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Consider a species in thermal equilibrium, with energy density and pressure depending on temperature and with vanishing chemical potential, as in the supplied thermodynamic relation. Put $U=\rho (T)V$. The first law of thermodynamics givesSince $dP/dT=(\rho +P)/T$,The differential is therefore exact:After fixing the irrelevant additive entropy zero, the entropy density at zero chemical potential isMore precisely the first law leaves an additive constant in $S$; extensivity fixes that constant for the usual entropy-density normalization. A nonzero chemical potential would instead require $Ts=\rho +P-\mu n$, so the stated formula is not a universal identity for a decoupled massive species with conserved particle number.

For a reversible thermodynamic process that is an adiabatic process in a comoving volume, $\dot{V}=3HV$ and the cosmological perfect-fluid continuity equation gives $\dot{\rho }=-3H(\rho +P)$. Differentiating the extensive expression explicitly, or using the exact differential above, givesThis is cosmological entropy conservation for the closed equilibrium gas with no entropy-producing energy injection.

For radiation in cosmology, $P=\rho /3$, so $s=4\rho /(3T)$. Comparing $s=c_1{g}_{\star }{T}^3$ and $\rho =c_2{g}_{\star }{T}^4$ yields the coefficient ratioWhen all relativistic species share the same temperature and count, the familiar normalizations are $c_1=2{\pi }^2/45$ and $c_2={\pi }^2/30$. More generally the energy and entropy effective counts can differ; the common $g_{\star }$ here uses the assumptions stated in the question.

Starting from a matter-antimatter symmetric ensemble, exact charge conjugation would pair every baryon-producing process with an equally probable antibaryon-producing process. Their contributions to the net baryon number cancel. Thus C violation is necessary. Violating charge conjugation alone is insufficient: if CP symmetry is exact, the parity-reflected conjugate channels still have equal probabilities and cancel in a CP symmetry invariant ensemble. One needs CP violation as well.

The remaining Sakharov conditions are baryon number violation and departure from thermal equilibrium. Conserving baryon number cannot turn an initially zero value into a nonzero one. In thermal equilibrium, under the usual CPT theorem hypotheses and with no imposed chemical potentials for charges odd under CPT symmetry, states and their conjugates have equal thermal weights, so the equilibrium expectation of net baryon number vanishes and thermal stationarity prevents its sustained production. A nonequilibrium history allows different forward and reverse populations to generate an asymmetry. These are the standard necessary conditions for dynamical baryogenesis from symmetric initial conditions, not a guarantee of a sufficiently large asymmetry.

Take $\gamma >1$ for the ideal gas. At fixed cylindrical radius, put $\mathcal{R}=k/(\mu m_p)$, where $k$ is the Boltzmann constant, $\mu$ the mean molecular weight and $m_p$ the proton mass. Also write $a=9\alpha \Omega /4$. Vertical hydrostatic equilibrium, thermal equilibrium and the ideal gas law requireConsequently, with $K=A/a$,Integrating from the midplane yieldsThe polytropic vertical structure in stellar gravity therefore has, for $|z|\le H$,whereThis is a polytropic equation of state $P\propto {\rho }^{1+1/\beta }$ with polytropic index $\beta$. For $|z|>H$, set $\rho =P=0$; temperature has no physical gas value in that vacuum. Here $H$ is the finite semi-thickness, rather than an exponential disk scale height.

At thermal equilibrium, microscopic reversibility equates the probabilities of a path and its reversed path when both include their Boltzmann distribution initial weights. Denote their endpoint states by $z_1,z_2$, including velocity if needed. Since the Hamiltonian is even under time reversal in classical mechanics,This detailed balance condition and the energy balance for an autonomous Lagrangian yieldThis is the fluctuation-dissipation relation for a Langevin particle: the strength of Gaussian white noise is fixed by the damping and temperature, with $k_B$ the Boltzmann constant.

For an equilibrium coarse-grained variable, the unresolved microscopic states contribute entropy; their statistical weight is encoded in the Helmholtz free energy, rather than in a single microscopic energy. Relative to the same reference measure, ${\rho }_{\mathrm{eq}}(x)\propto e^{-\beta F(x)}$, so microscopic reversibility becomesThis extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.

In the very early universe, assume species $i$ is in thermal equilibrium, is ultrarelativistic so that $kT\gg m_i{c}^2$ and $E(p)\simeq pc$, and has negligible chemical potential, ${\mu }_i\simeq 0$. Its occupation factor is then the Bose-Einstein distribution for a boson or the Fermi-Dirac distribution for a fermion.

Substitute $E=pc$ and set $y=pc/(kT)$. Since $p=(kT/c)y$ and $dp=(kT/c)dy$,

Let two systems exchange energy inside an otherwise isolated composite system, so $E_1+E_2=E$ is fixed. By the Second law of thermodynamics, equilibrium maximizes the total entropyAt an interior maximum,Thus $T_1=T_2$, which is entropy maximization under thermal contact and the condition for thermal equilibrium.

Stability requires this entropy maximum to be locally strict, so each ordinary subsystem has a concave entropy-energy relation. Sinceone has $\partial T/\partial E>0$, or equivalentlyThis is the entropy concavity and positive heat capacity criterion.

Now let $N_{\uparrow }=\alpha N$ and $N_{\downarrow }=(1-\alpha )N$. Because the spins are distinguishable, the number of microstates is the binomial coefficientThe Boltzmann entropy $S=k_B\log \Omega$ and the Stirling formula give, to leading order for large $N$,The total energy isThereforeSolving for the up-spin fraction gives the independent spin-one-half two-level system resultFor ordinary positive absolute temperature,with $\alpha \to 0$ as $T\to 0^{+}$ and $\alpha \to 1/2$ as $T\to +\infty$. Since the spectrum is bounded above, the population-inverted range $1/2<\alpha \le 1$ is also mathematically possible and corresponds to negative temperature; $\alpha =1/2$ is the infinite-temperature state.

Substitution givesand direct differentiation yieldsThus the energy increases with temperature throughout either finite-temperature branch, in particular throughout the required positive-temperature range.

When ${\mu }_{\nu }/T_{\nu }\gg 1$, the Fermi-Dirac distribution approaches the step function $f(p)=1$ for $p<{\mu }_{\nu }$ and zero for $p>{\mu }_{\nu }$. For one internal state and negligible mass, the degenerate relic neutrino energy density is thereforeso $\boxed{A=1/(8{\pi }^2)}$.

At thermal equilibrium, the antineutrino chemical potential is $-{\mu }_{\nu }$. Its occupation is approximately $e^{-(p+{\mu }_{\nu })/T_{\nu }}$, givingIt is exponentially smaller than the neutrino density; this particle-antiparticle imbalance is a cosmological lepton asymmetry.

In the nonrelativistic regime,The assumption $E-\mu \gg k_{B}T$ puts the Fermi-Dirac distribution in its Maxwell-Boltzmann limit, soTherefore the nonrelativistic Maxwell--Boltzmann number density is

For $p+e^{-}\leftrightarrow H+\gamma$, chemical equilibrium and the vanishing photon chemical potential implyApplying the number-density formula to each massive species and eliminating the chemical potentials givesUse charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_e{g}_p/g_H$. With the Hydrogen binding energywe obtain the Saha ionization equationThe assumptions are thermal equilibrium and chemical equilibrium before cosmological recombination, a nonrelativistic nondegenerate gas, zero photon chemical potential, charge neutrality, negligible proton-electron mass correction in the translational prefactor, and the stated degeneracy-factor approximation.

For photons in thermal equilibrium at temperature $T$, the number density per frequency interval is

A mean-field coupling of the Poisson equation to an ideal Boltzmann distribution of mobile ions. In Gaussian units with uniform dielectric constant $ϵ$, ${\nabla }^2\varphi =-(4\pi /ϵ)[{\rho }_f+{\sum }_i{q}_i{n}_{i,0}{e}^{-\beta q_i\varphi }]$. The potential reference and chemical potentials determine the normalization factors. It neglects discrete-ion correlations and assumes local thermal equilibrium. A counterion-only model is not the same equation as a symmetric salt reservoir.

The standard conditions for baryogenesis from a symmetric initial ensemble are baryon number violation, both C violation and CP violation, and departure from thermal equilibrium. Conserved baryon number cannot develop a net value from zero; an exact conjugation symmetry pairs production with equal antiproduction. In thermal equilibrium, CPT symmetry makes the equilibrium expectation of net baryon number vanish for zero imposed chemical potentials for charges odd under CPT symmetry. These conditions assume the usual local quantum-field framework and do not alone guarantee a large asymmetry.

In stellar structure, stellar thermal equilibrium means that internal heating balances the surface luminosity and any energy loss through neutrinos, with negligible secular storage of thermal energy on the timescale considered. A star can satisfy this balance while its interior is hotter than its surface and heat flows outward. It is therefore different from equal-temperature thermal equilibrium between systems. Neglecting neutrino losses, the balance is $L_{\mathrm{surface}}=L_{\mathrm{heat}}$. In a slowly mass-losing donor star, a stellar thermal equilibrium radius sequence applies only when thermal adjustment is fast compared with the mass-loss timescale; rapid loss instead requires the adiabatic stellar radius response exponent.

Thermal radiation is electromagnetic radiation emitted by matter because of its temperature. Matter in thermal equilibrium emits a spectrum governed by its absorptivity, approaching the Planck law for a blackbody.

In static thermal equilibrium, the local temperature times the stationary lapse is constant. Energy measured locally redshifts to conserved Killing energy by that same lapse; requiring a single equilibrium Boltzmann exponent gives the relation. The Killing vector field normalization fixes the constant but not the ratio of temperatures at different locations.

The first polar sine and cosine harmonics of a circular boundary represent displacements of its centre. A complete translation leaves perimeter and area unchanged, so line tension supplies zero stiffness to these modes. The centre of an unconfined domain has no normalizable positional thermal equilibrium in an infinite plane; recentering the boundary removes this freedom from a shape spectrum.