Equilibrium Thermodynamics of the Early Universe

For the first fractions of a second, the universe was a furnace so dense that every particle collided with its neighbours billions of times before the cosmos expanded appreciably. When reactions run that much faster than the Hubble expansion, thermodynamics wins: each species forgets its history and relaxes to the unique maximum-entropy state compatible with the conserved charges. That state is astonishingly economical to describe. A whole plasma of photons, electrons, positrons, neutrinos, quarks and gluons is captured by two universal distribution functions — Bose-Einstein for integer spin, Fermi-Dirac for half-integer spin — each pinned by one shared temperature $T$ and, per conserved charge, a chemical potential $\mu$ . This is the statistical-mechanical bedrock beneath the hot Big Bang. From those distributions three macroscopic numbers follow by integration over momentum: the number density $n$ , the energy density $\rho$ , and the pressure $P$ — the only microphysical inputs the expanding universe demands. The Friedmann equation asks for the total $\rho(T)$ ; entropy conservation asks for $s(T)$ ; the reaction networks of nucleosynthesis and recombination measure their departures from these equilibrium values. The bookkeeping of every relativistic species compresses into a single running number, the effective degrees of freedom $g_*(T)$ for energy and $g_{*s}(T)$ for entropy — a staircase that drops each time a species goes cold. This article derives $n$ , $\rho$ and $P$ from first principles in both limits, fixes the equation of state, and threads the results into photon counting, neutrino temperatures and the seeds of Big Bang Nucleosynthesis.

Key takeaways

In the hot early universe, rapid interactions enforce thermal equilibrium, so every species is a Bose-Einstein or Fermi-Dirac phase-space distribution set by one shared temperature $T$ and a chemical potential $\mu$ .
Integrating those distributions over momentum yields the number density $n$ , energy density $\rho$ and pressure $P$ — the sole microphysical inputs to the Friedmann equations and to the thermal history.
For relativistic species ( $T\gg m$ , $\mu\approx0$ ), $\rho\propto g\,T^4$ and $n\propto g\,T^3$ , with fermions suppressed by the exact factors $7/8$ (energy) and $3/4$ (number) relative to bosons.
For non-relativistic species ( $T\ll m$ ) the density is Boltzmann-suppressed, $n\propto(mT)^{3/2}e^{-m/T}$ , so heavy particles disappear from the plasma as it cools.
All relativistic species pack into $g_*(T)$ ; it falls from $106.75$ above the electroweak scale to $10.75$ at $T\approx1$ MeV to $3.36$ today, and it governs entropy conservation, neutrino decoupling and recombination.
Key idea: a species in thermal equilibrium is fixed entirely by $T$ , $\mu$ , its mass $m$ and its internal degrees of freedom $g$ ; integrating its Bose-Einstein or Fermi-Dirac distribution gives $n$ , $\rho$ and $P$ , which are everything the expanding universe needs to know about it.

The core insight

A particle species in thermal equilibrium is fully characterised by $T$ , $\mu$ , its mass $m$ and its internal degrees of freedom $g$ ; integrating the Bose-Einstein or Fermi-Dirac distribution over phase space gives $n$ , $\rho$ and $P$ — the only microphysical inputs the expanding universe requires.

Where this lives today

Equilibrium thermodynamics is the running engine of every precision-cosmology pipeline. The functions $g_*(T)$ and $g_{*s}(T)$ are tabulated to high accuracy — including the lattice-QCD treatment of the quark-hadron transition — because they set the expansion rate $H\propto\sqrt{g_*}\,T^2$ during nucleosynthesis, where a percent-level error shifts the predicted helium and deuterium abundances that BBN codes test against quasar-absorption measurements. The same entropy bookkeeping fixes the effective neutrino number $N_{\rm eff}$ , whose refined value $3.044$ is a target CMB experiments (Planck, and next CMB-S4 and Simons Observatory) measure to hunt for extra relativistic species. And the relic neutrino background at $1.95$ K, a direct consequence of the $(4/11)^{1/3}$ derived above, is the quarry of the PTOLEMY experiment. Every one of these frontier measurements rests on the equilibrium integrals in this article.

The physical picture

Picture the early universe as a box of gas so hot and dense that collisions randomise everything far faster than the box expands. Statistical mechanics then guarantees a unique end-state: all interacting species share one temperature, and the average occupation of each momentum state is fixed by nothing but spin statistics. Bosons are gregarious — they pile into low-energy states without limit. Fermions obey Pauli exclusion — no state holds more than one, so their occupation never exceeds unity. The first figure is the whole story of the $\pm1$ in the denominator: where the fermion factors $7/8$ and $3/4$ are secretly born, and where all three statistics merge into the classical Boltzmann tail once energies climb above $T$ .

Figure

The equilibrium occupation number f = 1/(exp(E/T) ± 1) at μ = 0. Bosons (Bose-Einstein, −1) pile up without bound as E/T → 0, fermions (Fermi-Dirac, +1) never exceed f = 1/2 there, and both collapse onto the classical Maxwell-Boltzmann tail e^(−E/T) once E ≳ 3T — the merging that lets one set of integrals serve every species

Whether a species counts as "relativistic" is decided by the single ratio $m/T$ . When $T\gg m$ its rest mass is an afterthought and it behaves like pure radiation, $\rho\propto T^4$ and $P=\rho/3$ . When $T\ll m$ it costs an energy $\sim m$ merely to conjure the particle, so its abundance is throttled by $e^{-m/T}$ — the species freezes out and dumps its entropy into whatever lighter particles remain. As the universe cools it steps through a sequence of thresholds, and at each one the count of relativistic degrees of freedom drops. The second figure is the quantity a cosmologist actually reaches for: the running staircase $g_*(T)$ , carrying the three canonical plateaus that anchor the thermal history.

Figure

The effective relativistic degrees of freedom g⋆(T) fall as species become non-relativistic and annihilate: from g⋆ = 106.75 above the electroweak scale, through the QCD transition and the pion-muon shoulder at g⋆ = 17.25, to g⋆ = 10.75 at T ≈ 1 MeV (photons, e±, three neutrinos), and down to g⋆ = 3.36 today after e⁺e⁻ annihilation reheats the photons (with g⋆s = 3.91)

The $7/8$ and $3/4$ that separate fermions from bosons are not fudge factors; they are the arithmetic fingerprint of the exclusion principle, the alternating Fermi-Dirac series trimming the sum by exactly $(1-2^{1-s})$ relative to the all-positive Bose-Einstein one. Chemical potentials, meanwhile, are almost always negligible: the cosmic particle-antiparticle asymmetry is tiny (baryon-to-photon ratio $\sim10^{-9}$ ), so setting $\mu=0$ turns the relativistic densities into pure functions of $T$ and $g$ .

The core equations

Everything begins with the average occupation of a single quantum state. Kinetic and chemical equilibrium fix it to depend on the state only through its energy $E(p)$ , with the sign in the denominator selecting the statistics: $+$ for fermions, $-$ for bosons. This is the one input from which $n$ , $\rho$ and $P$ are all built.

f(p)=\frac{1}{\exp\!\left(\dfrac{E(p)-\mu}{T}\right)\pm 1} \qquad (1)

where: $f(p)$ — mean number of particles per quantum state of momentum $p$ ; $E(p)=\sqrt{p^2+m^2}$ — relativistic energy (natural units $c=k_B=\hbar=1$ ); $\mu$ — chemical potential; $T$ — temperature shared by all interacting species; $+$ — Fermi-Dirac (fermions); $-$ — Bose-Einstein (bosons). Reading it: the whole thermodynamic state of a species lives in this one function of $p$ . Summing $f(p)$ over all momentum states with the phase-space measure $g\,\mathrm{d}^3p/(2\pi)^3$ gives the number density; the angular integral contributes $4\pi$ , leaving a one-dimensional integral over $p$ .

n=\frac{g}{2\pi^2}\int_0^\infty \frac{p^2\,\mathrm{d}p}{\exp\!\left(\frac{E-\mu}{T}\right)\pm 1} \qquad (2)

where: $n$ — particles per unit physical volume; $g$ — internal degrees of freedom (spin, colour, particle/antiparticle counted separately); the prefactor $g/2\pi^2$ is $g\int\mathrm{d}^3p/(2\pi)^3=(g/2\pi^2)\int p^2\,\mathrm{d}p$ . Reading it: weight each momentum shell by its occupation and its phase-space volume. Weighting the same integrand by each mode's energy gives the energy density; weighting by the momentum flux $p^2/3E$ gives the pressure. These two, with $n$ , are the complete macroscopic description.

\rho=\frac{g}{2\pi^2}\int_0^\infty \frac{E(p)\,p^2\,\mathrm{d}p}{\exp\!\left(\frac{E-\mu}{T}\right)\pm 1},\qquad P=\frac{g}{6\pi^2}\int_0^\infty \frac{p^2}{E(p)}\,\frac{p^2\,\mathrm{d}p}{\exp\!\left(\frac{E-\mu}{T}\right)\pm 1} \qquad (3)

where: $\rho$ — energy per unit volume; $P$ — isotropic pressure, the $\tfrac13\langle p^2/E\rangle$ momentum flux. Reading it: in the relativistic limit $E\to p$ , the pressure integrand becomes the energy integrand, so $P=\rho/3$ ; in the non-relativistic limit $p^2/E\to p^2/m\ll m$ , so $P\ll\rho$ . Carrying out the massless, $\mu=0$ integrals collapses the sum over all relativistic species into one number, the effective count of relativistic degrees of freedom — every boson with weight $1$ , every fermion with weight $7/8$ .

\rho_{\mathrm{rel}}=\frac{\pi^2}{30}\,g_*(T)\,T^4,\qquad g_*(T)=\sum_{\mathrm{bosons}}g_i\left(\frac{T_i}{T}\right)^4+\frac{7}{8}\sum_{\mathrm{fermions}}g_i\left(\frac{T_i}{T}\right)^4 \qquad (4)

where: $g_*(T)$ — effective relativistic degrees of freedom for energy; $7/8$ — the fermion suppression factor; $(T_i/T)^4$ — accounts for species (like neutrinos after decoupling) that no longer share the photon temperature. Reading it: the total radiation energy is $T^4$ times a running headcount of who is still relativistic. The entropy density follows from the Gibbs relation at $\mu=0$ and is what stays conserved in a comoving volume. Its headcount $g_{*s}$ uses cube-of-temperature weights.

s=\frac{\rho+P}{T}=\frac{2\pi^2}{45}\,g_{*s}(T)\,T^3,\qquad g_{*s}(T)=\sum_{\mathrm{bosons}}g_i\left(\frac{T_i}{T}\right)^3+\frac{7}{8}\sum_{\mathrm{fermions}}g_i\left(\frac{T_i}{T}\right)^3 \qquad (5)

where: $s$ — entropy per unit volume; $g_{*s}$ — effective degrees of freedom for entropy; $\rho+P=\tfrac43\rho$ for radiation. Reading it: conservation of $S=s\,a^3$ is the workhorse that relates temperatures across the expansion. For a non-relativistic species the quantum $\pm1$ is irrelevant and the density is exponentially suppressed — the seed of both the Saha equation and of freeze-out.

n=g\left(\frac{mT}{2\pi}\right)^{3/2}\exp\!\left(\frac{\mu-m}{T}\right),\qquad \rho\simeq m\,n,\qquad P=nT\ll\rho \qquad (6)

where: the Boltzmann factor $e^{-m/T}$ is the cost of the rest mass; $\rho\simeq mn$ because rest energy dominates; $P=nT$ recovers the ideal-gas law with $w=P/\rho\approx T/m\to0$ . Reading it: heavy particles are exponentially rare once $T$ falls below their mass. Collecting the two regimes gives the equation of state that feeds the Friedmann equation directly.

P=w\rho,\qquad w=\begin{cases}1/3 & \text{relativistic}\\[2pt] \approx 0 & \text{non-relativistic}\end{cases} \qquad (7)

where: $w$ — equation-of-state parameter. Reading it: radiation ( $w=1/3$ ) and dust ( $w\approx0$ ) are the two limits the equilibrium integrals interpolate between as a species crosses its mass threshold.

Derivation ARelativistic densities from the Bose and Fermi integrals

The goal is to turn the abstract phase-space integrals of equations (2) and (3) into the closed-form powers of temperature that every cosmology calculation quotes, and to expose where the celebrated coefficients $\zeta(3)/\pi^2$ , $\pi^2/30$ and the fermion factors $3/4$ and $7/8$ come from. We work in the ultra-relativistic, chemical-potential-free limit — set $m=0$ so $E=p$ , and $\mu=0$ — the regime in which photons and every hot species actually live.

Step 1. Change variables to the dimensionless $x=p/T$ in the number-density integral, $n=\frac{g}{2\pi^2}\int_0^\infty p^2\,\mathrm{d}p/(e^{p/T}\pm1)=\frac{gT^3}{2\pi^2}\int_0^\infty x^2\,\mathrm{d}x/(e^{x}\pm1)$ . Why: the substitution factors the entire temperature dependence out front as $T^3$ , leaving a pure number to evaluate once and reuse forever.

Step 2. Evaluate the bosonic case by expanding the denominator, $1/(e^x-1)=\sum_{k\ge1}e^{-kx}$ , so $\int_0^\infty x^2/(e^x-1)\,\mathrm{d}x=\Gamma(3)\sum_k k^{-3}=2\zeta(3)$ , giving $n_{\mathrm{boson}}=\frac{\zeta(3)}{\pi^2}gT^3$ . Why: the sum over $k$ is precisely the Riemann zeta function, which is how $\zeta(3)\approx1.202$ enters cosmology.

Step 3. The fermionic denominator expands with alternating signs, $1/(e^x+1)=\sum_{k\ge1}(-1)^{k+1}e^{-kx}$ , so the sum becomes the Dirichlet eta $(1-2^{1-s})\zeta(s)$ ; with $s=3$ this is $(1-2^{-2})=3/4$ , giving $n_{\mathrm{fermion}}=\tfrac34 n_{\mathrm{boson}}$ . Why: the sign alternation is the direct arithmetic imprint of Pauli exclusion — the entire origin of the $3/4$ .

Step 4. Repeat for the energy density, where the extra factor of $E=p=Tx$ raises the integrand to $x^3$ and the prefactor to $T^4$ : $\rho=\frac{gT^4}{2\pi^2}\int_0^\infty x^3\,\mathrm{d}x/(e^x\pm1)$ , with $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x=\Gamma(4)\zeta(4)=6\cdot\pi^4/90=\pi^4/15$ . Why: the same series machinery now lands on $\zeta(4)=\pi^4/90$ , which is what converts the messy integral into the clean $\pi^2/30$ .

Step 5. Collect: $\rho_{\mathrm{boson}}=\frac{g}{2\pi^2}T^4\cdot\frac{\pi^4}{15}=\frac{\pi^2}{30}gT^4$ , and the fermionic version carries the $s=4$ eta factor $(1-2^{-3})=7/8$ , giving $\rho_{\mathrm{fermion}}=\frac78\cdot\frac{\pi^2}{30}gT^4$ . Why: these are exactly the weights that build $g_*$ in equation (4), so this single derivation underwrites the whole staircase. *Note: the key identities used above are $\Gamma(4)\zeta(4)=6\cdot\pi^4/90=\pi^4/15$ and $(1-2^{-3})=7/8$ .*

Go deeper · Derivation B · 1 minThe non-relativistic limit and Boltzmann suppression

Derivation BThe non-relativistic limit and Boltzmann suppression

The aim is to show how the same phase-space integral that gave clean powers of $T$ for radiation instead produces an exponential $e^{-m/T}$ once a species goes cold, because this Maxwell-Boltzmann form is the microscopic seed of every abundance calculation downstream — the neutron-to-proton ratio, the Saha ionisation fraction, and dark-matter freeze-out. We take the regime $T\ll m$ and track how the quantum statistics quietly switch themselves off.

Step 1. When $T\ll m$ the exponent $(E-\mu)/T$ is large and positive for every momentum, so the occupation number is tiny and the $\pm1$ in the denominator is negligible: $e^{(E-\mu)/T}\pm1\approx e^{(E-\mu)/T}$ . The integral becomes $n\approx\frac{g}{2\pi^2}\int_0^\infty p^2 e^{-(E-\mu)/T}\mathrm{d}p$ . Why: with occupation numbers far below one there is never a contest for states, so Bose and Fermi statistics become indistinguishable from classical Boltzmann counting.

Step 2. Expand the energy for $p\ll m$ , $E=\sqrt{p^2+m^2}\approx m+p^2/2m$ , so the exponential factorises into a constant rest-mass piece and a Gaussian in momentum: $e^{-(E-\mu)/T}=e^{(\mu-m)/T}\,e^{-p^2/2mT}$ . Why: the rest energy $m$ pulls out as the temperature-independent suppression $e^{-m/T}$ , which is the whole physical content of the limit.

Step 3. Do the Gaussian moment $\int_0^\infty p^2 e^{-p^2/2mT}\mathrm{d}p=\frac{\sqrt\pi}{4}(2mT)^{3/2}$ , and assemble the constants to reach $n=g\left(\frac{mT}{2\pi}\right)^{3/2}e^{(\mu-m)/T}$ . Why: this is the classical ideal-gas density with a thermal de Broglie volume $(2\pi/mT)^{3/2}$ , now carrying the exponential mass penalty that empties the plasma of heavy particles.

Step 4. Finally, $\rho\approx m\,n$ because each particle's energy is dominated by its rest mass, while $P=nT$ from the equipartition of the Gaussian kinetic energy, so $w=P/\rho\approx T/m\to0$ . Why: a cold gas contributes rest-mass energy but almost no pressure, which is why non-relativistic matter behaves as pressureless dust in the Friedmann equation. *Note: the Gaussian moment $\int_0^\infty p^2 e^{-ap^2}\mathrm{d}p=\tfrac14\sqrt\pi\,a^{-3/2}$ , evaluated at $a=1/2mT$ , is what produces the $(mT/2\pi)^{3/2}$ prefactor in the non-relativistic number density.*

Go deeper · Derivation C · 1 minEntropy density and the conservation of comoving entropy

Derivation CEntropy density and the conservation of comoving entropy

The purpose is to establish why $S=s\,a^3$ is conserved and why the entropy headcount $g_{*s}$ — not the energy headcount $g_*$ — relates temperatures across mass thresholds, because this one conservation law is what lets us compute the neutrino temperature, the photon reheating from electron-positron annihilation, and the redshift-temperature relation of the microwave background. We start from the first law for a comoving volume at negligible chemical potential.

Step 1. For an adiabatically expanding fluid the first law reads $\mathrm{d}(\rho a^3)=-P\,\mathrm{d}(a^3)$ , which rearranges to $\mathrm{d}[(\rho+P)a^3/T]=0$ when $\mu=0$ , identifying the conserved quantity as $S=s\,a^3$ with $s=(\rho+P)/T$ . Why: expansion does no work on the surroundings and adds no heat, so the total entropy in a comoving box is fixed even as $\rho$ , $P$ and $T$ all change.

Step 2. For a relativistic gas $P=\rho/3$ , so $s=(\rho+\rho/3)/T=\tfrac43\rho/T$ ; inserting $\rho=\frac{\pi^2}{30}g_{*s}T^4$ gives $s=\frac{2\pi^2}{45}g_{*s}T^3$ . Why: the enthalpy density $\rho+P=\tfrac43\rho$ is what turns the $\pi^2/30$ of the energy density into the $2\pi^2/45$ of the entropy density.

Step 3. Conservation of $S=s\,a^3$ then reads $g_{*s}\,T^3 a^3=\text{const}$ , so $T\propto g_{*s}^{-1/3}/a$ . Why: whenever a species annihilates and pours its entropy into the remaining plasma, $g_{*s}$ drops and $T$ falls more slowly than $1/a$ for a moment — the photons are reheated relative to anything already decoupled.

Step 4. Apply this across electron-positron annihilation to the decoupled neutrinos, which carry their own conserved $g_{*s}$ : the photon bath is heated but the neutrinos are not, leaving $T_\nu/T_\gamma=(4/11)^{1/3}$ . Why: a bare bookkeeping identity fixes the relic neutrino temperature to $1.95$ K today, from the $7/8$ fermion factor alone. *Note: with pre-annihilation $g_{*s}=2+\tfrac78\cdot4=11/2$ and post-annihilation photon $g_{*s}=2$ , entropy conservation gives $(T_\nu/T_\gamma)^3=4/11$ ; today $g_{*s}=3.91$ and $g_*=3.36$ .*

Limiting cases

The phase-space integrals collapse into simple closed forms in two opposite regimes, governed by the ratio $x\equiv m/T$ . Between them, near a mass threshold, only a full numerical evaluation of the Fermi-Dirac and Bose-Einstein integrals will do — which is exactly what happens across electron-positron annihilation.

Case	Result	Meaning
Ultra-relativistic bosons ( $T\gg m$ , $\mu=0$ )	$n=\frac{\zeta(3)}{\pi^2}g\,T^3,\ \ \rho=\frac{\pi^2}{30}g\,T^4,\ \ P=\tfrac13\rho$	Photons ( $g=2$ ) are the archetype; $\zeta(3)\approx1.202$
Ultra-relativistic fermions ( $T\gg m$ , $\mu=0$ )	$n=\frac34\frac{\zeta(3)}{\pi^2}g\,T^3,\ \ \rho=\frac78\frac{\pi^2}{30}g\,T^4,\ \ P=\tfrac13\rho$	The $3/4$ and $7/8$ are the Pauli-exclusion signatures
Non-relativistic ( $T\ll m$ , either statistic)	$n=g\left(\frac{mT}{2\pi}\right)^{3/2}e^{-(m-\mu)/T},\ \ \rho\simeq mn,\ \ P=nT\ll\rho$	Boltzmann suppression; quantum statistics irrelevant since $f\ll1$
Degenerate fermions ( $T\to0$ , $\mu>0$ )	$f\to1$ below $p_F$ , $0$ above	Fermi sea; white dwarfs and neutron stars, not the hot early universe

Common trapUsing one temperature for every species at every epoch

The most common blunder is to assume a single global temperature holds forever. It does not. A common $T$ is enforced only while a species' interaction rate exceeds the Hubble rate; the moment it decouples, its distribution simply redshifts and its effective temperature scales as $1/a$ on its own, free to drift away from the photon temperature. This is precisely why $T_\nu\neq T_\gamma$ after electron-positron annihilation: the neutrinos had already decoupled, so the entropy released by the annihilating pairs heated the photons but not them, leaving $T_\nu=(4/11)^{1/3}T_\gamma$ . A second, quieter error is to write $E=p$ everywhere; the exact energy is $E=\sqrt{p^2+m^2}$ , and the massless form is legitimate only in the strict relativistic limit.

Worked examples

1 · The photon number and energy density today (fully worked)

The CMB is a relativistic Bose gas with $g=2$ polarisation states at $T_0=2.725$ K. Find the present photon number density and energy density.

Step 1. Convert the temperature to natural (energy) units: $k_B T_0=(8.617\times10^{-5}\ \mathrm{eV/K})(2.725\ \mathrm{K})=2.348\times10^{-4}$ eV. To turn an energy into an inverse length use $\hbar c=1.973\times10^{-5}\ \mathrm{eV\,cm}$ , so $T_0=k_BT_0/\hbar c=11.90\ \mathrm{cm^{-1}}$ .

Step 2. Apply the relativistic boson number density from the limiting-case table: $n_\gamma=\frac{\zeta(3)}{\pi^2}g\,T_0^3=\frac{2\zeta(3)}{\pi^2}T_0^3=\frac{2(1.202)}{9.870}(11.90\ \mathrm{cm^{-1}})^3$ . Cubing gives $(11.90)^3=1685\ \mathrm{cm^{-3}}$ , and $2(1.202)/9.870=0.2436$ , so $n_\gamma=0.2436\times1685\approx411\ \mathrm{cm^{-3}}$ — roughly four hundred CMB photons in every cubic centimetre of space.

Step 3. For the energy density use $\rho_\gamma=\frac{\pi^2}{30}g\,T_0^4=\frac{\pi^2}{15}T_0^4$ ; in the same units this is $0.260\ \mathrm{eV\,cm^{-3}}$ , and dividing by the critical density gives the radiation share $\Omega_\gamma h^2\approx2.47\times10^{-5}$ . These are the standard anchor values for the radiation content of the universe.

2 · The temperature of the relic neutrino background (fully worked)

Before electron-positron annihilation the plasma held photons, electrons, positrons and three neutrino species in equilibrium. The neutrinos decouple just before annihilation. Find the neutrino-to-photon temperature ratio, and $T_\nu$ today.

Step 1. Count the entropy degrees of freedom coupled to the photons just before annihilation. Photons contribute $g=2$ ; electrons and positrons contribute $\frac78\times4=3.5$ (two spin states each, particle and antiparticle). So $g_{*s}^{\text{before}}=2+3.5=\tfrac{11}{2}$ .

Step 2. After the pairs annihilate, only photons remain in that bath, so $g_{*s}^{\text{after}}=2$ . Comoving entropy conservation for the photon sector gives $g_{*s}(aT_\gamma)^3=\text{const}$ , while the decoupled neutrinos independently keep $aT_\nu=\text{const}$ ; before annihilation $T_\nu=T_\gamma$ .

Step 3. Taking the ratio, $\left(\frac{T_\nu}{T_\gamma}\right)^3=\frac{g_{*s}^{\text{after}}}{g_{*s}^{\text{before}}}=\frac{2}{11/2}=\frac{4}{11}$ , so $T_\nu/T_\gamma=(4/11)^{1/3}=0.7138$ , and with $T_\gamma=2.725$ K, $T_\nu\approx1.95\ \mathrm{K}$ . The same factor gives today's headcounts $g_*=2+\frac78\cdot6\cdot(4/11)^{4/3}=3.36$ and $g_{*s}=2+\frac78\cdot6\cdot(4/11)=3.91$ .

3 · The neutron-to-proton ratio before freeze-out

While weak interactions keep neutrons and protons in equilibrium, their abundance ratio is Boltzmann-controlled by the mass gap $\Delta m=m_n-m_p=1.293$ MeV. Estimate the ratio at $T=0.8$ MeV, just before weak freeze-out.

Step 1. Both are non-relativistic at these temperatures, so equation (6) applies to each. Their prefactors $(mT/2\pi)^{3/2}$ are nearly identical because $m_n\approx m_p$ , so the ratio reduces to the exponential of the mass difference: $n_n/n_p=e^{-(m_n-m_p)/T}=e^{-\Delta m/T}$ .

Step 2. Insert the numbers: $\Delta m/T=1.293/0.8=1.62$ , so $n_n/n_p=e^{-1.62}=0.20$ , about one neutron for every five protons.

Step 3. This equilibrium ratio is the input Big Bang Nucleosynthesis inherits. Once the weak rates fall below the Hubble rate near $T\approx0.8$ MeV the ratio freezes near this value (drifting only through neutron decay), and essentially all surviving neutrons end up bound in helium-4 — fixing the primordial helium mass fraction near $25\%$ . The Boltzmann suppression sets the starting line for the light-element abundances.

Physics of the portrait

Equilibrium cosmology is a fusion of two nineteenth- and twentieth-century revolutions — statistical mechanics and general relativity — and the person who first married them rigorously was Richard Tolman.

Origins · 1 minRichard Tolman and how this was discovered

Richard Tolman

1881–1948 · relativistic thermodynamics and the expanding universe

Richard Chace Tolman was a physical chemist turned mathematical physicist who, more than anyone, established that thermodynamics could be applied consistently to the universe as a whole. Working at Caltech through its golden age, he asked a question that sounds naive but is deep: does a box of blackbody radiation stay in thermal equilibrium as the space containing it expands? His answer, codified in the 1934 classic Relativity, Thermodynamics and Cosmology, was yes — provided the temperature redshifts as $T\propto1/a$ , exactly the scaling the equilibrium entropy density demands. This is why we can still speak of a photon "temperature" 13.8 billion years after the photons last interacted: Tolman proved the Planck spectrum is preserved by free expansion, with only its temperature sliding down. The distribution functions themselves were the gift of others — Bose and Einstein for integer spin in 1924, Fermi and Dirac for the exclusion principle in 1926 — and the numerical thermal history was later built by Gamow, Alpher and Herman, whose 1948 relic-background estimate pointed straight at the CMB that Penzias and Wilson found in 1965. But it was Tolman who made the thermodynamics of an expanding universe a rigorous subject rather than a plausibility argument.

History

1900

— Planck fixes the equilibrium spectrum of blackbody radiation

1924

— Bose and Einstein formulate the statistics of integer-spin particles

1926

— Fermi and Dirac formulate the statistics of the exclusion principle

1934

— Tolman treats the expanding universe as a thermodynamic system in Relativity, Thermodynamics and Cosmology

1948

— Alpher, Gamow and Herman use equilibrium densities to predict a relic radiation background of a few kelvin

Going further

Texts. Ryden, Introduction to Cosmology, 2nd ed., Ch. 9 · Dodelson, Modern Cosmology, 2nd ed., Ch. 2–3 · Kolb & Turner, The Early Universe, Ch. 3 · Carroll & Ostlie, An Introduction to Modern Astrophysics, Ch. 30 · Mo, van den Bosch & White, Galaxy Formation and Evolution, Ch. 3 · Weinberg, Cosmology, Ch. 3. The coefficients $\zeta(3)\approx1.202$ , $\pi^2/30$ , $2\pi^2/45$ , $7/8$ and $3/4$ all follow from the $\Gamma(s)\zeta(s)$ and Dirichlet-eta identities, with the key steps being $\int_0^\infty x^3/(e^x-1)\,\mathrm{d}x=\Gamma(4)\zeta(4)=\pi^4/15$ and the Gaussian moment $\int_0^\infty p^2 e^{-ap^2}\mathrm{d}p=\tfrac14\sqrt\pi\,a^{-3/2}$ . Numerical anchors: $n_\gamma\approx411\ \mathrm{cm^{-3}}$ and $\Omega_\gamma h^2\approx2.47\times10^{-5}$ at $T_0=2.725$ K; $g_*=106.75$ (electroweak), $17.25$ (post-QCD), $10.75$ (MeV era), $3.36$ (today) with $g_{*s}=3.91$ ; $T_\nu/T_\gamma=(4/11)^{1/3}=0.714$ ; $n_n/n_p=e^{-1.293/0.8}\approx0.20$ ; $\eta\approx6\times10^{-10}$ .

Prerequisites. Statistical Mechanics (Bose-Einstein / Fermi-Dirac distributions) · Special Relativity (energy-momentum relation) · Friedmann Equations and the Expanding Universe. Leads to. Thermal History of the Universe · Effective Number of Relativistic Degrees of Freedom · Entropy Conservation and Comoving Entropy · Decoupling and Freeze-Out · Neutrino Decoupling and the Cosmic Neutrino Background · Recombination and the Saha Equation.

Ask the tutor: Ask me to walk through how the $7/8$ fermion factor propagates from the Fermi-Dirac integral all the way to the $1.95$ K neutrino background, or to build the $g_*(T)$ staircase species-by-species for any epoch you name.

Why can a whole universe of particles be described by just T, μ and g?

Because thermal equilibrium erases microscopic history. When collisions are much faster than the expansion, a species reaches the unique maximum-entropy state consistent with its conserved charges, and the second law guarantees that state is the Bose-Einstein or Fermi-Dirac distribution. That distribution depends on each quantum state only through its energy, and on the species only through its temperature $T$ , chemical potential $\mu$ , mass $m$ and degrees of freedom $g$ . All the intractable many-body dynamics is thrown away, leaving four numbers per species — which is why cosmology can predict the CMB and the light-element abundances without ever solving the full equilibrium collision network.

What exactly is g, and how do I count it for a given particle?

The degrees of freedom $g$ count the distinct quantum states a particle can occupy at a given momentum. A photon has $g=2$ (two transverse polarisations). A massive spin- $\tfrac12$ electron has $g=2$ spin states, with the antiparticle counted separately, so an electron-positron pair contributes $g=4$ . Quarks carry an extra factor of $3$ for colour. Standard-Model neutrinos have one helicity each, so three flavours plus antiparticles give $g=6$ . In $g_*$ every bosonic degree of freedom enters with weight $1$ and every fermionic one with weight $7/8$ — getting $g$ right for each species is the whole of building the $g_*(T)$ staircase.

Where do the factors 7/8 and 3/4 actually come from?

They come from the difference between adding and alternating a series. The Bose-Einstein denominator $1/(e^x-1)$ expands into an all-positive series summing to the Riemann zeta $\zeta(s)$ ; the Fermi-Dirac denominator $1/(e^x+1)$ instead gives an alternating series summing to the Dirichlet eta $(1-2^{1-s})\zeta(s)$ . For the energy density $s=4$ , so the fermion-to-boson ratio is $1-2^{-3}=7/8$ ; for the number density $s=3$ , giving $1-2^{-2}=3/4$ . Both are exact imprints of Pauli exclusion: fermions cannot crowd the low-energy states the way bosons can, so they carry slightly less energy and fewer particles.

If neutrinos decoupled, how can we still assign them a temperature?

A decoupled species keeps a thermal-shaped distribution even without collisions, because free expansion redshifts every momentum by the same factor $1/a$ . A Fermi-Dirac distribution in $p/T$ simply continues to look Fermi-Dirac with $T\propto1/a$ — the functional form is frozen, only the scale slides. So the cosmic neutrino background is genuinely thermal at a well-defined $1.95$ K today, even though the neutrinos last scattered when the universe was one second old. This is Tolman's insight applied to fermions: expansion preserves the equilibrium spectrum and merely cools it — the same reason the photon background stays a perfect blackbody long after photons stopped interacting.

Why does the baryon-to-photon ratio η matter so much?

Because it is the one number the equilibrium formalism cannot supply, and everything else keys off it. The photon density $n_\gamma\propto T^3$ is fixed by equilibrium thermodynamics, while the net baryon number is a nearly conserved relic of an earlier asymmetry. Their ratio $\eta=n_b/n_\gamma\approx6\times10^{-10}$ is the single free parameter of Big Bang Nucleosynthesis: it sets how quickly protons and neutrons build deuterium and helium, and thus the primordial abundances. The same $\eta$ is independently pinned by the acoustic peaks of the CMB, and the two agree — one of the tightest consistency checks in cosmology.

When must I abandon the equilibrium formulas and solve the Boltzmann equation?

Precisely when a reaction rate falls below the Hubble rate — the condition for freeze-out. As long as $\Gamma\gg H$ the species tracks its equilibrium abundance and the algebraic formulas here are exact. Once $\Gamma\lesssim H$ it can no longer keep up: its abundance departs from equilibrium and freezes at a relic value the equilibrium formulas would badly misestimate. Computing that relic requires integrating the Boltzmann equation with a collision term, using the equilibrium densities as the reference the actual abundance is chasing. Dark-matter freeze-out, neutron-proton freeze-out, and recombination are all of this type — equilibrium sets the stage, but the final abundance is a non-equilibrium calculation.