(i) In equilibrium, the number density of a non-relativistic particle species is given by

$$n=g_{\mathrm{s}}\left(\frac{2 \pi m k T}{h^{2}}\right)^{3 / 2} e^{\left(\mu-m c^{2}\right) / k T}$$

where $m$ is the mass, $\mu$ is the chemical potential and $g_{\mathrm{s}}$ is the spin degeneracy. At around $t=100$ seconds, deuterium $D$ forms through the nuclear fusion of nonrelativistic protons $p$ and neutrons $n$ via the interaction:

$$p+n \leftrightarrow D$$

What is the relationship between the chemical potentials of the three species when they are in chemical equilibrium? Show that the ratio of their number densities can be expressed as

$$\frac{n_{D}}{n_{n} n_{p}} \approx\left(\frac{h^{2}}{\pi m_{p} k T}\right)^{3 / 2} e^{B_{D} / k T},$$

where the deuterium binding energy is $B_{D}=\left(m_{n}+m_{p}-m_{D}\right) c^{2}$ and you may take $g_{D}=4$. Now consider the fractional densities $X_{a}=n_{a} / n_{B}$, where $n_{B}$ is the baryon number of the universe, to re-express the ratio above in the form

$$\frac{X_{D}}{X_{n} X_{p}}$$

which incorporates the baryon-to-photon ratio $\eta$ of the universe. [You may assume that the photon density is $n_{\gamma}=\frac{16 \pi \zeta(3)}{(h c)^{3}}(k T)^{3}$.] From this expression, explain why deuterium does not form until well below the temperature $k T \approx B_{D}$.

(ii) The number density $n=N / V$ for a photon gas in equilibrium is given by the formula

$$n=\frac{8 \pi}{c^{3}} \int_{0}^{\infty} \frac{\nu^{2} d \nu}{e^{h \nu / k T}-1},$$

where $\nu$ is the photon frequency. By considering the substitution $x=h \nu / k T$, show that the photon number density can be expressed in the form

$$n=\alpha T^{3}$$

where the constant $\alpha$ need not be evaluated explicitly.

State the equation of state for a photon gas and explain why the chemical potential of the photon vanishes. Assuming that the photon energy density $\epsilon=E / V=(4 \sigma / c) T^{4}$, use the first law $d E=T d S-P d V+\mu d N$ to show that the entropy density is given by

$$s=S / V=\frac{16 \sigma}{3 c} T^{3}$$

Hence explain why, when photons are in equilibrium at early times in our universe, their temperature varies inversely with the scale factor: $T \propto a^{-1}$.