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Photon decoupling is a dynamical loss of frequent scattering, distinct from the chemical conversion of charged particles into neutral atoms. Before recombination, Thomson scattering on free Electrons keeps photons coupled to the baryon plasma. The physical scattering rate isin units with $c=1$. Define the approximate decoupling temperature byEquivalently the photon mean scattering time becomes comparable to an expansion time. A refined last-scattering definition uses the optical depth and the maximum of the visibility function; it need not coincide exactly with the local rate criterion. As $X_e$ falls, scattering becomes inefficient, ordinarily after the half-ionized recombination stage, so $T_{\mathrm{dec}}<T_{\mathrm{rec}}$.

Residual electron freeze-out concerns the remaining ionized fraction. The Saha equation presumes sufficiently rapid forward and reverse reactions to maintain chemical equilibrium. Real recombination is slowed by radiative bottlenecks and expansion; the free-electron fraction can therefore exceed its equilibrium value. Eventually the effective recombination rate per free Electron, roughly ${\alpha }_{\mathrm{eff}}(T)n_e$, falls below $H$. The remaining Electrons cannot all find and recombine with protons within an expansion time. A small nonzero residual fraction survives while the Saha prediction decreases exponentially toward zero.

The sketch shows this lag and residual floor. Its kinetic curve and the position of the decoupling marker are schematic: the data supplied determine the equilibrium curve, but not a quantitative recombination history or an exact $T_{\mathrm{dec}}$. Determining those needs the expansion history and atomic transition rates. The horizontal axis decreases toward the right to follow cosmological cooling.

For a massive particle of rest mass $m$, its four-momentum is $P^{\mu }=m{U}^{\mu }=(E/c,\mathbf{p})$, where the four-velocity is $U^{\mu }=\gamma (c,\mathbf{v})$, $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$. With metric signature $(-,+,+,+)$, $P\cdot P=-m^2{c}^2$. A photon has four-momentum $(E/c,(E/c)\hat{\mathbf{n}})$, a future-pointing null vector. Four-momentum conservation states that the total incoming and outgoing four-momenta agree for an isolated interaction; this includes both energy and momentum conservation in every inertial frame.

Take the incoming photon direction as $x$. Its energy is $\hslash \omega =\mu m{c}^2$, so the total initial four-momentum isThe invariant mass available after photon absorption is therefore $M=m\sqrt{1+2\mu }$. In the centre-of-momentum frame, the two equal daughters have opposite momenta. Their total energy is at least their combined rest energy $2\alpha m{c}^2$, with equality precisely when both daughters have zero momentum in that frame. Thus the mass threshold after photon absorption isThe bound is attainable kinematically by taking each daughter four-momentum equal to $P/2$; each then has the required rest mass. It immediately gives ${\alpha }_{\max }\to 1/2$ as $\mu \to 0$.

At this threshold both daughters move with the centre-of-momentum frame. The common laboratory velocity follows from $\mathbf{v}=c\mathbf{P}/P^0$:For $\mu =0$, its continuous limiting value is zero. At smaller allowed daughter rest masses, some of the centre-of-momentum energy appears as their relative kinetic energy, so their four-momenta need not be $P/2$.

Take the Minkowski metric with signature $(+---)$ and a coupling $e\ne 0$. For a charged scalar field use the gauge covariant derivativeA scalar electrodynamics Lagrangian isAny real potential bounded below and depending only on the modulus gives gauge invariance. Indeed,The derivative of the phase cancels the shifted gauge field in the first identity; commuting partial derivatives proves the second. These identities verify invariance of every term, including the interaction hidden in the kinetic term.

For an unbroken example choose $V=m_{\varphi }^2|\varphi {|}^2+\lambda |\varphi {|}^4$ with $m_{\varphi }^2>0$ and $\lambda >0$. The minimum is at zero. There is no vector mass term in the quadratic expansion and no Higgs mechanism. Pure electromagnetic theory has a neutral massless spin-one photon with two physical transverse polarizations, equivalently helicities $+1$ and $-1$; longitudinal and time-component polarizations are gauge redundancies. In the unbroken scalar theory, the same massless photon is accompanied by a spin-zero charged particle and its oppositely charged antiparticle, both of mass $m_{\varphi }$. A complex scalar has two real physical degrees of freedom, rather than two unrelated charged species.

For a Higgs example chooseThe vacuum modulus is nonzero. Around one vacuum representative, unitary gauge removes the phase and writes $\varphi =(v+h)/\sqrt{2}$. ThenThus the quadratic spectrum isThe gauge boson is a massive spin-one particle with three polarizations, and $h$ is a neutral massive spin-zero Higgs boson. The scalar phase supplies the longitudinal vector polarization; it is not an extra physical massless Goldstone boson. The degree count is unchanged: two massless-vector polarizations plus two scalar degrees become three massive-vector polarizations plus one radial scalar degree. The underlying gauge invariance remains a redundancy of the description.

For the two-charge theory, useand retain the same gauge transformation of $a_{\mu }$. Both derivatives transform with the phase of their own field. A manifestly stable two-charge scalar gauge potential is, for positive $m_{\varphi }^2,M^2,{\lambda }_{\varphi },{\lambda }_{\psi }$, $g\ge 0$ and a nonzero complex constant $b$,Since $\psi -b{\varphi }^2$ has charge $2e$, its modulus is gauge-invariant. Expanding its square exhibits the direct couplingIn particular the charge in ${\psi }^{\ast }{\varphi }^2$ is $-2e+2e=0$, which uses the stated charge ratio. The potential is real, bounded below and has the zero-field minimum, while the kinetic terms have the standard positive signs. A complete example is thereforeThere is genuine direct interaction even with $g=0$ because $b\ne 0$. In four spacetime dimensions $b$ has mass dimension $-1$, but the expanded potential contains only quadratic, cubic and quartic field monomials; the cubic coefficient $M^{2}b$ has dimension one. Thus the example is also power-counting renormalizable.

Take $D_{\mu }={\partial }_{\mu }+i{K}_{\mu }$, with the Pauli matrices acting on the Higgs doublet. For electroweak hypercharge $Y=1/2$,This specifies every component of the gauge covariant derivative for the electroweak interaction. Expanding the gauge-covariant kinetic term makes its derivative, trilinear and quartic interactions explicit:where the Pauli matrix multiplication law givesThe antisymmetric Pauli contribution vanishes because $W_{\mu }^a{W}^{b\mu }$ is symmetric in $a,b$. Reversing the sign convention for $D_{\mu }$ reverses the linear gauge interactions consistently, without changing the masses.

A nonzero vacuum expectation value requires ${\mu }^2<0$. Minimizing the Higgs potential gives $v^2=-{\mu }^2/\lambda$. By an gauge transformation choosein unitary gauge. The electroweak doublet gauge-boson mass matrix follows by inserting the vacuum expectation value in the gauge-covariant kinetic term:Define the charged electroweak gauge bosons and the neutral rotation through the Weinberg angle bywith inverse $W_{\mu }^3=c_W{Z}_{\mu }+s_W{A}_{\mu }$ and $B_{\mu }=-s_W{Z}_{\mu }+c_W{A}_{\mu }$. ThenThe factors differ because $W^{+}$ and $W^{-}$ are conjugate fields whereas $Z$ is real. The massless photon corresponds to the unbroken Lie algebra generator $Q=T_3+Y$, which annihilates $⟨\varphi ⟩$. Thus three of the four real gauge bosons acquire mass, with $m_W=m_Z{c}_W$. The three would-be Goldstone bosons provide their longitudinal polarizations; the remaining scalar $h$ is the Higgs boson.

Introduce the left-handed lepton doublet $L=({\nu }_{eL},e_L{)}^T$, with $Y_L=-1/2$, and the right-handed singlet $e_R$, with $Y_R=-1$. Use the chiral projectors $P_L=(1-{\gamma }^5)/2$ and $P_R=(1+{\gamma }^5)/2$ in the course's convention. The minimal Standard Model has no right-handed neutrino. The gauge-invariant fermion terms areThe doublet contraction in the Yukawa interaction is a singlet, and its total hypercharge is $+1/2+1/2-1=0$. A bare term ${\bar{e}}_L{e}_R$ would fail electroweak gauge invariance. The gauge covariant derivatives above already give all the requested fermion-gauge couplings. In terms of mass eigenstates, $e=g{s}_W=g^{\prime }{c}_W$ and they becomeThus the weak charged current is chiral and the neutrino has zero electric charge. The gauge-invariant electron Yukawa mass follows from the Yukawa interaction after electroweak symmetry breaking:Rephase $e_R$ to make $y_e$ real and positive. It givesThe electron Dirac mass is therefore compatible with the original gauge symmetry through the Higgs mechanism. The neutrino remains massless in this minimal renormalizable lepton sector.

Let $r_{\nu }=T_{\nu ,0}/T_0$, and retain the assumption that these neutrinos remain massless. Constancy of the neutrino degeneracy parameter gives $|{\mu }_{\nu ,0}|=|{\xi }_{\nu }|r_{\nu }{T}_0$. The condition on the critical density therefore yields the critical-density bound on massless neutrino degeneracyFor the standard instantaneous-neutrino decoupling approximation before electron-positron annihilation in cosmology, cosmological entropy conservation heats photons relative to the decoupled neutrinos. The electromagnetic entropy degrees of freedom change from $2+(7/8)4=11/2$ to $2$, givingThis numerical value requires that thermal-history assumption; the bound in terms of $r_{\nu }$ is the general answer if large degeneracy changes decoupling or subsequent entropy transfer. Setting $T_{\nu ,0}=T_0$ would instead give $26.2$. For $N$ equally degenerate flavors with the same temperature, the leading bound becomes $36.8N^{-1/4}$ under the standard temperature assumption. The neglected thermal corrections make a small change near this loose energy-density limit; this is not a detailed Big Bang nucleosynthesis bound.

A stellar equation of state supplies pressure and internal energy as functions of density, temperature and composition, together with thermodynamic derivatives needed for stability and transport. Hydrostatic equilibrium fixes the pressure gradient, but does not determine which microscopic components provide the pressure. In ordinary dense interiors local thermodynamic equilibrium is a useful starting point. A consistent mixture iswhere ions, Electrons, radiation and interaction corrections are distinguished. The classical Electron pressure and electron degeneracy pressure are two limits of the same Electron contribution, and must not be added as if they belonged to different particles. The finite-temperature electron equation of state interpolates between them.

In a fully ionized, nondegenerate, nonrelativistic gas, $P_g=\rho k_{B}T/(\mu m_u)$ and specific thermal energy is $u_g=3k_{B}T/(2\mu m_u)$. With nuclear mass fractions $X_A$, charges $Z_A$ and mass numbers $A_A$, the mean molecular weight satisfies ${\mu }^{-1}={\sum }_A{X}_A(1+Z_A)/A_A$, and ${\mu }_e^{-1}={\sum }_A{X}_A{Z}_A/A_A$ is the mean molecular weight per electron. This regime describes much of an ordinary main sequence interior. Toward cooler layers, ionization and molecular dissociation change particle numbers and consume heat. The Saha equation relates ionization to both temperature and Electron density: there is no universal horizontal ionization boundary. These regions have larger heat capacity and can have a reduced stellar adiabatic exponent. The simple fully ionized formula is then insufficient.

Equilibrium photons give radiation pressure $P_{\gamma }=a_r{T}^4/3$ and energy per volume $a_r{T}^4$, or specific energy $u_{\gamma }=a_r{T}^4/\rho$. In the nondegenerate gas regime, equality with gas pressure gives the radiation-to-gas pressure boundarya line of slope $1/3$ on a $(\log \rho ,\log T)$ plot. Higher temperatures at fixed mass density favor photon support. A monatomic gas has stellar adiabatic exponent $5/3$, while radiation alone has $4/3$; their adiabatic exponents of a monatomic gas-radiation mixture are not obtained by assuming a fixed pressure fraction during compression. Radiation support is particularly important in massive stars.

For Electrons, Pauli exclusion principle and the Fermi-Dirac distribution determine occupation numbers. The net Electron density is $n_e=\rho /({\mu }_e{m}_u)$ and the Fermi momentum is $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$. Define the kinetic electron Fermi temperatureFor $T\gg T_F$ the Electrons are nearly classical; for $T\ll T_F$ they are strongly degenerate and their pressure depends primarily on density. The intermediate region requires finite-temperature electron equation of state integrals, not a discontinuous switch of formulas.

The equation of state of a cold electron gas gives, in its two limits,These are respectively the $n=3/2$ and $n=3$ pressure-density powers, explaining the approximate white dwarf polytropic mass-radius relation sequence and the Chandrasekhar limit. The kinetic energy per volume is $3P_e/2$ in the nonrelativistic limit and $3P_e$ in the ultrarelativistic limit. Ions can still supply much of the heat capacity even when the Electron pressure supplies the mechanical support.

The Electron degeneracy crossover $T\sim T_F$ has slope $2/3$ at low mass density and $1/3$ at high mass density. The electron relativistic density threshold isa vertical marker where $p_F=m_{e}c$. It is different from the thermal electron relativistic threshold $k_{B}T\sim m_e{c}^2$, near $5.93\times 10^9\mathrm{K}$, a horizontal temperature scale. Hot dilute matter can have relativistic thermal Electrons without degeneracy; cold dense matter can have relativistic degenerate Electrons without reaching that temperature.

Electron degeneracy also does not automatically imply that Electrons dominate the total pressure. Comparing the cold Electron limit with radiation gives the radiation-to-degeneracy pressure boundaryIts logarithmic slopes are $5/12$ for nonrelativistic Electrons and $1/3$ for ultrarelativistic Electrons. One must compare the pressures separately from the degeneracy criterion; extrapolating the classical gas-radiation line into a degenerate region is incorrect.

The stellar equation-of-state regime diagram uses an illustrative fully ionized carbon composition, ${\mu }_e=2$, $\mu =12/7$. It displays the electron-degeneracy crossover, both radiation-pressure comparisons, and distinct thermal and density-driven relativity scales. The curves are limiting-model comparisons, not sharp phase boundaries or a calibrated complete equation of state. Partial ionization, molecular physics and interactions modify the low-temperature regions indicated on the plot.

At sufficiently high temperature, electron-positron thermal pair abundance can become important. The pair abundance depends on density and chemical potential as well as temperature; $k_{B}T=m_e{c}^2$ is not a universal onset line. In the dilute ultrarelativistic limit, both pair species together add energy density $7a_r{T}^4/4$ to the photons' $a_r{T}^4$, and have pressure one third of their energy density. While pairs are being created, thermal energy is spent on rest mass, which can reduce the stellar adiabatic exponent below $4/3$ and contribute to pair-instability supernova physics.

At high mass density and low temperature, Interactions governed by Coulomb's law invalidate the noninteracting-ion approximation. The ionic Coulomb coupling parameter $\Gamma =Z^2{e}^2/(4\pi {ϵ}_0{a}_i{k}_{B}T)$, where $a_i=(3/(4\pi n_i){)}^{1/3}$, grows as ${\rho }^{1/3}/T$. Corrections become significant when $\Gamma$ is of order one, and a sufficiently strongly coupled plasma can crystallize. At still greater mass density, electron capture alters ${\mu }_e$ and nuclear matter replaces the ideal electron-ion model; neutron star interiors require strong-interaction and relativistic equations of state. These further regimes lie beyond the simple pressure curves plotted here. A useful stellar EOS is thermodynamically consistent across the crossovers, rather than just the maximum of unrelated pressure laws.

Use the Minkowski metric convention $+---$. A particle's four-momentum is $P=(E/c,\mathbf{p})$, with $P\cdot P=m^2{c}^2$ for rest mass $m$. For a photon, its energy and wavelength obey $E=hc/\lambda$, and its four-momentum is null. Here $h$ is the Planck constant.

The initially stationary electron has $P=(mc,0,0,0)$. Choose the scattering plane as the $xy$ plane. The incoming and outgoing photon four-momenta areBy four-momentum conservation, the final electron has $P^{\prime }=P+K_1-K_2$. Its rest mass is unchanged, so the relativistic energy-momentum relation gives $P^{\prime 2}=P^2=m^2{c}^2$. Since $K_1^2=K_2^2=0$, expansion yieldsorMultiplying through gives the Compton scattering shift:It is nonnegative and vanishes for forward scattering, consistent with the recoil energy being supplied by the photon.

Use signature $(+---)$ and take the covariant spatial components $A_i$ as canonical coordinates. The canonical quantization of the electromagnetic field begins with the canonical momentaThus the primary momentum constraint of the electromagnetic potential is ${\Pi }^0=0$: $A_0$ has no independent velocity. The Hamiltonian obtained by the Legendre transform in mechanics, up to a boundary term, isPreserving the primary constraint requires the Gauss law constraint in gauge theory, ${\partial }_i{\Pi }^i=0$. It also follows by varying $A_0$. These are two first-class constraints; they generate the gauge freedom and remove two canonical pairs from the four potential components. The reduced phase space has four dimensions per spatial mode, hence two propagating photon degrees of freedom.

Impose Coulomb gauge, ${\partial }_i{A}_i=0$. With no charges, Gauss's law then gives ${\nabla }^2{A}_0=0$; vanishing boundary conditions set $A_0=0$. This is radiation gauge. The remaining components are transverse and obey the massless wave equation. Let ${ϵ}_i^{(r)}(\mathbf{k})$, $r=1,2$, be orthonormal transverse polarization vectors. Their photon polarization completeness relation isThe canonical transverse photon field is the Hermitian operatorwhereThe field and its conjugate momentum have the transverse equal-time commutatorThis is the quantized reduced bracket, or equivalently the Dirac bracket after imposing the constraints and gauge conditions. The normal-ordered Hamiltonian is ${\sum }_r\int d^{3}k{\omega }_{\mathbf{k}}{a}_r^{†}{a}_r/(2\pi {)}^3$. Its excitations are photons; circular combinations of the two transverse polarizations have helicity $+1$ and $-1$. The scalar and longitudinal potential components do not create additional physical photons.

The Feynman propagator is the vacuum expectation of a time-ordered product. The mode expansion directly gives the radiation-gauge photon propagator, with $z=x-y$:The first expression comes from the creation-annihilation commutator; the second is its contour-integral representation. The positive-energy pole lies below the real axis and the negative-energy pole above it. In this reduced free-field description, the temporal operator is zero. A photon propagator must specify its gauge; the spatial transverse propagator is not the same tensor as the covariant four-potential propagator.

For the commonly used covariant form, add the gauge fixing term $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$. The resulting Fourier-space kinetic operator isThe inversion of the gauge-fixed Maxwell kinetic operator gives $K^{\mu \nu }{D}_{\nu \rho }^F=i{\delta }^{\mu }{}_{\rho }$. In Feynman gauge, $\xi =1$, the photon propagator isEquivalently, $\square D_{\mu \nu }^F=i{\eta }_{\mu \nu }{\delta }^4(x-y)$ with the vacuum pole prescription. For general $\xi$, the momentum-space numerator is ${\eta }_{\mu \nu }-(1-\xi )k_{\mu }{k}_{\nu }/(k^2+i0)$.

A covariant canonical realization uses four polarization oscillators with $[a_r,a_s^{†}]=-(2\pi {)}^3{\eta }_{rs}{\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$. The resulting indefinite inner product is auxiliary. In Gupta-Bleuler quantization, impose $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. The scalar-longitudinal combination is thereby removed from the physical state space, leaving the same two transverse photon states. Thus the four-component Feynman-gauge numerator does not imply four physical polarization states.

A collisionless relativistic species redshifts with $T_{\nu }\propto a^{-1}$. The coupled plasma instead obeys cosmological entropy conservation, $g_{\ast s}{T}^3{a}^3=\mathrm{constant}$, and is heated relative to a decoupled species as massive particles annihilate. The effective number of relativistic energy degrees of freedom $g_{\ast }$ and effective number of relativistic entropy degrees of freedom $g_{\ast s}$ coincide while all species share a common relativistic temperature. Use $g_d:=g_{\ast s}(T_{\mathrm{dec}})\simeq 100$ for the Standard Model bath, excluding the already decoupled extra species.

Just before ordinary neutrino decoupling, the coupled bath contains two photon states, four Electron/Positron states and six ordinary neutrino/antineutrino states. ThereforeThe extra-neutrino temperature relative to that bath is then $(43/(4g_d){)}^{1/3}$. After the ordinary neutrinos decouple, Electron/Positron annihilation heats only the photon bath: its entropy count changes from $2+(7/8)4=11/2$ to $2$. Both neutrino populations miss that last heating, soThis is the two-stage entropy dilution of an early-decoupled neutrino. Numerically $T_{\nu ,0}\simeq 6.8\times 10^{-14}\mathrm{GeV}$. Relative to ordinary relic neutrinos it is cooler by $(43/(4g_d){)}^{1/3}\simeq 0.48$.

Let $g_{\nu }$ count the extra neutrino's populated helicity and antiparticle states. For one conventional neutrino plus antineutrino species $g_{\nu }=2$, the same state count as photons. The Fermi-Dirac distribution supplies the $3/4$ number-density factor, soThus a two-state species has about $2.9\%$ of the photon number density. Stating $g_{\nu }$ explicitly also covers a different choice of populated states.