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Having defined the phase space distribution function $f(\textbf{r},\textbf{p},t)$ in $\mu-$space, one can express the information that there are $N$ particles in the volume $V$ through the condition \begin{equation} \int f(\textbf{r},\textbf{p},t) d^3\textbf{r}d^3\textbf{p}=N \end{equation} as given in, for example, Kerson Huang, Statistical mechanics, second edition, Sec. 3.1, Eq. 3.4).

In equilibrium, the distribution function is independent of $t$ so that \begin{equation}\int f(\textbf{r},\textbf{p}) d^3\textbf{r}d^3\textbf{p}=N. \end{equation} If the particles are uniformly distributed in space, so that $f$ is independent of $\textbf{r}$, then the number density is given by \begin{equation} n=\frac{N}{V}=\int d^3\textbf{p}f(\textbf{p}) \end{equation}

  1. The expression for number density as given in Cosmology book by Kolb and Turner, they have an extra factor of $\frac{1}{(2\pi)^3}$: \begin{equation} n=\frac{N}{V}=\frac{g}{(2\pi)^3}\int d^3\textbf{p}f(\textbf{p}) \end{equation} where $g$ counts the number of internal degrees of freedom. But where does the factor $\frac{1}{(2\pi)^3}$ come from? Why is this factor missing in Huang's expression for number density?

  2. Kolb and Turner also writes an expression for the pressure for the relativistic particles satisfying $E^2=|\textbf{p}|^2+m^2$ (units in which $c=1$) in equilibrium as $$p=\frac{g}{(2\pi)^3}\int \frac{|\textbf{p}|^2}{3E}f(\textbf{p})d^3\textbf{p}.$$ What should the starting point in deriving this formula for pressure? In other words, what is the general formula for equilibrium pressure?

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  • $\begingroup$ This also has to do with how you do the Fourier transformation. $\endgroup$
    – Louis Yang
    Jun 22 '17 at 1:10
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  1. The extra factor of $\frac{1}{(2 \pi)^3}$ just comes from a different convention for the normalization of $f({\bf p}, {\bf r})$. The one in Kolb & Turner corresponds to Fermi-Dirac and Bose-Einstein distributions on the form: $$f({\bf p}) = \frac{1}{e^{(E({\bf p})-\mu)/T} \pm 1}.\tag{1}$$

  2. In general the energy-momentum tensor is given by: $$T^{\mu\nu} = \frac{g}{(2\pi)^3} \int d^3{\bf p} \frac{P^\mu P^\nu}{E({\bf p})} f({\bf p}). \tag{2}$$ For an ideal fluid, you can compare this to $$T^{\mu\nu} = (\rho + P)u^\mu u^\nu - g^{\mu \nu} P.\tag{3}$$ to check the expression for the pressure. (Note that I use the mostly minus metric convention here).

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  • $\begingroup$ How do you get the relation (2)? $\endgroup$
    – SRS
    May 9 '17 at 17:41
  • $\begingroup$ I did not actually derive it, but I guess you could start with the section on an isolated particle here : en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor, and generalize to the whole distribution f(p, x). Or at least convince yourself that (2) makes sense in light of that. $\endgroup$
    – Ihle
    May 10 '17 at 16:18

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