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?



1 Answer 1


Ok, I think I figured it out:

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.


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