Homework 2 solutions#

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 \(x \ll 1\):

  • \((1 + x^2)^{1/2} \sim 1 + \frac{x^2}{2} - \frac{x^4}{8} + \ldots\)

  • \(\sinh^{-1}(x) \sim x - \frac{x^3}{6} + \frac{3x^5}{40} + \ldots\)

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,

\[\eta = \frac{\mu - mc^2}{kT}\]

is not infinite.

a.#

We start by writing our number density as the integral of the Fermi-Dirac distribution over all momentum:

\[n_e = \frac{8\pi}{h^3} \int_0^\infty p^2 \frac{dp}{e^{\mathcal{E}/kT - \eta} + 1}\]

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

\[n_e = \frac{4\pi}{h^3} (2m)^{3/2} \int_0^\infty \frac{\mathcal{E}^{1/2} d\mathcal{E}}{e^{\mathcal{E}/kt - \eta} + 1}\]

Finally, defining \(\xi \equiv \mathcal{E}/(kT)\), we have:

\[n_e = \frac{4\pi}{h^3} (2 m k T)^{3/2} \int_0^\infty \frac{\xi^{1/2} d\xi}{e^{\xi - \eta} + 1} = \frac{4\pi}{h^3} (2mkT)^{3/2} F_{1/2} (\eta)\]

where we used the definition of the Fermi-Dirac integral given in the problem:

\[F_n(\eta) = \int_0^\infty \frac{\xi^n}{e^{\xi - \eta} + 1} d\xi\]

The pressure is done similarly:

\[P_e = \frac{8\pi}{3 h^3} \int_0^\infty p \frac{p}{m} \frac{p^2 dp}{e^{\mathcal{E}/kT - \eta} + 1}\]

where we used the non-relativistic velocity, \(v = p/m\). Doing the same substitutions, we have:

\[P_e = \frac{8\pi}{3h^2} (2m)^{3/2} \int_0^\infty \frac{\mathcal{E}^{3/2} d\mathcal{E}}{e^{\mathcal{E}/kT - \eta} + 1} = \frac{8\pi}{3h^3} (2mkT)^{3/2} k T F_{3/2}(\eta)\]

we see from these relations that:

\[\frac{P_e}{n_e} = \frac{2}{3} k T \frac{F_{3/2}(\eta)}{F_{1/2}(\eta)}\]

b.#

We now want to use the expansion for \(\eta \rightarrow \infty\):

\[F_n(\eta) = \frac{\eta^{n+1}}{n+1} \left [ 1 + \frac{\pi^2}{6} (n+1) n \eta^{-2} + \mathcal{O}(\eta^{-4}) \right ]\]

Keeping only the first term for number density, we have:

\[F_{1/2}(\eta) \sim \frac{2}{3} \eta^{3/2}\]

Note

The next term would be \(\propto \eta^{-1/2}\) which tends to \(0\) for \(\eta \rightarrow \infty\), which is why we can ignore it.

This gives us

\[n_e \sim \frac{8\pi}{2h^3} (2m kT)^{3/2} \eta^{3/2}\]

or

\[\eta \sim \left ( \frac{3h^3}{8\pi} \right )^{2/3} \frac{n_e^{2/3}}{2 m k T}\]

c.#

For the pressure integral, we use

\[F_{3/2}(\eta) \sim \frac{2}{5} \eta^{5/2} + \frac{\pi^2}{4} \eta^{1/2}\]

(again, the next terms tend to zero for \(\eta \rightarrow \infty\)).

This allows us to write our pressure as:

\[P_e = n_e k T \frac{2}{3} \frac{\frac{2}{5} \eta^{5/2} + \frac{\pi^2}{4} \eta^{1/2}}{\frac{2}{3}\eta^{3/2}}\]

or, substituting in our expression \(\eta(n_e)\),

\[\begin{align*} P_e &= n_e k T \frac{2}{3} \left [ \frac{3}{5} \left ( \frac{3h^3}{8\pi} \right )^{2/3} \frac{n_e^{2/3}}{2mkT} + \frac{3}{8} \pi^2 \left (\frac{3h^3}{8\pi} \right )^{-2/3} \frac{2mkT}{n_e^{2/3}} \right ] \\ &= \frac{h^2}{20m} \left ( \frac{3}{\pi} \right )^{2/3} n_e^{5/3} \left [ 1 + \frac{40\pi^2}{(3/\pi)^{4/3}} \frac{m^2}{h^4} \frac{(kT)^2}{n_e^{4/3}} \right ] \end{align*}\]

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. Adiabatic index#

We want to compute

\[\Gamma_1 = \left . \frac{d\log P}{d\log \rho} \right |_s = \frac{\rho}{P} \left . \frac{dP}{d\rho} \right |_s\]

for a gas composed of a mix of an ideal gas and radiation:

\[P = \frac{1}{3} a T^4 + \frac{\rho k T}{\mu m_u}\]

The corresponding specific energy is:

\[e = \frac{a T^4}{\rho} + \frac{3}{2} \frac{kT}{\mu m_u}\]

We start by writing our EOS as \(P = P(\rho, T(\rho, s))\) and then takking the derivative with respect to density:

\[\left . \frac{dP}{d\rho} \right |_s = \left . \frac{\partial P}{\partial \rho} \right |_T + \left . \frac{\partial P}{\partial T} \right |_\rho \left . \frac{dT}{d\rho} \right |_s \]

Now, from the first law of thermodynamics, we take entropy to be constant:

\[\begin{align*} dq = 0 &= de + P d\left ( \frac{1}{\rho} \right ) \\ &= \left . \frac{\partial e}{\partial T} \right |_\rho dT + \left . \frac{\partial e}{\partial \rho} \right |_T d\rho - \frac{P}{\rho^2} d\rho \end{align*}\]

where we expanded out \(de\) in terms of \(T\) and \(\rho\). This shows us that:

\[\left . \frac{dT}{d\rho} \right |_s = \left ( \left . \frac{\partial e}{\partial T} \right |_\rho \right )^{-1} \left (\frac{P}{\rho^2} - \left . \frac{\partial e}{\partial \rho} \right |_T \right )\]

Now we need to compute all the derivatives. From our equation of state, we have:

\[\left . \frac{\partial P}{\partial \rho} \right |_T = \frac{kT}{\mu m_i} = \frac{P_g}{\rho}\]
\[\left . \frac{\partial P}{\partial T} \right |_\rho = \frac{4}{3} a T^3 + \frac{\rho k}{\mu m_u} = \frac{1}{T} \left ( 4 P_\gamma + P_g \right )\]
\[\left . \frac{\partial e}{\partial \rho} \right |_T = - \frac{aT^4}{\rho^2} = -\frac{3P_\gamma}{\rho^2}\]
\[\left . \frac{\partial e}{\partial T} \right |_\rho = 4 \frac{aT^3}{\rho} + \frac{3}{2} \frac{k}{\mu m_u} = 12 \frac{P_\gamma}{\rho T} + \frac{3}{2} \frac{P_g}{\rho T}\]

Then inserting these into the above expression for \(dT/d\rho |_s\), we have:

\[\left . \frac{dT}{d\rho} \right |_s = \frac{T}{\rho} \frac{1 + 3 (1-\beta)}{12 (1- \beta) + \frac{3}{2}\beta} = 2 \frac{T}{\rho} \frac{4 - 3\beta}{24 - 21\beta}\]

and finally:

\[\begin{align*} \left . \frac{dP}{d\rho} \right |_s &= \frac{\beta P}{\rho} + \frac{P}{T} \left [ 4(1-\beta) + \beta\right ] 2 \frac{T}{\rho} \frac{4 - 3\beta}{24 - 21\beta} \\ &=\frac{P}{\rho} \frac{32 - 24\beta -3 \beta^2}{24 - 21\beta} \end{align*}\]

so

\[\Gamma_1 = \frac{32 - 24\beta -3 \beta^2}{24 - 21\beta}\]

We see that this has the proper limits:

  • Pure gas pressure: \(\beta = 1 \rightarrow \Gamma_1 = 5/3\)

  • Pure radiation pressure: \(\beta = 0 \rightarrow \Gamma_1 = 4/3\)