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

A white dwarf (WD) 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 \(v\).

We know from the general form of the equation for kinetic energy that:

\(K=\frac{1}{2} mv^2\)

We also know that momentum is given as:

\(p=mv\)

Rewriting kinetic energy to be in terms of momentum instead of speed, we get the equation:

\(K=\dfrac{p^2}{2m}\)

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?

Since this is a gravitationally bound system, we know that it is virtualized, which allows us to use Viral theorem.

\(K=-\frac{1}{2}U\)

Where \(U\) is the total gravitational energy. We can write an expression for the total kinetic energy and the total gravitational energy. In part (a), we already have an equation for the kinetic energy of a single particle of mass m. If we want to find the total kinetic energy of the electrons in the white dwarf, we need to multiply this by the number of particles. So we will get a total kinetic energy of:

\(K=\dfrac{p^2}{2m_e}N\)

Where N is the number of particles. For the total gravitational energy, we have already derived in a previous problem that the total gravitational potential energy for a sphere is given as:

\(U=-\dfrac{3GM^2}{5R}\)

Putting these expressions for total kinetic energy and potential energy into the Virial equation, we get:

\(\dfrac{p^2}{m_e}N=\dfrac{3GM^2}{5R}\)

c) 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 \(n_e\). (Hint: what is the relationship between an object’s kinetic energy and its momentum? From here, assume \(p\approx \Delta p\) and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume\(\sim (\Delta x)^3\).)

We begin with the expression for total kinetic energy (for \(N_e\) electrons) in part (b):

\(K=\dfrac{p^2}{2m_e}N_e\)

If we make the approximation that \(p\approx \Delta p\) for the electron, then this equation becomes:

\(K=\dfrac{\Delta p^2}{2m_e}N_e\)

Furthermore, to find an expression for the number of electrons, \(N_e\), we can assume that the number of protons, \(N_p\) in the white dwarf is equal to the number of electrons, \(N_e\). Since the mass of protons is much larger than the mass of electrons, we can make the approximation that \(N_p\approx N_e\approx \frac{M}{m_p}\), where M is the total mass of the white dwarf. Putting this into our equation, we get:

\(K=\dfrac{\Delta p^2}{2m_e}(\dfrac{M}{m_p})\)

From here, we can introduce the Heisenberg uncertainty principle. If we solve for \(\Delta p\), we find \(\Delta p\sim \frac{1}{\Delta x}\). Also, we know that number density is equal to number per volume. If we approximate the volume occupied by a single electron to be \(\Delta x^3\), then we find that \(n_e=\frac{1}{\Delta x^3}\), or \(\Delta x=\frac{1}{n_e^{1/3}}\). So this means that we have the relation \(\Delta p\sim n_e^{1/3}\). We can rewrite our expression for kinetic energy to be:

\(K=\dfrac{n_e^{2/3}}{2m_e}(\dfrac{M}{m_p})\)

Dropping constants to find a proportionality relationship between kinetic energy and number density, we get an answer of:

\(K\sim Mn_e^{2/3}\)

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

From Virial theorem, which says that the total kinetic energy is equal to half of the total potential energy, we can plug in the relation from part (c) to get:

\(Mn_e^{2/3}\sim \dfrac{3GM^2}{5R}\)

Which gives us a relationship between \(n_e\), M, and R of:

\(n_e^{2/3}\sim \dfrac{M}{R}\)

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

From part (d) we found a relationship between the number density, mass, and radius of a white dwarf. We also know, however, that the number density of electrons in the white dwarf is the number per volume:

\(n_e\sim \dfrac{N_e}{V}\)

In part (c), we made the claim that \(N_e\approx \frac{M}{m_p}\), and the volume of the white dwarf would be \(V=\frac{4}{3}\pi R^3\). So we can rewrite number density in terms of the mass and the radius of the white dwarf:

\(n_e\sim \dfrac{\frac{M}{m_p}}{\frac{4}{3}\pi R^3}\)

Simplifying this into a scaling relation:

\(n_e\sim \dfrac{M}{R^3}\)

Now plugging this into our relationship from part (d), we get:

\(\left(\dfrac{M}{R^3}\right)^{2/3}\sim \dfrac{M}{R}\)

Now, we can solve for a scaling relationship between the mass and the radius:

\(M\sim \dfrac{1}{R^3}\)

f) What would happen to the radius of a white dwarf if you add mass to it?

Interestingly, we found from part (e) that the mass is inversely proportional to the radius cubed, which means that a white dwarf with a larger mass will have a smaller radius. This makes sense because the mechanism which is keeping the white dwarf from collapsing on its own weight is the electron degeneracy pressure. As we increase mass, the gravitational force becomes stronger at a faster rate than the force due to pressure from electrons, so the electrons can't hold up the white dwarf as much, causing a decrease in size.


I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.