# Classical limit of non-interacting, relativistic quantum gas (Kapusta/Gale p.8)

I want to understand two equations in "Finite temperature field theory" by Kapusta and Gale on page 8. The partition function is $$\ln Z = V\int \frac{d^3 p}{(2\pi)^3}\;\ln\left(1\pm e^{-\beta(\omega-\mu)}\right)^{\pm 1},$$ where the upper sign is for fermions and the lower sign for bosons. For the dispersion relation $\omega=\sqrt{p^2+m^2}$ and in the classical limit $T\ll\omega-\mu$, I want to show $$P=\frac{T}{V}\ln Z=\frac{m^2 T^2}{2\pi^2}e^{\mu/T}K_2\left(\frac{m}{T}\right),$$ where $K_2$ is a modified Bessel function, which has (as one possible form) the integral representation (see NIST DLFM 10.32.8) $$K_2(z)=\frac{z^2}{3}\int_1^\infty dt\;e^{-zt}(t^2-1)^{3/2}.$$

In the limit $T\ll\omega-\mu$ we can use $$\ln\left(1+e^{-\beta(\omega-\mu)}\right)\approx e^{-\beta(\omega-\mu)}$$ and $$\ln\left(\frac{1}{1-e^{-\beta(\omega-\mu)}}\right)\approx e^{-\beta(\omega-\mu)},$$ i.e. fermions and bosons look the same in this limit.

I find something similar to the expression in the book, but not quite the right thing: $$P=T\int \frac{d^3 p}{(2\pi)^3}\;\ln\left(1\pm e^{-\beta(\omega-\mu)}\right)^{\pm 1}\\ =\frac{4\pi T}{(2\pi)^3}e^{\beta\mu}\int_0^\infty dp\;p^2 e^{-\beta\sqrt{p^2+m^2}}$$ Use the substitution $x=\sqrt{\frac{p^2}{m^2}+1} \Rightarrow p\,dp=m^2 x\, dx$ and $p=m\sqrt{x^2-1}$: $$P=\frac{Tm^3}{2\pi^2}e^{\frac{\mu}{T}}\int_1^\infty dx\;x\sqrt{x^2-1}e^{-\frac{m}{T}x}.$$ This does not look too far off, but it is not correct. Can someone help me?

Thanks!

Take the fermionic expression, first perform an integration by parts, then the substitution that I tried above and only then use the classical approximation: $$P=T\int\frac{d^3p}{(2\pi)^3}\ln\left[1+e^{-\beta(\omega-\mu)}\right]\\ =\frac{T}{2\pi^2}\int_0^\infty dp\;p^2\ln\left[1+e^{-\beta(\sqrt{p^2+m^2}-\mu)}\right]\\ =\frac{T}{2\pi^2}\left(\left.\frac{p^3}{3}\ln\left[1+e^{-\beta(\sqrt{p^2+m^2}-\mu)}\right]\right|_{p=0}^{p=\infty} - \int_0^{\infty}dp\;\frac{p^3}{3}\frac{e^{-\beta(\sqrt{p^2+m^2}-\mu)}}{1+e^{-\beta(\sqrt{p^2-m^2}-\mu)}}\left(-\beta\frac{p}{\sqrt{p^2+m^2}}\right)\right)\\ =\frac{1}{6\pi^2}\int_0^{\infty}dp\;\frac{p^4}{\sqrt{p^2+m^2}}\frac{1}{1+e^{\beta(\sqrt{p^2-m^2}-\mu)}}\\ =\frac{1}{6\pi^2}\int_1^\infty \frac{dx\;m^2 x\;m^3(x^2-1)^{3/2}}{mx}\frac{1}{e^{\frac{m}{T}x-\frac{\mu}{T}}+1}\\ \approx\frac{m^4}{6\pi^2}\int_1^\infty dx\;(x^2-1)^{3/2}\; e^{-\frac{m}{T}x+\frac{\mu}{T}}\\ =\frac{m^2 t^2}{2\pi^2}e^{\frac{\mu}{T}}\frac{1}{3}\left(\frac{m}{T}\right)^2\int_1^\infty dx\;(x^2-1)^{3/2}\; e^{-\frac{m}{T}x}\\ =\frac{m^2 T^2}{2\pi^2}e^{\frac{\mu}{T}}K_2\left(\frac{m}{T}\right)$$ The bosonic case is analogous.