Phys 132 Homework 25 Solutions
Professor Crystal Martin
TA: Ellie Hadjiyska
Problem 25: (Phillips Problems 3.7) (white dwarf cooling time) We found the cooling rate of the
white dwarf interior by setting the change in internal energy equal to the luminosity of the envelope
(Phillips Eq. (3.45), p. 101)
dTI
dt =−α·TI
7×107K¸7/2
where
α≈2
3k·12mH
M¯¸L¯≈6 K per year
We want to integrate this to find the cooling time of the white dwarf. Integrating from a very high
tempearature (T=∞) at time zero to a temperature TIat time tgives
ZTI
∞
T−7/2dT =−α(7 ×107K)−7/2Zt
0
dt
which gives
T−5/2
I=5α
2(7 ×107K)−7/2t=5
3k·12mH
M¯¸l¯(7 ×107K)−7/2t
Solving for tgives
t=3kTI
5
1
12mH
M¯
L¯·TI
7×107K¸−7/2
Using HSE and radiative transport with bound-free opacity (κbf ∝ρT −3.5) and applying appro-
priate boundary conditions (e.g. HW 3, problem 4), we have seen before that the pressure and
temperature of a white dwarf is related by
P2
2=CT8.5
I
8.5
where
C=16πacGk
3κ0¯m
M
L
Using the ideal gas law with 2/3 of the particles being electrons (e.g. for ionized helium), we find
the electron density to be
ne=2
3k·C
4.25¸1/2
T13/4
I
If we now assume that the interior temperature TIcorresponds to an electron number density of
ne= 10nQand use the Sun as a standard for mass and luminosity, the white dwarf mass-to-light
ratio is given by (Phillips Eq. (3.42) or (3.43))
M
L≈M¯
L¯·TI
7×107K¸−7/2
