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Use momentum conservation and conservation of energy in the laboratory inertial frame. The photon has momentum magnitude $E/c$, while the outgoing Electron satisfies the relativistic energy-momentum relation. Its total energy and momentum obeySubstitute these into $E_e^2=m^2{c}^4+c^2|{\mathbf{p}}_e{|}^2$. Cancelling $m^2{c}^4$ and $(E_1-E_2{)}^2$ gives $m{c}^2(E_1-E_2)=E_1{E}_2(1-\cos \theta )$. Both photon energies are positive, so division yieldsThis is Compton scattering. The energies include the Electron's rest energy; using only its kinetic energy in the mass-shell equation would give the wrong result. The outgoing photon has unchanged energy at $\theta =0$ and smaller energy for a nonzero scattering angle.

In the local orthonormal tetrad, the components of the photon number current areThey are respectively the photon number density and the photon number flux. Thus $J^{\mu }$ is the photon number four-current, not an energy current or an electric charge current. Photons carry no electric charge. For the isotropic background,Integrating the covariant Boltzmann equation over the invariant mass-shell measure, with vanishing momentum-boundary flux, giveswhere $C[f]$ is the collision operator in the covariant streaming equation. ThereforeCollisionless photon propagation satisfies this condition, as do Thomson scattering and ordinary Compton scattering when all participating photons are counted: each collision has one incoming and one outgoing photon. Photon emission, photon absorption, bremsstrahlung, double Compton scattering and electron-positron annihilation in cosmology or its inverse generally give a nonzero number source. A cancellation of net creation and destruction can nevertheless make the integrated source vanish even when individual number-changing reactions occur. Cosmological redshift itself changes photon energy, not photon number; in an expanding homogeneous universe a conserved photon number current gives ${\bar{n}}_{\gamma }\propto a^{-3}$.

The collisionless Boltzmann equation is conservation of $f$ along the photon trajectory. To linear order, use the unperturbed trajectory $d\mathbf{x}/d\eta =\mathbf{e}$ when differentiating $\Theta$. Deflection of $\mathbf{e}$ is first order and multiplies the first-order angular dependence of $\delta f$, while the background has no angular dependence; it therefore contributes only at second order. Similarly, the energy derivative of $\delta f$ times the perturbed energy change is second order. ThusFor a nontrivial spectrum this identifies the linear Free-streaming photon Boltzmann equation:In Newtonian gauge in cosmology, substitute the specified scalar gravitational-redshift source to writeMultiply its angular average by four. Since $⟨e^j{⟩}_{\Omega }=0$, the photon angular temperature moments imply the photon continuity equationFor the photon Euler equation, multiply by $3e^i$ and average. The source is $-3⟨e^i{e}^j{⟩}_{\Omega }{\partial }_j\psi =-{\partial }^i\psi$, while the second angular moment isIt follows thatThe background energy density is spatially homogeneous, so no spatial derivative acts on $\bar{\rho }$. The minus sign of its anisotropic stress term follows from the PDF's $\Pi =-\pi$ convention. Switching to the conventional trace-free stress $\pi$ changes that term to a plus sign.

Use a local orthonormal tetrad and units in which the speed of light is one. A particle has four-momentum $p^{\hat{a}}=E(1,\mathbf{e})$ and moves with velocity $\mathbf{e}$. Thus its energy density contributes $E$, its momentum density contributes $E{e}^i$, and its momentum flux contributes $E{e}^i{e}^j$. Integrating the phase-space distribution function gives the kinetic stress-energy tensor. The measure $d^{3}p/E$ is the future mass-shell Lorentz-invariant phase-space measure, so the same expression transforms as a spacetime stress-energy tensor; fixed state-counting or polarization factors are understood to be included in $f$.

Define $⟨F{⟩}_{\Omega }=(4\pi {)}^{-1}\int Fd\Omega$. The photon momentum measure is $d^{3}p=E^{2}dEd\Omega$, so every component of the kinetic stress-energy tensor has a radial factor $E^{3}dE=a^{-4}{ϵ}^{3}dϵ$. An isotropic background has $⟨e^i{⟩}_{\Omega }=0$ and $⟨e^i{e}^j{⟩}_{\Omega }={\delta }^{ij}/3$. HenceThis radiation pressure equation of state requires isotropy and masslessness, not a thermal spectrum.

The distribution perturbation is $\delta f=-ϵ{\bar{f}}^{\prime }(ϵ)\Theta$. Assume finite energy density and endpoint behavior ${ϵ}^4\bar{f}(ϵ)\to 0$ at zero and infinity, as for the Planck photon distribution. Integration by parts then givesConsequently the perturbed energy density and momentum density areSince $q^i=(\bar{\rho }+\bar{P})v^i=(4\bar{\rho }/3)v^i$, the photon angular temperature moments give $\delta =4⟨\Theta {⟩}_{\Omega }$ and $v^i=3⟨\Theta e^i{⟩}_{\Omega }$.

The sign of anisotropic stress must be specified. Direct kinetic integration gives the conventional trace-free spatial stress ${\pi }^{ij}=T^{ij}-P{\delta }^{ij}=+4\bar{\rho }⟨\Theta (e^i{e}^j-{\delta }^{ij}/3){⟩}_{\Omega }$. The PDF uses the opposite sign, ${\Pi }^{ij}=-{\pi }^{ij}$, equivalently $T^{ij}=P{\delta }^{ij}-{\Pi }^{ij}$. With that convention, all requested moments areTaking the spatial trace also givesThe angular temperature description assumes $\Theta$ is independent of photon energy; independent spectral distortions would require additional energy-dependent moments.

Evaluate the metric inner products of the proposed orthonormal tetrad to first order in the vector cosmological perturbation:The orthonormal tetrad therefore has metric signature $(-,+,+,+)$ to the required order. Put $q=ϵ/a^2$ and $b=\mathbf{e}\cdot \mathbf{B}$. The photon momentum components areThey satisfy the null vector condition to first order, with $|\mathbf{e}|=1$.

Use $\eta$ as parameter in the time component of the geodesic equation:The derivative of the direction is first order, since the background photon moves on a straight comoving line; $B_{i}d{e}^i/d\eta$ is consequently second order. With $d\mathbf{x}/d\eta =\mathbf{e}+O(B)$,The supplied Levi-Civita connection coefficients giveThe spatial gradients and all background expansion terms cancel. Since ${ϵ}^{\prime }/ϵ$ vanishes in the background, its product with $b$ is also second order. The photon energy redshift from a vector metric perturbation is thereforeAn independent check uses the covariant component $p_0=g_{0\mu }{p}^{\mu }=-ϵ+O(B^2)$. The covariant geodesic equation gives $d{p}_0/d\eta =(2p^0{)}^{-1}{\partial }_{\eta }{g}_{\alpha \beta }{p}^{\alpha }{p}^{\beta }=ϵe^i{\dot{B}}_i$; the conformal expansion term vanishes by the null vector condition and reproduces the same result.

A blackbody grain absorbs starlight of flux $L_{\star }/(4\pi r^2)$ and reradiates isotropically in its own rest frame. Aberration of light makes the incoming beam tilt against transverse motion. In the stellar frame, the reradiated photons carry momentum proportional to the grain's velocity, producing Poynting–Robertson drag; it does not require anisotropic emission in the grain's frame.

Define the radiation-pressure coefficientFor a spherical grain of radius $s$ and mass density $\rho$, $\beta =3L_{\star }/(16\pi cG{M}_{\star }\rho s)$, assuming unit radiation-pressure efficiency. To first order in $v/c$, the first-order stellar radiation force on a blackbody grain isThe incoming absorbed photon momentum contains the relative-flux factor $1-\dot{r}/c$; reradiation supplies the additional $-\mathbf{v}/c$. Separating the velocity-independent radiation pressure leavesThus there are two radial first-order contributions and one tangential contribution. The approximation assumes $v\ll c$, steady absorption/reradiation and constant grain properties; a non-blackbody requires its actual optical efficiencies.

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.