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I am working through Mukhanov's book "Physical Foundations of Cosmology" and something has me puzzled in the section on evaluating integrals for number density, energy density and pressure.

The number density and energy density are given by

$n=\frac{g}{2\pi^2}\int_m^\infty \frac{\sqrt{\epsilon^2-m^2}}{e^{(\epsilon-\mu)/T}\mp 1}\epsilon\;d\epsilon$

$E=\frac{g}{2\pi^2}\int_m^\infty \frac{\sqrt{\epsilon^2-m^2}}{e^{(\epsilon-\mu)/T}\mp 1}\epsilon^2\;d\epsilon$

He then makes a change of variables from $\epsilon$ to $x=\epsilon/T$ and introduces the new variables $\alpha=m/T$ and $\beta=\mu/T$, and then indicates that the following integrals are relevant in calculating $n$ and $E$ for particles and antiparticles:

$J_{\mp}^{(\nu)}(\alpha,\beta)=\int_\alpha^\infty \frac{(x^2-\alpha^2)^{\nu/2}}{e^{x-\beta}\mp 1}\;dx+\int_\alpha^\infty \frac{(x^2-\alpha^2)^{\nu/2}}{e^{x+\beta}\mp 1}\;dx$

However, in changing variables, the factors of $\epsilon(\rightarrow x)$ and $\epsilon^2(\rightarrow x^2)$ seem to have gone missing! Can anyone tell me if they have been absorbed somehow in the move to considering particles and antiparticles, or if there is some other explanation? It is tempting to think that this is just a typo, but the whole section that follows is related to evaluating these integrals and nowhere do the factors of $x$ and $x^2$ appear.

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Roughly, in the integrands you have something like

$$J^{(3)}+\alpha^2J^{(1)}\sim (x^2-\alpha^2)^{3/2}+\alpha^2(x^2-\alpha^2)^{1/2}=(x^2-\alpha^2)^{1/2}(x^2-\alpha^2+\alpha^2)=(x^2-\alpha^2)^{1/2}x^2$$

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  • $\begingroup$ Thank you very much for your help. Is it also possible to express the number density, n, as a linear combination of $J^{(1)}$ and $J^{(3)}$ or only in terms of the differential of $J$'s? The factor of $x$ seems more problematic than $x^2$. $\endgroup$
    – dgwp
    Commented Feb 9, 2016 at 13:33
  • $\begingroup$ Well, as stated in problem 3.8, the best one can do is calculate the difference between $n_p$ and $n_\bar{p}$. I think, if you calculate the left-hand side and the right-hand side of (3.37), you will see how to express this difference through $J^{(3)}$ $\endgroup$
    – Photon
    Commented Feb 9, 2016 at 16:51
  • $\begingroup$ Can you explain why it is not possible to calculate, say, just $n_p$, rather than the difference between the two n's? After all, there is a simple relation between n and p ($n=\frac{dp}{d\mu}$). $\endgroup$
    – dgwp
    Commented Feb 9, 2016 at 17:21
  • $\begingroup$ I did this calculation a few years ago, so of course I don't remember the details, but I guess that if you calculate $n=dp/d\mu$, you will get different signs depending on whether you are considering particles or antiparticles (since $\mu$ changes sign if you pass from particles to antiparticles). Therefore you get a minus sign in front of the antiparticle part and will end up with a difference rather than a sum as in the case of the energy density. $\endgroup$
    – Photon
    Commented Feb 9, 2016 at 17:39
  • $\begingroup$ That makes perfect sense, thank you. I've been able to work through the whole section now, but there is one small thing that is troubling me, and that is: where does the extra condition $\alpha-\beta\gg 1$ come from in the nonrelativistic limit? $\endgroup$
    – dgwp
    Commented Feb 11, 2016 at 11:08

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