Past exam of the general relativity and cosmology course of the University of Oxford / 2012 / 4

For a separately conserved homogeneous component, energy conservation gives $d(\rho c^2{a}^3)=-pd(a^3)$, equivalently $\dot{\rho }+3H(\rho +p/c^2)=0$. Dust has negligible pressure, while isotropic radiation has $p_{\gamma }={\rho }_{\gamma }{c}^2/3$. HenceThe matter factor is volume dilution; photons additionally lose energy by cosmological redshift. Nonrelativistic baryons and cold dark matter both contribute to the matter term, so ${\Omega }_M={\Omega }_B+{\Omega }_C$. At equality ${\rho }_{M0}{a}_{\mathrm{eq}}^{-3}={\rho }_{\gamma 0}{a}_{\mathrm{eq}}^{-4}$, givingThis is equality for the source's matter-plus-photon model. Additional relativistic species would contribute to total radiation and shift the equality value. Although the source calls the $\rho$'s energy densities, its Friedmann equation and later ${\rho }_{\gamma }{c}^2$ show that their consistent SI meaning is mass-equivalent density.

For thermal photons, the available thermal length is ${\ell }_T=\hslash c/(k_{B}T)$. The number of occupied modes per volume scales as ${\ell }_T^{-3}$, while their characteristic energy is of order $k_{B}T=\hslash c/{\ell }_T$. Thus $n_{\gamma }\propto {\ell }_T^{-3}$ and $u_{\gamma }={\rho }_{\gamma }{c}^2\propto \hslash c{\ell }_T^{-4}$. More explicitly, the two photon polarizations and the Bose-Einstein distribution giveconsistent with the rounded coefficients supplied in the source.

Use its definition $s=n_{\gamma }/n_B$, with $n_B={\rho }_B/m_p={\Omega }_B{\rho }_c/m_p$ and ${\rho }_c$ the mass-equivalent cosmological critical density. Writing $Q=k_{B}T/(\hslash c)$, the photon energy formula gives $Q^4={\Omega }_{\gamma }{\rho }_{c}c/(0.66\hslash )$. ThereforeThis photon-to-baryon ratio from density parameters has leading coefficient $0.328\simeq 0.33$, as required. For an illustrative estimate, take the stated ${\rho }_c=10^{-26}\mathrm{kg}{\mathrm{m}}^{-3}$, ${\Omega }_{\gamma }=5\times 10^{-5}$ and ${\Omega }_B=0.05$. Then $\boxed{s\sim 1.4\times 10^9}$, i.e. of order a billion photons per baryon. The source alone supplies no numerical density parameters, so an estimate must state such additional inputs.

Strictly, this $s$ is the photon-to-baryon ratio, rather than the thermodynamic entropy per baryon itself. For a thermal photon gas, $S_{\gamma }/(k_B{N}_B)=[4u_{\gamma }/(3k_{B}T{n}_{\gamma })]s\simeq 3.6s$ with the exact blackbody coefficients. The ratio used in the source is proportional to the radiation entropy per baryon under adiabatic evolution.

Assume an ideal nondegenerate hydrogen gas in thermal equilibrium and chemical equilibrium, with ground-state atoms and charge neutrality. Let $n_p=n_e=X{n}_B$ and $n_H=(1-X)n_B$. The ground-state Saha equation, with the usual electron/proton spin factors cancelling the atomic ground-state degeneracy, isInvert it and use $n_B=n_{\gamma }/s=(0.24/s)(k_{B}T/(\hslash c){)}^3$. This givessoHere $m_e$ is the electron mass denoted $M_e$ in the source, and $0.24(2\pi {)}^{3/2}=3.78$. At $X=1/2$, the left side is two. At $k_{B}T$ of order $B$, the right side is extremely small because $1/s\sim 10^{-9}$ and $B/(m_e{c}^2)\ll 1$. A large exponential is needed to compensate, requiring $B/(k_{B}T)\gg 1$ rather than a temperature of order the binding energy.

Physically, there are enormously many photons per baryon, so even a small high-energy tail of the thermal photon distribution can maintain ionization. Recombination is delayed until that tail is sufficiently depleted. This explains the low recombination temperature without solving the equilibrium equation; actual cosmological recombination also eventually requires rate equations once equilibrium fails.

Big Bang nucleosynthesis is a kinetic process in an expanding universe. A common particle temperature does not imply chemical equilibrium for every nuclear reaction: reaction rates must also exceed the expansion rate and connect the required species efficiently. Around $10^9\mathrm{K}$, nuclear abundances retain the history of earlier freeze-out and cannot be inferred from equilibrium Boltzmann factors alone.

In particular, weak reactions cease to maintain neutron-proton chemical equilibrium. Subsequent free-neutron decay after freeze-out changes the surviving neutron inventory. The large photon-to-baryon ratio delays deuterium survival through photodissociation; once the deuterium bottleneck opens, reactions rapidly bind many surviving neutrons into helium-4. Coulomb barriers, decreasing density and further freeze-out prevent unrestricted equilibration of all nuclei. The final helium abundance therefore requires a reaction network and expansion history, not just the temperature and nuclear binding energies.

For illustration, if essentially all available neutrons end in helium-4 and the neutron-to-proton ratio at that stage is $r$, simple particle counting gives $Y_{\mathrm{He}}\simeq 2r/(1+r)$. The crucial input $r$ is itself set by out-of-equilibrium history. Kinetic thermalization of the plasma can remain effective while these composition-changing processes are out of equilibrium.