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At leading anisotropic order, perpendicular Alfvénic turbulence evolves independently of the compressive fluctuations. The slow modes and the entropy mode are transported by that turbulence. The fast magnetosonic frequency is of order $k_{\perp }(v_A^2+c_s^2{)}^{1/2}$, much larger than the reduced Alfvén frequency; the low-frequency reduced magnetohydrodynamics ordering filters out that fast-wave sector.

The passive evolution can be made explicit for a uniform ideal adiabatic plasma. Let $c_s^2=\gamma p_0/{\rho }_0$, $\delta =\delta \rho /{\rho }_0$, $b=\delta B_{\parallel }/B_0$, and define the normalized entropy perturbationUse $D={\partial }_t+{\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }$ and ${\nabla }_{\parallel }={\partial }_z+(\delta {\mathbf{B}}_{\perp }/B_0)\cdot {\nabla }_{\perp }$. Perpendicular force balance makes the leading total-pressure perturbation vanish:The ideal entropy equation gives $Ds=0$. The leading continuity and parallel-induction equations have the same compressive divergence term; subtracting them gives $D(\delta -b)=-{\nabla }_{\parallel }{u}_{\parallel }$. Parallel momentum gives $D{u}_{\parallel }=-{\nabla }_{\parallel }\delta p/{\rho }_0$. PutThe two compressive equations therefore becomeLet $c_T=v_A{c}_s/(v_A^2+c_s^2{)}^{1/2}$ be the tube speed and $h=c_s^{2}r/c_T$. ThenThus the slow and entropy fluctuations in reduced magnetohydrodynamics have characteristic amplitudes $f^{\pm }=u_{\parallel }\pm h$ obeyingThe two slow-wave populations propagate in opposite directions along the bent field lines while being mixed across them by ${\mathbf{u}}_{\perp }$. The entropy perturbation is carried by the fluid without a restoring-wave frequency. A pure entropy perturbation has $r=b=u_{\parallel }=0$ and $\delta =-s/\gamma$, so its density variation is accompanied by a temperature change with no leading pressure variation.

The coefficients and advecting fields in these three scalar equations come from the Alfvénic cascade; the scalars do not enter the leading perpendicular Alfvén equations. They therefore develop small perpendicular scales through passive mixing. If a scalar $f$ has constant variance flux ${\mathcal{F}}_f$ and is mixed on the Goldreich–Sridhar time ${\tau }_{\ell }\sim {\mathcal{E}}^{-1/3}{\ell }_{\perp }^{2/3}$, thenThe flux and amplitude need not equal the Alfvénic energy flux. This passive-spectrum statement uses the same locality and stationary inertial-range assumptions as part (a), and ignores additional damping. If a passive mode is absent initially and is not forced, its homogeneous leading equation keeps it absent; the Alfvénic cascade does not have to generate every compressive mode.

Use the material-first nominal stress tensor convention $P_{Ii}$, with $F_{iI}$ the deformation gradient. For any fixed reference material subvolume $V_0$ with outward unit normal $N$, conservation of energy isHere $b$ is body force per mass, $r$ is heat supply per mass and $q$ is nominal heat flux. Using momentum conservation to remove the kinetic energy balance gives the integral Lagrangian internal energy balance:The integral Lagrangian entropy inequality iswhere $\theta >0$ is temperature and $q/\theta$ is the thermodynamic entropy flux.

Since the subvolume is arbitrary, the divergence theorem givesExpand $\dot{U}=U_{,F_{iI}}{\dot{F}}_{iI}+U_{,\eta }\dot{\eta }+U_{,{\xi }_r}{\dot{\xi }}_r$ and eliminate the heat supply between these relations. The Clausius-Duhem inequality becomesUnder the usual local constitutive independence assumptions, the reversible mechanical and thermal variations can have either sign. The Coleman-Noll procedure therefore imposesDefine the thermodynamic force conjugate to an internal variable by $f_r=-{\rho }_0{U}_{,{\xi }_r}$. The energy balance and the remaining entropy constraint now give, respectively,

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.

During radiation domination, the effective number of relativistic energy degrees of freedom, denoted $g_{\ast }$, gives the energy densityThe paper's ${\mathcal{N}}_{\mathrm{eff}}$ is this $g_{\ast }$; it is not the conventionally normalized effective number of neutrino species. With the specified reduced Planck mass, the Friedmann equation reads $H^2={\rho }_R/(3M_{\mathrm{Pl}}^2)$, whenceWrite the weak interaction conversion rate as $\Gamma =A{G}_F^2{T}^5$. The ratio $\Gamma /H$ scales as $T^3$, so it falls during cooling: conversion eventually becomes too slow to maintain neutron-proton chemical equilibrium. Equating the conversion and expansion rates givesOn suppressing the numerical coefficient, the cosmological weak freeze-out temperature is consequentlyThe one-sixth power reflects a balance between a conversion rate proportional to $T^5$ and a Hubble parameter proportional to $\sqrt{g_{\ast }}{T}^2$.

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.

Extra relativistic neutrino species increase the effective number of relativistic energy degrees of freedom. At fixed temperature the Hubble parameter rises as $g_{\ast }^{1/2}$, while the standard conversion rate is unchanged. Consequently cosmological weak freeze-out moves to higher temperature, $T_F\propto g_{\ast }^{1/6}$, and the neutron-to-proton ratio $r_F=e^{-Q/T_F}$ increases. The initial neutron fraction $r_F/(1+r_F)$ therefore increases as well.

The quicker expansion also reduces the cosmic time required to cool to the deuterium bottleneck temperature: during radiation domination, $t\propto g_{\ast }^{-1/2}{T}^{-2}$ while the thermal count is roughly constant. Fewer neutrons decay before burning. Both factors in the neutron budget increase, giving a larger final helium-4 mass fraction. This is the primordial helium response to extra relativistic species, under the assumption that the additional species contribute mainly to expansion rather than independently changing the conversion rates.

A larger baryon-to-photon ratio favours bound nuclei at a fixed temperature through the equilibrium factor ${\eta }^{A-1}$. In particular it helps deuterium survive photodissociation sooner: there are fewer energetic photons per baryon, and nuclear reaction rates per baryon also rise. A schematic deuterium bottleneck condition is $\eta e^{B_D/T_B}\sim \mathrm{constant}$, suppressing slowly varying powers of temperature. It givesThus higher $\eta$ permits burning at a higher temperature. During radiation domination that means earlier cosmic time, $t_B\propto T_B^{-2}$, leaving less time for free-neutron decay after freeze-out. More neutrons enter helium, so the final helium-4 mass fraction increases, ordinarily rather weakly. This is the baryon-density effect on primordial helium.

The equilibrium helium factor ${\eta }^3$ indicates the direction of the change but is not a prediction that the final helium abundance grows without bound as ${\eta }^3$. Once almost all surviving neutrons are captured, the final primordial helium mass fraction is limited by their supply, as the shared neutron budget shows.

Let $e$ be the specific internal energy, $s$ the specific entropy, $T$ the absolute temperature, and $w=e+p/\rho$ the specific enthalpy. The first law of thermodynamics for a simple fluid of fixed composition is $de=Tds-pd(1/\rho )$, soThe unmagnetized Euler equations for an inviscid fluid are ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla \Phi -\nabla p/\rho$. Define the vorticity $\mathbf{\omega }=\nabla \times \mathbf{u}$ and use the vorticity cross-product identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\mathbf{\omega }\times \mathbf{u}+\nabla (u^2/2)$. Substitution gives the Crocco form of the unsteady Euler equation:Adiabaticity means $Ds/Dt=0$ for smooth flow without dissipative heating. It does not require $s$ to be uniform between different fluid elements, so the entropy-gradient term generally remains. For a perfect gas with constant adiabatic exponent $\gamma$, one may write $w=\gamma p/[(\gamma -1)\rho ]$, but the thermodynamic derivation does not require that particular equation of state.

Use a characteristic midplane temperature $T$, density $\rho$ and disk scale height $H$. For an ideal gas of fixed composition, dominant gas pressure gives $c_s^2\propto T$, where $c_s$ is the sound speed. Vertical hydrostatic equilibrium and the alpha disk prescription giveNumerical factors in the vertical averaging do not affect the exponents. The stipulated opacity becomesFor an optically thick, radiatively efficient disk, radiative diffusion gives a cooling rate per unit area proportional to $T^4/(\kappa \Sigma )$. The viscous heating rate is proportional to $\nu \Sigma {\Omega }^2$. ThusEquating heating and cooling gives $T\propto \alpha {\Sigma }^{1/2}{\Omega }^{-1/2}$, soAt fixed $\alpha$ and central mass, $\Omega \propto R^{-3/2}$. Therefore the gas-pressure alpha disk with inverse-density opacity satisfies

For thermal stability of an accretion disk, perturb $T$ at fixed $R$ and $\Sigma$, since the thermal adjustment is faster than radial mass redistribution. The logarithmic heating and cooling slopes are respectively one and two. At equilibrium, a fractional temperature increase therefore makes the excess heating $Q^{+}-Q^{-}$ decrease, and the disk cools back. Thus the local thermal equilibrium is stable within the assumed gas-pressure, radiative-diffusion closure.

For viscous stability of an accretion disk, use the thermally equilibrated relation between viscosity and surface density. It givesLocally freezing the background radius in the Keplerian viscous diffusion equation, a short radial Fourier mode has growth rate $-3k_R^2{\partial }_{\Sigma }(\nu \Sigma )<0$. Density enhancements diffuse rather than grow, so the disk is locally viscously stable as well.

For Planck law expressed per unit frequency, put $x=h\nu /(k_{B}T)$. At fixed temperature, the frequency dependence is $B_{\nu }\propto T^3{x}^3/(e^x-1)$. Differentiation givesThe positive interior maximum therefore satisfies $3(1-e^{-x_{\ast }})=x_{\ast }$, with $x_{\ast }=2.821439\dots$ independent of temperature. The zero solution of the rearranged equation is a boundary, not the interior maximum. Hence the frequency form of Wien's displacement law isThis frequency peak is not $c$ divided by the peak wavelength of $B_{\lambda }$: the two spectral densities differ by the frequency-to-wavelength Jacobian.

Motion toward incoming cosmic microwave background radiation produces a blueshift. The photon occupation function of a blackbody depends on $h\nu /(k_{B}T)$, so the same Doppler factor multiplies both frequency and effective temperature. Looking forward along the Sun's peculiar velocity, one obtainsTaking the mean cosmic microwave background temperature as about $2.73\mathrm{K}$ givesThe numerical total inherits the rounded mean temperature; the robust estimate is the positive millikelvin dipole increment.

The neutral-hydrogen column density is $N_{\mathrm{HI}}=\int n_{\mathrm{HI}}d\ell$, with units ${\mathrm{cm}}^{-2}$. Define the neutral-hydrogen column-density distribution byHere $\mathcal{N}$ counts absorption systems along a quasar sightline. A histogram per logarithmic column interval instead measures $N\ln (10)f(N,z)$; forgetting this factor changes the inferred slope. Another common convention uses the absorption distance $X$, with $dX/dz=H_0(1+z{)}^2/H(z)$, and $f_X=f/(dX/dz)$.

At a fixed epoch, weak Lyman-alpha forest lines have an approximately falling power-law distribution, $f(N,z)\simeq A(z)N^{-\beta }$ with $\beta \simeq 1.4$–$1.5$ over representative low-column ranges. Stronger absorbers deviate from one universal slope, and the normalization and fitted slopes depend on epoch and sample. The low-column fit is measured, for example, in Kim and collaborators' forest survey. The sketch distinguishes that empirical low-column trend from an illustrative higher-column continuation.

Ordinary Lyman-alpha forest lines cover roughly $10^{12}\lesssim N_{\mathrm{HI}}\lesssim 10^{17.2}{\mathrm{cm}}^{-2}$. They are numerous resonant Lyman-alpha absorption features from intervening gas at different cosmological redshifts, collectively resembling a forest in a quasar spectrum. Many are optically thin at the ionization edge even when their line centers are saturated.

A Lyman limit system becomes optically thick at the $91.2\mathrm{nm}$ hydrogen ionization edge. With ${\sigma }_{912}\simeq 6.3\times 10^{-18}{\mathrm{cm}}^2$, ${\tau }_{912}=N_{\mathrm{HI}}{\sigma }_{912}\gtrsim 1$ requiresIts spectrum consequently has a strong continuum decrement below the redshifted Lyman edge. In a classification separating the highest columns, Lyman limit systems extend up to the damped-system boundary.

A damped Lyman-alpha system conventionally has $N_{\mathrm{HI}}\gtrsim 2\times 10^{20}{\mathrm{cm}}^{-2}$. Its Lyman-alpha absorption is sufficiently strong that the Lorentzian wings of the Voigt profile are prominent. “Damped” refers to the finite radiative lifetime of the atomic excited state, not an inference that the gas has a high temperature. These systems trace substantial neutral reservoirs, usually associated with galaxies.

Use high-resolution quasar spectra to fit Voigt profiles to Lyman-alpha forest lines near $z=3$, where the transition appears near $121.6(1+3)\simeq 486.4\mathrm{nm}$. Each fit estimates the neutral-hydrogen column density and the velocity-width parameter $b$. The one-dimensional Maxwell-Boltzmann velocity distribution has variance $k_{B}T/m_{\mathrm{H}}$; consequently the thermal optical-depth profile is proportional to $\exp [-(v-v_0{)}^2/b^2]$, with the thermal cutoff of Lyman-alpha line widthsFor example, a purely thermal $b=20\mathrm{km}{\mathrm{s}}^{-1}$ corresponds to $T\simeq 2.4\times 10^4\mathrm{K}$.

An observed broad line is not by itself a thermometer: peculiar velocities, bulk gradients, blending and instrumental resolution also broaden it. For independent Gaussian contributions, $b^2=b_{\mathrm{th}}^2+b_{\mathrm{nonthermal}}^2+b_{\mathrm{inst}}^2$. After modeling the instrumental response, an uncorrected total width gives an upper bound on the thermal temperature, not a lower bound. The narrowest lines at each column density are least affected by additional broadening. Their lower envelope in the $b$–$N_{\mathrm{HI}}$ plane constrains the temperature-density relation of photoionized intergalactic gas, often written $T=T_0(1+\delta {)}^{\gamma -1}$.

Calibrate the conversion between column density, density and width with hydrodynamic models incorporating the cosmic ionizing background, pressure smoothing, resolution and selection. It estimates $T_0$ and the slope rather than assigning one temperature to every absorber. Where aligned metal lines trace the same gas, comparing species of different atomic mass can separately estimate a common nonthermal contribution because thermal $b^2$ scales as $m^{-1}$. The forest lower-envelope method and its numerical calibration are developed in Schaye and collaborators' temperature analysis.

A primordial light-element abundance determination seeks gas or stars with little subsequent chemical processing. The three abundances require different observations and corrections.

For helium, measure helium and hydrogen recombination emission in metal-poor extragalactic ionized regions. Correct line ratios for electron temperature, density, extinction, collisional excitation, underlying stellar absorption and unobserved ionization stages. Convert the number ratio $y=n_{\mathrm{He}}/n_{\mathrm{H}}$ to helium mass fraction $Y\simeq 4y/(1+4y)$, allowing for heavy elements when necessary. Since stars produce helium and metals, extrapolate a sample toward zero gas-phase metallicity to infer $Y_p$. The sensitivity to atomic data and these corrections is illustrated by Peimbert and collaborators' helium determination.

For deuterium, fit isotope-shifted hydrogen and deuterium absorption lines in relatively simple, metal-poor quasar absorption systems. The deuterium feature is displaced by about $-82\mathrm{km}{\mathrm{s}}^{-1}$ from the corresponding hydrogen line. Multiple transitions and velocity components help distinguish deuterium from unrelated hydrogen contamination. The fitted neutral-column ratio estimates D/H in predominantly neutral gas. Since stellar processing destroys deuterium, low-metallicity, minimally processed absorbers are favored; the technique is exemplified by Pettini and collaborators' deuterium measurement.

For lithium, measure the ${}^7\mathrm{Li}$ absorption feature near $670.8\mathrm{nm}$ in warm, old, metal-poor halo stars. Stellar-atmosphere models convert line strength to Li/H. The Spite plateau supplies a nearly constant low-metallicity level, but stellar metallicity trends, atmospheric temperature scales, departures from local thermal equilibrium and stellar lithium depletion must be considered. Thus the photospheric level is not automatically the initial abundance. Ryan and collaborators' halo lithium measurements illustrate the measured abundance and chemical-evolution corrections.

The agreement of the main light-element abundances with a hot early nuclear-burning phase, and the independent baryon-density consistency between nucleosynthesis and the microwave background, gives strong support to the hot Big Bang framework. These measurements connect the first minutes to recombination, despite their very different temperatures, densities and observables. In particular, the large primordial helium mass fraction and residual deuterium are natural outputs of Big Bang nucleosynthesis, rather than arbitrary products of subsequent star formation.

The cosmological lithium problem prevents claiming that the simplest abundance calculation explains every observation without qualification. One must distinguish nuclear-rate uncertainties, stellar depletion, atmosphere and abundance systematics, and possible new particle physics. Thus the broad hot Big Bang framework is strongly supported, while lithium tests the completeness of its simplest nucleosynthesis and stellar-abundance interpretation. An unresolved lithium discrepancy alone does not erase the independent successes of deuterium, helium and the cosmic microwave background.

Let the basic temperature be $\overline{T}=T_{\ast }-\beta z$ with $\beta >0$, the density be ${\rho }_0$, gravitational acceleration be $-g\hat{\mathbf{z}}$, and thermal expansion coefficient be $a_T$. Denote kinematic viscosity, thermal diffusivity and magnetic diffusivity by $\nu ,\kappa ,\eta$. Write the dimensional magnetic perturbation as ${\mathbf{b}}_d$, the temperature perturbation as ${\theta }_d$, and use constant permeability ${\mu }_0$. Linearizing the Boussinesq equations and the resistive induction equation about the static state givesThe Lorentz force identity is $(\nabla \times {\mathbf{b}}_d)\times B_0\mathbf{c}=B_0(\mathbf{c}\cdot \nabla ){\mathbf{b}}_d-\nabla (B_0\mathbf{c}\cdot {\mathbf{b}}_d)$. Its gradient term is absorbed into $P_d$, the perturbation magnetohydrodynamic total pressure. The temperature source follows from ${\mathbf{u}}_d\cdot \nabla \overline{T}=-\beta u_{d,z}$.

Choose any reference length $d$, since no vertical boundaries are prescribed. Scale length by $d$, time by $d^2/\kappa$, velocity by $\kappa /d$, temperature perturbation by $\beta d$, magnetic perturbation by $B_0$, and total pressure by ${\rho }_0\nu \kappa /d^2$. The dimensionless parameters areThey are the Prandtl number, magnetic-to-thermal diffusivity ratio, Rayleigh number, and Chandrasekhar number, respectively. Division by the chosen scales givesIn particular $Q\zeta =B_0^2{d}^2/({\mu }_0{\rho }_0\nu \kappa )$, so the diffusivity convention agrees with the momentum coefficient.

For a plane wave put $\mathbf{l}=(k,0,m)$, $q=|\mathbf{l}{|}^2=k^2+m^2$, and $d_{\parallel }=\mathbf{c}\cdot \mathbf{l}=k\sin \alpha +m\cos \alpha$. The induction and temperature equations imply $\hat{\mathbf{b}}=i{d}_{\parallel }\hat{\mathbf{u}}/(s+\zeta q)$ and $\hat{\theta }=\hat{w}/(s+q)$. The projection $P=I-\mathbf{l}{\mathbf{l}}^{\mathsf{T}}/q$ removes pressure. Because $(P\hat{\mathbf{z}}{)}_z=k^2/q$, its vertical momentum equation isThus the oblique-field plane-wave magnetoconvection dispersion relation isIf out-of-plane velocity is also allowed, its decoupled magnetic branch satisfies $(s/\sigma +q)(s+\zeta q)+Q\zeta d_{\parallel }^2=0$; it has no buoyant driving. At stationary marginality, for $k\ne 0$,These are stationary thresholds; an oscillatory marginal mode, if present for particular diffusivity ratios, is a different onset calculation.

For $m=1$, write $R_{mathrmstat}=F(k)+QG(k)$, where $F=(1+k^2{)}^3/k^2$ and $G=(1+k^2)(\cos \alpha +k\sin \alpha {)}^2/k^2$. Direct differentiation gives $F^{\prime }(k)=-2k^{-3}+6k+4k^3$, so its two minima are $\pm 1/\sqrt{2}$, with $F=27/4$ and $F^{\prime \prime }=36$. For $Q>0$ the negative sign has the smaller magnetic penalty: at every positive $k$, $G(-k)<G(k)$ because both sine and cosine of the inclination are positive. The wavenumber selection in oblique-field magnetoconvection therefore starts from $a=-1/\sqrt{2}$. Expanding $F^{\prime }(k_c)+Q{G}^{\prime }(k_c)=0$ gives $k_c=a+C_{1}Q+O(Q^2)$, where $C_1=-G^{\prime }(a)/F^{\prime \prime }(a)$. Taylor expansion of the value then givesThis proves the small-field forms, including the order of the correction.

For large $Q$, $G$ is nonnegative and vanishes only at $b=-\cot \alpha$. The candidate $k=b$ has finite threshold $F(b)$, so a global minimizer must satisfy $QG(k_c)\le F(b)$ and hence tend to $b$. Here $G^{\prime \prime }(b)=2(1+b^2){\sin }^2\alpha /b^2>0$. Expanding the stationarity condition at $b$ yieldsThe value expansion isThe leading coefficient follows from $1+{\cot }^2\alpha ={\csc }^2\alpha$. The expansion holds at fixed $0<\alpha <\pi /2$.

Finally choose $\cot \alpha =1/\sqrt{2}$. Then the nonmagnetic minimizing wavevector also has $d_{\parallel }=0$, and its magnetic penalty vanishes. Since $F(k)\ge 27/4$ and $QG(k)\ge 0$, this is the global minimum for every nonnegative field strength:This field-independent stationary threshold in oblique magnetoconvection has a simple physical explanation: the disturbance is constant along the imposed field, so it does not bend field lines.

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.

For an isolated collisionless stellar system, define the scalar second mass moment and total kinetic energy byTake moments of the Collisionless Boltzmann equation, with vanishing surface fluxes. For any time-independent phase-space function $A$, integration by parts givesApply this first to $A=r^2$ and then to $A=\mathbf{r}\cdot \mathbf{v}$. The result isThe potential is self-consistent Newtonian gravity, soSymmetrize the force integral under $\mathbf{r}\leftrightarrow {\mathbf{r}}^{\prime }$:Thus the dynamical virial theorem is $\ddot{I}/2=2T+W$. For a steady system, $\ddot{I}=0$, and consequentlyFor bounded statistically steady motion, averaging instead gives $2⟨T⟩+⟨W⟩=0$ provided $[\dot{I}(\tau )-\dot{I}(0)]/\tau \to 0$. Collisionlessness alone does not imply instantaneous virial equilibrium: a collapsing system retains the $\ddot{I}$ term. The displayed derivation assumes a finite second mass moment, or a limiting version with vanishing boundary terms, and no confining boundary-pressure contribution.

In virial equilibrium, the total energy is $E_{\mathrm{tot}}=T+W=-T$. Define the kinetic temperature $\Theta$ at fixed particle number $N$ by $T=3N{k}_B\Theta /2$. Along an equilibrium sequence,This is the negative heat capacity of a virialized gravitational system: removing total energy makes the kinetic temperature rise. A collisionless stellar system need not be in thermal equilibrium, so without that additional assumption this is a kinetic-temperature interpretation of heat capacity, rather than an assertion that every stellar distribution has a thermodynamic temperature.

To distinguish the moment quantity from the gravitational constant, write it as $\mathcal{G}={\sum }_i{\mathbf{p}}_i\cdot {\mathbf{r}}_i$. Newton's equations give ${\dot{\mathbf{p}}}_i={\mathbf{F}}_i$ and ${\dot{\mathbf{r}}}_i={\mathbf{p}}_i/m_i$. Differentiating provesAverage from $0$ to $\tau$, divide by $\tau$, and apply the supplied long-time boundary condition. The result isThe PDF's intermediate formula omits this minus sign. It cannot follow from the preceding derivative identity: an attractive inverse-square circular orbit already has $\mathbf{F}\cdot \mathbf{r}<0$ and $T>0$, contradicting the printed positive-sign version. The later homogeneous-potential relation is consistent with the corrected sign.

Interpret $W$ as the total pair potential energy, with each pair counted once, for example $W={\sum }_{i<j}C|{\mathbf{r}}_i-{\mathbf{r}}_j{|}^{p+1}$. The printed pair sum leaves one particle index unspecified; an ordered-pair convention with the corresponding factor absorbed into $C$ has the same degree of homogeneity. Under a common dilation of all positions,Differentiating with respect to $\lambda$ at one is the Euler theorem for homogeneous functions. Since ${\mathbf{F}}_i=-{\nabla }_{i}W$, it givesSubstitution into the correctly signed averaged identity proves the virial theorem for a homogeneous pair potential:For $p\ne -1$, the conserved total energy can be evaluated using these averages:Use the kinetic temperature convention $⟨T⟩=\kappa \Theta$, where $\kappa =3N{k}_B/2>0$ for three-dimensional thermalized particles, or the same positive coefficient as a kinetic-temperature definition. At fixed particle number and interaction parameters,Thus the equilibrium heat capacity is negative precisely whensubject to existence of the bounded equilibrium sequence assumed by the virial theorem. At $p=-3$ the virial energy is zero and this derivative vanishes. At $p=-1$ the written power-law potential is constant, its force vanishes, and the bounded-motion virial condition excludes nonzero kinetic energy; division by $p+1$ is not permissible. The Newtonian case $p=-2$ recovers $E_{\mathrm{tot}}=-⟨T⟩$.

Use stellar homology: dimensionless density, temperature and luminosity profiles are taken to have the same shape, so the dimensional scale relations follow from the structure equations. Mass conservation and hydrostatic support giveThe ideal gas equation then gives $T_c\propto \mu M/R$, with the dimensional constants held fixed. The specific stellar energy-generation rate has the prescribed hydrogen factor and temperature exponent, soIndependently, radiative diffusion in a star gives $T_c/R\propto \kappa {\rho }_{c}L/(R^2{T}_c^3)$, henceEquating the nuclear supply and radiative transport establishes composition-dependent CNO homology:At one fixed composition, these reduce to $R\propto M^{3/4}$ and $L\propto M^3$ as required.