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?

  • $\begingroup$ This also has to do with how you do the Fourier transformation. $\endgroup$
    – Louis Yang
    Jun 22 '17 at 1:10
  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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.