Electron-Positron Annihilation and Photon Reheating

For a few seconds when the universe was about one second old, its entire radiation budget was rewritten by a single reaction. Photons, electrons, positrons, and neutrinos had until then shared one temperature, locked together by fast electromagnetic and weak collisions. Then two things happened in close succession. First, near $T\sim 1\;\mathrm{MeV}$ the weak reactions that tied neutrinos to the plasma fell behind the expansion, and the neutrinos decoupled — from that moment they free-streamed, their temperature simply redshifting as $T_\nu\propto a^{-1}$ . Second, the temperature dropped below the electron rest energy $m_e c^2=0.511\;\mathrm{MeV}$ , so electron–positron pairs could no longer be replenished and annihilated wholesale, $e^+e^-\to 2\gamma$ . Because the neutrinos had already left, all of that rest-mass energy and entropy poured into the photon gas and nowhere else. The photons were reheated relative to the neutrinos. The size of the effect is fixed exactly by the conservation of comoving entropy. Counting the entropy-weighted degrees of freedom before annihilation (photons plus $e^\pm$ ) and after (photons only) forces the photon comoving temperature up by $(11/4)^{1/3}$ , leaving the sharp, parameter-free prediction $T_\nu/T_\gamma=(4/11)^{1/3}\approx 0.714$ . That one number is why the relic neutrino sea today is colder than the microwave background, why the radiation energy density is written through $N_{\rm eff}$ , and why the expansion rate during nucleosynthesis — and hence the primordial helium abundance — comes out as it does.

Key takeaways

After neutrinos decouple near $T\sim 1\;\mathrm{MeV}$ , the temperature falls below $m_ec^2=0.511\;\mathrm{MeV}$ and $e^+e^-$ pairs annihilate to photons, dumping entropy into the photon bath but not into the already-decoupled neutrinos.
Comoving-entropy conservation in the coupled photon– $e^\pm$ fluid fixes the transfer exactly: the entropy degrees of freedom drop from $g_{*S}=11/2$ to $2$ , boosting the photon comoving temperature by $(11/4)^{1/3}\approx 1.401$ .
The result is the parameter-free ratio $T_\nu/T_\gamma=(4/11)^{1/3}\approx 0.7138$ , frozen for all later times.
With the measured CMB at $T_\gamma=2.7255\;\mathrm{K}$ this predicts a relic neutrino background at $T_\nu=1.945\;\mathrm{K}$ and $n_\nu\approx 336\;\mathrm{cm^{-3}}$ (about $112\;\mathrm{cm^{-3}}$ per flavour).
Across annihilation the total entropy count steps from $g_{*S}=10.75$ down to $g_{*S}=3.909$ ; the radiation energy is then bookkept through the effective neutrino number, with $N_{\rm eff}=3.044$ once non-instantaneous decoupling and QED corrections are included.
This epoch sets the radiation content that governs the expansion rate during Big Bang Nucleosynthesis, and thereby the light-element yields.
Key idea: entropy released by $e^+e^-$ annihilation goes only into the photon bath because the neutrinos have already decoupled; comoving-entropy conservation then locks in $T_\nu/T_\gamma=(4/11)^{1/3}$ .

The core insight

Entropy released by $e^+e^-$ annihilation goes only into the photon bath because neutrinos have already decoupled; comoving-entropy conservation then locks in $T_\nu/T_\gamma=(4/11)^{1/3}$ .

Where this lives today

The $(4/11)^{1/3}$ ratio is both an active input and an active target across cosmology. Every Big Bang Nucleosynthesis code uses the reheated photon temperature and the colder neutrino background to set the expansion rate, and the tight agreement between predicted and observed primordial deuterium and helium confirms that the thermal history is understood. Planck measures $N_{\rm eff}=2.99\pm 0.17$ from the damping tail and phase of the CMB acoustic peaks, consistent with the theoretical $3.044$ ; any excess would signal extra light relics — a sterile neutrino, a thermalized axion, or a dark-sector particle present at annihilation — and CMB-S4 aims to reach $\sigma(N_{\rm eff})\approx 0.03$ , enough to detect or exclude whole classes of them. Most ambitiously, the PTOLEMY experiment is attempting the direct capture of relic neutrinos on tritium, $\nu_e+\,^3\mathrm{H}\to\,^3\mathrm{He}+e^-$ , which would turn the predicted $1.95\;\mathrm{K}$ , $336\;\mathrm{cm^{-3}}$ neutrino sea into a measured background — the first direct detection of matter left from the universe's first second.

The physical picture

Think of two gases once thermally welded together and now disconnected. One is the photon gas, still tied to $e^\pm$ by rapid scattering; the other is the neutrino gas, just decoupled and coasting freely. While both were coupled they cooled in lockstep at a shared temperature. Once the neutrinos let go, the only thing that changes their temperature is the stretching of space — they redshift as $a^{-1}$ and nothing else touches them. The photons, by contrast, are about to receive a windfall. That windfall is the annihilation of the $e^\pm$ pairs. As the temperature slides below the electron mass, pair creation can no longer keep up with annihilation, and the electrons and positrons convert almost entirely into photons. Their rest-mass energy and, more to the point, their entropy have one open channel: the photon bath they are still coupled to. The neutrinos, thermally severed, get none of it. The photon gas therefore cools more slowly than a freely redshifting gas would during the annihilation, and emerges warmer than the neutrinos by a fixed factor.

Figure

Two gases that were once locked together drift apart: neutrinos decouple near T ≈ 1 MeV, then e⁺e⁻ pairs annihilate into photons near kT ≈ mₑc² = 0.511 MeV, dumping their entropy into the photon bath alone, so the photons end up hotter than the neutrinos by the factor (11/4)^(1/3) ≈ 1.401 while the neutrinos keep cooling as a⁻¹

Entropy — not energy — is the right ledger, because comoving entropy is conserved exactly in the adiabatic expansion even while energy is shuffled between species. The entropy of a relativistic bath scales as $g_{*S}(aT)^3$ , where $g_{*S}$ counts the entropy-weighted degrees of freedom. Before annihilation the coupled fluid holds photons ( $g_{*S}=2$ ) plus $e^\pm$ , which as fermions carry a factor $7/8$ over four spin states: $g_{*S}=2+\tfrac{7}{8}(4)=\tfrac{11}{2}$ . Afterward only photons remain, $g_{*S}=2$ . Demanding $g_{*S}(aT_\gamma)^3$ be unchanged forces $aT_\gamma$ up by $(11/4)^{1/3}$ — and since $aT_\nu$ never jumped, $T_\nu/T_\gamma$ is pinned at $(4/11)^{1/3}$ . Applied to the universe as a whole, this appears as a step in the $g_{*S}(T)$ curve: from $g_{*S}=10.75$ (photons, $e^\pm$ , and three neutrino flavours at one temperature) down to $g_{*S}=3.909$ once the pairs annihilate and the neutrinos are left colder. That drop is the signature of the reheating, and every later step of the thermal history inherits it.

The core equations

The whole result is packed into one ratio. It is exact in the instantaneous-decoupling limit, carries no free parameters, and is the sharpest handle we have on the physics of the first second.

\frac{T_\nu}{T_\gamma}=\left(\frac{4}{11}\right)^{1/3}\approx 0.7138 \qquad (1)

where: $T_\nu$ — temperature of the decoupled, free-streaming neutrino background; $T_\gamma$ — photon (CMB) temperature after reheating. Reading: the neutrinos end up cooler than the photons by a fixed factor a little under three-quarters, and because both later redshift identically as $a^{-1}$ , that factor is frozen for all time. Everything traces back to a single conservation law. The entropy density of a relativistic bath is set by its temperature and its degree-of-freedom count, and the total comoving entropy is invariant under adiabatic expansion.

s=\frac{2\pi^2}{45}\,g_{*S}\,T^3,\qquad S=s\,a^3=\mathrm{const} \qquad (2)

where: $s$ — entropy density; $g_{*S}$ — effective entropy degrees of freedom; $a$ — scale factor; $S$ — comoving entropy, conserved adiabatically. Reading: since $S\propto g_{*S}(aT)^3$ , any drop in $g_{*S}$ is paid for by a rise in $aT$ — the temperature falls more slowly than $a^{-1}$ through a threshold. The counter $g_{*S}$ sums each species weighted by its spin states, its temperature relative to the photons, and a factor $7/8$ for fermions from Fermi–Dirac versus Bose–Einstein statistics.

g_{*S}=\sum_{\rm bosons} g_i\left(\frac{T_i}{T}\right)^3+\frac{7}{8}\sum_{\rm fermions} g_i\left(\frac{T_i}{T}\right)^3 \qquad (3)

where: $g_i$ — internal (spin) degrees of freedom of species $i$ ; $T_i$ — its temperature, which may differ from the photon temperature $T$ after decoupling; $7/8$ — the fermion statistical factor. Reading: after neutrinos and photons part ways their $T_i$ differ, so the $(T_i/T)^3$ weights matter and $g_{*S}$ is no longer a simple integer. Applied across the annihilation, the coupled fluid's counter drops from $11/2$ to $2$ . This is the whole engine of the effect.

g_{*S}^{\rm before}=2+\frac{7}{8}(2+2)=\frac{11}{2}=5.5\quad\longrightarrow\quad g_{*S}^{\rm after}=2 \qquad (4)

where: the leading $2$ is the two photon polarizations; $2+2$ are the spin states of $e^-$ and $e^+$ ; after annihilation only the photons remain. Reading: the ratio of after to before is $2/(11/2)=4/11$ , and this ratio of counters becomes the cube of the temperature ratio. Once the ratio is fixed, the neutrino energy density is written relative to the photons through the effective neutrino number, which absorbs both the temperature ratio and small corrections to instantaneous decoupling.

\rho_\nu=\frac{7}{8}\left(\frac{4}{11}\right)^{4/3}N_{\rm eff}\,\rho_\gamma \qquad (5)

where: $\rho_\gamma=\tfrac{\pi^2}{15}T_\gamma^4$ is the photon energy density (natural units); $N_{\rm eff}=3.044$ in the Standard Model; the quartic power $(4/11)^{4/3}$ appears because energy scales as $T^4$ . Reading: numerically $\rho_\nu\approx 0.69\,\rho_\gamma$ , so the neutrinos still carry a comparable share of the radiation budget despite being colder. Collecting photons and neutrinos gives the total relativistic energy count that governs the expansion rate long after annihilation.

g_*^{\rm rad}=2+\frac{7}{8}\cdot 2\cdot N_{\rm eff}\left(\frac{4}{11}\right)^{4/3}\approx 3.38 \qquad (6)

where: the $2$ is the photon contribution and the second term the three cooled neutrino species with $N_{\rm eff}\approx 3.044$ . Reading: this multiplies $\tfrac{\pi^2}{30}T_\gamma^4$ to give the radiation density that enters $H^2\propto\rho$ during and after nucleosynthesis. Finally, the ratio plus the measured CMB temperature predicts the relic neutrino temperature today, a genuine number-out-of-the-theory.

T_\nu(t_0)=\left(\frac{4}{11}\right)^{1/3}T_\gamma(t_0)=1.945\;\mathrm{K} \qquad (7)

where: $T_\gamma(t_0)=2.7255\;\mathrm{K}$ is the COBE/FIRAS-measured CMB temperature. Reading: the cosmic neutrino background is real and predicted to sub-percent precision, even though its energy is far too low for present detectors to register directly.

Derivation AWhy comoving entropy is conserved and gives a fixed degree-of-freedom product

The goal is to establish the master relation on which everything rests: the product $g_{*S}(aT)^3$ is constant through the expansion, so that whenever the count $g_{*S}$ drops across a mass threshold, the temperature must rise above pure redshifting to compensate. We build it from the first law applied to a comoving volume together with the relativistic-gas relation between entropy, energy, and pressure.

Step 1. Take a comoving volume $V\propto a^3$ of the plasma and apply the first law in the form $\mathrm{d}(\rho a^3)=-p\,\mathrm{d}(a^3)$ , which states that the change in internal energy equals the work done by expansion, with no heat entering or leaving because the whole universe expands adiabatically. Why: there is no "outside" to exchange heat with, so the expansion is adiabatic by construction and the first law reduces to an energy–work balance.

Step 2. For a relativistic gas with negligible chemical potential the entropy density obeys $s=(\rho+p)/T$ . Differentiate the comoving entropy $S=s a^3=(\rho+p)a^3/T$ and use $\mathrm{d}p/\mathrm{d}T=(\rho+p)/T$ (the relativistic Gibbs–Duhem relation) together with the first law from Step 1. Why: this pair of thermodynamic identities is exactly what is needed to turn the energy balance into a statement about entropy.

Step 3. The algebra collapses to $\mathrm{d}(sa^3)=0$ , i.e. $S=\mathrm{const}$ . Substituting $s=\tfrac{2\pi^2}{45}g_{*S}T^3$ gives $g_{*S}(aT)^3=\mathrm{const}$ . Why: the entropy density carries all its temperature dependence in $T^3$ and all its species content in $g_{*S}$ , so conservation of $sa^3$ is precisely conservation of $g_{*S}(aT)^3$ .

Step 4. Read off the consequence at a threshold. If $g_{*S}$ falls from $g_1$ to $g_2$ while the fluid stays in equilibrium, then $g_1(a_1T_1)^3=g_2(a_2T_2)^3$ , so $a_2T_2=(g_1/g_2)^{1/3}a_1T_1>a_1T_1$ : the comoving temperature is boosted by $(g_1/g_2)^{1/3}$ . Why: this is the single lever that converts an entropy-count change into a temperature jump, and it is what heats the photons at annihilation. *The key checks: $\mathrm{d}(sa^3)=0$ follows directly from the first law together with $s=(\rho+p)/T$ ; the entropy density of a relativistic bath is $s=\tfrac{2\pi^2}{45}g_{*S}T^3$ ; and the comoving-temperature boost $(g_1/g_2)^{1/3}$ at a threshold is the immediate consequence.*

Go deeper · Derivation B · 2 minDeriving the neutrino-to-photon temperature ratio

Derivation BDeriving the neutrino-to-photon temperature ratio

The aim is the exact value $T_\nu/T_\gamma=(4/11)^{1/3}$ , obtained by tracking two separately conserved comoving entropies through the $e^+e^-$ annihilation — the neutrinos', which is untouched, and the photon– $e^\pm$ fluid's, which sheds the pairs into photons. We hold the neutrino sector fixed, boost the photon sector by degree-of-freedom counting, and divide the two.

Step 1. The neutrinos decoupled before annihilation, so their comoving entropy is conserved on its own: $g_{*S}^\nu(aT_\nu)^3=\mathrm{const}$ . Since no neutrino species appears or disappears here, $g_{*S}^\nu$ is fixed and therefore $aT_\nu=\mathrm{const}$ , i.e. $T_\nu\propto a^{-1}$ — pure redshifting. Why: a decoupled, free-streaming gas has nothing to exchange entropy with, so its only temperature change is the geometric dilution of expansion.

Step 2. The electromagnetically coupled fluid conserves its own comoving entropy across the annihilation, with the counter dropping from $g_{*S}^{\rm before}=\tfrac{11}{2}$ to $g_{*S}^{\rm after}=2$ :

\frac{11}{2}(a_1T_1)^3=2\,(a_2T_{\gamma,2})^3

Why: photons and $e^\pm$ stay in mutual equilibrium right through annihilation, so their combined comoving entropy — and only that — is the conserved quantity, and the count of relativistic species in it changes.

Step 3. Solve for the photon comoving temperature after annihilation: $(a_2T_{\gamma,2})^3=\tfrac{11}{4}(a_1T_1)^3$ , hence $a_2T_{\gamma,2}=(11/4)^{1/3}a_1T_1$ . The photon $aT$ is boosted by $(11/4)^{1/3}\approx 1.401$ relative to what pure redshifting would have given. Why: this is the master lever of Derivation A with $g_1/g_2=11/4$ ; the "reheating" is exactly this boost of the comoving photon temperature.

Step 4. Combine the two sectors. Before annihilation $T_1=T_{\nu,1}$ , and Step 1 gives $a_2T_{\nu,2}=a_1T_1$ . Dividing the neutrino relation by the photon relation, the scale factors and the common $a_1T_1$ cancel:

\frac{T_{\nu,2}}{T_{\gamma,2}}=\frac{a_2T_{\nu,2}}{a_2T_{\gamma,2}}=\frac{a_1T_1}{(11/4)^{1/3}a_1T_1}=\left(\frac{4}{11}\right)^{1/3}

Why: expressing both sectors through the common pre-annihilation value $a_1T_1$ is what makes the model-independent ratio drop out, with no reaction rates or timescales entering.

Step 5. Evaluate: $(4/11)^{1/3}=0.71377\ldots$ . The result is exact in the instantaneous-decoupling approximation and depends on nothing but the counting of degrees of freedom. Why: because the derivation used only conserved entropies and species counts, the number is a clean prediction of the standard thermal history rather than a fit. *The key steps check out explicitly: $g_{*S}^{\rm before}=2+\tfrac{7}{8}(4)=\tfrac{11}{2}$ ; the ratio of after to before is $2/(11/2)=4/11$ ; and $(4/11)^{1/3}=0.7138$ .*

Go deeper · Derivation C · 1 minThe relic neutrino background temperature and number density today

Derivation CThe relic neutrino background temperature and number density today

The purpose is to turn the frozen ratio into concrete present-day predictions: the temperature of the cosmic neutrino background and how many relic neutrinos fill each cubic centimetre right now. After annihilation both temperatures redshift identically, so the ratio survives to today; we combine it with the measured CMB temperature and the equilibrium number-density formulae for bosons and fermions.

Step 1. After annihilation the photon and neutrino temperatures both scale as $a^{-1}$ , so $T_\nu/T_\gamma$ is constant from then until now. With $T_\gamma(t_0)=2.7255\;\mathrm{K}$ ,

T_\nu(t_0)=\left(\frac{4}{11}\right)^{1/3}\times 2.7255\;\mathrm{K}=1.945\;\mathrm{K}

Why: the whole point of the reheating result is that it is a ratio of comoving temperatures, which is redshift-invariant, so measuring $T_\gamma$ today fixes $T_\nu$ today.

Step 2. The photon number density follows from the Bose–Einstein relation $n_\gamma=\tfrac{2\zeta(3)}{\pi^2}T_\gamma^3$ , which at $2.7255\;\mathrm{K}$ evaluates to $n_\gamma\approx 411\;\mathrm{cm^{-3}}$ . Why: photons are the one species whose absolute abundance we know precisely, so all relic-neutrino counts are best expressed as ratios to it.

Step 3. A fermionic species carries an extra factor $3/4$ relative to a boson with the same degrees of freedom, and scales as $T^3$ . Per flavour (neutrino plus antineutrino), $n_\nu=\tfrac{3}{4}\left(\tfrac{4}{11}\right)n_\gamma=\tfrac{3}{11}n_\gamma\approx 112\;\mathrm{cm^{-3}}$ . Why: the $3/4$ is the Fermi–Dirac versus Bose–Einstein number ratio, and $(T_\nu/T_\gamma)^3=4/11$ converts photon counts to neutrino counts.

Step 4. Summing the three flavours gives the total relic neutrino density,

n_\nu^{\rm tot}=3\cdot\frac{3}{11}\,n_\gamma=\frac{9}{11}\,n_\gamma\approx 336\;\mathrm{cm^{-3}}

so roughly three hundred relic neutrinos occupy every cubic centimetre around us. Why: this is the concrete, if undetectable, population predicted by the reheating physics — a direct, countable consequence of the $(4/11)^{1/3}$ ratio. *The concrete numbers are: $T_\nu\approx 1.95\;\mathrm{K}$ , $n_\gamma\approx 411\;\mathrm{cm^{-3}}$ , $n_\nu\approx 112\;\mathrm{cm^{-3}}$ per flavour, and $\approx 336\;\mathrm{cm^{-3}}$ total.*

Limiting cases

The temperature ratio and the radiation content take clean limiting forms on either side of the annihilation and at the level of precision cosmology.

Case	Result	Meaning
Well before annihilation ( $T\gg m_e$ )	$T_\nu=T_\gamma,\quad g_{*S}=10.75$	Photons, $e^\pm$ , and three neutrino species share one temperature.
Well after annihilation ( $T\ll m_e$ )	$\dfrac{T_\nu}{T_\gamma}=\left(\dfrac{4}{11}\right)^{1/3}\approx 0.7138$	Photons reheated, neutrinos left colder; $g_{*S}=2+\tfrac{7}{8}\cdot 6\cdot\tfrac{4}{11}=3.909$ .
Today	$T_\nu=1.945\;\mathrm{K}\ (T_\gamma=2.7255\;\mathrm{K})$	The relic neutrino temperature from the frozen ratio and the measured CMB.
Precision (non-instantaneous decoupling)	$N_{\rm eff}=3.044$	Slight residual heating of neutrinos plus finite- $T$ QED raises $N_{\rm eff}$ above 3.

Common trapThe photons are "reheated" — so their temperature went up

No temperature ever rises here; the universe cools monotonically throughout. "Reheating" is relative bookkeeping, not an absolute rise: the entropy dumped by the annihilating pairs makes the photon temperature fall more slowly than $a^{-1}$ during annihilation, so afterward the photons are warmer than the neutrinos than they otherwise would have been. A related trap: track this with the entropy count $g_{*S}$ , not the energy count $g_*$ . The two are equal only while everything shares one temperature; once neutrinos and photons reach different temperatures they diverge ( $g_{*S}=3.909$ versus $g_*=3.38$ today), and only the entropy count is conserved as $g_{*S}(aT)^3$ . Deriving the ratio from $g_*$ gives the wrong answer.

Worked examples

1 · The relic neutrino background today (fully worked)

Given the measured CMB temperature $T_\gamma=2.7255\;\mathrm{K}$ and photon density $n_\gamma=411\;\mathrm{cm^{-3}}$ , predict the cosmic neutrino background temperature, the number of relic neutrinos per cubic centimetre, and their typical energy scale today.

Step 1. Apply the frozen ratio: $T_\nu=(4/11)^{1/3}\,T_\gamma=0.71377\times 2.7255\;\mathrm{K}=\mathbf{1.945\;K}$ . This is the temperature the neutrino sea has cooled to, colder than the CMB purely because it missed the annihilation entropy.

Step 2. Count the neutrinos. Per flavour (neutrino plus antineutrino) the fermionic number density is $n_\nu=\tfrac{3}{11}n_\gamma=\tfrac{3}{11}\times 411=\mathbf{112\;cm^{-3}}$ . Over the three flavours, $n_\nu^{\rm tot}=\tfrac{9}{11}\times 411=\mathbf{336\;cm^{-3}}$ .

Step 3. Sanity check against the photons: $n_\nu^{\rm tot}/n_\gamma=9/11=0.818$ , so there are about four relic neutrinos for every five CMB photons — a population comparable to the CMB, yet at a typical energy of only a fraction of a milli-electron-volt, far below any detector threshold. The theory predicts a 1.945 K neutrino sea at 336 cm⁻³ that pervades all of space.

2 · The radiation energy budget after annihilation

Using $N_{\rm eff}=3.044$ , find the neutrino-to-photon energy ratio, the total relativistic energy count $g_*^{\rm rad}$ , and the present radiation density parameter given $\Omega_\gamma h^2=2.47\times 10^{-5}$ for the photons alone.

Step 1. The energy ratio comes straight from equation (5): $\rho_\nu/\rho_\gamma=\tfrac{7}{8}(4/11)^{4/3}N_{\rm eff}$ . With $(4/11)^{4/3}=0.2596$ , this is $\tfrac{7}{8}\times 0.2596\times 3.044=\mathbf{0.691}$ . Despite being colder, the neutrinos hold roughly $69\%$ as much energy as the photons because energy scales as $T^4$ but there are six neutrino degrees of freedom against two photon polarizations.

Step 2. Assemble the total relativistic energy count: $g_*^{\rm rad}=2+\tfrac{7}{8}\cdot 2\cdot N_{\rm eff}(4/11)^{4/3}=2+1.75\times 3.044\times 0.2596=2+1.383=\mathbf{3.38}$ . Note this differs from the entropy count $g_{*S}=3.909$ — a direct illustration of why the two must not be confused.

Step 3. Scale up the radiation density: $\Omega_r h^2=\Omega_\gamma h^2\,(1+\rho_\nu/\rho_\gamma)=2.47\times 10^{-5}\times 1.691=\mathbf{4.18\times 10^{-5}}$ . The neutrinos nearly double the radiation budget over photons alone.

Step 4. This $\Omega_r$ fixes the redshift of matter–radiation equality, $1+z_{\rm eq}=\Omega_m/\Omega_r$ , and the expansion rate $H\propto\sqrt{\rho}$ during nucleosynthesis. The cooled neutrinos raise the radiation density about 69% over the photons alone — the number that enters the Friedmann equation for the early universe.

Physics of the portrait

The reheating factor $(4/11)^{1/3}$ was worked out in full in the early 1950s, by the group that first took the hot early universe seriously enough to compute its consequences.

Origins · 1 minRalph Alpher and how this was discovered

Ralph Alpher

1921–2007 · physical cosmology and the thermal history of the early universe

Ralph Alpher was, with George Gamow and Robert Herman, one of the architects of the hot Big Bang. His 1948 doctoral work — the famous " $\alpha\beta\gamma$ " paper with Bethe and Gamow — argued that the light elements were forged in a hot, dense early universe, and in the same year he and Herman predicted that a relic radiation background should survive today at a few kelvin, roughly a decade and a half before Penzias and Wilson found exactly such a background at three degrees. It is the follow-up work that owns this article: in 1953 Alpher, James Follin, and Herman published a detailed calculation of conditions in the first minutes, and it was there that the difference between the neutrino and photon temperatures produced by $e^+e^-$ annihilation was first computed carefully — the $(4/11)^{1/3}$ factor in all but name. Their treatment already recognized that neutrinos decouple before the pairs annihilate, that the annihilation entropy therefore heats only the photons, and that the surviving neutrino background must be correspondingly colder. Alpher spent decades underappreciated as the CMB prediction was, for a time, credited elsewhere; he received the National Medal of Science only in 2005. The modern precision version — $N_{\rm eff}=3.044$ from non-instantaneous decoupling and finite-temperature QED — is the direct descendant of their program of actually calculating, rather than sketching, the thermal history.

History

1948

— Alpher and Herman predict a surviving relic radiation background of a few kelvin

1953

— Alpher, Follin and Herman compute the neutrino-to-photon temperature difference from e⁺e⁻ annihilation

1965

— Penzias and Wilson detect the cosmic microwave background near 3 K

1990

— COBE FIRAS measures the CMB as a 2.7255 K blackbody

2020

— Precision QED and non-instantaneous decoupling fix N_eff = 3.044, confirmed by Planck's N_eff = 2.99 ± 0.17

Going further

Texts. Ryden, Introduction to Cosmology, 2nd ed., ch. 9–10 · Dodelson & Schmidt, Modern Cosmology, 2nd ed., ch. 2–3 · Kolb & Turner, The Early Universe, ch. 3 · Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., ch. 30 · Weinberg, Cosmology (2008), ch. 3 · Mo, van den Bosch & White, Galaxy Formation and Evolution, ch. 3. The key numerical results follow from comoving-entropy conservation and degree-of-freedom counting: $g_{*S}^{\rm before}=2+\tfrac{7}{8}(4)=5.5$ ; ratio $2/5.5=4/11$ ; boost $(11/4)^{1/3}=1.401$ ; $T_\nu=(4/11)^{1/3}\times 2.7255\;\mathrm{K}=1.945\;\mathrm{K}$ ; $n_\gamma\approx 411\;\mathrm{cm^{-3}}$ and $n_\nu^{\rm tot}=\tfrac{9}{11}n_\gamma\approx 336\;\mathrm{cm^{-3}}$ ; $\rho_\nu/\rho_\gamma=\tfrac{7}{8}(4/11)^{4/3}N_{\rm eff}\approx 0.69$ ; $g_*^{\rm rad}\approx 3.38$ . The value $N_{\rm eff}=3.044$ comes from precision non-instantaneous-decoupling and finite-temperature QED calculations; $m_ec^2=0.511\;\mathrm{MeV}$ and $T_\gamma=2.7255\;\mathrm{K}$ are measured constants. The historical calculation is Alpher, Follin & Herman (1953).

Prerequisites. Equilibrium Thermodynamics of the Early Universe · Entropy Conservation and Comoving Entropy · Neutrino Decoupling and the Cosmic Neutrino Background · Effective Number of Relativistic Degrees of Freedom. Leads to. Big Bang Nucleosynthesis · Primordial Helium and the Deuterium Bottleneck · Last Scattering Surface and Photon Decoupling · Thermal History of the Universe.

Ask the tutor: Ask me to walk through why the entropy count drops from 11/2 to 2, or to show how $N_{\rm eff}$ turns the reheating ratio into a measured constraint on light relics.

Are the photons actually heated up during e⁺e⁻ annihilation?

Not in absolute terms. The universe is cooling the entire time, and the photon temperature never rises — it only falls more slowly than $a^{-1}$ for the brief span while the pairs are annihilating. "Reheating" is a relative statement: after annihilation the photons are warmer than the freely redshifting neutrinos than they otherwise would have been. Equivalently, the entropy released by the pairs is added to the photon bath, so the same comoving entropy is now shared among fewer degrees of freedom, holding the photon temperature up. It is best to think of the neutrinos as the fixed reference that keeps redshifting, against which the photons end up boosted.

Why does the number 11 appear in the ratio?

It is the entropy degree-of-freedom count of the photon–electron–positron fluid before annihilation, times two. Photons contribute $g_{*S}=2$ from their two polarizations. Electrons and positrons are fermions, so they carry the statistical factor $7/8$ , and between them they have four spin states, contributing $\tfrac{7}{8}\times 4=\tfrac{7}{2}$ . The total is $2+\tfrac{7}{2}=\tfrac{11}{2}$ . After annihilation only the photons' $g_{*S}=2$ remains. The ratio of after to before is $2/(11/2)=4/11$ , and because entropy conservation makes this the cube of the temperature ratio, you get $T_\nu/T_\gamma=(4/11)^{1/3}$ .

Why must I use the entropy count gS rather than the energy count g?

Because the conserved quantity in the adiabatic expansion is comoving entropy, $g_{*S}(aT)^3$ , not comoving energy. As long as every species shares a single temperature, $g_*$ and $g_{*S}$ happen to be equal, and it does not matter which you use. But the whole point of reheating is that neutrinos and photons end up at different temperatures, and once that happens the two counts diverge — today $g_{*S}=3.909$ while $g_*=3.38$ . Only the entropy version is conserved through the threshold, so only it gives the correct temperature ratio. Using $g_*$ would be tracking a quantity that is not actually constant.

How many relic neutrinos are around us, and can we detect them?

The theory predicts about $336\;\mathrm{cm^{-3}}$ in total — roughly $112$ per flavour — pervading all of space, comparable to the $411\;\mathrm{cm^{-3}}$ of CMB photons. Detecting them directly is extraordinarily hard because their temperature is only $1.945\;\mathrm{K}$ , giving each a typical energy of a fraction of a milli-electron-volt, far below the threshold of any neutrino detector built for solar or reactor neutrinos. The PTOLEMY experiment aims to capture them on tritium via neutrinoless-threshold beta decay, which has no energy threshold. Indirectly, though, the neutrino background is already "seen": its gravitational effect on the CMB and on structure formation is exactly what the $N_{\rm eff}$ measurement registers.

Why is N_eff = 3.044 instead of exactly 3?

Because the instantaneous-decoupling picture is an idealization. In reality neutrino decoupling is not perfectly complete when $e^+e^-$ annihilation begins, so the higher-energy tail of the neutrino distribution is still weakly coupled and catches a small share of the annihilation entropy. This slightly warms the neutrinos above the naive $(4/11)^{1/3}$ prediction. Adding finite-temperature QED corrections to the electromagnetic plasma's equation of state shifts things a touch more. The net effect raises the effective neutrino number from $3$ to $3.044$ . It is a small correction, but it is a genuine, calculable piece of physics — and precision CMB measurements are now sensitive to it.

How does this epoch connect to Big Bang Nucleosynthesis?

Nucleosynthesis happens just after reheating and is sensitive to the expansion rate, set by the total radiation density $\rho\propto g_*^{\rm rad}T_\gamma^4$ — which the reheated photons and colder neutrinos fix through $g_*^{\rm rad}\approx 3.38$ . A faster expansion (larger $N_{\rm eff}$ ) freezes out the neutron-to-proton ratio earlier and higher, leaving more neutrons to be locked into helium and raising the primordial helium fraction. This is why light-element abundances constrain $N_{\rm eff}$ , and why reheating physics, though it plays out in a fraction of a second, leaves a fingerprint on the universe's composition we can still read today.