I am completely at a loss in deriving the equation 2.62 given in Scott Dodelson's Modern Cosmology book. That is the following expression for the pressure of Fermi-Dirac and Bose-Einstein distributions,

$$P_i= g_i\int \frac{d^3p}{(2\pi)^3} f_i(\vec{x},\vec{p}) \frac{p^2}{3E(p)},$$


$$ f_i(\vec{x},\vec{p}) = \frac{1}{e^{(E-\mu)/T}\pm 1}. $$ I am rather weak in statistical physics and to derive this expression I don't even have clue for where to begin. So please answer on how to derive this. A full derivation may not be needed, hints on where to begin and what quantities to compute will also work. Or even a reference to books, papers are also welcome.


1 Answer 1


Think of the energy density and the pressure as the diagonal elements of the stress-energy tensor of an isotropic gas, for an observer locally at rest. Each element $T^{\alpha\alpha}$ represents the flux of the $\alpha$-component of the energy-momentum vectors of all particles in the $\alpha$-direction (see also this question). In other words, we can write

$$ T^{\alpha\alpha}(\vec{x}) = \int v^\alpha p^\alpha\, n(\vec{x},\vec{p})\,\text{d}^3\vec{p}, $$ where $v^\alpha = (c,\vec{v})$ is the velocity vector, $p^\alpha = (E/c,\vec{p})$ is the energy-momentum vector, and

$$n(\vec{x},\vec{p}) = \frac{g}{(2\pi\hbar)^3}\frac{1}{e^{[E(p)-\mu]/kT}\pm 1} = \frac{g}{(2\pi\hbar)^3}f(\vec{x},p)$$

is the phase-space number density. Therefore,

$$\begin{align} T^{00} &= \rho c^2 = \frac{g}{(2\pi\hbar)^3}\int E(p)\, f(\vec{x},p)\,\text{d}^3\vec{p},\\ T^{11} &= \frac{g}{(2\pi\hbar)^3}\int v_xp_x\, f(\vec{x},p)\,\text{d}^3\vec{p},\\ T^{22} &= \frac{g}{(2\pi\hbar)^3}\int v_yp_y\, f(\vec{x},p)\,\text{d}^3\vec{p},\\ T^{33} &= \frac{g}{(2\pi\hbar)^3}\int v_zp_z\, f(\vec{x},p)\,\text{d}^3\vec{p}. \end{align}$$

Since the distribution is isotropic in $p$, we have $T^{11} = T^{22} = T^{33} = P$, so that

$$ P = \frac{1}{3}\left(T^{11} + T^{22} + T^{33} \right) = \frac{g}{(2\pi\hbar)^3}\int \frac{1}{3}\vec{v}\cdot\vec{p}\; f(\vec{x},p)\,\text{d}^3\vec{p}. $$ Finally, from

$$ E = \frac{mc^2}{\sqrt{1-v^2/c^2}},\qquad \vec{p} = \frac{m\vec{v}}{\sqrt{1-v^2/c^2}}, $$ we find $\vec{v} = \vec{p}c^2/E$, and

$$ P = \frac{g}{(2\pi\hbar)^3}\int \frac{p^2c^2}{3E(p)}\, f(\vec{x},p)\,\text{d}^3\vec{p}. $$

  • $\begingroup$ Your derivation of $\vec{v}=\vec{p} c^{2} / E$ is specific to special relativity. While the question was regarding the pressure in a FRW Universe. So I think a general relativistic motivation needs to be given. $\endgroup$
    – Virgo
    Apr 1 at 0:57

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