Baryon Asymmetry and Baryogenesis

Every direct observation says the Universe is built of matter, not antimatter. Cosmic rays carry antiprotons only at the trickle expected from secondary collisions; the sky shows no annihilation γ-rays from seams where matter and antimatter domains would meet; and no antihelium has ever been found. Yet the Standard Model is very nearly matter–antimatter symmetric, and thermal equilibrium in the hot early Universe would have enforced equal abundances. So there is a genuine cosmological puzzle: why is there something — galaxies, stars, us — rather than the nothing of a pure photon gas left after everything annihilated its partner? The imbalance is compressed into a single pure number, the baryon-to-photon ratio $\eta = n_B/n_\gamma \approx 6\times10^{-10}$ : roughly one surviving baryon for every billion-and-a-half photons. That number is measured two independent ways — from primordial deuterium forged in Big Bang Nucleosynthesis at redshift $z\sim10^9$ , and from the acoustic peaks of the microwave background at $z\approx1100$ — and the two agree at $\eta_{10}\approx6.1$ (equivalently $\Omega_b h^2\approx0.0224$ ), one of the great consistency checks of modern cosmology. In 1967 Andrei Sakharov distilled the problem into three conditions any theory must satisfy to grow such an asymmetry from a symmetric start: baryon-number violation, violation of both C and CP, and a departure from thermal equilibrium. These are logical necessities under CPT, not features of a model. Remarkably the Standard Model contains all three qualitatively — sphalerons, the CKM phase, the electroweak transition — yet fails quantitatively, so the measured $\eta$ is concrete evidence for new physics and points toward leptogenesis.

Key takeaways

The Universe is observed to be almost pure matter; the imbalance is the baryon-to-photon ratio $\eta \equiv n_B/n_\gamma \approx 6\times10^{-10}$ , about one baryon per billion-and-a-half photons.
This tiny number is pinned two independent ways — primordial deuterium from BBN and the acoustic peaks of the CMB — and both give $\eta_{10}\approx6.1$ , i.e. $\Omega_b h^2\approx0.0224$ .
A symmetric start ( $n_B=n_{\bar B}$ ) cannot make it: freeze-out of nucleon–antinucleon annihilation leaves only $\eta\sim10^{-18}$ , eight orders of magnitude too small, so a dynamical mechanism is forced.
Sakharov (1967) proved that generating the asymmetry needs three ingredients at once: baryon violation, C and CP violation, and departure from thermal equilibrium.
The Standard Model has all three in principle (sphalerons violate $B+L$ , the CKM phase violates CP, the electroweak transition can fall out of equilibrium) but is far too weak on the last two, so baryogenesis is physics beyond the Standard Model.
The invariant to track is the entropy-normalized $Y_B = (n_B-n_{\bar B})/s \approx 8.7\times10^{-11}$ , not $\eta$ itself, because photon reheating changes $\eta$ but not $Y_B$ .
Key idea: A dynamically generated baryon asymmetry requires all three Sakharov conditions simultaneously — baryon violation, C and CP violation, and out-of-equilibrium dynamics — and the measured $\eta \approx 6\times10^{-10}$ (confirmed by both BBN and the CMB) is too large to be an initial condition and too subtle for the Standard Model, pointing to new physics.

Where this lives today

Baryogenesis is one of the sharpest open problems in physics, and the hunt for the missing new physics is fully experimental. If the asymmetry was made at the electroweak scale, a strongly first-order transition is needed — which the Standard Model cannot supply for a 125 GeV Higgs — so extensions with extra Higgs doublets or scalar singlets are sought at the LHC; their new CP-violating phases would generate electric dipole moments, and ACME and neutron-EDM searches have pushed the electron and neutron EDMs to record limits, squeezing electroweak baryogenesis hard. If instead the seed is leptonic, leptogenesis ties the asymmetry to the Majorana nature of neutrinos: neutrinoless double-beta decay (KamLAND-Zen, LEGEND, nEXO) would establish lepton-number violation, and the CP phase measured in neutrino oscillations (T2K, NOvA, DUNE, Hyper-Kamiokande) probes the low-energy shadow of the CP violation leptogenesis needs. Meanwhile AMS-02 keeps scanning cosmic rays for a single antihelium nucleus that would upend the matter-dominated picture.

The physical picture

Picture the quark epoch as a near-perfect matter–antimatter mirror: for every billion antiquarks there are a billion-and-one quarks. As the Universe cools below the QCD scale, quark–antiquark pairs annihilate into radiation. The billion antiquarks pair off with a billion quarks and vanish into photons; the single leftover quark per billion survives. Every atom in every star is built from that one-in-a-billion residue. Baryogenesis is the theory of where that lopsided extra digit came from — and the first figure makes the bookkeeping literal.

Figure

For every billion antiquarks the early Universe held a billion-and-one quarks; the pairs annihilate to photons and the lone survivor per billion becomes every atom of matter, leaving today n_B/n_γ = η ≈ 6×10⁻¹⁰

Why can equilibrium never do this on its own? In thermal equilibrium a particle and its antiparticle have identical mass — guaranteed by CPT — so at zero chemical potential their number densities are exactly equal. Any reaction that makes a baryon runs equally fast in reverse, and by C and CP symmetry equally fast for the antibaryon, so the books balance to zero. Only if the Universe falls out of equilibrium, so forward and reverse rates temporarily differ while B, C, and CP are all violated, can a net excess be frozen in. That is the logical skeleton of Sakharov's three conditions. The second figure is the payoff: the same number $\eta$ read off two experiments that share no physics except the baryon density itself. Deuterium abundance in distant gas clouds fixes $\eta$ from nuclear reactions when the Universe was seconds old; the ratio of odd to even acoustic peaks in the microwave background fixes it from photon–baryon fluid dynamics 380 000 years later. That they land on the same value across nine decades of redshift is the observational bedrock the whole subject stands on.

Figure

Two independent baryometers agree: primordial deuterium from BBN at z ∼ 10⁹ and the CMB acoustic peaks at z ≈ 1100 both give η_10 ≈ 6.1, equivalently the baryon density Ω_b h² ≈ 0.0224

The core equations

The whole subject hangs on one dimensionless number and the conditions needed to produce it. We start with the definition, build the machinery to convert it into the density astronomers quote, and end with the three Sakharov conditions cast as sharp inequalities on rates. The asymmetry is defined as the ratio of net baryon number to photon number. Today antibaryons are utterly negligible, so the two forms below coincide, but keeping $n_{\bar B}$ explicit is what makes $\eta$ the asymmetry rather than merely the baryon abundance.

\eta \equiv \frac{n_B}{n_\gamma}, \qquad \eta_B \equiv \frac{n_B - n_{\bar B}}{n_\gamma} \qquad (1)

where: $n_B$ — number density of baryons; $n_{\bar B}$ — number density of antibaryons (negligible today, so $\eta_B\approx\eta$ ); $n_\gamma$ — number density of CMB photons, $\approx411\ \mathrm{cm^{-3}}$ now. Reading: $\eta$ counts surviving baryons per photon, and its measured value $\approx6\times10^{-10}$ is the residue the mirror in the first figure left behind. The denominator is a blackbody quantity fixed by temperature alone. Integrating the Bose–Einstein distribution over momentum for the two photon polarizations gives a clean cube law, evaluated below at the present CMB temperature.

n_\gamma = \frac{2\zeta(3)}{\pi^2}\,T^3 \approx 411\left(\frac{T}{2.725\ \mathrm{K}}\right)^3\ \mathrm{cm^{-3}} \qquad (2)

where: $\zeta(3)\approx1.202$ — Riemann zeta value; $T$ — photon temperature. Reading: because $n_\gamma\propto T^3$ tracks the entropy, using it as the yardstick for baryons builds in the expansion history automatically. To compare $\eta$ with the density astronomers report, multiply by the mean nucleon mass and divide by the critical density. The result is the "baryometer" relation — note the coefficient multiplies $\eta$ itself, not $\eta_{10}$ .

\Omega_b h^2 = 3.66\times10^{7}\,\eta = 3.66\times10^{-3}\,\eta_{10}, \qquad \eta_{10}\equiv10^{10}\eta \qquad (3)

where: $\Omega_b$ — present baryon density parameter; $h$ — Hubble constant in units of $100\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$ ; $\eta_{10}$ — baryon-to-photon ratio in units of $10^{-10}$ . Reading: $\eta_{10}=6.1$ gives $\Omega_b h^2=0.0223$ , matching Planck's $0.0224$ . Because photon number is not conserved when species annihilate, $\eta$ drifts through cosmic history. The clean invariant is the entropy-normalized ratio, since comoving entropy and comoving baryon number are each separately conserved after baryogenesis.

s = \frac{2\pi^2}{45}\,g_{*s}\,T^3, \qquad Y_B = \frac{n_B-n_{\bar B}}{s} \qquad (4)

where: $s$ — entropy density; $g_{*s}$ — effective entropic degrees of freedom; $Y_B$ — comoving baryon asymmetry, constant after baryogenesis. Reading: today $s/n_\gamma\approx7.04$ , so $Y_B\approx\eta/7.04\approx8.7\times10^{-11}$ — the number a theorist actually computes. The engine of the subject is Sakharov's theorem: three conditions that are jointly necessary and (with dynamics) sufficient to build $B\neq0$ from $B=0$ .

\text{Sakharov:}\quad (1)\ \Delta B \neq 0;\quad (2)\ C,\ CP\ \text{violated};\quad (3)\ \text{out of equilibrium} \qquad (5)

where: $\Delta B$ — change in baryon number by microphysics; $C$ — charge conjugation; $CP$ — combined charge–parity. Reading: drop any one and the asymmetry either cannot change, cancels between B and $\bar B$ , or is erased back to zero. Condition (3) has teeth because of the equilibrium form of the net density. For a relativistic species the particle-minus-antiparticle density depends only on the chemical potential, and B-violating reactions in equilibrium drive that potential to zero.

n - \bar n = \frac{g}{6\pi^2}\,T^3\left[\pi^2\,\frac{\mu}{T} + \left(\frac{\mu}{T}\right)^3\right] \;\xrightarrow{\ \mu\to0\ }\; 0 \qquad (6)

where: $\mu$ — chemical potential (the antiparticle carries $-\mu$ ); $g$ — internal degrees of freedom. Reading: fast B-violation forces $\mu_B\to0$ and erases the asymmetry, so it must survive only after those reactions freeze out. The Standard Model's own B-violation is the electroweak sphaleron, a thermally activated field configuration that changes $B+L$ but leaves $B-L$ untouched. This is why a high-scale $B-L$ asymmetry is the robust seed.

\Gamma_{\rm sph}\propto e^{-E_{\rm sph}/T}\ (T<T_{\rm EW}), \qquad \Delta(B+L)\neq0,\quad \Delta(B-L)=0 \qquad (7)

where: $E_{\rm sph}\sim8\pi v/g\sim10\ \mathrm{TeV}$ — sphaleron barrier; $T_{\rm EW}\sim100\ \mathrm{GeV}$ — electroweak scale; $B,L$ — baryon and lepton number. Reading: above $T_{\rm EW}$ the barrier vanishes and sphalerons are in full swing; below it they shut off exponentially. Finally, "out of equilibrium" is made quantitative by comparing an interaction rate with the expansion rate — the same freeze-out criterion used throughout the thermal history.

\Gamma < H \quad\Longleftrightarrow\quad \text{decoupled / out of equilibrium} \qquad (8)

where: $\Gamma$ — rate of the B-violating (or CP-violating) process; $H$ — Hubble rate. Reading: when a species' key reaction falls below $H$ , the reaction can no longer track equilibrium and whatever asymmetry exists is frozen in.

Derivation AWhy a symmetric universe cannot make the observed asymmetry

The goal is to show that if the Universe began with exactly equal baryons and antibaryons and no process ever violated baryon number, the residue left after annihilation is catastrophically smaller than what we measure — roughly eight orders of magnitude short — so a symmetric initial condition is excluded and a dynamical mechanism generating $\Delta B\neq0$ is mandatory. We estimate the freeze-out abundance of protons annihilating against antiprotons and compare the leftover $n_p/n_\gamma$ with the observed $\eta$ .

Step 1. Nucleons and antinucleons annihilate through $p\bar p\to\gamma\gamma$ (and pions) as long as the annihilation rate $n_p\langle\sigma v\rangle$ exceeds the expansion rate $H$ ; annihilation effectively ceases when $n_p\langle\sigma v\rangle\sim H$ . For strong-interaction cross sections this freeze-out lands near $T_f\sim20\ \mathrm{MeV}$ . Why: once the partners can no longer find each other within a Hubble time, whatever abundance remains is frozen — this is the same logic as WIMP freeze-out applied to baryons.

Step 2. In equilibrium down to $T_f$ the abundance is Boltzmann-suppressed by the nucleon rest mass, $n_p/n_\gamma\sim(m_p/T_f)^{3/2}\,e^{-m_p/T_f}$ , with $m_p=938\ \mathrm{MeV}$ . The exponent $m_p/T_f\approx938/20\approx47$ is enormous. Why: a symmetric gas has no protected excess, so the only survivors are the exponentially rare pairs that failed to find a partner before freeze-out.

Step 3. Evaluate: $e^{-47}\approx2.4\times10^{-21}$ and the prefactor $(47)^{3/2}\approx322\approx3.2\times10^{2}$ , giving $n_p/n_\gamma\sim10^{-18}$ . This sits about eight orders of magnitude below the observed $\eta\approx6\times10^{-10}$ . Why: the gap is far too large to blame on our crude prefactor — no reasonable freeze-out temperature rescues a symmetric universe.

\left.\frac{n_p}{n_\gamma}\right|_{\rm sym}\sim10^{-18}\ \ll\ \eta_{\rm obs}\approx6\times10^{-10} \qquad\text{(symmetric residual)}

This order-of-magnitude relic from nucleon freeze-out at $T_f\sim20$ MeV gives $\eta_{\rm sym}\sim10^{-18}$ .

Go deeper · Derivation B · 1 minFrom the baryon-to-photon ratio to the cosmic baryon density

Derivation BFrom the baryon-to-photon ratio to the cosmic baryon density

The goal is to convert the microscopic ratio $\eta$ into the macroscopic density parameter $\Omega_b h^2$ that BBN codes and CMB fits actually report, so that a single measured number can be quoted in either language and cross-checked. We build the baryon mass density from $\eta$ and the photon density, divide by the critical density, and read off the coefficient — confirming that $\Omega_b h^2 = 3.66\times10^{7}\,\eta$ , with the factor multiplying $\eta$ and not $\eta_{10}$ .

Step 1. The baryon mass density is the number density times the mean nucleon mass: $\rho_b = m_N\,n_B = m_N\,\eta\,n_\gamma$ , with $m_N\approx939\ \mathrm{MeV}=1.674\times10^{-24}\ \mathrm{g}$ and $n_\gamma=411\ \mathrm{cm^{-3}}$ today. Why: baryonic mass is carried by nucleons, so counting nucleons and weighting by their mass gives the density that gravitates.

Step 2. The critical density is $\rho_{\rm crit}=3H_0^2/8\pi G=1.878\times10^{-29}\,h^2\ \mathrm{g\,cm^{-3}}$ . Forming $\Omega_b h^2 = \rho_b/(\rho_{\rm crit}/h^2)$ isolates the $h$ -independent combination. Why: $\Omega_b h^2$ is what data constrain because the $h^2$ cancels the poorly known Hubble factor, leaving a clean physical density.

Step 3. Insert the numbers: $\Omega_b h^2 = (1.674\times10^{-24}\times411\,/\,1.878\times10^{-29})\,\eta = 3.66\times10^{7}\,\eta$ . Writing $\eta=10^{-10}\eta_{10}$ gives the equivalent $\Omega_b h^2=3.66\times10^{-3}\,\eta_{10}=\eta_{10}/273$ . Why: the two forms are identical; the common slip is to multiply the large coefficient by $\eta_{10}$ instead of $\eta$ , which is off by $10^{10}$ .

\Omega_b h^2 = 3.66\times10^{7}\,\eta = 3.66\times10^{-3}\,\eta_{10} \qquad\text{(baryometer relation)}

As a check: Planck's $\Omega_b h^2=0.0224$ inverts to $\eta_{10}\approx6.1$ , i.e. $\eta\approx6.1\times10^{-10}$ .

Go deeper · Derivation C · 1 minEntropy dilution and the invariance of the comoving asymmetry

Derivation CEntropy dilution and the invariance of the comoving asymmetry

The goal is to show that $\eta=n_B/n_\gamma$ is not constant in time because photon number jumps whenever a species annihilates and dumps its entropy into the photon bath, whereas $Y_B=n_B/s$ is constant, so baryogenesis calculations quote $Y_B$ and only convert to $\eta$ at the end. We compute the entropy-to-photon ratio, evaluate it after electron–positron annihilation, and use it to relate the early and late values of $\eta$ .

Step 1. Divide the entropy density by the photon density: $s/n_\gamma = (2\pi^2 g_{*s}/45)\,/\,(2\zeta(3)/\pi^2) = (\pi^4/45\zeta(3))\,g_{*s}\approx1.80\,g_{*s}$ . It depends only on the entropic degrees of freedom at that epoch. Why: since both $s$ and $n_\gamma$ scale as $T^3$ , their ratio is pure counting — it tells you how many entropy units each photon carries.

Step 2. After $e^+e^-$ annihilation the bath is photons plus decoupled neutrinos: $g_{*s}=2+\tfrac{7}{8}\cdot2\cdot3\cdot(4/11)\approx3.91$ , so $s/n_\gamma\approx1.80\times3.91\approx7.04$ . Hence $Y_B\approx\eta/7.04$ . Why: the $4/11$ factor is the neutrino-to-photon temperature ratio cubed — the neutrinos never shared in the reheating, so they count with a reduced weight.

Step 3. Because comoving entropy $S=s\,a^3$ and comoving baryon number $n_B\,a^3$ are each conserved, $Y_B$ is fixed once and for all, while $\eta=Y_B\,(s/n_\gamma)$ tracks $g_{*s}$ . Before $e^+e^-$ annihilation the electrons and positrons still counted, so photons were later reheated and $\eta$ fell by a factor $\approx11/4$ . Why: annihilation adds photons per comoving volume without adding baryons, so any ratio to $n_\gamma$ must drop — only the ratio to $s$ is protected.

Y_B=\frac{n_B}{s}=\text{const},\qquad \eta=\frac{n_B}{n_\gamma}=Y_B\,\frac{s}{n_\gamma} \qquad\text{(invariant vs. observable)}

As a check: $s/n_\gamma=1.80\times3.91=7.04$ today, giving $Y_B=8.7\times10^{-11}$ .

Limiting cases

The one asymmetry admits several equivalent measures; the table collects the conversions and the regimes where each is the natural variable.

Case	Result	Meaning
Today, baryon-to-photon	$\eta_0=\dfrac{n_B-n_{\bar B}}{n_\gamma}\approx6.1\times10^{-10}$	the directly quoted value; $n_{\bar B}\approx0$ now, so $\eta_0\approx n_B/n_\gamma$
Entropy-normalized (invariant)	$Y_B=\dfrac{n_B-n_{\bar B}}{s}\approx\dfrac{\eta_0}{7.04}\approx8.7\times10^{-11}$	constant through thermal history; the quantity theory computes
Pre- $e^+e^-$ annihilation	$\eta_{\rm early}\approx\dfrac{11}{4}\,\eta_0$	photon reheating dilutes $\eta$ by $\approx11/4$ ; $Y_B$ is untouched
Symmetric-universe residual	$\eta_{\rm sym}\sim10^{-18}\ll10^{-10}$	freeze-out of $p\bar p$ annihilation leaves far too little — a mechanism is required
Standard Model output	$\eta_{\rm SM}\lll\eta_{\rm obs}$	CP too small and EW transition a crossover, so no surviving asymmetry

Common trapConfusing η with Y_B, and treating η as constant in time

Two traps travel together. First, do not conflate $\eta=n_B/n_\gamma$ with $Y_B=n_B/s$ : they differ by the entropy-to-photon factor, today $s/n_\gamma\approx7.04$ , so $Y_B\approx\eta/7.04\approx8.7\times10^{-11}$ . Quoting $Y_B$ where $\eta$ is meant (or vice versa) is a factor-of-7 error. Second, $\eta$ is not constant back in time: when $e^+e^-$ pairs annihilate they pour entropy into the photons, raising $n_\gamma$ per comoving volume by about $11/4$ and diluting $\eta$ . Track the entropy-normalized $Y_B$ through the thermal history — that is the invariant, and it is why baryogenesis results are always stated as $Y_B$ and only converted to today's $\eta$ at the very end.

Worked examples

1 · From the CMB baryon density to η and Y_B (confidence builder)

Planck reports $\Omega_b h^2 = 0.02237$ . Convert this to the baryon-to-photon ratio $\eta$ and to the entropy-normalized $Y_B$ , and check consistency with the BBN determination.

Step 1. Invert the baryometer relation from Derivation 2, $\Omega_b h^2 = 3.66\times10^{7}\,\eta$ , so $\eta = (0.02237)/(3.66\times10^{7}) = 6.11\times10^{-10}$ , i.e. $\eta_{10}=\mathbf{6.11}$ . This is the value read from acoustic-peak physics at $z\approx1100$ .

Step 2. Convert to the invariant using $s/n_\gamma\approx7.04$ : $Y_B = \eta/7.04 = 6.11\times10^{-10}/7.04 = \mathbf{8.7\times10^{-11}}$ . This is the number any baryogenesis model must reproduce as its final output.

Step 3. Compare with BBN: deuterium in low-metallicity quasar absorbers gives $\eta_{10}=6.0\pm0.3$ , overlapping $6.11$ within errors. The two probes sit at redshifts differing by six orders of magnitude yet agree, confirming a single primordial asymmetry. η = 6.1×10⁻¹⁰, Y_B = 8.7×10⁻¹¹, consistent with BBN's η_10 = 6.0 ± 0.3.

2 · The symmetric-universe residue, worked to a number

Suppose baryon number were exactly conserved and the Universe began symmetric. Estimate the surviving $n_p/n_\gamma$ if nucleon–antinucleon annihilation freezes out at $T_f\approx20\ \mathrm{MeV}$ , and compare with the observed $\eta$ .

Step 1. The Boltzmann-suppressed abundance is $n_p/n_\gamma\sim(m_p/T_f)^{3/2}e^{-m_p/T_f}$ . Form the exponent: $m_p/T_f = 938/20 = \mathbf{46.9}$ .

Step 2. Evaluate the pieces: $e^{-46.9} = 2.4\times10^{-21}$ , and $(46.9)^{3/2} = 321 = 3.2\times10^{2}$ . Multiply: $n_p/n_\gamma \sim 3.2\times10^{2}\times2.4\times10^{-21} \approx \mathbf{7.7\times10^{-19}}$ , i.e. of order $10^{-18}$ .

Step 3. Compare: the observed $\eta\approx6\times10^{-10}$ exceeds this by a factor $\sim10^{9}$ . No plausible shift in $T_f$ closes a nine-order-of-magnitude gap, so the symmetric universe is excluded and a dynamical origin — a mechanism satisfying all three Sakharov conditions — is unavoidable. Symmetric residue ∼ 10⁻¹⁸, about a billion times too small.

Physics of the portrait

Baryogenesis has a single founding document: a three-page 1967 note by a Soviet physicist better known to the public as a dissident than as the man who framed why the Universe contains matter.

Origins · 1 minAndrei Sakharov and how this was discovered

Andrei Sakharov

1921–1989 · the three conditions for baryogenesis and the origin of matter

Andrei Dmitrievich Sakharov was the theoretical architect of the Soviet hydrogen bomb before he became the Soviet Union's most prominent human-rights dissident, awarded the Nobel Peace Prize in 1975. Between those two lives he wrote, in 1967, a short paper — "Violation of CP Invariance, C Asymmetry, and Baryon Asymmetry of the Universe" — that founded the entire field. At a time when baryon number was assumed to be exactly conserved and the newly discovered CP violation in kaons (Cronin and Fitch, 1964) was a curiosity, Sakharov argued that these ingredients, plus a departure from thermal equilibrium in an expanding Universe, could dynamically generate the observed excess of matter starting from a symmetric Big Bang. The paper was almost entirely ignored for a decade. Its vindication came in stages: 't Hooft's 1976 discovery that the electroweak vacuum violates $B+L$ supplied the Standard Model's own baryon-number violation, and Fukugita and Yanagida's 1986 leptogenesis tied the asymmetry to neutrino masses. Today the "Sakharov conditions" open every early-Universe course — a rare case of a whole subfield resting on one physicist's three-line insight.

History

1928

— Dirac's relativistic equation predicts antimatter

1932

— Anderson discovers the positron in cosmic rays

1964

— Cronin and Fitch observe CP violation in kaon decay

1967

— Sakharov states the three conditions for baryogenesis

1976

— 't Hooft shows the electroweak vacuum violates B + L

1985

— Kuzmin, Rubakov and Shaposhnikov show hot sphalerons reprocess B + L

1986

— Fukugita and Yanagida propose leptogenesis

Going further

Texts. Ryden, Introduction to Cosmology, 2nd ed., ch. 9 (baryon-to-photon ratio, BBN) · Dodelson, Modern Cosmology, 2nd ed., ch. 3 (early-Universe thermodynamics) · Kolb & Turner, The Early Universe (1990), ch. 6 (baryogenesis and the Sakharov conditions) · Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., ch. 30 · Mo, van den Bosch & White, Galaxy Formation and Evolution, ch. 3 (thermal history). Key numerical anchors: (1) $n_\gamma=(2\zeta(3)/\pi^2)T^3\approx411\ \mathrm{cm^{-3}}$ at $T=2.725\ \mathrm{K}$ ; (2) $s/n_\gamma=(\pi^4/45\zeta(3))g_{*s}\approx1.80\times3.91\approx7.04$ today; (3) $Y_B=\eta_0/7.04\approx8.7\times10^{-11}$ ; (4) the baryometer $\Omega_b h^2=3.66\times10^{7}\,\eta$ , inverting Planck's $0.0224$ to $\eta_{10}\approx6.1$ ; (5) the symmetric residue $n_p/n_\gamma\sim10^{-18}$ from nucleon freeze-out at $T_f\sim20\ \mathrm{MeV}$ . The sphaleron conversion $B=\tfrac{28}{79}(B-L)$ is the standard three-generation, one-Higgs-doublet Standard Model value.

Prerequisites. Thermal History of the Universe · Equilibrium Thermodynamics of the Early Universe · Entropy Conservation and Comoving Entropy · Decoupling and Freeze-Out. Leads to. Big Bang Nucleosynthesis · Primordial Helium Abundance and the Deuterium Bottleneck · Last Scattering Surface and Photon Decoupling.

Ask the tutor: Ask me to walk through any Sakharov condition, convert between η, Y_B and Ω_b h², or contrast electroweak baryogenesis with leptogenesis step by step.

Is there really no antimatter anywhere in the Universe?

None in any significant amount within our horizon. If galaxies or clusters of antimatter existed, their boundaries with ordinary matter would light up in annihilation γ-rays that we do not see, and the diffuse γ-ray background limits any antimatter domain to be larger than the visible Universe. Cosmic rays contain a few antiprotons and positrons, but exactly at the level expected from ordinary collisions of cosmic rays with interstellar gas — secondary production, not primordial antimatter. AMS-02 keeps searching for a single antihelium nucleus, which could not be made secondarily and would be a bombshell; so far, nothing. The working conclusion is a globally matter-dominated Universe, which is exactly what baryogenesis must explain.

Why is η so small — is a part-per-billion asymmetry not suspiciously tiny?

It is small as a ratio but it is the whole reason anything exists. The number $\eta\sim10^{-9}$ says that in the early quark soup, matter beat antimatter by only about one part in a billion. Everything material today is that leftover billionth; the other billion parts annihilated into CMB photons, which is why photons outnumber baryons by roughly $10^{9}$ to one. So the smallness of $\eta$ is not a fine-tuning embarrassment — it measures how mild the CP-violating, out-of-equilibrium push was. The real challenge is the reverse of "why so small": most naive mechanisms predict either zero or far too little, so producing even $10^{-10}$ is hard.

What is a sphaleron and why does it matter for baryogenesis?

A sphaleron is a non-perturbative Standard Model process — a thermally activated "over the barrier" transition of the electroweak gauge and Higgs fields — that changes $B+L$ while conserving $B-L$ . Above the electroweak scale ( $\sim100$ GeV) the barrier effectively disappears and sphalerons freely reshuffle $B+L$ ; below it they are exponentially suppressed and switch off. Their importance is twofold. They give the Standard Model its own baryon-number violation (Sakharov condition 1). And because they conserve $B-L$ , any asymmetry stored in $B-L$ at high energy survives, and sphalerons redistribute it into a fixed mix of baryon and lepton number — the standard $B=\tfrac{28}{79}(B-L)$ . This is the bridge that lets a lepton asymmetry become the baryon asymmetry we observe.

How can BBN and the CMB measure the same η if they probe such different epochs?

Because both are sensitive to the baryon density, just through different physics. BBN operates when the Universe is seconds to minutes old ( $T\sim$ MeV): a higher $\eta$ means more baryons per photon, which speeds up the nuclear reactions that burn deuterium into helium, so the surviving D/H is a steep, monotonic function of $\eta$ . Measuring D/H in pristine gas thus reads off $\eta$ . The CMB, emitted 380 000 years later, is sensitive to $\eta$ through the baryon loading of the photon–baryon fluid, which enhances the odd (compression) acoustic peaks over the even (rarefaction) ones. Two completely different handles — nuclear reaction rates versus acoustic oscillations — both point to $\eta_{10}\approx6.1$ . Their agreement is strong evidence that a single, universal baryon asymmetry was in place before nucleosynthesis.

What is leptogenesis, and why is it currently the favored mechanism?

Leptogenesis makes the asymmetry in the lepton sector first and lets sphalerons convert part of it to baryons. Heavy right-handed (Majorana) neutrinos — the same particles that make ordinary neutrinos light through the seesaw mechanism — decay out of equilibrium; if their decays violate CP, they produce slightly more leptons than antileptons. Electroweak sphalerons then reprocess this $L$ asymmetry into a $B$ asymmetry via the $B-L$ they preserve. It is favored because it is economical: it reuses physics we independently believe (neutrinos have mass, and the seesaw explains why they are so light), it works at a high scale where the required CP violation is easy to arrange, and it ties a cosmological number to laboratory neutrino physics.

Does the Standard Model really contain all three Sakharov conditions?

Qualitatively yes, quantitatively no — and that mismatch is the whole point. Baryon violation: sphalerons supply it. C and CP violation: C is maximally violated by the weak interaction, and the CKM quark-mixing matrix carries a CP-violating phase. Out of equilibrium: the electroweak phase transition could in principle provide it. But two of the three fail numerically. The CKM CP violation, once you fold in the small quark-mass splittings that must enter, is something like $10^{-20}$ in the relevant measure — astronomically short of $10^{-10}$ . And for the measured 125 GeV Higgs the electroweak transition is a smooth crossover, not the strongly first-order transition needed to protect any asymmetry. So the Standard Model has the right cast of characters but cannot deliver the performance, which is why the observed $\eta$ is direct evidence for physics beyond it.