# How does Dodelson calculate the number density of a particle species?

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)?

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, $$m\gg \frac{k_B T}{c^2}$$.

• 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. Commented Mar 19, 2021 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 $\int_{-1}^1 d(\cos(\theta)) = \int_0^\pi \sin(\theta)d\theta = 2$. Commented Mar 19, 2021 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? Commented Mar 19, 2021 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. Commented Mar 19, 2021 at 22:18
• @GluonSoup I can't imagine how, that's just $\int_0^{2\pi} d\phi = \phi\big|^{2\pi}_0 = 2\pi-0 = 2\pi$. Commented Mar 21, 2021 at 0:36