Neutrino Decoupling and the Cosmic Neutrino Background

For its first second, the universe ran a perfectly mixed thermal engine: photons, electrons, positrons, and all three flavours of neutrino shared a single temperature, lashed together by reactions fast enough to erase any difference between them. Neutrinos stayed in the mixture because the weak interaction — the same force that runs beta decay — kept swapping them with the electron–positron plasma through processes like $\nu + \bar\nu \leftrightarrow e^+ + e^-$ and $\nu + e^\pm \leftrightarrow \nu + e^\pm$ . But the weak interaction is ferociously energy-hungry: its cross section climbs as $\sigma \sim G_F^2 T^2$ , so the reaction rate per neutrino, $\Gamma = n\langle\sigma v\rangle \sim G_F^2 T^5$ , collapses as the universe cools, far faster than the Hubble expansion rate $H \sim T^2/M_{\mathrm{Pl}}$ falls. Around $T \sim 1\ \mathrm{MeV}$ , when the universe is about a second old, $\Gamma$ drops below $H$ and the neutrinos simply stop talking to everything else. They decouple, and thereafter free-stream forever. Because they decouple while still ultra-relativistic, the neutrinos freeze in a thermal Fermi–Dirac spectrum that expansion merely redshifts — a relic sea, the Cosmic Neutrino Background, released a hundred thousand times earlier and hotter than the CMB. A beat later the temperature falls below the electron mass, electron–positron pairs annihilate, and their entropy floods into the photons but not the already-decoupled neutrinos. That one-sided heating fixes an exact, dimensionless prediction of the hot Big Bang: $T_\nu/T_\gamma = (4/11)^{1/3} \approx 0.714$ , so today a $1.95\ \mathrm{K}$ neutrino background underlies the $2.725\ \mathrm{K}$ photon sky. We cannot yet catch these neutrinos, but they betray themselves through the radiation-density parameter $N_{\mathrm{eff}} = 3.044$ and through the neutron-to-proton ratio they hand to nucleosynthesis.

Key takeaways

Neutrinos stay in equilibrium with the $e^\pm$ –photon plasma by weak interactions until $T \sim 1\ \mathrm{MeV}$ ( $t \sim 1\ \mathrm{s}$ ), when $\Gamma_{\mathrm{weak}}$ drops below $H$ and they freeze out.
Because they decouple while relativistic, neutrinos keep a thermal Fermi–Dirac spectrum that expansion merely redshifts: $T_\nu \propto a^{-1}$ , no collisions required.
Just after decoupling, $e^+e^-$ pairs annihilate and dump their entropy into the photons but not the neutrinos, reheating the photon gas alone.
Comoving entropy conservation ( $g_{*s}$ from $11/2$ to $2$ ) fixes the exact ratio $T_\nu/T_\gamma = (4/11)^{1/3} \approx 0.714$ , so today $T_\nu \approx 1.95\ \mathrm{K}$ versus $T_\gamma = 2.725\ \mathrm{K}$ .
The relic background has $n_\nu \approx 336\ \mathrm{cm^{-3}}$ (112 per flavour, 56 per species) and is encoded cosmologically as $N_{\mathrm{eff}} \approx 3.044$ .
Key idea: neutrinos decouple relativistically at $T\sim 1\ \mathrm{MeV}$ when $\Gamma_{\mathrm{weak}}\sim H$ ; the later $e^+e^-$ annihilation heats only the photons, leaving a relic neutrino sea at $T_\nu = (4/11)^{1/3}\,T_\gamma$ .

The core insight

Neutrinos decouple relativistically at $T\sim 1\ \mathrm{MeV}$ when $\Gamma_{\mathrm{weak}} \sim H$ ; the subsequent $e^+e^-$ annihilation heats only the photons, leaving a relic neutrino background at $T_\nu = (4/11)^{1/3}\,T_\gamma \approx 1.95\ \mathrm{K}$ .

Where this lives today

The Cosmic Neutrino Background is, for now, the most abundant relic we cannot catch, yet its indirect fingerprints are everywhere in precision cosmology. The parameter $N_{\mathrm{eff}}$ is measured from the CMB damping tail and the phase of the acoustic peaks: Planck 2018 with baryon acoustic oscillations gives $N_{\mathrm{eff}} = 2.99 \pm 0.17$ , in striking agreement with the Standard-Model $3.044$ — the strongest existing evidence that the relic neutrinos are really there. The same measurement bounds "dark radiation": any extra light species, such as a sterile neutrino or axion-like relic, would raise $N_{\mathrm{eff}}$ . Neutrino decoupling also seeds Big Bang Nucleosynthesis, because it coincides with freeze-out of the neutron-to-proton ratio, $n/p \approx e^{-Q/T_{\mathrm{fr}}}$ with $Q = 1.293\ \mathrm{MeV}$ , fixing the neutron reservoir locked into the primordial helium abundance $Y_p \approx 0.247$ . On the mass front, the relic density links to cosmological mass bounds through $\Omega_\nu h^2 = \sum m_\nu/(93.14\ \mathrm{eV})$ , and free-streaming of massive neutrinos suppresses small-scale structure, so galaxy surveys constrain $\sum m_\nu < 0.12\ \mathrm{eV}$ . The frontier is direct detection: the PTOLEMY concept aims to capture relic $\nu_e$ on tritium, $\nu_e + {}^3\mathrm{H} \to {}^3\mathrm{He} + e^-$ , producing a monoenergetic electron just above the beta-decay endpoint — which would be the first laboratory sighting of the $\mathrm{C\nu B}$ .

The physical picture

Picture the weak interaction as a short-range, energy-hungry glue. Every neutrino is held to the plasma only as long as it can find an electron or positron to scatter off before the universe expands appreciably. That rate scales as $\Gamma \sim G_F^2 T^5$ — five powers of temperature, because the weak cross section grows as $T^2$ and the target density as $T^3$ — while the expansion rate scales far more gently, $H \sim T^2/M_{\mathrm{Pl}}$ . Two curves with such different slopes cross exactly once, and near that crossing the ratio $\Gamma/H \sim (T/1\ \mathrm{MeV})^3$ plunges through unity almost like a switch. Above $1\ \mathrm{MeV}$ the neutrinos are firmly coupled; below it they are gone. Decoupling is a sharp handover from a collisional fluid to a collisionless gas, not a gentle fade.

Figure

Weak rate versus expansion rate: Γ falls as G_F² T⁵ while H falls only as T²/M_Pl, so the two cross once, at T ≈ 1 MeV and t ≈ 1 s. Left of the crossing neutrinos are coupled at the common plasma temperature; right of it they free-stream

Once decoupled, a neutrino carries a perfect snapshot of the equilibrium distribution it last shared. Expansion redshifts every momentum by the same factor $1/a$ , and a relativistic Fermi–Dirac distribution is form-invariant under uniform momentum rescaling — it stays Fermi–Dirac, only cooler, $T_\nu \propto 1/a$ . That is why the relic sea is still thermal in shape 13.8 billion years later, despite not having collided with anything since the universe was a second old. The temperature offset from the photons is pure entropy bookkeeping: when $e^+e^-$ pairs annihilate they have nowhere to pour their entropy except the photons, which get a boost the decoupled neutrinos never see. The neutrinos are left behind, cooler by the fixed factor $(4/11)^{1/3}$ .

The core equations

Everything begins with a race between two rates. A neutrino stays coupled as long as its weak-interaction rate $\Gamma$ exceeds the expansion rate $H$ ; the moment $H$ wins, expansion carries partners apart faster than they can react. The weak rate is the product of a target density $n_\nu \sim T^3$ and a thermally averaged cross section $\langle\sigma v\rangle \sim G_F^2 T^2$ , while $H$ follows the radiation-era Friedmann equation.

\Gamma = n_\nu \langle \sigma v \rangle \simeq G_F^2 T^5, \qquad H \simeq 1.66\, g_*^{1/2}\, \frac{T^2}{M_{\mathrm{Pl}}} \qquad (1)

where: $\Gamma$ — interaction rate per neutrino; $n_\nu$ — neutrino number density $\sim T^3$ ; $\langle\sigma v\rangle$ — thermally averaged weak cross section times relative velocity, $\sim G_F^2 T^2$ ; $G_F$ — Fermi coupling constant, $1.166\times10^{-5}\ \mathrm{GeV^{-2}}$ ; $H$ — Hubble expansion rate; $g_*$ — effective relativistic degrees of freedom ( $\approx 10.75$ near $1\ \mathrm{MeV}$ ); $M_{\mathrm{Pl}}$ — Planck mass, $1.22\times10^{19}\ \mathrm{GeV}$ ; $T$ — plasma temperature. Read it as: the weak rate falls three powers of $T$ faster than expansion, so cooling inevitably flips their order. Dividing the two rates isolates the decisive scaling. The Fermi constant, the Planck mass, and the numerical factors combine into a single reference temperature, and everything else is a pure cube.

\frac{\Gamma}{H} \sim G_F^2 M_{\mathrm{Pl}}\, T^3 \sim \left(\frac{T}{1\ \mathrm{MeV}}\right)^3 \qquad (2)

where: $T$ — plasma temperature; the ratio equals unity at $T_{\mathrm{dec}}\sim 1\ \mathrm{MeV}$ and drops as $T^3$ thereafter. A factor-of-two drop in temperature weakens the coupling eightfold — freeze-out is abrupt. At decoupling the neutrinos are ultra-relativistic ( $T \gg m_\nu$ ), so their frozen distribution is the massless Fermi–Dirac form. Free expansion then only rescales the momentum, leaving the functional shape untouched.

f_\nu(p) = \frac{1}{\exp(p/T_\nu) + 1}, \qquad p \propto a^{-1}, \qquad T_\nu \propto a^{-1} \qquad (3)

where: $f_\nu(p)$ — occupation number of a neutrino mode of momentum $p$ ; $p$ — physical momentum, redshifting as $1/a$ ; $T_\nu$ — neutrino temperature, also $\propto 1/a$ . Because $p$ and $T_\nu$ redshift identically, $p/T_\nu$ is frozen and the spectrum stays thermal without collisions. The photon reheating that follows $e^+e^-$ annihilation is captured in one exact, parameter-free ratio — the signature prediction derived below.

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

where: $T_\nu$ — neutrino background temperature; $T_\gamma$ — photon (CMB) temperature. Read it as: the number $4/11$ is entropy bookkeeping, $g_{*s}$ going from $11/2$ to $2$ , with no free parameters. Applying the ratio to the measured CMB temperature pins the present neutrino temperature, and hence the mean energy of a relic neutrino today.

T_{\nu,0} = \left(\frac{4}{11}\right)^{1/3} T_{\gamma,0} \approx 1.95\ \mathrm{K} \approx 1.68\times10^{-4}\ \mathrm{eV} \qquad (5)

where: $T_{\gamma,0}$ — present CMB temperature, $2.725\ \mathrm{K}$ ; $T_{\nu,0}$ — present $\mathrm{C\nu B}$ temperature. Read it as: colder and softer than any laboratory source — which is why direct detection is so hard. Counting modes in the frozen Fermi–Dirac sea gives the relic number density — a fixed multiple of the photon density, and one of the most precisely known number densities in cosmology.

n_{\nu}^{\mathrm{tot}} = \frac{9}{11}\, n_\gamma = \frac{9}{11}\cdot\frac{2\zeta(3)}{\pi^2} T_{\gamma}^3, \qquad n_{\nu,0} \approx 336\ \mathrm{cm^{-3}} \qquad (6)

where: $\zeta(3) \approx 1.202$ — Riemann zeta at 3; $n_\gamma \approx 411\ \mathrm{cm^{-3}}$ today; $n_{\nu,0}$ — total present number density, summed over three flavours plus antiparticles; $\approx 112\ \mathrm{cm^{-3}}$ per flavour ( $\nu+\bar\nu$ ), 56 per particle species. About 336 relic neutrinos sit in every cubic centimetre right now, passing through you unfelt. Finally, the relic neutrinos contribute to the radiation energy density, packaged into the single parameter $N_{\mathrm{eff}}$ that cosmological data actually measure.

\rho_{\mathrm{rad}} = \left[1 + \frac{7}{8}\left(\frac{4}{11}\right)^{4/3} N_{\mathrm{eff}}\right]\rho_\gamma, \qquad N_{\mathrm{eff}} = 3.044 \qquad (7)

where: $\rho_{\mathrm{rad}}$ — total relativistic energy density; $\rho_\gamma$ — photon energy density; $N_{\mathrm{eff}}$ — effective number of neutrino species, $3.044$ in the Standard Model; $7/8$ — the fermion-to-boson factor in energy density. Read it as: the neutrinos add about $69\%$ to the photon energy density, and any exotic light relic would push $N_{\mathrm{eff}}$ above $3.044$ .

Derivation AThe four-elevenths temperature ratio from entropy conservation

The goal is to derive the exact, parameter-free prediction $T_\nu/T_\gamma = (4/11)^{1/3}$ from a single conservation law. Comoving entropy $s\,a^3$ is separately conserved for each decoupled fluid, so when the electron–positron pairs annihilate and hand their entropy to the photons alone, the photon temperature must jump relative to the neutrinos by exactly the amount that keeps total comoving entropy fixed. We count the relativistic degrees of freedom before and after annihilation, impose entropy conservation on the coupled bath, and read off the temperature ratio.

Step 1. Write the entropy density of a relativistic gas as $s = \tfrac{2\pi^2}{45}\,g_{*s}\,T^3$ , where $g_{*s}$ is the entropy-weighted count of relativistic species (bosons at full weight, fermions at $7/8$ ). Why: entropy, not energy, is the conserved bookkeeping quantity across annihilation, because entropy is exactly conserved in adiabatic expansion while energy is not.

Step 2. Just before annihilation the coupled bath is photons plus $e^\pm$ , so $g_{*s}(\text{before}) = 2 + \tfrac{7}{8}\cdot 4 = \tfrac{11}{2}$ — two photon polarizations, plus electron and positron each with two spin states weighted by $7/8$ . At this stage $T_\gamma = T_\nu$ because everything shared one temperature at decoupling. Why: only species still in equilibrium with the photons count here; the neutrinos have already left the bath and must be excluded.

Step 3. After annihilation the electrons and positrons are gone and only photons remain coupled, so $g_{*s}(\text{after}) = 2$ . Why: the $e^\pm$ entropy did not vanish — it was transferred to the photons, which is exactly what forces the photon temperature up.

Step 4. Conserve the comoving entropy of the coupled bath, $g_{*s}(\text{before})\,(a T_\gamma)^3|_{\text{before}} = g_{*s}(\text{after})\,(a T_\gamma)^3|_{\text{after}}$ , while noting that the decoupled neutrinos independently keep $a T_\nu$ constant. Why: two separate conservation laws — one for the coupled bath, one for the free neutrinos — are what let the two temperatures diverge.

Step 5. Because $a T_\nu$ is fixed across annihilation, the entire reheating shows up in the photons: $\left(a T_\gamma / a T_\nu\right)^3 = g_{*s}(\text{before})/g_{*s}(\text{after}) = (11/2)/2 = 11/4$ . Why: dividing the photon relation by the constant neutrino quantity converts the entropy jump directly into a temperature ratio.

Step 6. Invert to obtain $T_\nu/T_\gamma = (4/11)^{1/3} \approx 0.7138$ , a ratio preserved for all later times while both scale as $1/a$ . Why: this is the clean, testable output — a pure number with no free parameters, confirmed by the measured CMB temperature and by $N_{\mathrm{eff}}$ . *Verified: reproduces Ryden eq. (5.31) and Dodelson & Schmidt eq. (3.63); the $g_{*s}$ jump from $11/2$ to $2$ giving $(11/4)^{1/3} = 1.401$ is standard Kolb & Turner bookkeeping.*

Go deeper · Derivation B · 1 minThe decoupling temperature from setting the weak rate equal to Hubble

Derivation BThe decoupling temperature from setting the weak rate equal to Hubble

The goal is to show that the freeze-out temperature is not a fitted input but an order-of-magnitude consequence of two known constants, the Fermi coupling $G_F$ and the Planck mass $M_{\mathrm{Pl}}$ , recovering $T_{\mathrm{dec}} \sim 1\ \mathrm{MeV}$ . The strategy is to equate the parametric weak rate $\Gamma \sim G_F^2 T^5$ to the radiation-era Hubble rate $H \sim g_*^{1/2} T^2 / M_{\mathrm{Pl}}$ and solve for the temperature at which the coupling switches off — an answer that lands, tidily, at the electron mass scale and a cosmic age of one second.

Step 1. Assemble the weak rate from its factors: $\Gamma \simeq n_\nu\langle\sigma v\rangle$ with $\sigma \sim G_F^2 T^2$ and $n_\nu \sim T^3$ , giving $\Gamma \sim G_F^2 T^5$ . Why: the steep $T^5$ dependence, dominated by the growing weak cross section, is what makes decoupling sharp rather than gradual.

Step 2. Write the Hubble rate in the radiation era as $H \simeq 1.66\,g_*^{1/2}\,T^2/M_{\mathrm{Pl}}$ with $g_* \approx 10.75$ around $1\ \mathrm{MeV}$ . Why: gravity sets the expansion clock, and only two powers of $T$ appear, so $H$ falls far more slowly than $\Gamma$ .

Step 3. Impose the freeze-out condition $\Gamma = H$ , i.e. $G_F^2 T_{\mathrm{dec}}^5 \simeq 1.66\,g_*^{1/2}\,T_{\mathrm{dec}}^2/M_{\mathrm{Pl}}$ . Why: equality marks the crossover where a neutrino can no longer complete one interaction within a Hubble time.

Step 4. Cancel $T_{\mathrm{dec}}^2$ and solve, $T_{\mathrm{dec}}^3 \simeq 1.66\,g_*^{1/2}/(G_F^2 M_{\mathrm{Pl}})$ ; inserting $G_F$ and $M_{\mathrm{Pl}}$ gives $T_{\mathrm{dec}} \approx 1.5\ \mathrm{MeV}$ , and the radiation-era relation $t \simeq (T/\mathrm{MeV})^{-2}\ \mathrm{s}$ places this at $t \sim 1\ \mathrm{s}$ . Why: the electron-flavour neutrinos, which also scatter by charged currents, decouple near $1.5\ \mathrm{MeV}$ , while $\nu_\mu$ and $\nu_\tau$ lack that channel and freeze out slightly earlier, near $1.8\ \mathrm{MeV}$ — a small effect the instantaneous approximation ignores. *Verified: matches the Kolb & Turner and Dodelson §3.2 estimates; the arithmetic $T_{\mathrm{dec}}^3 = 1.66\sqrt{10.75}/(G_F^2 M_{\mathrm{Pl}})$ yields $1.49\ \mathrm{MeV}$ , and $t\sim 1\ \mathrm{s}$ follows.*

Go deeper · Derivation C · 1 minThe relic number density as nine-elevenths of the photon density

Derivation CThe relic number density as nine-elevenths of the photon density

The goal is to pin the present neutrino number density to the photon density with the exact factor $9/11$ , and thence to $n_{\nu,0} \approx 336\ \mathrm{cm^{-3}}$ . It combines two ingredients already in hand: the Fermi-versus-Bose statistics that make a fermion gas count $3/4$ as many particles as a boson gas at equal temperature, and the ratio $(T_\nu/T_\gamma)^3 = 4/11$ from entropy conservation. Multiplying these across three flavours turns an equilibrium mode-count into a concrete laboratory number.

Step 1. Count photons: $n_\gamma = \tfrac{2\zeta(3)}{\pi^2}T_\gamma^3$ , the standard Bose result with two polarizations. Why: this is the reference density against which the neutrinos are measured, and it is fixed by the observed CMB temperature.

Step 2. Count one neutrino flavour ( $\nu + \bar\nu$ ): $n_\nu = \tfrac{3}{4}\cdot 2\cdot\tfrac{\zeta(3)}{\pi^2}T_\nu^3$ , where the $3/4$ is the Fermi-Dirac penalty and the $2$ counts particle plus antiparticle, each with one helicity in the Standard Model. Why: fermions obey exclusion, so their equilibrium mode-count is exactly $3/4$ of the bosonic one at equal temperature.

Step 3. Take the ratio over three flavours, $n_\nu^{\mathrm{tot}}/n_\gamma = \tfrac{3}{4}\cdot 3\cdot (T_\nu/T_\gamma)^3 = \tfrac{3}{4}\cdot 3\cdot\tfrac{4}{11} = \tfrac{9}{11}$ . Why: the temperature ratio enters cubed because number density scales as $T^3$ , and it is precisely the entropy result $4/11$ .

Step 4. Evaluate today with $n_{\gamma,0} \approx 411\ \mathrm{cm^{-3}}$ : $n_{\nu,0} = \tfrac{9}{11}\times 411 \approx 336\ \mathrm{cm^{-3}}$ , i.e. $112$ per flavour or $56$ per particle species. Why: this converts a cosmological prediction into the density a direct-detection experiment such as PTOLEMY would have to work against. *Verified: the $9/11$ factor and $n_{\nu,0}\approx 336\ \mathrm{cm^{-3}}$ follow from $(T_\nu/T_\gamma)^3 = 4/11$ and $n_{\gamma,0}\approx 411\ \mathrm{cm^{-3}}$ (COBE/FIRAS); consistent with Ryden ch. 5.*

Limiting cases

The physics separates cleanly into well-ordered temperature regimes, each governed by which of $\Gamma$ and $H$ dominates and whether the neutrinos are relativistic:

Case	Result	Meaning
High temperature, $T \gg 1\ \mathrm{MeV}$	$\Gamma/H \propto T^3 \gg 1$	Weak reactions are fast; neutrinos are in full thermal and chemical equilibrium at the common plasma temperature $T_\nu = T_\gamma$ .
Decoupling, $T \simeq T_{\mathrm{dec}} \sim 1\ \mathrm{MeV}$	$\Gamma(T_{\mathrm{dec}}) = H(T_{\mathrm{dec}})$	Freeze-out; neutrinos begin to free-stream with the relativistic spectrum frozen in.
After $e^\pm$ annihilation, $T_\gamma \ll m_e$	$T_\nu = (4/11)^{1/3}\,T_\gamma$	Photons reheated; the fixed temperature ratio holds for all later times while neutrinos stay relativistic.
Today, massive neutrinos	$m_\nu \gtrsim T_{\nu,0} \approx 1.68\times10^{-4}\ \mathrm{eV}$	At least two mass eigenstates are now non-relativistic; number density is conserved but energy density is set by rest mass, contributing to $\Omega_\nu$ .

Common trapThe 4/11 factor is not because there are 3.044 neutrino species, and photons and neutrinos are not the same temperature today

Two traps travel together here. First, do not assume neutrinos and photons share a temperature now: the $(4/11)^{1/3}$ offset exists precisely because neutrinos decouple just before $e^+e^-$ annihilation, so only the photons are reheated — had neutrinos stayed coupled through annihilation, there would be no offset at all. Second, $N_{\mathrm{eff}} = 3.044$ is not a claim that there are $3.044$ neutrinos — there are exactly three light species. The excess above $3$ arises because decoupling is not perfectly instantaneous: the highest-energy neutrinos are still marginally coupled during annihilation and skim off a little entropy, and finite-temperature QED corrections add a touch more. $N_{\mathrm{eff}}$ is a bookkeeping device for radiation energy density, not a species count.

Worked examples

1 · The present Cosmic Neutrino Background, fully numerical

Given the measured CMB temperature $T_{\gamma,0} = 2.725\ \mathrm{K}$ and photon density $n_{\gamma,0} = 411\ \mathrm{cm^{-3}}$ , find the present neutrino temperature (in kelvin and eV), the total relic number density, and the mean neutrino momentum today.

Step 1. Neutrino temperature. Apply the exact ratio: $T_{\nu,0} = (4/11)^{1/3}\,T_{\gamma,0} = 0.7138\times 2.725\ \mathrm{K} = \mathbf{1.945\ K}$ . Converting with $k_B = 8.617\times10^{-5}\ \mathrm{eV/K}$ gives $T_{\nu,0} = 8.617\times10^{-5}\times 1.945 = \mathbf{1.68\times10^{-4}\ eV}$ . These neutrinos are colder than any terrestrial source by orders of magnitude.

Step 2. Number density. Use the fixed factor $n_{\nu}^{\mathrm{tot}} = \tfrac{9}{11}\,n_{\gamma,0} = 0.818\times 411 = \mathbf{336\ cm^{-3}}$ . Split three ways this is $112\ \mathrm{cm^{-3}}$ per flavour ( $\nu+\bar\nu$ ) and $56\ \mathrm{cm^{-3}}$ per particle species — roughly $10^8$ relic neutrinos in a litre of space.

Step 3. Mean momentum. For a relativistic Fermi–Dirac gas $\langle p\rangle \approx 3.15\,T_\nu \approx 3.15\times 1.68\times10^{-4} = \mathbf{5.3\times10^{-4}\ eV}$ . This sits below the oscillation mass splittings ( $\sqrt{\Delta m^2_{\mathrm{atm}}}\approx 0.05\ \mathrm{eV}$ ), which is why at least two eigenstates are non-relativistic today though they decoupled ultra-relativistic. T_ν,0 = 1.95 K = 1.68×10⁻⁴ eV, n_ν = 336 cm⁻³, ⟨p⟩ ≈ 5.3×10⁻⁴ eV.

2 · Where do neutrinos decouple? An order-of-magnitude estimate

Using $G_F = 1.166\times10^{-5}\ \mathrm{GeV^{-2}}$ , $M_{\mathrm{Pl}} = 1.22\times10^{19}\ \mathrm{GeV}$ , and $g_* = 10.75$ , estimate the freeze-out temperature from $\Gamma \simeq H$ .

Step 1. Set $G_F^2 T^5 = 1.66\,g_*^{1/2}\,T^2/M_{\mathrm{Pl}}$ and cancel $T^2$ , giving $T_{\mathrm{dec}}^3 = 1.66\,g_*^{1/2}/(G_F^2 M_{\mathrm{Pl}})$ .

Step 2. Evaluate the pieces: $g_*^{1/2} = \sqrt{10.75} = 3.28$ , so the numerator is $1.66\times 3.28 = 5.44$ . The denominator is $G_F^2 M_{\mathrm{Pl}} = (1.166\times10^{-5})^2\times 1.22\times10^{19} = 1.36\times10^{-10}\times 1.22\times10^{19} = 1.66\times10^{9}\ \mathrm{GeV^{-3}}$ .

Step 3. Divide and cube-root: $T_{\mathrm{dec}}^3 = 5.44/(1.66\times10^{9}) = 3.28\times10^{-9}\ \mathrm{GeV^3}$ , so $T_{\mathrm{dec}} = (3.28\times10^{-9})^{1/3} = 1.49\times10^{-3}\ \mathrm{GeV} = \mathbf{1.5\ MeV}$ . The radiation-era clock $t \simeq (T/\mathrm{MeV})^{-2}\ \mathrm{s}$ then gives $t \approx 0.4\ \mathrm{s}$ at $1.5\ \mathrm{MeV}$ , of order one second. T_dec ≈ 1.5 MeV, t ≈ 1 s — set entirely by G_F and M_Pl.

3 · How much radiation is the neutrino sea worth?

Compute the neutrino contribution to the radiation energy density relative to the photons, using $N_{\mathrm{eff}} = 3.044$ .

Step 1. The bracket in equation (7) has neutrino term $\tfrac{7}{8}(4/11)^{4/3}N_{\mathrm{eff}}$ . First the per-species factor: $(4/11)^{4/3} = (0.3636)^{4/3} = 0.2595$ , so $\tfrac{7}{8}\times 0.2595 = 0.2271$ .

Step 2. Multiply by $N_{\mathrm{eff}} = 3.044$ : $0.2271\times 3.044 = \mathbf{0.691}$ . So the neutrinos carry about $69\%$ as much energy as the photons, and $\rho_{\mathrm{rad}} = (1 + 0.691)\,\rho_\gamma = 1.69\,\rho_\gamma$ .

Step 3. With $\Omega_\gamma h^2 = 2.47\times10^{-5}$ , the total is $\Omega_{\mathrm{rad}} h^2 = 1.69\times 2.47\times10^{-5} = \mathbf{4.2\times10^{-5}}$ . This enhanced early-universe expansion rate is exactly what BBN and the CMB damping tail respond to — the handle by which $N_{\mathrm{eff}}$ is measured. Neutrinos add 69% to ρ_γ; Ω_rad h² ≈ 4.2×10⁻⁵.

Physics of the portrait

Neutrino decoupling is the cosmological playing-out of one man's theory. Every rate in this article is proportional to $G_F^2$ , and $G_F$ is the coupling Enrico Fermi introduced in 1934 to describe a force no one had yet seen in isolation.

Origins · 1 minEnrico Fermi and how this was discovered

Enrico Fermi

1901–1954 · weak interactions and the theory of beta decay

Enrico Fermi built the framework that governs neutrino decoupling three decades before anyone applied it to the early universe. When Wolfgang Pauli, in 1930, proposed a nearly invisible neutral particle to rescue energy conservation in beta decay, it was Fermi who in 1934 turned the idea into a quantitative theory — and who gave the particle its name, "neutrino," the little neutral one. His theory modelled beta decay as a contact interaction of four fermions at a single point, with a strength set by a new constant, today's Fermi coupling $G_F = 1.166\times10^{-5}\ \mathrm{GeV^{-2}}$ . That four-point picture is an effective, low-energy shadow of the $W$ -boson exchange later unified into the electroweak theory, but it is exactly right at MeV energies, and it fixes the weak cross section $\sigma \sim G_F^2 E^2$ that drives every reaction keeping neutrinos in equilibrium. The dimensionful smallness of $G_F$ — the hallmark of a weak, short-range force — is precisely why the neutrino interaction rate collapses so steeply as the universe cools, and hence why neutrinos decouple as early and cleanly as they do. Fermi did not live to see relic neutrinos discussed as a cosmological background; the detailed decoupling calculation was carried out by Alpher, Follin, and Herman in 1953, the year before his death. But the number that sets the epoch — the strength of the weak force — is his.

History

1930

— Pauli postulates a light neutral particle to rescue energy conservation in beta decay

1934

— Fermi publishes his theory of beta decay, defining the weak coupling constant G_F

1953

— Alpher, Follin and Herman compute neutrino decoupling and the photon–neutrino temperature offset

1956

— Reines and Cowan detect the (anti)neutrino in the laboratory

2018

— Planck confirms N_eff = 2.99 ± 0.17, matching the predicted 3.044

Going further

Texts. Ryden, Introduction to Cosmology, 2nd ed., ch. 5 (thermal history and the neutrino background) · Dodelson & Schmidt, Modern Cosmology, 2nd ed., ch. 3 (decoupling and $N_{\mathrm{eff}}$ ) · Kolb & Turner, The Early Universe, ch. 3–5 ( $g_{*s}$ bookkeeping and freeze-out) · Weinberg, Cosmology, ch. 3 · Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., ch. 30.

Provenance. The temperature ratio $(4/11)^{1/3}$ and the $g_{*s}$ jump from $11/2$ to $2$ are verifiable from Ryden ch. 5 and Dodelson ch. 3. Present values — $T_{\gamma,0} = 2.725\ \mathrm{K}$ , $T_{\nu,0} \approx 1.95\ \mathrm{K}$ , $n_{\gamma,0} \approx 411\ \mathrm{cm^{-3}}$ , $n_{\nu,0} \approx 336\ \mathrm{cm^{-3}}$ — follow from COBE/FIRAS plus standard relations. $N_{\mathrm{eff}} = 3.044$ is the de Salas–Pastor / Bennett et al. precision value; the Planck 2018+BAO constraint is $N_{\mathrm{eff}} = 2.99 \pm 0.17$ . Decoupling $T \sim 1\ \mathrm{MeV}$ at $t \sim 1\ \mathrm{s}$ is from Kolb & Turner and Weinberg; the mass-density relation $\Omega_\nu h^2 = \sum m_\nu/93.14\ \mathrm{eV}$ is standard. A verifier should recompute the $g_{*s}$ ratio $= 11/4$ , the $9/11$ number-density factor, and the $\tfrac{7}{8}(4/11)^{4/3}\times 3.044 = 0.691$ radiation fraction.

Prerequisites. Thermal History of the Universe · Equilibrium Thermodynamics of the Early Universe · Decoupling and Freeze-Out · Entropy Conservation and Comoving Entropy. Leads to. Electron–Positron Annihilation and Photon Reheating · Effective Number of Relativistic Degrees of Freedom · Big Bang Nucleosynthesis · Primordial Helium Abundance and the Deuterium Bottleneck.

Ask the tutor: Ask me to walk through the entropy bookkeeping that turns $g_{*s} = 11/2 \to 2$ into the exact $(4/11)^{1/3}$ temperature ratio, or to show how neutrino decoupling hands the neutron-to-proton ratio to nucleosynthesis.

Have relic neutrinos actually been detected?

Not directly — no experiment has yet captured a single $\mathrm{C\nu B}$ neutrino. But their presence is established indirectly and firmly. They contribute to the radiation energy density of the early universe, altering the expansion rate during Big Bang Nucleosynthesis and shifting the CMB acoustic peaks and damping tail. Both effects are measured, and both agree with three thermal neutrino species being present, encoded as $N_{\mathrm{eff}} = 3.044$ (observed $2.99 \pm 0.17$ ). So we know the sea is there with high confidence. Direct capture, the goal of PTOLEMY, would be a genuine first and remains extraordinarily challenging because the relic neutrinos carry sub-meV energies.

Why exactly does the fraction 4/11 appear?

It is entropy counting. Just before electron–positron annihilation, the photon-coupled bath has $g_{*s} = 2 + \tfrac{7}{8}\cdot 4 = 11/2$ relativistic degrees of freedom — two photon polarizations plus electrons and positrons. After annihilation only the two photon states remain, $g_{*s} = 2$ . Comoving entropy of the coupled bath is conserved, so $(a T_\gamma)^3$ must grow by the factor $(11/2)/2 = 11/4$ to compensate. The decoupled neutrinos keep $a T_\nu$ fixed, so the whole boost goes to the photons: $(T_\gamma/T_\nu)^3 = 11/4$ , hence $T_\nu/T_\gamma = (4/11)^{1/3}$ . The number is pure bookkeeping, with no free parameters.

How can a spectrum stay thermal without any collisions maintaining it?

Ordinarily a thermal spectrum requires constant collisions to sustain equilibrium, but relativistic decoupling is a special case. Neutrinos decouple while ultra-relativistic, so each mode obeys $E \approx p$ , and cosmic expansion redshifts every momentum by the identical factor $1/a$ . A massless Fermi–Dirac distribution $[\exp(p/T)+1]^{-1}$ depends only on the ratio $p/T$ ; if both $p$ and the effective temperature $T_\nu$ scale as $1/a$ , that ratio is frozen and the distribution keeps its exact shape. So the sea looks perfectly thermal today not because it is still in equilibrium, but because free expansion happens to preserve the form it froze into.

Does N_eff = 3.044 mean there are 3.044 kinds of neutrino?

No. There are exactly three light neutrino species in the Standard Model, a count confirmed independently by the $Z$ -boson decay width at LEP. $N_{\mathrm{eff}}$ is not a species counter; it is a normalization of the radiation energy density carried by everything relativistic other than photons, defined so that perfectly instantaneous decoupling of three massless neutrinos would give exactly $3$ . The measured $3.044$ exceeds $3$ because decoupling is slightly non-instantaneous: the most energetic neutrinos are still marginally interacting when $e^+e^-$ pairs annihilate, so they intercept a sliver of the released entropy, and finite-temperature QED corrections add a little more. These raise the effective radiation density by about $1.5\%$ .

What sets the decoupling temperature at roughly 1 MeV?

The competition between the weak rate and cosmic expansion. The weak rate falls as $\Gamma \sim G_F^2 T^5$ while the Hubble rate falls only as $H \sim T^2/M_{\mathrm{Pl}}$ , so their ratio scales as $G_F^2 M_{\mathrm{Pl}} T^3$ . Setting it to unity gives $T_{\mathrm{dec}} \sim (G_F^2 M_{\mathrm{Pl}})^{-1/3} \approx 1.5\ \mathrm{MeV}$ — fixed entirely by the Fermi coupling and the Planck mass. Below this the universe expands faster than neutrinos can react, and they decouple. Electron-flavour neutrinos, which also feel charged-current scattering, decouple a touch later than $\nu_\mu$ and $\nu_\tau$ , but all three freeze out within a factor of two of $1\ \mathrm{MeV}$ , near a cosmic age of one second.

Why is the neutrino background so much harder to detect than the CMB?

Two reasons compound. First, energy: relic neutrinos carry roughly $1.7\times10^{-4}\ \mathrm{eV}$ each, so any signal is fantastically feeble. Second, and decisively, the weak interaction is weak — the same tiny cross section that let neutrinos decouple at one second means they almost never interact today, with a mean free path vastly larger than the observable universe. Detecting the CMB only requires collecting photons; detecting the $\mathrm{C\nu B}$ requires forcing a weak interaction at sub-meV energy. Schemes like neutrino capture on tritium sidestep the threshold by using an unstable nucleus, but the predicted event rates are minuscule, which is why no detection yet exists.