Past exam of the general relativity and cosmology course of the University of Oxford / 2017 / 4

Normalize the present scale factor to $a_0=1$. Conservation of the number of electrons associated with the baryons makes their physical number density scale as $a^{-3}$. At this early epoch they are free rather than bound in neutral atoms, soHere the specified present density is interpreted as the conserved total electron abundance, including electrons that might be bound today. A present free-electron density would require an ionization-fraction correction.

Density alone does not determine whether a gas is relativistic; its random kinetic energy must be compared with the rest energy. In the usual thermal Hot Big Bang model, use the measured cosmic microwave background temperature $T_0\simeq 2.7\mathrm{K}$ and the temperature scaling of thermal photon number and energy densities, $T_{\gamma }\propto a^{-1}$. Tight coupling then givesSince $m_e{c}^2\simeq 5.1\times 10^5\mathrm{eV}$,Their thermal speed is of order $\sqrt{3k_B{T}_e/m_e}\simeq 0.04c$. The gas is fully ionized because its temperature is far above the recombination temperature, but ionization does not imply relativistic electrons. Thermal electron-positron pairs are exponentially suppressed at this temperature.^[ref]

The given age and scale-factor ratio should be treated as independent supplied estimates in the following rate comparison. They do not describe the same epoch in standard radiation-era expansion: $a/a_0=10^{-6}$ normally corresponds to an age of order a year, not $10^4$ years. The density and temperature estimates above depend on the scale-factor ratio, not on assigning it the stated age.

For statistically independent Thomson scattering events, the probability of no scattering over a proper distance $s$ is $\exp (-n_e{\sigma }_{e}s)$. The photon mean free path in an ionized universe is thereforeThis is about half an astronomical unit. It is a proper distance at that epoch, not a present-day comoving distance. The Thomson scattering approximation is appropriate because the electrons are nonrelativistic and typical photon energies are much smaller than $m_e{c}^2$. The source's spelling of the process is a typographical error; the physical process is Thomson scattering.

A photon travels at $c$, so its typical scattering interval isabout four minutes. The supplied universe age isThusA photon scatters extremely frequently compared with the expansion timescale, of order the cosmic age. The plasma is opaque and its photons are tightly coupled to the electrons rather than freely streaming over a cosmological distance. In rate language ${\Gamma }_{\gamma }=n_e{\sigma }_{e}c\gg H$, where $H$ is the Hubble parameter.^[ref]

This is a statement about transport and momentum coupling. Elastic Thomson scattering alone does not create or destroy photons and hence is not, by itself, proof that an arbitrary photon spectrum reaches complete blackbody thermal equilibrium. The interaction interval calculated here is the interval experienced by a photon; the interval per electron differs because there are many more photons than electrons. Even replacing the inconsistent supplied age by the standard age of order a year at this scale factor would leave scattering much faster than expansion.

As the temperature falls, electrons and protons form neutral hydrogen through cosmological recombination. The free electron fraction drops sharply, and so does the scattering rate ${\Gamma }_{\gamma }=n_e{\sigma }_{T}c$. When this rate becomes comparable to or smaller than the Hubble parameter, the plasma becomes transparent and photon decoupling occurs. The Thomson scattering cross-section itself does not abruptly vanish. The rapid change is in the abundance of free targets.

The Hydrogen binding energy is $13.6\mathrm{eV}$, but recombination occurs at a much smaller thermal energy, of order $0.3\mathrm{eV}$, or a few thousand kelvin. There are roughly a billion photons per baryon, so the high-energy tail can keep hydrogen ionized long after $k_{B}T$ drops below the binding energy. The falling exponential ionizing tail then makes the transition comparatively sharp. Last scattering follows when sufficiently few electrons remain free.^[ref]

To estimate a horizon, specify an expansion model as well as the redshift. The proper particle horizon at emission isAs a transparent analytic estimate, approximate a spatially flat matter-dominated universe by $a(t)=(t/t_{\mathrm{now}}{)}^{2/3}$ throughout. ThenTaking the illustrative present value $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, this model has $t_{\mathrm{now}}=2/(3H_0)\simeq 9.3\times 10^9\mathrm{yr}$. At $z_{\ast }=10^3$,This is a proper horizon radius at emission; its corresponding present comoving length is about $270\mathrm{Mpc}$.

For the same model the comoving distance to the Cosmic microwave background last-scattering surface and its angular diameter distance areTherefore the angular extent associated with one radial horizon length isThis is an order-one-degree estimate. Including the earlier radiation domination and late dark energy changes the coefficient; it does not make the horizon span the sky. A horizon patch diameter has twice this angle, whereas the acoustic scale uses a sound horizon and differs again. The redshift by itself, without an expansion model or normalization, is insufficient for a unique horizon distance.

The horizon problem is that the observed cosmic microwave background has nearly the same mean temperature across angular separations vastly greater than this scale. In the purely decelerating Hot Big Bang model, sufficiently separated last-scattering regions had neither time to exchange signals nor a common post-Big-Bang causal region that could establish their observed thermal uniformity. The puzzle concerns the missing causal explanation of those initial correlations; it is not an inconsistency of specifying uniform initial conditions mathematically. An earlier period of cosmic inflation can enlarge a formerly causally connected region beyond the later observable horizon.