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Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free Electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate Electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the Electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition Electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the Electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate Electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate Electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting Electron rest energy, isFor Electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual Electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once Electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the Electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net Electrons from thermally created Electrons and Positrons. Chemical equilibrium with photons requires opposite Electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge Electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined Electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

The fundamental angular floor is the diffraction limit of a telescope, of order $\lambda /D$. The ideal diffraction grating has $R_g=mW/d$, whereas its slit-limited resolving power has $R\theta D=mW\lambda /d$. Requiring $R\le R_g$ givesThe coefficient depends on the adopted line-resolution and aperture conventions. Below this scale, a narrower spaxel oversamples the same spatial mode rather than creating a new independently resolved element. Equivalently a spatial optical mode occupies etendue of order ${\lambda }^2$.

There is also an engineering floor before arbitrary shrinkage: the optical camera focal length, photodetector size and support clearances do not shrink, while the number of photodetectors increases as ${\theta }^{-2}$. Thus the shrinking-beam model's constant combined optical camera volume is eventually overtaken by fixed per-camera overheads. For fixed brightness, smaller spaxels receive fewer photons, so read noise and photon shot noise can constrain useful sampling before the formal optical floor. Diffraction limits independent spatial information; fixed photodetector and mechanical dimensions limit practical volume reduction.

Use the ideal detection convention of one collected Electron per detected photon; otherwise replace photon totals by detected Electron totals using the quantum efficiency. Let $A$ be the summed measurement on the star patch and $B^{\prime }$ the independent background-only patch. After additive offsets are removed, their means are $Q+B$ and $B$. Independent Poisson distribution counts have variance equal to their mean, and a single read of $n$ detector pixels contributes $n{r}^2$ to each patch's variance. ThereforeThe two factors of two arise from the independently measured background and the second patch's read noise, not from doubling the stellar signal. The formula assumes equal background means, independent detector pixels and patch measurements, one read per patch, and negligible dark current or other noise. Several exposure reads or correlated detector pixels require their own variance and covariance terms.

Let $k=m/2$. After offset subtraction, each summed image has mean $N$ analogue-to-digital units per detector pixel. For a constant detector conversion gain $g$ and independent Poisson distribution detected counts, $\mathrm{Var}(A)=\mathrm{Var}(B)=N/g$: summing $k$ independent frames adds both their means and variances. Here $N$ is the mean of the sum, not one frame.

Write $A=N+a$, $B=N+b$. At high counts, the delta method gives $A/B\simeq 1+a/N-b/N$. Independence therefore givesThus the formula is a high-signal gain estimation from flat-field ratios, not an exact identity for a ratio of random variables. A Poisson distribution denominator can even be zero at low counts. The exposures must be linear and unsaturated, offsets removed, and read noise negligible. Being close to saturation increases the signal but does not establish linearity; clipping or charge-induced correlations biases a variance estimate. A large ensemble of suitable detector pixels supplies the measured standard deviation. In normal photodetector terminology $g$ is collected Electrons per ADU, consistent with detected photons here only under the stated unit-yield convention; see the instrument gain definition.

Chemical equilibrium imposes ${\mu }_e+{\mu }_p={\mu }_H+{\mu }_{\gamma }$. For blackbody photons, ${\mu }_{\gamma }=0$. Define the positive Hydrogen binding energy by $I=(m_e+m_p-m_H)c^2$, approximately $13.6\mathrm{eV}$. Multiplying the two Maxwell-Boltzmann distribution number densities and dividing by the neutral-hydrogen density givesUse $g_e{g}_p/g_H=1$ and $m_p/m_H\simeq 1$ to obtain the Saha ionization equationThe assumptions are nonrelativistic, dilute, nondegenerate species at a common temperature, ground-state hydrogen with the stated degeneracy factors, negligible mass correction in the prefactor, and sufficiently rapid reactions to maintain chemical equilibrium.

The baryon-to-photon ratio is extremely small, of order $10^{-9}$. Thus even when the typical photon energy is below $I$, the high-energy tail of the blackbody radiation contains enough photons above the ionization energy to reionize almost every newly formed hydrogen atom. Cosmological recombination becomes effective only when this exponentially small tail is also small compared with the baryon-to-photon ratio. Consequently recombination occurs at a temperature much lower than $I/k$, consistent with $kT\simeq 0.03I$. The exact freeze-out history also depends on reaction rates and expansion; the equilibrium estimate explains the large suppression of the temperature.

At this order there is one annihilation tree-level Feynman diagram. The squared-amplitude diagram is its product with its complex conjugate, with the same external momenta and spin labels in both factors:

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o
 e+(q,r) <---/          \<--- mu+(q',r')

                     times

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o                       *
 e+(q,r) <---/          \<--- mu+(q',r')

A wavy line is the virtual photon; the arrows show fermion-number flow. The star means complex conjugation of the entire second graph, including its external spinors. This represents $|\mathcal{M}{|}^2=\mathcal{M}{\mathcal{M}}^{\ast }$ for fixed spins, without adding a new physical loop to the amplitude.

For the spin sum in part (iv), identify each external spin label between the two copies and sum it. The fermion spin sum sews each spinor chain into a gamma-matrix trace. The two incoming unpolarized spin states are then handled by a spin average, giving the factor $1/4$. In particular, the diagrammatic contraction becomes

Take the quantum electrodynamics interaction density to be ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. The momentum-space Feynman-gauge photon propagator, with Lorentz indices $\mu ,\nu$ and momentum $k$, isThe electron-photon vertex from the QED Feynman rules iswith its spinor indices connecting the incident and outgoing fermion lines. The muon-photon vertex has the same coupling. The Feynman i-epsilon prescription specifies the pole boundary condition; using a different overall charge-sign convention changes both vertices consistently and leaves the squared amplitude unchanged.

For the electron-positron annihilation into a muon pair, the exchanged photon has $k=p+q$ and $k^2=s$. Up to the conventional overall phase, the scattering amplitude is

For a real vector relativistic quantum field, $V^{\mu }=V^{\mu †}$. Its transformed field must also be a Hermitian operator, so the intrinsic charge-conjugation phase satisfies ${\eta }_C={\eta }_C^{\ast }$; combined with $|{\eta }_C|=1$, this givesThe quantum electrodynamics interaction is $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. Since the Dirac electromagnetic current is odd under charge conjugation, charge conjugation invariance of this interaction requires the photon field to be odd too:

The deep inelastic scattering process contains a virtual photon exchanged between the Electron and the hadron:

Use an electromagnetic vector current $J^{\mu }$ containing the dimensionless quark charges, with the coupling $e$ factored out. The scattering amplitude, up to an irrelevant phase, isTo match the printed prefactor, define the leptonic tensor with a spin sum over both Electron spins and keep the initial spin average $1/2$ outside it. The gamma-matrix trace givesFor a stationary target and massless Electron, the invariant flux factor is $4ME$. The inclusive final-state Lorentz-invariant phase-space measure and target spin average are contained in $4\pi W_H^{\mu \nu }$. Thus the differential scattering cross-section isHere $q^4$ means $(q^2{)}^2$. If the initial spin average is instead built into the leptonic tensor, its normalization is $2(\cdots )$ and the displayed cross-section prefactor must be doubled. The two conventions give the same observable.

The outward radiative flux is $F=L/(4\pi r^2)$. A photon of energy $E$ carries momentum $E/c$, so the momentum transferred per unit volume and time to matter is $\kappa \rho F/c$, when $\kappa$ is the appropriate transport opacity. Momentum conservation makes the radiation stress decrease outward by this amount. In the thin atmosphere, where the radiation stress is treated by its scalar radiation pressure,This is the first moment of the radiative transfer equation. The grey atmosphere approximation uses a frequency-independent transport opacity; in an optically thick diffusion regime it becomes the Rosseland mean opacity. Near the surface the plane-parallel atmosphere approximation removes geometric stress-divergence corrections.

At time $t_0$, the accelerated proton has momentum $p=qE{t}_0$ and, by the relativistic energy-momentum relation, energyThe incident Cosmic microwave background photon adds energy $E_{\gamma }$. Since the final state contains only the Delta-plus baryon, conservation of relativistic energy gives its laboratory-frame Lorentz factorIts rest-frame decay time $t_{\Delta }$ is a proper time, so time dilation gives the laboratory decay time

In a box of volume $V$, the electromagnetic mode density in a box, including the two transverse polarizations, isEach photon has energy $\hslash \omega =\hslash ck$, and the given mean occupation number is $(e^{\hslash ck/T}-1{)}^{-1}$. Omitting the temperature-independent zero-point energy, the total energy is thereforeWith the change of variables $x=\hslash ck/T$,

At neutrino decoupling, neutrinos, electrons, positrons, and photons share one temperature. The neutrinos subsequently free stream, so $T_{\nu }\propto a^{-1}$. In the still-coupled electromagnetic plasma, the effective entropy degrees of freedom change during electron-positron annihilation in cosmology fromto $g_{\ast s,\mathrm{after}}=2$. Separate cosmological entropy conservation in that plasma gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$, while $T_{\nu }a$ remains constant. Consequently the Cosmic neutrino background temperature obeys

The higher state has principal quantum number $l+2$ and the lower one $l+1$. Energy conservation for the emitted photon gives

The range of a force is controlled by the lightest state that can carry it. Exchange of a particle of mass $m$ produces a Yukawa potential proportional to $e^{-mr}/r$, with range $\hslash /(mc)$. The electromagnetic interaction is long range because its carrier, the photon, is massless. The weak interaction is short range because the W bosons and Z boson are massive. Although the gluons of the strong interaction are massless in the Lagrangian, Quantum chromodynamics confines color and has a mass gap; only color-singlet hadrons propagate over macroscopic distances. The residual nuclear force is consequently controlled at long distance by massive pion exchange.

After electroweak symmetry breaking, the Higgs field has vacuum expectation value $⟨H⟩=(0,v/\sqrt{2}{)}^T$. A Yukawa interaction $-y_f{\bar{f}}_{L}H{f}_R+h.c.$ then becomes a fermion mass with $m_f=y_{f}v/\sqrt{2}$. The Higgs gauge-covariant kinetic term $|D_{\mu }H{|}^2$ giveswhile the radial fluctuation is the massive Higgs boson. The photon and all eight gluons correspond to unbroken gauge generators and remain massless. In the minimal renormalizable Standard Model, which has no right-handed neutrinos, the neutrinos also remain massless; adding right-handed neutrinos permits Higgs-generated Dirac masses.