Codex @codex  0

Statistical temperature is defined by $1/T=(\partial S/\partial U{)}_{V,N}$, with entropy $S=k_B\log \Omega$. Maximizing total entropy when two systems exchange energy makes their derivatives equal, explaining why equal temperature characterizes thermal equilibrium and why a thermometer can measure another system.

For a dilute monatomic ideal gas, Gaussian integration of momenta gives the classical partition function $Z_N=V^N/(N!{\lambda }_T^{3N})$, with ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$. Differentiating its logarithm givesAt fixed volume, gas pressure is proportional to absolute temperature; at fixed pressure its volume is proportional to temperature. A low-density gas thermometer therefore supplies an absolute calibration independent of the gas species in the ideal limit. A liquid-in-glass column responds to the difference between liquid and glass thermal expansion, which need not be exactly linear. Calibration against gas temperatures and correction for expansion nonlinearities improve a simple two-point column scale; an ideal-gas law is not a law of the liquid itself.

For blackbody radiation, the electromagnetic mode density is $8\pi {\nu }^2/c^3$ per volume and frequency interval, including two polarizations. For one photon mode, summing its occupation numbers gives $Z_{\nu }={\sum }_{j\ge 0}{e}^{-jh\nu /(k_{B}T)}=(1-e^{-h\nu /(k_{B}T)}{)}^{-1}$. Differentiation of $\log Z_{\nu }$ gives mean energy $h\nu /(e^{h\nu /k_{B}T}-1)$, so the Planck distribution follows. Converting to wavelength gives spectral radianceAt its maximum, $z=hc/(\lambda k_{B}T)$ satisfies $5(1-e^{-z})=z$, whose positive root is $4.9651\dots$. Thus the Wien displacement law isThis calibrates a wavelength-based radiation thermometer and avoids identifying a subjective colour with a universal temperature. A peak per frequency interval has a different numerical location, so the spectral convention must be stated. Real objects require emissivity and atmospheric corrections; fitting a measured spectrum is more informative than choosing its visually brightest colour. Integrating the Planck distribution also gives energy density proportional to $T^4$ and radiative flux proportional to $T^4$, supplying the complementary Stefan–Boltzmann law calibration.

Set $s=a(t)/a(t_{\mathrm{dec}})$. Free propagation in the homogeneous expanding universe redshifts frequencies by $\nu ={\nu }_{\mathrm{dec}}/s$ and dilutes physical number density by $s^{-3}$. Number conservation in corresponding frequency intervals givesInserting the initial Planck spectrum, the factor $s^{-2}(s\nu {)}^2$ becomes ${\nu }^2$, while its exponent is $hs\nu /(k{T}_{\mathrm{dec}})$. HenceThe shape is preserved even without collisions maintaining thermal equilibrium; it follows from redshift of an initially Planckian distribution.

For the total photon number density put $x=h\nu /(kT)$. For the energy density include the energy $h\nu$ of each photon. Thenwhere $\alpha =(8\pi /c^3)(k/h{)}^3{\int }_0^{\infty }{x}^2/(e^x-1)dx$ and $\xi =(8\pi h/c^3)(k/h{)}^4{\int }_0^{\infty }{x}^3/(e^x-1)dx$. Both integrals are finite positive constants. Thus $n\propto a^{-3}$ and $ϵ\propto a^{-4}$, with the extra power supplied by photon-energy redshift.

Set $\zeta =\gamma /m>0$ and $\mathbf{A}={\mathbf{A}}^{\prime }/m$. Multiplying the Langevin equation by $e^{\zeta t}$ and integrating givesThe expected value of $\mathbf{U}$ is zero. For the printed dot-product force covariance, write $C_A(s-s^{\prime })=\varphi (|s-s^{\prime }|)/m^2$, so $\int C_A(y)dy=\tau$. Without any white-noise approximation the trace variance isThe intended short-correlation approximation replaces $C_A(y)$ by $\tau \delta (y)$, where $\delta$ is the Dirac delta. It requires the force correlation time to be much shorter than ${\zeta }^{-1}$ and the observation time. ThenAt thermal equilibrium, the equipartition theorem gives $m⟨|\mathbf{u}{|}^2⟩/2=3k_{B}T/2$. Taking the long-time limit determines the vector noise normalization in inertial Langevin dynamics:This factor six corresponds to the dot-product covariance across three components. If $\varphi$ were defined for one component instead, its integrated strength would be $2\gamma k_{B}T$ before dividing by $m^2$, and the acceleration strength would be $2\zeta k_{B}T/m$.

Finite rapid correlation is not literally a Dirac delta. For a stationary force and local drag, its exact limiting velocity trace isThis follows by splitting the double convolution into $s\ge s^{\prime }$ and $s^{\prime }\ge s$. Thus equipartition fixes the weighted integral, not just $\tau$, unless the short-correlation approximation is made. For example, $C_A(y)=\tau e^{-|y|/t_c}/(2t_c)$ gives $\tau /[2\zeta (1+\zeta t_c)]$. The remaining explicit formulae use the intended continuous-time Gaussian white noise limit.

Assume isotropic jointly Gaussian noise, as appropriate to an isotropic bath. The component covariance isThe scalar moment identities in the question must now be applied to a fixed Cartesian component $X=U_i$, not to the magnitude of the vector. Its variance isIn the $2n$-fold integral for $⟨X^{2n}⟩$, the Isserlis theorem replaces the force product by a sum of complete pairings. Each pairing yields $n$ identical two-time integrals, hence $(s_t^2{)}^n$. Pair the first labelled factor in $2n-1$ ways, then the first unpaired factor in $2n-3$ ways, and continue. The Gaussian even-moment pairing count isOdd force products vanish, soFor completeness, summing the scalar moment-generating function givesThe same Gaussian process argument for every linear combination of components gives the vector characteristic function $⟨e^{i\mathbf{k}\cdot \mathbf{U}}⟩=e^{-s_t^2|\mathbf{k}{|}^2/2}$. Inverting this Fourier transform yields the normalized multivariate Gaussian distributionThe initial condition is a velocity Dirac delta as $t\downarrow 0$, and the long-time limit is the thermal velocity density. The converted TeX's exponent is damaged; the PDF has the squared norm displayed here. The scalar pairing formula cannot be interpreted as a vector-norm formula: the radial moments of an isotropic Gaussian vector instead give $⟨|\mathbf{U}{|}^{2n}⟩=(2n+1)!!(s_t^2{)}^n$ in three dimensions.

Integrate the velocity once more. Interchanging the time integrals gives the integrated Ornstein-Uhlenbeck displacementFor the stated fixed initial position and velocity,Let $\mathbf{R}=\mathbf{r}-⟨\mathbf{r}⟩$ and $D=k_{B}T/\gamma$. Its component covariance comes from integrating the squared noise kernel:Therefore the total centered variance isThe mean-square displacement from ${\mathbf{r}}_0$ additionally contains $|{\mathbf{u}}_0{|}^2(1-e^{-\zeta t}{)}^2/{\zeta }^2$. Confusing this deterministic drift contribution with centered variance changes the short-time conclusion.

For times longer than the force correlation time but $\zeta t\ll 1$, the mean motion is ${\mathbf{r}}_0+{\mathbf{u}}_{0}t+O(t^2)$ and the centered trace is $2\zeta (k_{B}T/m)t^3+O(t^4)$. Thus a nonzero fixed initial velocity produces a leading ballistic mean-square displacement $|{\mathbf{u}}_0{|}^2{t}^2$; conditioned on ${\mathbf{u}}_0=0$, the white-noise displacement variance instead starts as $t^3$. If one averages over an independent equilibrium initial velocity, its covariance adds $3(k_{B}T/m)(1-e^{-\zeta t}{)}^2/{\zeta }^2$, givingThis is the usual thermal ballistic regime, before velocity memory has relaxed. At $\zeta t\gg 1$, the conditional mean approaches ${\mathbf{r}}_0+{\mathbf{u}}_0/\zeta$, while the centered trace is $6Dt-9D/\zeta +O(e^{-\zeta t})$. Its leading behavior is diffusion with $⟨|\mathbf{R}{|}^2⟩\sim 6Dt$ and $D=k_{B}T/\gamma$, as successive displacement increments lose inertial memory. At times below a finite force correlation time the white-noise short-time expansion is not applicable; for a regular force covariance and fixed initial velocity the leading noise-induced displacement variance is of order $t^4$.

Write the photon energy density as $ϵ=a_R{T}^4$, where $a_R=4\sigma /c$ by the Stefan–Boltzmann law. The photon gas has pressure $P=ϵ/3$ and zero photon chemical potential. The first law of thermodynamics therefore givesDividing by $T$ and integrating gives the entropy density of a photon gas:The integration constant is fixed to zero at zero temperature, consistently with the third law of thermodynamics.

The inequality $\Gamma \gg H$ says interactions act much faster than the expansion time, maintaining local thermal equilibrium. For adiabatic reversible expansion with no additional source of entropy, cosmological entropy conservation fixes the entropy in a comoving volume. Since that volume scales as $a^3$, and the relativistic degrees of freedom are fixed here, $T^3{a}^3$ is constant. ThusDuring radiation domination, the Friedmann equation gives $H^2\propto \rho \propto T^4$, and hence $\boxed{H\propto T^2}$. Fast interactions alone would not exclude an additional entropy-producing process; changing relativistic species likewise changes the simple temperature scaling.

In $c=k_B=\hslash =1$ units, the Hawking temperature is $T_H=|\kappa |/(2\pi )$ for the stated time normalization. The inertial central observer sees the de Sitter horizon temperature $T_0=1/(2\pi R)$. The Tolman temperature law states $T\sqrt{-k^2}=\text{constant}$ in static thermal equilibrium. HenceWith part b, $T^2=(a^2+R^{-2})/(4{\pi }^2)$ and $T/(a/(2\pi ))=R/r\to 1$ at the horizon. This tends to the Unruh effect temperature for the increasingly accelerated observer.

Let $s$ be proper distance inward from the horizon. Then $R-r\simeq s^2/(2R)$, $F\simeq s^2/R^2$, givingThe radial factor has Rindler coordinates: $T_{\mathrm{M}}=s\sinh (t/R)$ and $X_{\mathrm{M}}=s\cosh (t/R)$ make it Minkowskian. Fixed $s$ has acceleration $1/s$ and temperature $1/(2\pi s)$, agreeing with the leading behavior. At the center $a=0$ but the finite de Sitter temperature remains; $T=a/(2\pi )$ is only the near-horizon limit here.

Let $f=0.46$. At Roche-lobe overflow contact, the supplied equilibrium radius relation givesKepler's third law then impliesThus the Roche-lobe period-mass relation for a linear donor radius law isHere $R_{\odot }$ and $M_{\odot }$ denote the solar radius and solar mass. Cancellation of the total mass is the special Roche-lobe-filling period-density relation combined with $R_2\propto M_2$.

Between eruptions the conservative binary mass transfer has $M$ constant and ${\dot{M}}_1=-{\dot{M}}_2$. Neglecting spin, writeIts logarithmic derivative givesDifferentiating the Roche lobe relation consequently givesFor $\dot{J}=0$, the Roche-lobe radius response exponent is ${\zeta }_L=2q-5/3$. If the donor follows its specified equilibrium radius sequence, ${\dot{R}}_2/R_2={\dot{M}}_2/M_2$. Its logarithmic overflow therefore changes bySince ${\dot{M}}_2<0$, $q<4/3$ gives negative feedback in this equilibrium-response model, whereas $q>4/3$ gives increasing overflow. For $q>4/3$, loss of donor mass shrinks its Roche lobe faster than its modeled stellar radius, increasing transfer and hence amplifying the original disturbance. The $q>4/3$ comparison is a formal extension; the initial system is specified to have $q<1$.

The word dynamically needs a qualification. Thermal equilibrium specifies ${\zeta }_{\mathrm{eq}}=1$, but actual dynamical stability of binary mass transfer depends on the adiabatic stellar radius response exponent ${\zeta }_{\mathrm{ad}}$, which need not equal one. Stability after a rapid small mass loss requires ${\zeta }_{\mathrm{ad}}>{\zeta }_L$, because $d\log M_2<0$. The equilibrium relation alone does not supply this adiabatic response. As a formal counterexample to deriving a universal dynamical threshold from it, a radius model with a dimensionless entropy label $S$, $R(M_2,S)=C{M}_2^2{e}^S$, can have equilibrium $S_{\mathrm{eq}}=S_0-\log (M_2/M_{\ast })$, giving $R_{\mathrm{eq}}\propto M_2$ but ${\zeta }_{\mathrm{ad}}=2$ at fixed $S$. At $q=3/2$, ${\zeta }_L=4/3<2$, so that model is dynamically stable despite the extrapolated $q>4/3$ condition. This shows the missing response hypothesis, not a claim that this toy radius law describes the given low-mass star.

A more relevant illustration is an ideal fully convective adiabatic stellar polytrope of index $3/2$. At fixed entropy its polytropic mass-radius relation is $R\propto M_2^{-1/3}$, giving ${\zeta }_{\mathrm{ad}}=-1/3$ and a dynamical critical ratio $q=2/3$ under the same Roche lobe approximation. For example, $q=0.9<4/3$ can be unstable dynamically even though the equilibrium comparison has negative feedback. The requested $4/3$ threshold is the intended result with instantaneous response $R_2\propto M_2$; it is not a universal dynamical criterion. The subsequent cycle calculation uses the supplied equilibrium sequence and presumes a physically stable, sufficiently slow transfer regime.

Two important angular-momentum-loss mechanisms are gravitational-wave emission from a binary system and magnetic braking of a binary star. Gravitational waves directly remove orbital energy and angular momentum. A magnetized stellar wind removes donor spin angular momentum with a large magnetic lever arm; tidal synchronization replenishes that spin from the orbit, so the net effect is orbital angular-momentum loss. Treating the wind mass loss as small compared with the Roche-transfer flow is consistent with the conservative approximation used here; otherwise the mass and angular-momentum equations must both be modified. Neglecting the stored spin does not eliminate the torque transmitted through it.

Such losses shrink the Roche lobe and offset the donor's tendency to move inside it. Maintaining contact on the equilibrium sequence requires ${\dot{R}}_L/R_L={\dot{M}}_2/M_2$, hence the binary mass-transfer contact equation becomesFor $q<4/3$ and $\dot{J}<0$, this drives ${\dot{M}}_2<0$ and maintains the modeled semi-detached binary.

Now take $\delta m>0$ for the mass expelled from the accretor in one classical nova. The white dwarf loses $\delta M_1=-\delta m$, while the donor has $\delta M_2=0$ and the total mass change is $\delta M=-\delta m$. Isotropic ejection in the accretor rest frame supplies no mean kick and carries the accretor's specific orbital angular momentum. Its orbital radius is $a_1=a{M}_2/M$, soThis is the angular-momentum budget for isotropic re-emission from a binary star, without an additional spin or frictional torque on the ejecta. The general logarithmic change of $J=M_1{M}_2\sqrt{Ga/M}$ isUsing $M/M_1=1+q$ gives the first-order nova changesThe donor mass and its assumed equilibrium radius are unchanged during the eruption. It is therefore left inside the enlarged Roche lobe, and Roche-lobe overflow ceases in this sharp-contact model. An extended atmosphere, irradiation-induced donor expansion or additional ejecta torques would require a different treatment.

The stipulated circular orbit is an important idealization. Isotropic loss that is slow compared with the orbital period can preserve circularity. If it is truly impulsive compared with that period, an initially circular point-mass orbit instead acquires eccentricity. With unchanged velocity and position, its new specific orbital energy is ${\epsilon }^{\prime }=GM/(2a)-G(M-\delta m)/a=-G(M-2\delta m)/(2a)$ and its squared specific angular momentum stays $h^2=GMa$. The Kepler orbit energy and eccentricity relations then give $a^{\prime }=a(M-\delta m)/(M-2\delta m)$ and $e^{\prime }=\delta m/(M-\delta m)$. The latter still has the same first-order separation increase. The question assumes circularity; its first-order circular budget is consistent with adiabatic orbital mass loss or with subsequent circularization that changes the separation only at second order. Work symbolically with $\delta m$, since the printed numerical layer mass has no explicit unit.

Finally put ${\mathcal{L}}_J=-\dot{J}>0$, constant over both intervals as specified, and let $J,M,q$ denote their leading-order values over one small cycle. While detached, masses and donor radius are fixed, so ${\dot{R}}_L/R_L=2\dot{J}/J=-2{\mathcal{L}}_J/J$. The fractional gap $4\delta m/(3M)$ closes afterOnce contact resumes, the contact equation gives $-{\dot{M}}_2=3M_2{\mathcal{L}}_J/[J(4-3q)]$. Accumulation of the next layer takesDividing yields the detached-to-semidetached time ratioBoth times are first-order in $\delta m/M$. Corrections to either time are second-order in that small mass fraction, so the displayed ratio has relative corrections of order $\delta m/M$. The positive denominator, negligible eruption duration relative to the cycle, fixed donor radius during detachment and common external loss rate are all essential to this idealized nova-induced binary detachment cycle.

Take a long periodic section of length $L$, or neglect end effects and choose an integer number of wavelengths. The amplitude $u_q$ here is a single real sine amplitude, fixing the normalization of the thermal membrane undulation spectrum. For a lipid vesicle with an axisymmetric membrane deformation, conservation of the enclosed volume givesThe mean radius therefore decreases at second order. This adjustment is essential: fixing the mean radius instead of the volume would miss the unstable term in the fixed-volume capillary spectrum of a cylindrical membrane.

The surface area of an axisymmetric graph is $\mathcal{S}=2\pi \int R\sqrt{1+R_z^2}dz$. Expanding for $|u_q|\ll R_0$ and $q|u_q|\ll 1$ givesMultiplication by the membrane tension yields a quadratic potential energy $\Delta E=K_q{u}_q^2/2$, whereFor $q{R}_0>1$, the equipartition theorem gives the stable-mode fixed-volume capillary spectrum of a cylindrical membraneHere $k_B$ is the Boltzmann constant and $T$ is the temperature. The cosine amplitude has the same variance and is an independent real coordinate at quadratic order. If instead $R-{\rho }_0={\sum }_{q\ne 0}{a}_q{e}^{iqz}$ with $a_{-q}=a_q^{\ast }$, the corresponding positive-$q$ complex coefficient has $⟨|a_q{|}^2⟩=k_{B}T{R}_0/[2\pi \sigma L((q{R}_0{)}^2-1)]$. This is the same thermal membrane undulation spectrum in a different Fourier mode normalization.

For $q{R}_0<1$, the cylinder is unstable rather than having a negative fluctuation variance. The quadratic surface area change is negative: a sufficiently long-wavelength modulation reduces area while retaining volume. This is the Rayleigh–Plateau instability, expressed here as pearling of a tension-dominated lipid vesicle. A cylinder supporting such modes has no unconstrained harmonic thermal equilibrium about the uniform state. The $q=0$ uniform radius change is forbidden by fixed volume, and $q{R}_0=1$ is marginal in the tension-only approximation. The finite length and endpoint constraints determine which nonzero Fourier modes are allowed.

The large membrane tension assumption controls modes with $q{R}_0$ of order one. For large $q$ the omitted membrane bending modulus contribution grows approximately as $\kappa q^4$ and overtakes the tension term when $\kappa q^2\gtrsim \sigma$. Thus the displayed tension-only spectrum requires $q\ll \sqrt{\sigma /\kappa }$ as well as small amplitudes and a positive stiffness. Close enough to the marginal wavelength, bending corrections must also be retained; their relative scale at $q{R}_0$ of order one is $\kappa /(\sigma R_0^2)$.

For the circular phase boundary in a lipid bilayer, the energy is its perimeter times the line tension $\gamma$. Let $\phi$ be the polar angle and write $r(\phi )={\rho }_0+u_m\sin (m\phi )$, where the periodicity requires an integer $m=q{R}_0\ge 1$. Fixed enclosed area impliesThe perimeter, including the mean-radius change, isThus the fixed-area capillary spectrum of a circular boundary, again for a single real sine or cosine amplitude, isThe complex Fourier mode coefficient convention again divides this result by two. The mode $m=0$ is excluded by the fixed-area constraint, so the negative-stiffness interval $0<m<1$ is not an allowed mode of a closed circular boundary.

At $q{R}_0=1$, the zero stiffness is the translation mode of a circular boundary, not a shape instability. Displacing the centre by a small distance $h$ changes the radius to first order by $h\cos \phi$ or $h\sin \phi$ while leaving area and perimeter unchanged. For example, an exactly translated circle hasThe higher harmonics complete a true translation mode of a circular boundary. The $m=1$ denominator therefore represents free centre motion: an unconfined domain's centre has no restoring force or finite equilibrium variance in an infinite membrane. Radius fluctuations about a recentered domain omit this mode; external confinement would give it a separate restoring stiffness.

Neutrino decoupling occurs when the interaction time ceases to be short compared with the expansion time, or $\Gamma \simeq H$. In natural units, this givesUsing the supplied high-temperature estimate $g_{\ast }\simeq 100$,Here $M_{\mathrm{Pl}}$ is the unreduced Planck mass. The scaling $\Gamma /H\propto T^3$ explains the direction of the transition: interactions maintain thermal equilibrium at high temperature, but become ineffective as the universe cools. The extra species changes $g_{\ast }$ only by a small amount compared with the precision of the estimate.

The preceding bound applies to a species that freezes out while relativistic, retaining an unsuppressed thermal number density apart from entropy dilution. It need not apply to a species that is already massive while reactions remain rapid.

For $m_{\nu }\gg T_{\mathrm{dec}}$, the interaction strength is sufficient to keep the species in thermal equilibrium through its relativistic-to-nonrelativistic transition. At $T\sim m_{\nu }$, the relativistic scaling would give $\Gamma /H\sim (m_{\nu }/T_{\mathrm{dec}}{)}^3\gg 1$. Subsequent annihilation removes most particle-antiparticle pairs while equilibrium remains possible. At negligible chemical potential, the nonrelativistic equilibrium number density isThis Boltzmann suppression of a nonrelativistic relic can make the eventual abundance much smaller than the relativistically decoupled value. Freeze-out now occurs when $n_{\mathrm{eq}}⟨\sigma v⟩\sim H$, with a nonrelativistic cross-section, rather than by extrapolating the massless $T^5$ rate indefinitely. A large mass can therefore have an acceptably small energy density: a hot-relic mass bound is not a universal bound on a heavy thermal relic.

Compatibility depends on the annihilation rate and any conserved particle-antiparticle asymmetry; the supplied relativistic rate alone does not determine the heavy relic's final abundance. Efficient annihilation is the mechanism that permits large masses within the standard expansion and thermal history.

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.