Past exam of the general relativity and cosmology course of the University of Oxford / 2018 / 4

For a photon gas in thermal and chemical equilibrium, photon number is not a conserved thermodynamic charge. Number-changing processes can create and absorb photons without a corresponding conserved-number constraint, so the equilibrium photon chemical potential is zero.

At $T=5\times 10^9\mathrm{K}$, $k_{\mathrm{B}}T\simeq 0.431\mathrm{MeV}$. The other light species are electrons, Positrons, three neutrino species and their antiparticles. Neutrinos are ultrarelativistic; electrons and Positrons are only moderately relativistic at this particular temperature because $m_e{c}^2=0.511\mathrm{MeV}$. The later simple counting formulas use the intended ultrarelativistic approximation for them. Baryons, muons and heavier particles are not an abundant relativistic bath at this temperature.

Pair equilibrium implies ${\mu }_{e^{-}}+{\mu }_{e^{+}}=2{\mu }_{\gamma }=0$. Approximate charge symmetry, with the net electron excess tied to the much smaller baryon asymmetry, then makes both electron and Positron chemical potentials small compared with $k_{\mathrm{B}}T$. For neutrinos one additionally assumes negligible lepton asymmetry. Their near-zero chemical potentials are not implied merely by their statistics. After neutrino decoupling, a distribution initially having negligible chemical potential retains a redshifted Fermi-Dirac distribution, with its own temperature.

Let $V$ be proper volume, let $g_i$ count the populated internal states, and take ${\mu }_i=0$ and $ϵ=pc$. Spherical integration in momentum space givesUsing the fermion-to-boson thermal integral ratio, the minus-denominator boson integral is $2\zeta (3)$ and the plus-denominator fermion integral is $3\zeta (3)/2$. HenceThe energy integral has one extra factor $pc=k_{\mathrm{B}}Tx$:The supplied distribution identity gives the fermion integral as $(1-2^{-3}){\pi }^4/15=7{\pi }^4/120$. ThereforeFor an isotropic ultrarelativistic gas, momentum flux gives $P_i=\frac{1}{3}\int pv{f}_i{d}^{3}p$. Since $v=c$ and $pc=ϵ$,Photons have $g_{\gamma }=2$, electrons and Positrons each have $g=2$, and each neutrino or antineutrino species has one populated helicity in the standard thermal bath.

For zero chemical potential, each species contributes $S_i=(U_i+P_{i}V)/T=4U_i/(3T)$. ThusEntropy density scales as $T^3$ at fixed relativistic particle content; total entropy also includes the changing proper volume. An isolated adiabatically expanding comoving gas has conserved total entropy and $aT$ constant, rather than its total entropy falling as $T^3$.

At equal temperature, single electrons and photons each have two spin/polarization states. In the ultrarelativistic approximation,Positrons are a separate species with the same contribution, so $(S_{e^{-}}+S_{e^{+}})/S_{\gamma }=7/4$. This distinction is needed for the heating count in part d. At the exact supplied temperature the electron rest mass gives corrections to the ideal $7/8$ ratio; it is the limiting relativistic value requested by the problem.

There is also a sign convention issue in the introductory entropy formula. With the chemical-potential convention used in the supplied distribution, the Euler relation is $U=TS-PV+{\sum }_i{\mu }_i{N}_i$, so $S=(U+PV-{\sum }_i{\mu }_i{N}_i)/T$. Its chemical-potential term has the opposite sign to the printed one. For a concrete counterexample, a dilute monatomic ideal gas with one internal state and $n{\lambda }_T^3=e^{-1}$, where ${\lambda }_T$ is the thermal de Broglie wavelength, has $\mu =-k_{\mathrm{B}}T$, $U=3N{k}_{\mathrm{B}}T/2$ and $PV=N{k}_{\mathrm{B}}T$. Its entropy is $7N{k}_{\mathrm{B}}/2$; the printed plus sign would instead give $3N{k}_{\mathrm{B}}/2$. All chemical potentials have been neglected here, so this source error does not change any of the above results.

As the plasma cools through electron-positron annihilation, the pairs transfer their entropy to the coupled photons. Neutrinos have largely decoupled and do not share that heating in the instantaneous-decoupling approximation. Before annihilation the electromagnetic entropy count iswhile after it $g_{\ast s,\mathrm{after}}=2$. Conservation of electromagnetic comoving entropy, $g_{\ast s}{T}_{\gamma }^3{a}^3=\text{constant}$, gives the relative photon heating factorEquivalently $T_{\gamma }/T_{\nu }=(11/4{)}^{1/3}$ afterward. The exact $11/4$ count compares an asymptotic relativistic pair bath with the pair-free bath; at the supplied temperature some finite-electron-mass entropy loss has already begun.

The word “rises” requires care. The photons become hotter than an otherwise freely redshifting bath at the same scale factor; their absolute temperature need not rise during the expansion. In the smooth equilibrium process,since the pair contribution grows with temperature. This monotonic photon cooling during pair annihilation is slower than $a^{-1}$ while $a{T}_{\gamma }$ increases. The factor $1.401$ compares the photon temperature with the freely redshifting neutrino bath, not two consecutive absolute temperatures along the expanding history.

Use the observed present photon temperature $T_{\gamma 0}\simeq 2.725\mathrm{K}$. This is an additional observational input, consistent with ^[ref]. Standard neutrino decoupling and electron-positron entropy transfer giveFor nonrelativistic relic neutrinos this is the temperature parameter of the redshifted momentum distribution, not the temperature of a newly thermalized gas of massive particles.

The photon number density from part b isOne neutrino mass eigenstate and its antineutrino have two populated helicity states. ThusThe total over three species is about $336\mathrm{c}{\mathrm{m}}^{-3}$, including antineutrinos. Comoving number conservation preserves these abundances after the particles become nonrelativistic.

For the supplied $H_0=72\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}=2.33\times 10^{-18}{\mathrm{s}}^{-1}$,Assuming nonnegative other energy components, relic neutrino rest-mass density cannot exceed the near-critical total density. With the common per-species abundance $n_{\ast }=n_{{\nu }_i}+n_{{\bar{\nu }}_i}$,Using the total $336\mathrm{c}{\mathrm{m}}^{-3}$ would instead bound the mean mass, not the sum. This critical-density bound on the sum of relic neutrino masses is deliberately conservative: baryons, dark matter and dark energy leave less density available. It assumes the standard thermal abundance, negligible chemical potentials, no later dilution or extra populated states, and the idealized decoupling count. Small non-instantaneous-decoupling corrections do not matter for this estimate.