Heading Inside a White Dwarf: Stellar Scaling Relationships, Part III

A white dwarf can be considered a gravitationally bound system of massive particles.

a) What is the relationship between the total kinetic energy of the electrons that are supplying the pressure in a white dwarf, and the total gravitational energy of the WD?

What we have here is an application of our old friend the Virial Theorem, $1/2U=KE$. The total kinetic energy of all of the electrons in the white dwarf is therefore the sum of all of the electrons' individual kinetic energies (bear in mind that these are not all of the particles in the WD, and thus their total kinetic energy is not the total kinetic energy within the WD--this'll be important to remember later). If we assume $N$ electrons, this gives us $$-\frac{3GM^2}{5R} = Nm_{e^-}v_{e^-}^2$$

b) According to the Heisenberg uncertainty Principle, one cannot know both the momentum and position of an election such that $\Delta p \Delta x > \frac{h}{4\pi}$. Use this to express the relationship between the kinetic energy of electrons and their number density $n_{e^-}$.

Our first step is going to be to write the kinetic energy of the electrons in terms of their momentum, similar to what we did in part c) of the last post. This gives us $$KE = \frac{Np_{e^-}^2}{2m_{e^-}^2}$$ as the kinetic energy of all of the electrons. We can then substitute in $\Delta p_{e^-}$ for $p_{e^-}$, and now we'll stop worrying too much about constants and equations and move into the realm of scaling relationships. Assuming we are approaching the limit where $\Delta p_{e^-} \Delta x = \frac{h}{4\pi}$, we know that $\Delta p_{e^-} \sim \frac{1}{\Delta x}$ (everything else in there is just a constant). Next, we know that number density $n_{e^-}$ is equal to $\frac{N}{V}$, and can approximate $V$ as $\Delta x ^3$, so $n_{e^-} \sim \Delta x ^{-3}$. We can then relate $n_{e^-}$ and $\Delta p_{e^-}^2$ as they appear in our equation as $\Delta p_{e^-} ^2 \sim n_{e^-}^{2/3}$. Finally, we can substitute $\frac{M}{m_{e^-}}$ in for $N$, giving us a final relation between the kinetic energy of electrons and their number density: $$KE=\frac{Mn_{e^-}^{2/3}}{m_{e^-}^2}$$ Since $m_{e^-}$ is a constant, the scaling relationship here is $KE \sim Mn_{e^-}^{2/3}$.

c) What is the relationship between $n_{e^-}$ and the mass $M$ and radius $R$ of a WD?

Once again, we know that $n_{e^-} = \frac{N}{V}$, and that $N = \frac{M}{m_{e^-}}$. Volume, of course, is $4/3\pi R^3$. Dropping the constants $m_{e^-}$ and $4/3\pi$, we have $$n_{e^-} \sim \frac{M}{R^3}$$

d) Substitute back into your Virial energy statement, aggressively yet carefully drop constants, and relate the mass and radius of a WD.

Remembering from part b) that $KE \sim Mn_{e^-}^{2/3}$, and from part c) that n_{e^-} \sim \frac{M}{R^3}, we know that $$KE \sim \frac{M^{5/3}}{R^2}$$ We can plug this into the right side of our Virial statement, giving us $$-\frac{3GM^2}{5R} \sim \frac{M^{5/3}}{R^2}$$ As for the left side, it is a simple matter to start aggressively dropping constants ($3/5$ and $G$) and reveal that the left side scales as $\frac{M^2}{R}$. If we solve for $M$ in the resulting scaling relation, we find that $$M\sim\frac{1}{R^3}$$ This is actually a pretty fascinating result, which tells us that the more massive a WD gets, the smaller it is. While initially extremely unintuitive, this actually makes a fair amount of sense when we consider the process of gravitational contraction that has made the WD so small to begin with, and the electron degeneracy pressure that is preventing it from collapsing into a neutron star. The heavier it is, the harder it will be for degenerate electrons to hold the WD up, and the more it will contract under the inexorable force of gravity.

Credit and thanks to Anne Madoff, Scott Zhuge, Louise Decoppet, and Jennifer Shi to working through to this surprising and cool result with me.