1. Degeneracy
In class, we derived:
\[\begin{align*}
P_e &= A f(x) \\
\rho e_e &= A g(x)
\end{align*}\]
with
\[A = \frac{\pi}{3} \left ( \frac{m_e c}{h} \right )^3 m_e c^2\]
and
\[\begin{align*}
f(x) &= x (2x^2 - 3) (1 + x^2)^{1/2} + 3 \sinh^{-1}(x) \\
g(x) &= 8x^3 \left [ (1+ x^2)^{1/2} - 1 \right ] - f(x)
\end{align*}\]
where \(x = p/(m_e c)\) is the Fermi momentum.
a.
We want to expand in the non-relativistic limit (\(x \ll 1\)).
Let’s look at the terms in \(f(x)\) for \(x \ll 1\):
then
\[\begin{align*}
f(x) &\sim (2x^3 - 3x) \left (1 + \frac{x^2}{2} - \frac{x^4}{8} \right ) + 3x - \frac{1}{2} x^3 + \frac{9}{40}x^5 + \ldots \\
&\sim \frac{8}{5}x^5
\end{align*}\]
Note
You need to carry up to \(x^5\) in your expansions because of all of the
cancellations that occur. Otherwise you will miss some factors.
Now, \(g(x)\) expands as:
\[\begin{align*}
g(x) &= 8x^3 \left [ (1+x^2)^{1/2} - 1 \right ] - f(x) \\
&\sim 8x^3 \left [1 + \frac{1}{2}x^2 - 1 \right ] - f(x) \\
&\sim \frac{12}{5} x^5
\end{align*}\]
This gives:
\[\begin{align*}
P_e &\sim A \frac{8}{5} x^5 \\
\rho e_e &\sim A \frac{12}{5} x^5
\end{align*}\]
Now to put in numbers. We know that
\[\frac{\rho}{\mu_e} = \frac{8\pi}{3} m_u \left (\frac{m_e c}{h} \right )^3 x^3\]
or
\[x = \left ( \frac{3}{8\pi m_u} \right )^{1/3} \frac{h}{m_e c} \left ( \frac{\rho}{\mu_e} \right )^{1/3}\]
then
\[P_e = \frac{8\pi}{15} \left ( \frac{3}{8\pi} \right )^{5/3} \frac{h^2}{m_e} \left (\frac{\rho}{\mu_e m_u} \right )^{5/3}\]
putting in numbers, we have:
\[P_e \sim 9.6\times 10^{12} \left ( \frac{\rho / 1~\mathrm{g~cm^{-3}}}{\mu_e} \right )^{5/3} \mathrm{erg~cm^{-3}}\]
and
\[\rho e_e = \frac{3}{2} P_e\]
b.
Now we want the relativistic limit, \(x \gg 1\).
We first note that \(\sinh^{-1}(x)\) is a very slowing varying function
of \(x\) (plot it and you will see). In particular, the leading term of
an expansion of \(\sinh^{-1}(x)\) for \(|x| \gg 1\) is \(\log(2|x|)\) (you
can find this in a table of series expansions). This grows much
slower than the polynomial growth of the other terms, so we can ignore
this.
Then we have (for \(|x| \gg 1\)):
\[f(x) \sim x \cdot \underbrace{(2 x^2)}_{2x^2 -3} \cdot \underbrace{x}_{(1+x^2)^{1/2}} \sim 2x^4\]
and
\[g(x) \sim 8x^3 \cdot \underbrace{x}_{[(1+x^2)^{1/2} - 1]} - f(x) \sim 6 x^4\]
Putting these into our expressions,
\[P_e \sim A 2x^4 = \frac{\pi}{3} \left (\frac{m_e c}{h} \right )^3 m_e c^2 \cdot 2 x^4
= \frac{2\pi}{3} \left ( \frac{3}{8\pi} \right )^{4/3} h c \left ( \frac{\rho}{\mu_e m_u} \right )^{4/3}\]
and putting in numbers
\[P_e \sim 1.2\times 10^{15} \left ( \frac{\rho / 1~\mathrm{g~cm^{-3}}}{\mu_e} \right )^{5/3} \mathrm{erg~cm^{-3}}\]
finally,
\[\rho e_e = \frac{g(x)}{f(x)} P_e = 3 P_e\]
2. Non-zero temperature degeneracy
We want to consider finite-temperature, but non-relativistic degeneracy.
We can do this analytically.
Finite temperature means that the degeneracy parameter,
\[\Psi = \frac{\mu - mc^2}{kT}\]
is not infinite.
a.
We start by writing our number density as the integral of the Fermi-Dirac distribution
over all momentum:
\[n_e = \frac{8\pi}{h^3} \int_0^\infty p^2 \frac{dp}{e^{\mathcal{E}/kT - \Psi} + 1}\]
We then change variable in terms of \(\mathcal{E}\):
\[p = \sqrt{2 m \mathcal{E}} \rightarrow dp = \sqrt{\frac{m}{2\mathcal{E}}} d\mathcal{E}\]
giving
\[n_e = \frac{4\pi}{h^3} (2m)^{3/2} \int_0^\infty \frac{\mathcal{E}^{1/2} d\mathcal{E}}{e^{\mathcal{E}/kt - \Psi} + 1}\]
Finally, defining \(\xi \equiv \mathcal{E}/(kT)\), we have:
\[n_e = \frac{4\pi}{h^3} (2 m k T)^{3/2} \int_0^\infty \frac{\xi^{1/2} d\xi}{e^{\xi - \Psi} + 1} = \frac{4\pi}{h^3} (2mkT)^{3/2} F_{1/2} (\Psi)\]
where we used the definition of the Fermi-Dirac integral given in the problem:
\[F_n(\Psi) = \int_0^\infty \frac{\xi_n}{e^{\xi - \Psi} + 1} d\xi\]
The pressure is done similarly:
\[P_e = \frac{8\pi}{3 h^3} \int_0^\infty p \frac{p}{m} \frac{p^2 dp}{e^{\mathcal{E}/kT - \Psi} + 1}\]
where we used the non-relativistic velocity, \(v = p/m\). Doing the same substitutions, we have:
\[P_e = \frac{8\pi}{3h^2} (2m)^{3/2} \int_0^\infty \frac{\mathcal{E}^{3/2} d\mathcal{E}}{e^{\mathcal{E}/kT - \Psi} + 1} = \frac{8\pi}{3h^3} (2mkT)^{3/2} k T F_{3/2}(\Psi)\]
we see from these relations that:
\[\frac{P_e}{n_e} = \frac{2}{3} k T \frac{F_{3/2}(\Psi)}{F_{1/2}(\Psi)}\]
b.
We now want to use the expansion for \(\Psi \rightarrow \infty\):
\[F_n(\Psi) = \frac{\Psi^{n+1}}{n+1} \left [ 1 + \frac{\pi^2}{6} (n+1) n \Psi^{-2} + \mathcal{O}(\Psi^{-4}) \right ]\]
Keeping only the first term for number density, we have:
\[F_{1/2}(\Psi) \sim \frac{2}{3} \Psi^{3/2}\]
Note
The next term would be \(\propto \Psi^{-1/2}\) which tends to \(0\) for \(\Psi \rightarrow \infty\), which
is why we can ignore it.
This gives us
\[n_e \sim \frac{8\pi}{2h^3} (2m kT)^{3/2} \Psi^{3/2}\]
or
\[\Psi \sim \left ( \frac{3h^3}{8\pi} \right )^{2/3} \frac{n_e^{2/3}}{2 m k T}\]
c.
For the pressure integral, we use
\[F_{3/2}(\Psi) \sim \frac{2}{5} \Psi^{5/2} + \frac{\pi^2}{4} \Psi^{1/2}\]
(again, the next terms tend to zero for \(\Psi \rightarrow \infty\)).
This allows us to write our pressure as:
\[P_e = n_e k T \frac{2}{3} \frac{\frac{2}{5} \Psi^{5/2} + \frac{\pi^2}{4} \Psi^{1/2}}{\frac{2}{3}\Psi^{3/2}}\]
or, substituting in our expression \(\Psi(n_e)\),
\[\begin{align*}
P_e &= n_e k T \frac{2}{3} \left [ \frac{3}{5} \left ( \frac{3h^3}{8\pi} \right )^{2/3} \frac{n_e^{2/3}}{2mkT} +
\frac{3}{8} \pi^2 \left (\frac{3h^3}{8\pi} \right )^{-2/3} \frac{2mkT}{n_e^{2/3}} \right ] \\
&= \frac{h^2}{20m} \left ( \frac{3}{\pi} \right )^{2/3} n_e^{5/3} \left [ 1 + \frac{40\pi^2}{(3/\pi)^{4/3}} \frac{m^2}{h^4} \frac{(kT)^2}{n_e^{4/3}} \right ]
\end{align*}\]
We see that the first term is the zero-temperature expression we derived in class for non-relativistic electron degeneracy. The second term is the finite-temperature
correction.
3. Intensity vs. flux
a.
We want to compare the intensity leaving a source to that received by a detector.
The energy leaving a source with area \(dA\) in direction \(\theta\) into a cone \(d\Omega\) is:
\[dE^\mathrm{emit} = I \cos \theta dA d\Omega dt\]
this is received by a detector a distance \(r\) away, with a detector area \(dA^\prime\).
That means the solid angle of the detector as seen by the source is:
\[d\Omega = \frac{dA^\prime \cos\theta^\prime}{r^2}\]
where the \(\cos\theta^\prime\) is the projection of the detector’s \(dA^\prime\) onto the
line of sight. This means that
\[dE^\mathrm{emit} = I \cos \theta dA \frac{dA^\prime \cos\theta^\prime}{r^2} dt\]
Now the detector receives an energy
\[dE^\mathrm{recv} = I^\prime \cos\theta^\prime dA^\prime d\Omega^\prime dt\]
where \(d\Omega^\prime\) is the solid angle subtended by the emitter as seen by the
detector. This is just
\[d\Omega^\prime = \frac{dA \cos \theta}{r^2}\]
so
\[dE^\mathrm{recv} = I^\prime \cos\theta^\prime dA^\prime \frac{dA \cos\theta}{r^2} dt\]
Finally, since \(dE^\mathrm{emit} = dE^\mathrm{recv}\), we see that \(I = I^\prime\).
b.
The flux emitted by our source is:
\[f = \int_\Omega I\cos\theta d\Omega = B \int \cos \theta \sin\theta d\theta d\phi\]
where we used \(I = B\) is constant.
We need to figure out our integration limits. The sphere will subtend an angle \(\theta_c\)
as seen some distance \(d\) away, with
\[\theta_c = \tan^{-1} \frac{R}{d} \sim \sin^{-1} \frac{R}{d}\]
then the integration limits are:
\[f = B \int_{\theta=0}^{\theta=\theta_c} \int_{\phi=0}^{\phi=2\pi} \cos\theta \sin\theta d\theta d\phi\]
changing variables, \(\xi = \sin\theta \rightarrow d\xi = cos\theta d\theta\), we have:
\[f = 2\pi B \int_0^{R/d} \xi d\xi = \pi B \left ( \frac{R}{d} \right )^2\]
So we see that the flux falls off as \(\sim 1/d^2\).