pervect said:
I did some more looking, and finally did run across a webpage which talked about the white dwarf star radius vs mass relation.
As I remembered, it was weird. Larger white dwarfs have a smaller radius!
It's also not dependent on temperature.
Yes, in fact, the mass-radius relationship of white dwarfs is an excellent instructive tool because it can be very approximately derived by combining a basic knowledge of several areas of physics. In order to determine it, one must first consider the equation of state of degenerate matter. The pressure will depend only on the density of the material as long as:
kT << E_F
Why? Think crudely about the electrons as "trying" to fit themselves into a Maxwell-Boltzmann distribution, but failing because there are only so many states available in position-momentum space. Specifically, the exclusion principle limits them on the low-momentum end, so a degenerate gas will tend to fill up all of the states available between zero momentum and the fermi momentum. In practice, there will always be a high-energy tail, but one can approximately think of it as a filled sphere in momentum space; that is, the density is given by:
n_e=\int_0^{p_F}f(p)dp\propto p_F^3
Likewise, I can find the energy density of degenerate gas with simply:
U_e=\int_0^{p_F}E(p)f(p)dp
In general, the pressure and energy density are non-trivially related, but to a rough approximation, one can usually say
P_e \propto U_e
Given these things, we now have the tools necessary to derive a scaling relation for the equation of state; that is:
P_e \propto \int_0^{p_F}E(p)f(p)dp
There are two limits that are of interest: relativistic and non-relativistc. In the non-relativistic limit, one gets
E(p)=\frac{p^2}{2m}\propto p^2
In the relativistic limit, it is instead:
E(p)=pc\propto p
Substituting these into my above equation and combing with my first equation, we get:
P_e\propto p_F^5 \propto n_e^{5/3} Non-relativistic degeneracy pressure
P_e \propto p_F^4 \propto n_e^{4/3} Relativistic degeneracy pressure
What does all of that have to do with the mass-radius relationship? Well, imagine we combine this with some elementary gravitational physics. That is, let's recall hydrostatic equilibrium:
\frac{dP}{dR} \sim \frac{P}{R} = -\rho_e g = -\rho_e \frac{GM}{R^2}
P \propto \frac{n_eM}{R}
Plugging the equations of state into this and considering that
n_e \propto \frac{M}{R^3},
we finally have the mass-radius relationships for non-relativistic and relativistic degeneracy pressure:
R \propto M^{-1/3} Non-relativistic
M \propto constant Relativistic
The first is the mass-radius relation you noted, and the radius does indeed decrease with mass. Notice that for the relativistic case, however, the mass/radius go to a constant. If derived in detail, it turns out that this will give you the famous Chandrasekhar mass!
