Worksheet 14.1, Problem 2: The Mass-Radius Relation for White Dwarfs

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

(a)  Express the kinetic energy of a particle of mass m in terms of its momentum p instead of the usual notation using its speed v.
(b)  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?
(c)  According to the Heisenberg uncertainty Principle, one cannot know both the momentum and position of an election such that \(\Delta x \Delta p >\frac{h}{4\pi}\). Use this to express the relationship between the  kinetic energy of electrons and their number density ne (Hint: what is the relationship between an object’s kinetic energy and its momentum? From here, assume p = ∆p and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume ~ (∆x)3.)
(d)  Substitute back into your Virial energy statement. What is the relationship between ne and the mass M and radius R of a WD?
(e)  Now, aggressively yet carefully drop constants, and relate the mass and radius of a WD.
(f)  What would happen to the radius of a white dwarf if you add mass to it? 

(a)  Express the kinetic energy of a particle of mass m in terms of its momentum p instead of the usual notation using its speed v.

Let's start with kinetic energy, where \(v\) is velocity, \(K\) is kinetic energy, and \(m\) is mass.

\[K=\frac{1}{2}mv^2\]

Momentum (\(p\)) is \(p=mv\). So:

\[\boxed{K=\frac{p^2}{2m}}\]

(b)  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?

We are back to once again, you guessed it, the one and only: Virial Theorem.

\[K=-\frac{1}{2}U\]

\[\frac{1}{2}Mv^2=\frac{GM^2}{2R}\]

\[N_e m_e v^2=\frac{GM^2}{R}\]

Or:

\[\frac{N_e p^2}{2m_e}=\frac{GM^2}{2R}\]

(c)  According to the Heisenberg uncertainty Principle, one cannot know both the momentum and position of an election such that \(\Delta x \Delta p >\frac{h}{4\pi}\). Use this to express the relationship between the  kinetic energy of electrons and their number density ne (Hint: what is the relationship between an object’s kinetic energy and its momentum? From here, assume p = ∆p and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume ~ (∆x)3.)

We will use the cutoff case for this problem, so:

\[\Delta x \Delta p =\frac{h}{4\pi}\]

We are told that \(\Delta p \approx p\), so:

\[p\Delta x =\frac{h}{4\pi}\]

We are also told that an election takes up the space \(\Delta x^3\), so the number density of electrons can be shown as:

\[n_e=\frac{1}{\Delta x^3}\]
We can solve this for \(\Delta x\) and insert it into our Heisenberg equation:

\[\frac{p}{n_e^{\frac{1}{3}}} =\frac{h}{4\pi}\]

Earlier we established that \(K=\frac{N_e p^2}{2m_e}\)

We can solve this for \(p\).

\[p=\sqrt{\frac{2m_eK}{N_e}}\]

Now we can throw this into the Heisenberg:

\[\frac{\sqrt{\frac{2m_eK}{N_e}}}{n_e^{\frac{1}{3}}} =\frac{h}{4\pi}\]

Let's clean up some exponents:

\[\frac{2m_eK}{N_e n_e^{\frac{2}{3}}} =\frac{h^2}{16\pi^2}\]
Finally let's solve for K.

\[K=\frac{n_e^{\frac{2}{3}}h^2 N_e}{32m_e\pi^2}\]

(d)  Substitute back into your Virial energy statement. What is the relationship between ne and the mass M and radius R of a WD?

Our statements is that:

\[K=-\frac{1}{2}U\]

\[\frac{n_e^{2/3}h^2 N_e}{32m_e\pi^2}=\frac{GM^2}{2R}\]

Let's get rid of the fractions and simplify:

\[n_e^{2/3}h^2 N_e R=16GM^2 m_e\pi^2\]

This is truly messy, but it's the best we can get.

(e)  Now, aggressively yet carefully drop constants, and relate the mass and radius of a WD.

Now here is the fun part.

\[n_e^{2/3}h^2 N_e R=16GM^2 m_e\pi^2\]

\[n_e^{2/3} N_e R=M^2\]

We need to get rid of \(n_e\), so we should remember that for every electron in the star, there is a proton with mass.  So:

\[n_e \sim \frac{M}{R^3}\]

This makes sense because a number density whose item has mass is pretty much regular density with the mass ignored.  Luckily the mass is just a constant, so we can drop that.

\[n_e^{2/3} \sim \frac{M^{2/3}}{R^2}\]

Let's substitute this in:

\[\frac{M^{2/3}N_e}{R^2} R\sim M^2\]

\[\frac{N_e}{R} \sim M^{4/3}\]

We already established that proton number is tied to election count, so \(N_e \sim M\).

\[\frac{M}{R} \sim M^{4/3}\]

\[R \sim M^{-1/3}\]

Or:

\[\boxed{M \sim \frac{1}{R^3}}\]

(f)  What would happen to the radius of a white dwarf if you add mass to it? 
Since the Mass and Radius are inversely proportionate (with cube thrown in),  if you added mass, the radius would actually decrease.

I worked with G. Grell and N. James on this problem.