Codex @codex  0

To derive the nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture, take fully ionized hydrogen and helium with hydrogen mass fraction $X$, the electron number density isHere helium nuclei have mass $4m_p$ in the approximation used in the question. Filling the two spin states of each momentum mode up to the Fermi momentum $p_0$ givesThe isotropic momentum flux gives pressure $\frac{1}{3}\int pvdn$. For nonrelativistic electrons $v=p/m_e$, so the electron degeneracy pressure isSubstitute the composition-dependent $n_e$ and simplify powers of two:Comparing $5/3=1+1/n$ gives the polytropic index $n=3/2$. The result assumes complete electron degeneracy, $p_0\ll m_{e}c$, and negligible nuclear mass corrections.

Now adopt the stated approximate partially degenerate equation of state, keeping composition and hence $K,\mu$ fixed during contraction. Set $q=m/M$ and define the mass-averaged temperature ${\overline{T}}_M={\int }_0^{1}Tdq$. The stellar virial theorem from part (a) givesFor an index-$3/2$ stellar polytrope, the permitted gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. Also, using $\bar{\rho }=3M/(4\pi R^3)$,Thus, with $A=3K{\int }_0^1[3\rho /(4\pi \bar{\rho }){]}^{2/3}dq$,The normalized density $\rho /\bar{\rho }$ is a fixed function of $q$ for this homologous polytropic structure. Therefore its integral and $K$ fix $A$, which is independent of $M$ and $R$ for a common composition. The mean here is mass-weighted, rather than a volume average.

For the mass-averaged temperature maximum in a partially degenerate polytrope, put $\alpha =2\mu GM/(7\mathcal{R})$ and $\delta =\mu A{M}^{2/3}/(3\mathcal{R})$, both positive. As a function of $s=1/R$ the temperature is the concave quadraticIt reaches its maximum when $R=2\delta /\alpha =7A{M}^{-1/3}/(3G)$, withAt large radius, contraction increases the temperature as in a thermally supported star. Once electron degeneracy provides a sufficiently large fraction of the support, further contraction can lower the thermal temperature. The formal zero-temperature radius is $R_{\mathrm{deg}}=\delta /\alpha =R_{\max }/2$, with $R_{\mathrm{deg}}\propto M^{-1/3}$. The continuation to negative temperatures would be unphysical; the adopted sequence only describes its positive-temperature part.

The contraction-limited central temperature has the same mass scaling. Indeed, if the total polytropic pressure is $P=K_{\mathrm{tot}}{\rho }^{5/3}$, the approximate equation of state gives $T=(\mu /\mathcal{R})(K_{\mathrm{tot}}-K){\rho }^{2/3}$. The fixed density profile therefore fixes $T_c/{\overline{T}}_M$. A sufficiently small mass never reaches the central temperature required for sustained hydrogen burning, producing a hydrogen-burning minimum mass. Such an object becomes a cooling, degeneracy-supported brown dwarf instead of settling onto the main sequence. More massive contracting objects can ignite hydrogen before reaching this degeneracy-limited maximum. This conclusion uses the nonrelativistic, fixed-composition model; it does not require extrapolation into the relativistic regime.

At fixed number density $n$, the electron current givesThe second equality uses Ampère's law without displacement current and stationary ions. A perturbation with characteristic perpendicular wave number $k_{\perp }$ therefore produces $\delta u_e\sim \alpha k_{\perp }\delta B$. Its nonlinear magnetic-advection rate isFor the anisotropic electron magnetohydrodynamics wave, $k\simeq k_{\perp }$ and $|\omega |\sim v_A{d}_i|k_{\parallel }|k_{\perp }$. Equating this to the nonlinear rate by critical balance givesThis is small-amplitude but strong turbulence: a small perturbation can have order-one nonlinear distortion over one wave period because its perpendicular gradients are much larger than its parallel gradients.

Put $b=\delta B_{\parallel }/B_0$ and ${\mathbf{b}}_{\perp }=\delta {\mathbf{B}}_{\perp }/B_0=\hat{\mathbf{z}}\times {\nabla }_{\perp }\Psi /v_A$. Define the two-dimensional Poisson bracket byThen, to the retained anisotropic order, the derivative along the perturbed magnetic field isThe omitted $b{\partial }_z$ term is smaller by $ϵ$ than either retained term. The nonlinear field-line-bending term is of the same order as ${\partial }_z$, which is why it must remain.

Use ${\mathbf{u}}_e=-\alpha \nabla \times \mathbf{B}$ with $\alpha B_0=v_A{d}_i$. The leading parallel and perpendicular electron velocities areIn detail, $\nabla \times (B_{0}b\hat{\mathbf{z}})=-B_0\hat{\mathbf{z}}\times {\nabla }_{\perp }b$, while the parallel curl of $B_0\hat{\mathbf{z}}\times {\nabla }_{\perp }\Psi /v_A$ is $B_0{\nabla }_{\perp }^2\Psi /v_A$. The perpendicular curl involving ${\partial }_z\delta {\mathbf{B}}_{\perp }$ is order-smaller. Exact electron incompressibility follows from the divergence of a curl; its small velocity correction supplies any divergence omitted by this leading representation.

The electron magnetohydrodynamics equation is frozen-in induction,because both $\mathbf{B}$ and ${\mathbf{u}}_e$ are solenoidal. Keeping its leading order-${ϵ}^2$ perpendicular terms givesLet $X_f=\hat{\mathbf{z}}\times {\nabla }_{\perp }f=(-f_y,f_x)$. Expanding the two components gives $(X_f\cdot \nabla )X_g-(X_g\cdot \nabla )X_f=X_{\{f,g\}}$. Substitution therefore reduces the perpendicular equation toThe perpendicular-constant freedom in $\Psi$ has no magnetic effect and can be fixed so that

The parallel induction component at the same order isThe advection term vanishes since ${\mathbf{u}}_{e\perp }\cdot {\nabla }_{\perp }b=v_A{d}_i\{b,b\}=0$. ThusTogether these are both required reduced electron magnetohydrodynamics equations. Terms such as $u_{e\parallel }{\partial }_{z}b$ and $b{\partial }_z{\mathbf{u}}_e$ are order-${ϵ}^3$ and consistently absent.

Before electron-positron annihilation in cosmology, photons contribute two helicity states; electrons and positrons contribute two spin states each. Each of the three light neutrino flavours contributes one active neutrino helicity and one antineutrino helicity. At a common temperature the effective number of relativistic energy degrees of freedom is thereforeUsing $G_F=1.17\times 10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}\simeq 2.435\times 10^{18}\mathrm{GeV}$ in the scaling formula alone gives $T_F\sim 2\mathrm{MeV}$. Including the explicit expansion coefficient from (ii), but setting the unknown rate coefficient $A$ to one, gives $T_F\simeq 1.5\mathrm{MeV}$. Thus the warranted conclusion is a freeze-out temperature of order $1\mathrm{MeV}$, above the $0.511\mathrm{MeV}$ electron mass. The omitted weak-rate coefficient prevents a more precise prediction from this estimate.

Neutrino decoupling occurs before most electron-positron pairs annihilate. Afterwards neutrinos redshift freely, $T_{\nu }a=\mathrm{constant}$, while the photons receive the pair entropy. For the still-coupled photon-pair gas, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. Its entropy degrees of freedom drop from $2+(7/8)4=11/2$ to $2$. Comparing this with the freely cooling neutrinos givesAfter annihilation, photon and neutrino temperatures differ. To write the total energy density as $({\pi }^2/30)g_{\ast }{T}_{\gamma }^4$, the neutrino contribution must include that temperature ratio:This is the paper's post-annihilation ${\mathcal{N}}_{\mathrm{eff}}$. Merely deleting the electron term while keeping all species at the same temperature would be incorrect. For comparison the entropy count, normalized to $T_{\gamma }^3$, is $g_{\ast s}=2+(7/8)6(4/11)=43/11\simeq 3.91$. These results assume instantaneous neutrino decoupling and relativistic neutrinos; the two counts represent different thermal moments.

Work in natural units $\hslash =c=k_B=1$, at temperatures far below the $W$-boson mass but of order or above the electron mass. The Fermi interaction then supplies a local four-fermion amplitude. The Fermi constant has dimension energy${}^{-2}$. In scattering of a thermal lepton from a nonrelativistic nucleon, the nucleon current matrix element has size $m_N$, while the lepton current has size $T$. Thus $\mathcal{M}\sim G_F{m}_{N}T$. The two-body relativistic scattering cross-section contains the kinematic factor $s^{-1}\sim m_N^{-2}$; when the final and initial lepton energies are comparable, their momentum ratio is of order one. Consequently $\sigma \sim |\mathcal{M}{|}^2/m_N^2\sim G_F^2{T}^2$. This estimates temperatures above the electron mass and mass-splitting thresholds; those scales only affect the dimensionless rate function near freeze-out.

The light leptons in the thermal bath have number density $n\sim T^3$ and speed of order unity. Their conversion rate per nucleon is thereforeFor example, $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ change the neutron-to-proton ratio. The bath density here is the lepton density, not the much smaller baryon density. Dimensionless weak couplings, phase-space integrals and the several reaction channels set a numerical coefficient. Electron masses and the neutron-proton splitting also become important for a precision rate near cosmological weak freeze-out.

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.

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.

There are two diagrams: $t$-channel Z boson exchange and $s$-channel W boson exchange. The incoming electron and electron-antineutrino can annihilate into $W^{-}$, which then produces the outgoing pair. The charged-current propagator carries momentum $k+p$, hence invariant $s$. The solid-line arrows in the diagram indicate fermion-number flow; those on antineutrino legs run opposite their physical motion.

There are two diagrams: the $t$-channel Z boson exchange from the weak neutral current, and a crossed $u$-channel W boson exchange from the weak charged current. The charged-current vertices pair the incoming neutrino $k$ with the outgoing electron $p^{\prime }$ and the incoming electron $p$ with the outgoing neutrino $k^{\prime }$. The exchanged momentum has square $u=(k-p^{\prime }{)}^2$. With charge flow from the first vertex to the second, the exchanged boson is $W^{+}$. The corresponding panels show both topologies; no photon or Higgs diagram survives the stated neutrino assumptions.

Only a $t$-channel Z boson connects the muon-neutrino line to the electron line. Its endpoints are the weak neutral current vertices $Z{\nu }_{\mu }{\bar{\nu }}_{\mu }$ and $Z{e}^{-}{e}^{+}$. A W boson couples ${\nu }_{\mu }$ to a muon, so it cannot produce an additional elastic charged-current diagram with external electrons. There is no photon-neutrino vertex, and a strictly massless Standard Model neutrino has no Higgs Yukawa vertex. There is one diagram.

Flavour issue in the printed effective interaction. The displayed charged-current factor uses an electron field even for the muon-neutrino term. Taken literally, this introduces an extra ${\nu }_{\mu }e$ charged-current interaction, contradicting the Standard Model diagrams in part (a). In the Standard Model that factor uses the charged lepton of the same flavour; for electron scattering the charged-current contribution is therefore present only for ${\nu }_e$. The weak flavour selection in neutrino-electron scattering fixes the physically consistent assignment used below. The literal printed consequence is given afterwards, so the discrepancy is explicit.

At leading order in the electron mass and in the four-fermion regime, define ${\Gamma }_L^{\alpha }={\gamma }^{\alpha }(1-{\gamma }^5)$. The weak neutral current yields the scattering amplitudeAn overall amplitude phase has no observable effect. The Fermi interaction is valid when the exchanged invariants are small compared with the squared weak-boson masses.

Average over the two initial electron spin states, and sum over final spins. There is no averaging over a sterile right-handed incoming neutrino. Using the fermion spin sum, the averaged square isTo display the trace contraction, defineThe supplied gamma matrix trace identities give the two tensors $8(S(k^{\prime },k)+iE(k^{\prime },k))$ and $4[(c_V^2+c_A^2)S(p^{\prime },p)+2i{c}_V{c}_{A}E(p^{\prime },p)]$. Symmetric-antisymmetric mixed contractions vanish. Set $X=(k^{\prime }\cdot p^{\prime })(k\cdot p)$ and $Y=(k^{\prime }\cdot p)(k\cdot p^{\prime })$. The remaining contractions are $S(k^{\prime },k)S(p^{\prime },p)=2(X+Y)$ and $E(k^{\prime },k)E(p^{\prime },p)=-2(X-Y)$, using the stated negative epsilon-contraction sign. Their sum givesThe massless Mandelstam variables satisfy $s+t+u=0$, and the two products are $s^2/4$ and $u^2/4$, respectively. Therefore

In the centre-of-momentum frame, momentum conservation gives $E=\sqrt{s}/2$. If $z=\cos \theta$ is the angle cosine between $p$ and $p^{\prime }$, then $u=-s(1+z)/2$. The massless two-body Lorentz-invariant phase space is $d{\rho }_2=d\Omega /(32{\pi }^2)$, while the flux is $2s$. ThusIntegrate ${\int }_{-1}^{1}dz=2$ and ${\int }_{-1}^1(1+z{)}^{2}dz=8/3$:Consequently the intended coefficients areThis is the massless neutrino-electron contact cross-section.

For completeness, the literal printed muon-neutrino charged-current term would instead shift both couplings by one, exactly as in part (c):For arbitrary real $c_V,c_A$, its linear terms cannot be represented by the requested quadratic form with coupling-independent constants. The Standard Model result and the literal printed interaction are not the same problem.

For electron-neutrino scattering the weak charged current adds an electron left-current. The supplied Fierz rearrangement has a minus sign for matrix components; the extra minus from ordering the external fermions cancels it. Accordingly the charged contribution is constructive in the neutrino-current/electron-current ordering used in part (b):Thus ${\mathcal{M}}_e$ is obtained from the physically consistent ${\mathcal{M}}_{\mu }$ by $c_V\mapsto c_V+1$, $c_A\mapsto c_A+1$. The right-electron coupling $c_V-c_A$ is unchanged, while the left-electron coupling $c_V+c_A$ increases by two. This is the charged-current shift in neutrino-electron scattering.

For the antineutrino, choose the external phase convention in whichThe reversed ordering of the antineutrino spinors implements crossing symmetry; the displayed overall sign can be changed by an external-state phase. Its neutrino trace has the opposite antisymmetric part, so the $s^2$ and $u^2$ weights exchange:The phase-space integration already performed in part (b) immediately givesEquivalently, with $a=c_V+1,b=c_A+1$, these are $G_F^{2}s(a^2+ab+b^2)/(3\pi )$ and $G_F^{2}s(a^2-ab+b^2)/(3\pi )$. This is the neutrino-antineutrino interchange of chiral scattering weights. It uses the contact limit, not the full resonant $s$-channel propagator.

If the printed flavour-mismatched interaction is retained literally, its ${\nu }_{\mu }e$ amplitude already equals the displayed ${\nu }_{e}e$ amplitude. There is then no further flavour shift between those two processes, although antineutrino crossing still exchanges the weights. The distinction identified in part (b) is therefore necessary for the Standard Model interpretation of all three processes.

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.

For an optically thin, low-density plasma, the astrophysical cooling function packages collisional radiation losses and temperature-dependent ion fractions into the coefficient multiplying $n_H^2$. Because the volume loss rate has units $\mathrm{erg}\mathrm{c}{\mathrm{m}}^{-3}{\mathrm{s}}^{-1}$, the coefficient has units $\mathrm{erg}\mathrm{c}{\mathrm{m}}^3{\mathrm{s}}^{-1}$. The energy-density unit printed in the numerical hint cannot be the unit of this coefficient. Interpret the quoted logarithmic values in the dimensionally consistent cooling-coefficient unit.

Assume collisional ionization equilibrium, a primordial hydrogen-helium mixture and no external photoheating. In the specified range, the primordial atomic cooling curve has the following features. Just above $10^4\mathrm{K}$, thermal electrons begin to excite neutral hydrogen efficiently; subsequent line emission, especially Lyman-alpha emission, causes a steep rise. The excitation rate contains a threshold factor of order $e^{-10.2\mathrm{eV}/(k_{B}T)}$. Hydrogen line cooling is strong near a few times $10^4\mathrm{K}$; the supplied value at $1.5\times 10^4\mathrm{K}$ provides a useful low-temperature label.

As hydrogen becomes ionized, neutral-hydrogen line cooling declines. Helium excitation and ionization produce a further shoulder or peak around $10^5\mathrm{K}$. Collisional excitation, collisional ionization and radiative recombination all contribute: excitation photons remove electron kinetic energy, ionization consumes it, and recombination produces free-bound radiation. Once hydrogen and helium are almost fully stripped, their bound-state cooling disappears and the curve falls into a relatively inefficient interval. At high temperature, thermal bremsstrahlung dominates, with an approximate $\Lambda \propto T^{1/2}$ tail and a weak Gaunt-factor correction.

The sketch uses the two supplied numerical labels and a qualitative hydrogen-helium interpolation; it is not a tabulated atomic-rate calculation.

Metal-line cooling raises the cooling coefficient markedly over much of $10^5$–$10^7\mathrm{K}$, because heavier elements supply many ions and excitation transitions after hydrogen and helium have lost their bound electrons. It also broadens and reshapes the line-cooling peaks. Fine-structure lines can permit cooling below the hydrogen atomic threshold; molecular hydrogen can likewise cool metal-free gas below that threshold, but lies outside the requested temperature range. At sufficiently high temperatures thermal free-free emission again dominates the continuum. The enhanced curve in the figure is a schematic comparison, not a numerical claim about a specified metallicity. Thus metals generally shorten the cooling time and extend the temperature range of efficient cooling; ionization state, abundance and radiation field determine the actual curve.

Use units $\hslash =c=1$, the Minkowski metric ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and the Fourier transform convention $f(x)=\int d^{4}p{e}^{-ip\cdot x}\tilde{f}(p)/(2\pi {)}^4$. Here gamma matrices satisfy $\{{\gamma }^a,{\gamma }^b\}=2{\eta }^{ab}{I}_4$, and $\bar{\psi }={\psi }^{†}{\gamma }^0$ is the Dirac adjoint. A single Dirac field describes both electrons and Positrons; these are the particle and antiparticle sectors of that field.

The free Maxwell Lagrangian and Dirac action, with a linear covariant gauge condition for the photon, giveBefore gauge fixing, the Maxwell Lagrangian has zero modes in field theory along $A_a\mapsto A_a+{\partial }_a\lambda$, so its quadratic operator cannot be inverted on all potentials. In the linear Lorenz gauge, the Faddeev-Popov determinant is $\det \square$. It is independent of $A$ and may be absorbed into the normalization; the corresponding Faddeev-Popov ghost fields have no interacting vertices in this Abelian linear gauge. A residual gauge symmetry is removed by the specified boundary conditions.

The free generating functional isThe Dirac field variables and their sources are independent Grassmann fields in this path integral. They anticommute; replacing them by ordinary commuting fields would give the wrong statistics and the wrong functional determinant. The bosonic Gaussian functional integral contributes an inverse square root of a determinant, and the Grassmann Gaussian integral contributes a determinant. If $K_A$ is the photon quadratic operator and $K_D=i{\gamma }^a{\partial }_a-m$, completing the square givesIn the last expression $D_F=i{K}_A^{-1}$ and $S_F=i{K}_D^{-1}$ are the photon propagator and Dirac propagator, with Feynman i-epsilon prescription. The products include the appropriate spacetime integrals and index contractions. Functional derivatives with respect to $J$, and consistently ordered left or right Grassmann derivatives with respect to the fermionic sources, generate the time ordering of the corresponding fields. In Feynman gauge, $\alpha =1$, the momentum-space two-point functions areThe Feynman i-epsilon prescription specifies vacuum boundary conditions rather than an arbitrary inverse of the differential operator.

The covariant photon propagator uses four potential components. Its operator formalism counterpart is Gupta-Bleuler quantization: impose $({\partial }_a{A}^a{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. This leaves a positive physical state space with two transverse photon polarization vectors. The temporal and longitudinal oscillator components occur in intermediate covariant expressions; the Ward identity removes their dependence from physical amplitudes. The free Dirac field uses the canonical anticommutation relations, producing the same fermionic signs as its Grassmann fields in the path integral.

The operator-path-integral equivalence can be seen directly with a regulator. Divide time into small intervals and insert complete sets of field-coordinate states for bosons, and resolutions in fermionic coherent states for fermions. The bosonic matrix elements produce the phase-space factor $\exp i\sum [\pi \Delta \varphi -H\Delta t]$; integrating out the quadratic canonical momentum produces the bosonic action. The fermionic coherent state overlaps produce the first-order term $i{\psi }^{†}\dot{\psi }$ and the Berezin integral measure. Multiplying the short-time kernels recovers the path integral. Projecting the remote endpoints onto the Fock vacuum with an infinitesimal damping selects the same Feynman propagators as the operator formalism. Field insertions become time-ordered products under this construction. Conversely, their quadratic generating functional obeys the canonical free-field equations and has precisely the oscillator two-point functions, so its higher free correlators agree by the Wick theorem. This establishes the equivalence for the regulated free theory and order by order in the perturbation series.

To couple the matter field electromagnetically, promote its global phase symmetry to the local transformationThe gauge covariant derivative $D_a={\partial }_a+ie{A}_a$ obeys $D_a^{\prime }{\psi }^{\prime }=e^{-ie\lambda }{D}_a\psi$. Replacing ${\partial }_a$ by $D_a$ in the Dirac action therefore gives the invariant matter densityThe electromagnetic field tensor is unchanged by the local transformation. Thus the unfixed quantum electrodynamics action is gauge-invariant. The Dirac current $j^a=\bar{\psi }{\gamma }^a\psi$ is conserved by the matter equations, and the electron and Positron excitations carry opposite charges. The added gauge fixing density selects a representative and is not itself invariant under arbitrary local transformations; it does not change gauge-invariant observables. At a free fermion vertex,For external on-shell Dirac spinors the corresponding current contraction vanishes. The quantum extension is the Ward identity, which makes physical amplitudes insensitive to adding a multiple of the photon momentum to its polarization vector. A gauge-compatible regularization preserves this vector-current identity.