ρc
11
dρ
dr
= 0 .
r=0
Lane &amp; Emden found a useful change of variables:
s
ρ = ρc θ
n
r = αξ
α=
where n = polytropic index
n=
(1/n)−1
(n + 1)K ρc
4πG
1
γ−1
γ =1+
1
n
A lot will cancel out &amp; simplify.
n
d
ρ
dθ
4πG
1
n
c
ρc(1/n)−1 θ1−n ξ 2α2
= −
2
2
n
ξ α ρc θ α dξ
α dξ
K
n+1
Above: α2 ρc cancels.
(1/n)−1
ρc
d 2 1−n n−1 dθ
4πG
n
ξ θ
nθ
= −
α2 ξ 2 θn dξ
dξ
K
n+1
Above: n cancels, and θ1−nθn−1 = 1.
4πG α2
1 d 2 dθ
ξ
= −
(1/n)−1
ξ 2 θn dξ
dξ
K(n + 1)ρc
Recalling definition of α, the whole RHS = −1.
Thus, the standard form of the Lane-Emden equation is
d 2 dθ
ξ
= −ξ 2 θn
dξ
dξ
θ = 1
with boundary conditions
at ξ = 0 .
dθ/dξ = 0
The free parameter n determines the shape of the solution θ(ξ).
Next time we’ll discuss analytic &amp; numerical solutions for θ(ξ), and how they
are useful in specifying stellar parameters.
12
```