Worksheet 14.1: Problem 2: 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. 


Kinetic Energy is equal to \(\frac{1}{2}mv^{2}\).
Momentum, \(p\), is equal to \(mv\).
We can then rewrite Kinetic Energy as:
\[K\:=\:\frac{p^{2}}{2m}\]

(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? 


White Dwarfs can be assumed to be perfect blackbodies in this problem, and thus we can use the Viral Theorem to relate kinetic and potential energy. The Virial Theorem tells us that:
\[K\:=\:-\frac{1}{2}U\]

(c)  According to the Heisenberg uncertainty Principle, one cannot know both the momentum and the position of an election such that ∆p∆x > h/4π . 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 ≈ ∆p and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume ~ (∆x)³.) 


 

From the Heisenberg uncertainty Principle, we can assume the following expression:

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

We can then rearrange that to solve for momentum and see that:

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

Now that we have another equation that involves momentum, we can set it equal to the first relationship that we obtained in part a:

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

\[p\:=\:\sqrt{2Km}\]

We can now set our two expressions equal to each other:

\[\frac{h}{4\pi \Delta x}\:=\:\sqrt{2Km}\]

Solving for K, we find that:

\[K\:=\:\frac{h^{2}}{32 \pi^{2} \Delta x^{2} m}\].

Now, we have to express this relationship in terms of the number density in the region of \(\Delta x\_, which seems tricky at first because we don't have an explicit term for the number density given to us. Number Density is equal to:

\[n_{e}\:=\:\frac{Number\:of\:Electrons}{Volume}\]

We know that volume scales like so:

\[Volume\:\sim\:\Delta x^{3}\].

We can then substitute this into our expression for the number density and get:

\[n_{e}\:=\:\frac{Number\:of\:Electrons}{\Delta x^{3}}\]

\[n_{e}\:=\:\frac{1}{\Delta x^{3}}\]

 Next, we can rearrange our expression for K:

\[K\:=\:\frac{h^{2}}{32 \pi^{2} \Delta x^{2} m}\]

\[\Delta x^{2}\:\sim\:n_{e}^{-\frac{2}{3}}\]

\[K\:=\:\frac{h^{2}n_{e}^{\frac{2}{3}}}{32 \pi^{2}m}\]

Since we used terms pertaining to a single electron, this is the expression for the Kinetic Energy of one electron. In order to find the energy of the entire White Dwarf, we will need to know the number of electrons. The number of protons and electrons is about equal in the star, so the number of electrons is as follows:

\[N_{e}\:=\:N_{p}\:=\:\frac{M}{m_{p}}\]

We can multiply the kinetic energy of one electron by this expression in order to get the total kinetic energy of the white dwarf. Thus:

\[K_{tot}\:=\:\frac{h^{2}n_{e}^{\frac{2}{3}}M}{32 \pi^{2}m_{e}m_{p}}\]

 

(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? 


The Virial Theorem is as follows:

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

We know from the previous part that the total kinetic energy of the star is equal to:

\[K_{tot}\:=\:\frac{h^{2}n_{e}^{\frac{2}{3}}M}{32 \pi^{2}m_{e}m_{p}}\]

We can then set these equal to each other and see that:

 \[\frac{h^{2}n_{e}^{\frac{2}{3}}M}{32 \pi^{2}m_{e}m_{p}}\:=\:\frac{GMm_{e}}{2R}\]

Next, we can solve for number density and be left with:

 \[n_{e}\:=\:(\frac{16\pi^{2}Mm_{e}m_{p}}{Rh^{2}})^{\frac{3}{2}}\]

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


Starting with the expression for number density we got in the last part, we can start aggressively dropping constant to get:

\[n_{e}\:=\:(\frac{16\pi^{2}Mm_{e}m_{p}}{Rh^{2}})^{\frac{3}{2}}\]

\[n_{e}\:\sim\:(\frac{M}{R})^{\frac{3}{2}}\]

Then, if we assume that the number density of electrons is equal to the number density of protons, we can arrive at the following relationship:

\[n_{e}\:=\:\frac{Mass}{Volume}\:\sim\:\frac{M}{R^{3}}\]

Thus:

\[\frac{M}{R^{3}}\:\sim\:(\frac{M}{R})^{\frac{3}{2}}\]

Finally, by rearranging, we see that:

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

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


Strangely enough, the above relationship implies that the radius would decrease if you were to add more mass to it. This is contrary to what we have seen in larger, main sequence stars.