Kolb and Turner's expresion for number density Eqn. (3.46) and pressure in Eqn. (3.48)

Question

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?

Answer 1

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).

Answer 2

I think relation (2) can be derived the following way. Consider an element of surface $\delta S$ with its unit vector $\vec n$. The particles that will collide with it during a time ${\rm d} t$ lie in a tilted cylinder of volume $$ V = v \;{\rm d} t\; \cos \theta \; \delta S \;. $$ Now, the quantity ${\rm d}N$ defined as $$ {\rm d} N = \frac{g}{h^3} f(p)\;{\rm d}^3 p $$ gives the number of particles with momentum within ${\rm d}^3 p$ over the whole $4\pi$ solid angle. Here I use physical units (not natural) so that is why the Planck's constant $h$ appears. The fraction of those particles that have the right $\theta$ and $\phi$ is therefore given by the ratio $$ \frac{\sin \theta\; {\rm d}\theta \;{\rm d}\phi}{4\pi}\;. $$ Let us now select those particles that will effectively collide with our surface $\delta S$: $$ {\rm d} N_{coll} = \frac{g}{h^3} f(p)\;{\rm d}^3 p\; \frac{\sin \theta\; {\rm d}\theta \;{\rm d}\phi}{4\pi}\; v \cos \theta \; \delta S \;{\rm d} t\;. $$ We can assume that each particle transfers a momentum $2\vec p \cdot \vec n = 2\;p\cos\theta$ to the elementary surface, so the force is that value divided by ${\rm d}t$ and the pressure is obtained by further dividing by $\delta S$. Putting all of this together one has the pressure $$ dP = {\rm d} N_{coll}\;\frac{2\;\vec p \cdot \vec n}{{\rm d}t \;\delta S} =\frac{g}{2\pi h^3}f(p)\;p\;v\;\cos^2\theta \sin\theta \;{\rm d}\theta \; {\rm d}\phi\;{\rm d}^3 p\;. $$ The integral over $\phi$ gives $2\pi$ and $\theta$ shall be integrated between $0$ and $\pi/2$ to account for a positive pressure (from only one side of the surface). In the end the integral over $\theta$ gives a factor $1/3$, and we end up with $$ {\rm d} P = \frac{g}{h^3}\;f(p)\; \frac{p\;v}{3}\;{\rm d}^3p\;. $$ To finish we need to evaluate the product $p\times v$. Recall the definitions $$ \vec p = \gamma m \vec v $$ and $$ E = \gamma m c^2\;, $$ where $\gamma$ is the Lorentz factor, and from which we have $$ p\times v = \frac{p^2}{\gamma m} = \frac{p^2 c^2}{E}\;. $$ So in the end we have $$ {\rm d} P = \frac{g}{h^3}\;f(p)\; \frac{p^2 c^2}{3E} {\rm d}^3p\;. $$ To compare with the original post, we switch to natural units $\hbar = c =1$, in which case $h=2\pi$ and we find $$ P = \frac{g}{(2\pi)^3}\int \frac{|{\bf p}|^2}{3E} \;f({\bf p})\; {\rm d}^3{\bf p}\;. $$