Dodelson - Problem 2.14

Chapter 2 - The Smooth, Expanding Universe

Exercise 14
(a) Compute the pressure of a relativistic species in equilibrium with temperature $T$. Show that $\mathcal{P} = \frac{\rho}{3}$ for both Fermi-Dirac and Bose-Einstein statistics.
(b) Suppose the distribution function depends only on E/T as it does in equilibrium. Find $d\mathcal{P}/dT$. A simple way to do this is to rewrite df/dT in the integral as -(E/T)df/dE and then integrate Eq. (2.62) by parts.

Solution
(a) Recall from Special Relativity that $pc = \sqrt{E^2 - m^2c^4}$. Thus in the relativistic limit where $E >> mc^2$ we have that $pc \approx E$. Since we are employing units where $c = 1$, we have that $$\mathcal{P} = g_i \int \frac{d^3p}{(2\pi)^3}f_i(\vec{x},\vec{p}) \frac{p^2}{3E(p)}$$
$$ = g_i \int \frac{d^3p}{(2\pi)^3}f_i(\vec{x},\vec{p}) \frac{E(p)}{3}$$
$$ = \frac{\rho}{3}$$
as required.

(b) If the distribution function is function of E/T only, we get
$$\mathcal{P} = g_i \int \frac{d^3p}{(2\pi)^3}f_i(E/T) \frac{p^2}{3E(p)}$$.
Since there is no angular dependence we can easily transform to spherical coordinates and write
$$\mathcal{P} = g_i \int \frac{4\pi}{(2\pi)^3}f_i(E/T) \frac{p^4}{3E(p)}dp$$,
where we are integrating over all space. Changing the variable of integration to the energy
$$E^2 = p^2 +m^2$$
$$dp = \frac{E}{p}dE$$
and noting that we are in the relativistic limit where $p \approx E$, we obtain
$$\mathcal{P} = g_i \int \frac{4\pi}{(2\pi)^3}f_i(E/T) \frac{E^3}{3}dE$$.

Differentiating with respect to temperature under the integral sign yields
$$\frac{d\mathcal{P}}{dT} = g_i \int \frac{4\pi}{(2\pi)^3}\frac{\delta f_i(E/T)}{\delta T} \frac{E^3}{3}dE$$
$$ =  -g_i \int \frac{4\pi}{(2\pi)^3}\frac{\delta f_i(E/T)}{\delta E} \frac{E^4}{3T}dE$$
where we made use of the fact that $\frac{df}{dT} = -(E/T)\frac{df}{dE}$.

Now we integrate by parts. Letting
$$u = E^4 -> du = 4E^3 dE$$
$$dv =  \frac{df}{dE} dE -> v = f$$
we obtain
$$\frac{d\mathcal{P}}{dT} = \frac{g_i}{3T} \frac{4\pi}{(2\pi)^3} \left( E^4f \big{|}_{0}^{\infty} - \int_{0}^{\infty} E^4 f dE \right)$$

Notice that  $E^4f \big{|}_{0}^{\infty} = 0$ and so we simply get
$$\frac{d\mathcal{P}}{dT} = \frac{1}{T}\frac{4}{3}\rho$$
$$= \frac{\rho + \mathcal{P}}{T}$$