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Thermal photon number density is proportional to $T^3$. A sudden photon heating factor $(1+r{)}^{1/4}$ therefore raises photon number by $(1+r{)}^{3/4}$ and dilutes the baryon-to-photon ratio by its inverse when baryon number is conserved. Comparing Big Bang nucleosynthesis and cosmic microwave background baryon-density determinations can test entropy injection between the two epochs, subject to changes in the expansion rate and nuclear abundances.

The Compton y parameter is the Thomson scattering optical depth weighted by the dimensionless electron temperature. It measures the integrated electron pressure along a line of sight. When hot, nonrelativistic electrons scatter much colder photons, the first-order radiation energy gain is $4y$ times the incident energy. The spectrum is redistributed while the total photon number density is conserved locally by scattering.

The cosmological baryon density is the mass density in baryonic matter, including hydrogen and helium. Up to nuclear binding-energy and composition corrections it is the proton mass times the baryon number density. The baryon-to-photon ratio links that number density to the photon number density. Larger baryon density increases nuclear reaction rates and permits the deuterium bottleneck to end earlier.

A standard thermal Cosmic neutrino background with one populated neutrino and one antineutrino helicity per species has $n_{\nu }+n_{\bar{\nu }}=3n_{\gamma }/11$ after electron-positron annihilation in cosmology. If this species is nonrelativistic today and dominates a spatially flat matter universe, equate $m_{\nu }(n_{\nu }+n_{\bar{\nu }})$ to the critical density $3H_0^2{M}_P^2/(8\pi )$, using the unreduced Planck mass. The photon number density gives the displayed conditional mass. Extra populated states or several massive species change the abundance count; vacuum-dominated cosmology changes the density available for matter.

Use natural units and a positive nuclear binding energy ${\epsilon }_D=m_n+m_p-m_D$. Chemical equilibrium of $n+p\leftrightarrow D+\gamma$ requires ${\mu }_D={\mu }_n+{\mu }_p$. The nonrelativistic Maxwell--Boltzmann number density givesHere $g_D=3$ and $g_n=g_p=2$ count the spin states. Taking $m_n\simeq m_p\simeq m_N$ and $m_D\simeq 2m_N$, then writing $X_i=n_i/n_B$, $n_B=\eta n_{\gamma }$ and using the photon number density yields the displayed abundance. The coefficient is approximately $8.14$. Omitting the spin-state ratio $3/4$ gives the rougher prefactor $16\zeta (3)/\sqrt{\pi }$. A convention $B=m_D-m_n-m_p<0$ instead writes the exponential as $e^{-B/T}$. The small baryon-to-photon ratio suppresses stable deuterium until the temperature is well below the binding scale, producing the deuterium bottleneck.

For $T\ll m_e$ and negligible chemical potentials, the combined equilibrium density of electrons and positrons divided by the photon number density isThe exponential suppression eventually makes pair equilibrium incompatible with the residual electrons required by charge neutrality.

The entropy density of a photon gas is its entropy per physical volume, $s=S/V$. For homogeneous equilibrium vacuum radiation with two states of electromagnetic polarization and zero photon chemical potential,The first equality is the entropy density at zero chemical potential, obtained from the thermodynamic Euler relation $U=TS-pV+{\mu }_{\gamma }{N}_{\gamma }$ with internal energy $U=uV$. The second uses the photon gas equation of state, $u=a_R{T}^4$ and $p=u/3$. If an independently constrained photon number gives a nonzero chemical potential, the general entropy density is $(u+p-{\mu }_{\gamma }{n}_{\gamma })/T$ and $u=a_R{T}^4$ need not hold. This distinction explains why the absence of a conserved number matters for entropy as well as for the spectral shape. ^[ref]

A short first law of thermodynamics check, for neighboring equilibrium states with ${\mu }_{\gamma }=0$, isThe thermodynamic Euler relation fixes the additive normalization. Equivalently, this radiation entropy tends to zero with temperature at fixed volume; no contribution from the containing walls is included. Interpreting $TdS$ as transferred heat and $pdV$ as boundary work requires a reversible thermodynamic process. For a reversible thermodynamic process that is also an adiabatic process, there is no entropy production or entropy exchange, soAn irreversible adiabatic process can produce entropy even without heat exchange, so the word adiabatic alone is insufficient to infer $T^{3}V=\text{constant}$.

For an analytic example, expand an initially equilibrated photon gas reversibly from $V_0$ to $8V_0$, keeping only the same two photon states of electromagnetic polarization. Its temperature halves, its energy density becomes $u_0/16$, its entropy density of a photon gas becomes $s_0/8$, and its total entropy stays $S_0$. The internal energy halves rather than remaining fixed: the missing half leaves as work against radiation pressure,This is a stipulated reversible radiation model, not a report of a piston experiment. Maintaining a real cavity along this path requires controlling wall heat exchange and including the wall heat capacity in any actual apparatus balance.

For a computed scale at stipulated $T=2.725\mathrm{K}$, $s\simeq 2.04122\times 10^{-14}\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{m}}^{-3}$. Combining this with the photon number density givesThis is the total equilibrium radiation entropy divided by its mean photon number, not a separately assigned entropy carried by each labeled particle. During the ideal expansion, $n_{\gamma }\propto T^3$ and $T^{3}V$ is fixed, so the mean number $N_{\gamma }=n_{\gamma }V$ also stays constant. This follows from the path; it was not imposed as the independent number constraint used to establish chemical potential.

The beautiful connection is that expansion lowers each typical photon energy and dilutes the radiation in just the proportions needed to retain its equilibrium spectral shape and total entropy. For the cosmic microwave background, homogeneous expansion of the universe has physical volume proportional to the cube of the scale factor; fixed-species cosmological entropy conservation therefore gives $T\propto a^{-1}$. This describes preservation of an already established Planck photon distribution and connects the FIRAS spectral observation in photon gas to an entropy prediction; FIRAS did not perform a direct entropy measurement. When other equilibrated species disappear and transfer energy into radiation, use the effective number of relativistic entropy degrees of freedom, $g_{\ast s}$: the conserved quantity for the combined sector is $g_{\ast s}{T}^{3}V$, under the corresponding equilibrium and no-entropy production assumptions. A changing species population, separate decoupled sectors or entropy production requires an appropriate entropy balance rather than the pure photon gas formula. ^[ref]

The Kompaneets equation describes frequency diffusion of a mean photon occupation number $n$ through repeated nonrelativistic Compton scattering. For thermal electrons, with $x_e=h\nu /(k_B{T}_e)$ and $dy=(k_B{T}_e/m_e{c}^2)n_e{\sigma }_{T}cdt$, its full homogeneous form is ${\partial }_{y}n=x_e^{-2}{\partial }_{x_e}[x_e^4({\partial }_{x_e}n+n+n^2)]$. The drift and stimulated terms permit a thermal stationary spectrum. In the hot-electron limit $T_e\gg T_{\gamma }$, with low photon energies, they can be neglected relative to the diffusion term; any fixed linear frequency variable $x$ then gives ${\partial }_{y}n=x^{-2}{\partial }_x(x^4{\partial }_{x}n)$. The quantity $n$ is not the frequency-integrated photon number density.

The official notes, PDF pp. 24–27, handwritten pp. 17–20 obtain the photon number density by integrating the Planck distribution, explain the temperature and dilution scalings, and compare the interaction time with the expansion time for thermal equilibrium. They also correct the missing cube and factor-of-two error in the paper's printed photon number density.

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.

The intended quantity is $\boxed{N_{\mathrm{th}}^{\ast }}$, the threshold population inversion density of the cavity after switching to low loss. At this value stimulated gain exactly balances the photon-loss rate:Above threshold a nonzero seed grows; below threshold it decays. The unstarred $N_{\mathrm{th}}$ in the subpart is shorthand for the starred threshold quantity in the displayed equations, rather than a different total atomic population. The photon lifetime ${\tau }_c$ fixes the photon-density loss rate $n/{\tau }_c$.

At the end of the pulse, the photon density returns to negligible size. Put $u=N_f^{\ast }/N_i^{\ast }$ and $r=N_i^{\ast }/N_{\mathrm{th}}^{\ast }$. The result of i givesUsing $x=\beta \eta =1-u$, the nonzero-pulse solution obeysFor $r>1$, select $0<x<1$, corresponding to depletion below threshold. The root $x=0$ is also an algebraic solution for every $r$ but represents no macroscopic pulse. A positive nonzero root exists only above threshold.

For a nonzero photon density, divide the laser rate equations to remove time:Integrating from the initial inversion and retaining a small seed density $n_i$ givesThe source's pulse approximation neglects $n_i$ compared with the macroscopic pulse density, givingA positive seed, or a spontaneous-emission source omitted from these equations, is nevertheless needed to start the pulse. Setting $n(0)=0$ exactly in the unforced equations gives the counterexample $n(t)=0$, $N^{\ast }(t)=N_i^{\ast }$ for all time, even above threshold. The displayed formula is therefore used in the negligible-seed limit, rather than as an assertion that a pulse starts from an identically zero field.

During the pulse, the photon density reaches its maximum when $\dot{n}=0$ with $n>0$, so $N^{\ast }=N_{\mathrm{th}}^{\ast }$. Evaluating i at this inversion givesThe ideal instantaneous output power is $P(t)=V_c\hslash {\omega }_{L}n(t)/{\tau }_c$. ThereforeFor $r\gg 1$, almost all the extractable energy is stored briefly as intracavity photons, and $P_{\max }\simeq E_{\max }/{\tau }_c$. The leading rise while the inversion remains near its initial value is $n\propto e^{(r-1)t/{\tau }_c}$, with short growth time ${\tau }_c/(r-1)$. Depletion then takes the inversion below threshold. At late times $N^{\ast }\simeq N_f^{\ast }$ andwhich approaches a tail with decay time ${\tau }_c$ at large $r$. The pulse is consequently asymmetric, with a rapid rise and a slower approximately exponential trailing edge. The negligible-seed build-up delay depends on the seed strength. The pumping and spontaneous population decay have been neglected over the short pulse; parasitic cavity loss multiplies useful power by the output fraction described in ii.

Use $X_i=n_i/n_B$, so $X_D$ counts deuterium nuclei per baryon, not baryons bound into deuterium. For the radiative capture reaction, chemical equilibrium and the zero photon chemical potential give ${\mu }_D={\mu }_n+{\mu }_p$. Dividing the three nonrelativistic Maxwell--Boltzmann number densities removes all chemical potentials:Here $g_D=3$ and $g_n=g_p=2$. The binding convention in this PDF is a negative mass defect, $B_D=m_D-m_n-m_p<0$. The positive nuclear binding energy is ${\epsilon }_D=-B_D$, so the abundance is enhanced by $e^{-B_D/T}$, not by $e^{B_D/T}$.

Since $X_D/(X_n{X}_p)=n_B{n}_D/(n_n{n}_p)$, insert the baryon-to-photon ratio $n_B=\eta n_{\gamma }$ and the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. This givesAt equal nucleon mass $m_N$ in the prefactors, $m_D\simeq 2m_N$, the deuterium equilibrium abundance simplifies toThe dimensionless prefactor is about $8.14$. These expressions describe thermal equilibrium; residual primordial deuterium after nuclear reactions cease must instead be obtained from the reaction kinetics.

The perturbations are density contrasts, ${\delta }_c=\delta {\rho }_c/{\bar{\rho }}_c$ and ${\delta }_r=\delta {\rho }_r/{\bar{\rho }}_r$, evaluated in the common comoving synchronous gauge. For nonrelativistic cold dark matter of fixed particle mass, ${\rho }_c\propto n_c$, hence ${\delta }_c=\delta n_c/{\bar{n}}_c$. Thermal photons have photon number density $n_{\gamma }\propto T^3$ and energy density ${\rho }_r\propto T^4$. A small local temperature perturbation therefore givesFor adiabatic initial conditions, the ratio $n_{\gamma }/n_c$ is unperturbed, so $\delta n_{\gamma }/{\bar{n}}_{\gamma }=\delta n_c/{\bar{n}}_c$. Combining these equations provesEquivalently ${\delta }_r/(1+w_r)={\delta }_c/(1+w_c)$, with $w_r=1/3$ and $w_c=0$. The relation applies before causal pressure gradients can separate the two species on superhorizon scales; it need not remain true after cosmological horizon crossing.

At neutrino decoupling, the interaction rate is comparable with the Hubble parameter. Equating the supplied rates givesHere the Planck mass is the unreduced $m_P=G^{-1/2}\simeq 1.22\times 10^{19}\mathrm{GeV}$ in natural units, $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$, and the effective number of relativistic energy degrees of freedom is $g_{\ast }=10.75$. The supplied rate gives an order-of-magnitude neutrino decoupling estimate; a precise neutron-proton conversion temperature uses its own weak-rate coefficients.

The chemical equilibrium condition for $n+{\nu }_e\leftrightarrow p+e^{-}$ is ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. With negligible lepton chemical potentials, ${\mu }_n={\mu }_p$. The nonrelativistic Maxwell--Boltzmann number density therefore givesbecause $g_n=g_p=2$ and the small mass difference can be neglected in the prefactor. This is the neutron-proton chemical equilibrium relation.

For the deuterium equilibrium abundance, ${\mu }_D={\mu }_n+{\mu }_p$ because photons have zero chemical potential. The sign convention in the paper is important: $B=m_D-m_n-m_p<0$, while the positive nuclear binding energy is ${\Delta }_D=-B$. Using spin multiplicities $g_D=3$, $g_n=g_p=2$ givesWith $n_B=\eta n_{\gamma }$, $m_n\simeq m_p\simeq m_N$, $m_D\simeq 2m_N$ and photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, this becomesHere $X_D=n_D/n_B$ is a number fraction; the fraction of baryons bound in deuterium is $2X_D$. The formula estimates the onset of the deuterium bottleneck opening while depletion of free neutrons and protons remains small. Because $\eta \ll 1$, stable deuterium becomes abundant only far below the positive nuclear binding energy, not simply at $T={\Delta }_D$.

Let $T_f$ denote cosmological weak freeze-out of the neutron-to-proton ratio, $r_f=e^{-Q/T_f}$, and let $t_N$ mark efficient nuclear burning. Free-neutron decay after freeze-out givesAlmost every surviving neutron enters helium-4, whose atomic nucleus contains two neutrons and four baryons. Thus neutron-limited helium synthesis givesA longer neutron half-life leaves more neutrons available, increasing the primordial helium mass fraction. If the longer lifetime reflects weaker weak interactions, the accompanying earlier cosmological weak freeze-out can reinforce that effect.

During radiation domination, $H\propto \sqrt{G{g}_{\ast }}{T}^2$ whereas the weak conversion rate is proportional to $T^5$. Hence $T_f\propto (G{g}_{\ast }{)}^{1/6}$. Increasing the effective number of relativistic energy degrees of freedom raises $T_f$, increasing $r_f$, and shortens the interval for free-neutron decay after freeze-out. Both increase the primordial helium mass fraction while nuclear burning remains efficient. The same reasoning gives the primordial helium response to a varying gravitational constant: larger $G$ over this interval produces more helium, with fixed microscopic rates and the approximate Friedmann equation unchanged. Smaller $G$ has the opposite effect. A particular varying-$G$ model must specify its background expansion and time history; the simple trend must not be extrapolated to changes so large that nuclear burning itself fails.

In the Maxwell-Boltzmann velocity distribution, each Cartesian component has mean-square velocity $k_B{T}_e/m_e$. Thus $⟨v^2⟩=3k_B{T}_e/m_e$. The original PDF calls this quantity an rms velocity, but the rms speed is its square root. Also, its $n_{\gamma }$ must mean the dimensionless mean occupation number $n$, since its equilibrium value is a Planck photon distribution; it is not the frequency-integrated photon number density.

Take $x=h\nu /(k_B{T}_{\gamma })$ at a fixed local radiation temperature. Because $x$ is proportional to frequency,The hot-electron Kompaneets equation therefore becomesFor traversal length $d\ell =cdt$, the Compton y parameter isIt measures electron pressure integrated along the line of sight, or the Thomson scattering optical depth weighted by $k_B{T}_e/(m_e{c}^2)$. In this hot-electron limit the mean fractional energy gain is approximately $4y$. The derivation assumes nonrelativistic electrons, Thomson-limit photon energies and $T_e\gg T_{\gamma }$. The original PDF uses $h\nu$, as required for ordinary frequency; the converted TeX incorrectly substitutes $\hslash \nu$.

Use natural units $\hslash =c=k_B=1$ and the unreduced Planck mass $M_P=G^{-1/2}$. Neutrino decoupling occurs when the weak interaction rate falls below the Hubble parameter. The weak-decoupling temperature estimate equates the given rate with the radiation-era expansion rate:Here $g_{\ast }$ counts the effective number of relativistic energy degrees of freedom, and the Fermi constant is $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$. The omitted reaction-rate coefficients make this an order-of-magnitude estimate: decoupling occurs at a temperature of order one MeV.

The charge-conserving channel is $n+{\nu }_e\leftrightarrow p+e^{-}$, together with $p+{\bar{\nu }}_e\leftrightarrow n+e^{+}$. The positron sign in the printed neutron-capture channel violates electric charge conservation. Chemical equilibrium requires ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Negligible lepton chemical potentials therefore give ${\mu }_n\simeq {\mu }_p$. Since neutrons and protons are nonrelativistic particles, their Maxwell-Boltzmann distribution givesBoth nucleons have two spin states, and their masses agree closely in the prefactor. Thus neutron-proton chemical equilibrium impliesThe heavier neutron is suppressed by a Boltzmann factor.

For the deuterium equilibrium abundance, retain the paper's convention $B=m_D-m_n-m_p<0$; the positive nuclear binding energy is ${\epsilon }_D=-B\simeq 2.22\mathrm{MeV}$. Chemical equilibrium of $n+p\leftrightarrow D+\gamma$ gives ${\mu }_D={\mu }_n+{\mu }_p$ because the equilibrium photon chemical potential vanishes. Using the nonrelativistic Maxwell--Boltzmann number density for each species and dividing givesHere $g_D=3$ and $g_n=g_p=2$. With the baryon-to-photon ratio $\eta =n_B/n_{\gamma }$, the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, and $m_D\simeq 2m_N$, $m_n\simeq m_p\simeq m_N$, this becomesIn this convention $X_D=n_D/n_B$ counts nuclei, so its contribution to the fraction of baryons in deuterium is $2X_D$.

The deuterium bottleneck explains why helium-4 formation waits until $T$ is far below ${\epsilon }_D$. At higher temperature, there are vastly more photons than baryons; even the dilute energetic tail can destroy newly formed deuterium. A crude estimate balances ${\eta }^{-1}{e}^{-{\epsilon }_D/T}$ against one, giving $T\sim {\epsilon }_D/\log ({\eta }^{-1})\sim 0.1\mathrm{MeV}$ for $\eta$ of order $10^{-9}$. The power-law factors in the deuterium equilibrium abundance and the nuclear reaction rates refine this estimate. Once stable deuterium accumulates, further reactions can efficiently place nearly all surviving neutrons into helium-4.

Write $r_N=n_n/n_p$ at the onset of nuclear burning. Each helium-4 nucleus requires two neutrons and contains four baryons. Neutron-limited helium synthesis therefore gives the primordial helium mass fractionFor a realistic rough cosmological weak freeze-out estimate, $r_f$ is about $1/6$ and free-neutron decay after freeze-out reduces it to about $1/7$, giving roughly one quarter of the baryonic mass in helium-4. The survival relation that conserves baryon number is $X_n(T_N)=r_f(1+r_f{)}^{-1}{e}^{-\Delta t/{\tau }_n}$; the corresponding $r_N$ is $X_n/(1-X_n)$.

The rough neutrino decoupling rate by itself does not determine that one-quarter value. Substituting $T_d=1.8\mathrm{MeV}$ literally into $e^{-Q/T_d}$ would give $r\simeq 0.49$ and $Y_4\simeq 0.65$ before neutron decay. Neutron-proton chemical equilibrium persists to a lower temperature, approximately $0.7$--$0.8\mathrm{MeV}$ once the actual conversion rates are included. The rate scaling fixes the MeV scale; accurate abundances need the conversion coefficients and the delay through the deuterium bottleneck. This sequence is described in David Tong's discussion of nucleosynthesis.

Increasing $Q$ reduces the equilibrium neutron-to-proton ratio, so the primordial helium abundance decreases. Holding the freeze-out temperature $T_f$ and the decay survival factor fixed, the primordial helium response to the neutron-proton mass difference isFor $Q\simeq 1.29\mathrm{MeV}$ and $T_f=0.7$--$0.8\mathrm{MeV}$ this is about $0.85$--$0.87$, a reduction of roughly fifteen percent. A larger $Q$ also changes neutron decay and detailed weak rates; their quantitative effects require more microscopic information. The sign from the Boltzmann factor is already clear.

Write $n_b=n_p+n_H$ for the cosmological baryon density in this hydrogen-only model, and $x_e=n_p/n_b$ for the hydrogen ionization fraction. By charge neutrality, $n_e=n_p$, so the Saha ionization equation becomesThe baryon-to-photon ratio is $\eta =n_b/n_{\gamma }$, and the equilibrium photon number density is $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. At an order-one hydrogen ionization fraction, neglecting factors of order one therefore givesReplacing $T$ inside the slowly varying logarithm by $B$ givesWith $B=13.6\mathrm{eV}$, $M_e=5\times 10^5\mathrm{eV}$ and $\eta =10^{-10}$, this is $T\simeq 0.351\mathrm{eV}\simeq 4.1\times 10^3\mathrm{K}$. Solving the preceding implicit approximation instead gives $T\simeq 0.306\mathrm{eV}\simeq 3.5\times 10^3\mathrm{K}$. Both are reasonable estimates of the recombination temperature at the accuracy requested. The small baryon abundance delays hydrogen recombination: even when $T\ll B$, the high-energy tail of the abundant photons can still ionize the much rarer hydrogen atoms.