Download PS #1 Solutions - Stars and Stellar Explosions 1. Opacity sources

PS #1 Solutions - Stars and Stellar Explosions
1. Opacity sources and L-M relationship [35 points]
In class I derived the relationship between the luminosity and mass of stars under the
assumption that energy is transported by radiative diffusion and that the opacity is due
to Thompson scattering. We will carry out many related estimates during this course
so it is important to become familiar with this process. Consider a star in hydrostatic
equilibrium in which energy transport is by radiative diffusion. The star is composed of
ionized hydrogen and is supported primarily by gas pressure.
a) Derive an order of magnitude estimate of the luminosity L of a star of mass M and
radius R if the opacity is due to free-free absorption, for which κ ≈ 1023 ρT −7/2 cm2 g−1
(ρ is in cgs).
b) If all stars have roughly the same central temperature, and are supported by gas
pressure, what is the mass-luminosity scaling (proportionality) relationship for stars?
c) Give a quantitative argument as to whether free-free opacity dominates electron scattering opacity in stars more massive than the sun or in stars less massive than the sun.
SOLUTION:
We know that the radiative diffusion equation gives us
4caT 3 dT
.
F~ = −
3κρ dr
Since everything is radially symmetric, we switch to radial coordinates. We can set the
total flux leaving the star as F = L/4πR2 , and approximate dT /dr ∼ −T /r. The average
density is given by ρ = 3M/4πR3 . We can use the expression for κ given in the question.
Thus we have
L
4πR2
∼
L ∼
4caT 4 4πR3
3κR 3M
2
16πRcaT 15/2 4πR3
.
3 × 1023
3M
A star in hydrostatic equilibrium obeys the virial theorem which implies kT ∼ GM µmp /3R.
Substituting this (i.e. T ∼ GM µmp /3Rk where µ ∼ 0.5, and specifically µ = 0.6 for solar
metallicity) into the above expression gives
L=
43 π 3 ac
3
3 × 1023
Gmp µ
3k
15/2
M 11/2 R−1/2 ∼ 1035
M
M
11/2 R
R
−1/2
ergs s−1
Our estimate is about 10 x too high for the sun but that’s reasonable given the simplicity
of the estimate. Note that only a factor of 2 difference in the estimate of temperature
(e.g., how to estimate the potential energy) give rise to a factor of 215/2 = 181 difference
in L, thus, I gave full credit for any reasonable values for L between 1033 and 1035
(b)
(a) If the temperature is constant and pressure is dominated by gas pressure, then hydrostatic
equilibrium gives us
dP
dr
nkT
R
=
∼
1
GM ρ
r2
GM ρ
R2
−
R
∼
R
∝
GM µmp
kT
M.
Plugging this proportionality in our expression for L above we get
L ∝ M 5.
(c) Since R ∝ M (temperature is constant assumption), then density scales like ρ ∝ M/R3 ∝
M −2 . Since the free-free opacity scales like κff ∝ ρ at constant temperature, we get
κff ∝ M −2 . Because of this inverse proportionality, we expect free-free absorption to get
more and more important in lower mass stars.
2. Verifying Ideal Gas Law Assumption [35 points]
The density of the interior of the sun is significantly larger than that of water, for example,
the density is ∼ 150 g/cm3 near the center of the sun. Why, then, do we treat stellar interiors
as an ideal gas? The important physics here is as follows: in a gas, interparticle forces are
typically unimportant (unless particles happen to get very close to each other) while in a liquid,
interparticle forces are important even for when particles are at a “typical” separation from
each other. For an ionized plasma, the relevant interparticle force is the Coulomb electric force.
So what this problem is asking you to consider is whether, given the density and temperature
at the center of the sun, the Coulomb force between particles is important or whether the ideal
gas law is a reasonable assumption. Note that you do not need any quantum mechanics for
this problem. It’s purely classical.
a) Provide a quantitative relation between the temperature and density of a star which indicates
when we can treat it as a gas (rather than a liquid) throughout its interior, in spite of the very
high densities. Is our assumption valid at the center of the sun?
b) If all stars have roughly the same central temperature (that of the sun), use a scaling
argument to estimate the stellar mass at which the simple non-interacting ideal gas assumption
breaks down.
SOLUTION:
(a) An ideal gas is one in which the particles are far apart from one another, and one in which
the thermal energy far out weighs any Coulomb potential energy between particles. Thus,
we require:
kT |q 2 /r|.
Since we are assuming ionized hydrogen, q is either the charge of a proton or an electron
(±4.8 × 10−10 esu), and the interparticle distance r can be thought of as n−1/3 , where n
is the number density. Therefore, we require:
T 1.7 × 10−3 n1/3 .
For typical solar values at the center, ρc ∼ 150 g/cm3 . Assuming ionized hydrogen,
m ' mp /2, this translates into n ∼ 1.5 × 1026 cm−3 . Thus we require
T 106 K.
We know, however, that the solar interior has T ∼ 107 K, and so our condition is satisfied
and we can treat it as an ideal gas.
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(b) Let’s assume something very simple like n ∝ M/R3 . If the temperature doesn’t change
in the star as a function of mass, then from hydrostatic equilibrium we get
dP
dr
P
R
ρT
MT
R3
R
GMr ρ
r2
2
M
R5
M2
R4
M2
R4
M.
= −
∝
∝
∝
∝
Therefore, n ∝ M −2 , and thus to get comparable thermal and Coulomb energies, you
need a mass of about 0.03M , which is getting close to dwarf stars or giant planets.
You could also use a rough mass-radius relation for main sequence stars, R ∝ M 0.75 ,
to get n ∝ M −1.25 . To get kT ∼ q 2 /r, you would need a mass of ∼ 4 × 10−3 M , which
is, again, giant planet size. Of course, the mass-radius relation for main sequence stars
doesn’t extend to such low masses, and in reality main sequence stars have slightly varying compositions and temperatures, so this isn’t fully accurate. But the idea is the same.
The assumption of an ideal gas breaks down for very low mass stars and giant planets.
There’s yet another way to solve this, not directly assuming that the central temperature is constant, but just that ideal gas pressure is dominant. We determined that the
ideal condition will breakdown when kT ∼ q 2 /r. If we multiply both sides by the number
of particles in the star N ∼ M/mp , then we have N kT ∼ (M/mp )q 2 /r. The LHS is the
thermal energy of the star, which from the virial theorem is ∼ GM 2 /R, the gravitational
potential energy. So then, since r ∼ n−1/3 ∼ R (M/mp )−1/3
GM 2
(M/mp )4/3 q 2
∼
R
R
So the radius cancels, and we can write
−3/2
M ∼ q 3 m−2
∼ 10−3 M
p G
which nice since it only involves fundamental constants.
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