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In chapter three of Modern Cosmology, Dodelson models the evolution of a particle plasma as the universe expands. On page 61, the author gives the formula for the species-dependent equilibrium density as:$$n_i=g_i\space e^{\mu_i/T}\int\frac{d^3p}{(2\pi)^3}e^{-E_1/T}\tag1$$For equilibrium (when a species is created as often as it's annihilated), we have:$$n_i^{(0)}=g_i\space\int\frac{d^3p}{(2\pi)^3}e^{-E_1/T}\tag2$$Now this is the part I don't follow. How does Dodelson go from Eq. (2) to this:$$n_i^{(0)}=g_i\left(\frac{m_iT}{2\pi}\right)^{3/2} e^{-m_1/T}\tag3$$I get that $\int d^3 p=\frac{p^3}{({2\pi})^3\space6}$, and I get that $E_1=m_1$, and I get that $p=\sqrt{E^2-m^2}$, but I can't put them together. How do we go from Eq. (2) to Eq. (3)?

Quark Soup
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I'm afraid your primary error is one of elementary calculus. $\int d^3p$ is not equal to $\frac{p^3}{(2\pi)^3 6}$. Secondly, $E\neq m$ (how could it, since as you correctly say in your next words, $p=\sqrt{E^2-m^2}$??).

To evaluate this integral we go to spherical coordinates, in which $\int d^3 p = \int_0^\infty p^2 dp \int_{0}^{\pi} sin(\theta)\space d\theta \int _0^{2\pi} d\phi = 4\pi \int_0^\infty p^2 dp$, where we've used the spherical symmetry of the problem to evaluate the angular integrals. From there, we have

$$n_i^{(0)} = g_i \frac{4\pi}{(2\pi)^3} \int_0^\infty p^2 e^{-\sqrt{p^2+m_i^2}/T} dp$$ This integral doesn't have a nice form. However, if we assume that $m_i \gg p$, we can approximate $\sqrt{p^2+m_i^2} \approx m_i + \frac{p^2}{2m_i}$ (a non-relativistic approximation), which allows us to simplify to

$$n_i^{(0)} \approx g_i \frac{4\pi}{(2\pi)^3}e^{-m_i/T} \int_0^\infty p^2 e^{-p^2/2m_iT} dp$$ $$ = g_i\frac{4\pi}{(2\pi)^3} e^{-m_i/T} \frac{\sqrt{\pi}}{4}(2m_iT)^{3/2}$$ which simplifies after a bit of algebra to $$n_i^{(0)} \approx g_i e^{-m_i/T} \left(\frac{m_iT}{2\pi}\right)^{3/2}$$

The non-relativistic approximation is valid as long as $m\gg T$ or, in SI units, $m\gg \frac{k_B T}{c^2}$.

J. Murray
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  • I don't understand the terminology of $\int_{-1}^1 d(\cos(\theta))$. Is this supposed to be $\int_{-1}^{1} \cos(\theta) d\theta$? Obviously it's supposed to result in $2\pi$, but I don't see this step completely because the notation is unfamiliar. – Quark Soup Mar 19 '21 at 21:46
  • @GluonSoup It’s a fairly common notation which is useful when only cosine (or no angular dependence at all) shows up in the integrand. In this case we have the latter, so $\int_{-1}^1 d(\cos(\theta)) = \int_0^\pi \sin(\theta)d\theta = 2$. – J. Murray Mar 19 '21 at 21:50
  • Following your assumption that $m_i\gg p$, I would like to understand how I compare momentum to mass. If, say, the mass of a proton, $m_P$, was $1.67\times 10^{-27}\space kg$, then to what, exactly, am I comparing this? – Quark Soup Mar 19 '21 at 22:14
  • @GluonSoup Everything here is taking place in natural units, where the Boltzmann constant $k_B$ and the speed of light $c$ are set to 1. If you want to use SI units, then the assumption would be that $m_i \gg p/c$. In different terms, the particle's kinetic energy is much less than its mass energy, or its velocity is much less than $c$. All of these approximations are equivalent. – J. Murray Mar 19 '21 at 22:18
  • So then, we're assuming that the motion of particles in this 1 MeV soup is non-relativistic? That' they're basically at rest compared to $c$? – Quark Soup Mar 19 '21 at 22:23
  • @GluonSoup Yes. 1 MeV is highly non-relativistic for protons, since $m_P c^2 \approx 1$ GeV. It is, however, highly relativistic for electrons, sincd $m_e c^2 \approx 0.5$ MeV. – J. Murray Mar 19 '21 at 22:53
  • I've been over this several times in MMa and I even cross referenced Wiki on the spherical coordinates. This is what MMa gives for an integral of the above formula:$$\int d^3 p = \int_0^\infty p^2 dp \int_{0}^{\pi} sin(\theta)\space d\theta \int _0^{2\pi} d\phi = 4\pi^2 \int_0^\infty p^2 dp$$That is, the $\pi$ term is squared. Was this a misprint on your side or is this really what you get when you take the integral? – Quark Soup Mar 21 '21 at 00:26
  • @GluonSoup It’s really what you get. The theta integral gives you $2$, not $2\pi$. – J. Murray Mar 21 '21 at 00:28
  • I agree that that $\int_{0}^{\pi} sin(\theta)\space d\theta$ is 2, but I'm getting $2\pi^2$ for $\int _0^{2\pi} d\phi$ – Quark Soup Mar 21 '21 at 00:34
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    @GluonSoup I can't imagine how, that's just $\int_0^{2\pi} d\phi = \phi\big|^{2\pi}_0 = 2\pi-0 = 2\pi$. – J. Murray Mar 21 '21 at 00:36
  • Thank you. That was it. Mistake in the transcription. – Quark Soup Mar 21 '21 at 00:37