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Strict positivity throughout phase space is incompatible with compact support. The following calculation uses positive smooth rapidly decaying solutions with all displayed integrals justified; nonnegative cases use the corresponding entropy limits where justified.

Write $B=|{\mathbb{S}}^2|$ and use integration over $(v,v_{\ast },\sigma )$. The two symmetrized weak collision identities areandThey follow from particle interchange and the angular exchange for elastic collisions. They are the weak identities needed here; no invalid fixed-angle Jacobian is used.

The collision invariants $1$, $v$ and $|v{|}^2$ obey ${\varphi }^{\prime }+{\varphi }_{\ast }^{\prime }=\varphi +{\varphi }_{\ast }$. Momentum follows from $v^{\prime }+v_{\ast }^{\prime }=v+v_{\ast }$, and energy from $|v^{\prime }{|}^2+|v_{\ast }^{\prime }{|}^2=|v{|}^2+|v_{\ast }{|}^2$. HenceFor the Boltzmann H functional $H=\int \int f\log fdxdv$, the spatial transport contribution is a boundary divergence and the term $\int Q$ is zero. Set $a=f^{\prime }{f}_{\ast }^{\prime }$, $b=f{f}_{\ast }>0$. The second weak identity with $\varphi =\log f$ gives the Boltzmann H theoremIndeed this integrand with its negative sign is $b(1-X)\log X$ for $X=a/b>0$, which is nonpositive. The kinetic H functional decreases; the physical entropy with opposite sign increases.

A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as the entropy or order parameter. It can have latent heat when the entropy jumps. A continuous phase transition has a continuously vanishing order parameter and no latent heat, with singular higher derivatives and a diverging correlation length. The LG theory compares global minima, not merely the points where a local minimum loses stability.

For the uniform quartic Landau free energythe equation of state is $rm+u{m}^3=h$. At zero field the stable minimum is $m=0$ for $r>0$, and $m=\pm \sqrt{-r/u}$ for $r<0$. Thus tuning $r$ through zero gives a continuous transition. For a fixed $r<0$, varying $h$ through zero instead switches between the two ordered minima and makes $m$ jump, giving a first-order phase transition in the conjugate field.

A temperature-like first-order transition can occur at zero field when $u<0$ and a positive sextic term $v{m}^6/6$ stabilizes the potential. Write $q=m^2$. Stationarity of a nonzero phase gives $r+uq+v{q}^2=0$, while equality with $V(0)=0$ gives $rq/2+u{q}^2/4+v{q}^3/6=0$. Solving these two conditions yieldsThe order parameter jumps from zero to $\pm \sqrt{q_{\mathrm{coex}}}$. At this pointso the competing stationary points are genuinely global minima. This phase coexistence condition differs from the spinodal points $r=0$ and $r=u^2/(4v)$, which mark loss or creation of local stability, not equilibrium coexistence.

If the symmetry permits a cubic term $w{m}^3/3$, a positive quartic coefficient does not preclude a first-order transition. For $V=r{m}^2/2+w{m}^3/3+u{m}^4/4$ with $w\ne 0,u>0$, stationarity and coexistence give $m=-2w/(3u)$ and $r=2w^2/(9u)$. This is the first-order transition in a cubic-quartic Landau potential; the symmetry restriction on the expansion is therefore part of the prediction.

Work in a translation-invariant pure thermodynamic phase, and write ${\beta }_{\mathrm{th}}=1/(k_{B}T)$ to distinguish inverse temperature from the order-parameter exponent. The connected correlation function isAway from criticality in the massive scalar order channel, it decays exponentially, possibly multiplied by an algebraic factor. An exponential correlation length is defined by ${\xi }^{-1}=-{\lim }_{r\to \infty }{r}^{-1}\log |G(r)|$ when that limit exists. At a continuous transition $\xi$ diverges and the critical decay is algebraic. In an ordered symmetric mixture, a nondecaying contribution can remain; selecting a pure branch prevents confusing it with connected critical fluctuations.

Let $M={\sum }_{\mathbf{n}}{\sigma }_{\mathbf{n}}$ and $m=⟨M⟩/N$. Differentiating the partition function with the energy term $-hM$ gives the correlation-function susceptibility sum ruleIf the source is instead the dimensionless ${\beta }_{\mathrm{th}}h$, the explicit inverse-temperature factor is absent. In continuum physical coordinates the lattice sum is approximately $a^{-D}\int d^{D}rG(\mathbf{r})$.

Choose a normalized blocking kernel $B_b({\sigma }^{\prime },\sigma )\ge 0$ such that ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )=1$, replacing the sum by an integral for continuous variables. A deterministic example is a product of delta functions imposing that each coarse variable equals the average of spins in a block of $b^D$ sites, with its field normalization included consistently. Define the effective Boltzmann weight byKernel normalization makes the blocked partition function exactly equal to the original. Its coarse-grained variables retain the long-distance observables through their defining relation to the original fields. An exact RG step generally generates all symmetry-allowed operators; keeping only a small coupling set is an approximation, not an exact closure assumption. The geometrical coarse lattice hasso its physical volume is unchanged. A subsequent coordinate rescaling may restore the numerical lattice spacing to its initial value, which is the equivalent rescaled-coordinate convention.

Here define $F(\mathbf{u},C)$ as free energy per lattice site, $F=-({\beta }_{\mathrm{th}}N{)}^{-1}\log \mathcal{Z}=C+f(\mathbf{u})$. Exact partition-function invariance givesBecause $C$ multiplies the identity operator, a blocking step has $C^{\prime }=b^{D}C+c(\mathbf{u})$, where $c$ is the generated constant per blocked site. With $g(\mathbf{u})=b^{-D}c(\mathbf{u})$ this impliesThe identity-operator contribution to renormalization-group free energy includes the eliminated modes' entropy, connected vacuum terms and field-measure normalization. It cannot be discarded when calculating an absolute free energy, even though it cancels out of normalized correlation functions.

There is a useful convention behind the word “singular” in this inhomogeneous equation. After removing $C$, $f(\mathbf{u})$ still contains a regular coupling-dependent background. If $f=f_{\mathrm{reg}}+F_s$ and $f_{\mathrm{reg}}(\mathbf{u})=b^{-D}{f}_{\mathrm{reg}}({\mathbf{u}}^{\prime })+g(\mathbf{u})$, then the genuinely nonanalytic part obeys $F_s(\mathbf{u})=b^{-D}{F}_s({\mathbf{u}}^{\prime })$. Thus the printed inhomogeneous representative and homogeneous singular scaling below are consistent after background subtraction. At resonances or marginal points this subtraction may leave additive or multiplicative logarithms; a strictly homogeneous pure-power form is then qualified accordingly.

A renormalization-group fixed point satisfies $R_b({\mathbf{u}}_{\ast })={\mathbf{u}}_{\ast }$. Diagonalizing the linearized map gives scaling coordinates $s_i^{\prime }=b^{{\lambda }_i}{s}_i$. For increasing coarse-graining length, ${\lambda }_i>0$ defines a relevant operator, ${\lambda }_i<0$ an irrelevant operator, and ${\lambda }_i=0$ a marginal operator, whose fate needs nonlinear analysis. Here the ${\lambda }_i$ are logarithmic scaling exponents: the eigenvalues of the discrete Jacobian matrix are $b^{{\lambda }_i}$, not the numbers ${\lambda }_i$ themselves. The critical surface is the stable manifold flowing into the critical fixed point after every relevant scaling field is tuned to zero. A repulsive renormalization-group trajectory leaves the fixed point along relevant directions and flows toward a different long-distance phase.

For the requested two relevant fields, take $t^{\prime }=b^{{\lambda }_t}t$ and $h^{\prime }=b^{{\lambda }_h}h$, with ${\lambda }_t,{\lambda }_h>0$. Ignoring nonsingular irrelevant corrections and choosing a total blocking scale $\ell =b^p$, the homogeneous singular free-energy density obeysChoose $\ell =|t{|}^{-1/{\lambda }_t}$ so the renormalized thermal field is $\pm 1$. This provesMetric factors and regular analytic redefinitions of the scaling fields only change amplitudes. Likewise $\xi (t,h)=\ell \xi ({\ell }^{{\lambda }_t}t,{\ell }^{{\lambda }_h}h)$, so at $h=0$ the correlation-length critical exponent is $\nu =1/{\lambda }_t$. Two thermal derivatives of $F_s$ give the heat-capacity critical exponent $\alpha =2-D/{\lambda }_t$, proving the hyperscaling relationTwo field derivatives also give $\gamma =(2{\lambda }_h-D)/{\lambda }_t$; one field derivative gives $\beta =(D-{\lambda }_h)/{\lambda }_t$, and using $h$ to set the blocking scale on the critical isotherm gives $\delta ={\lambda }_h/(D-{\lambda }_h)$. These statements assume that no dangerously irrelevant coupling makes the scaling function singular as it is removed. In particular the naive hyperscaling form need not describe the quartic ordered phase above four dimensions, where its stabilizing quartic coupling is dangerously irrelevant. That restriction reconciles this derivation with the mean-field exponents and Question 3.

The scaling form of the connected correlation function is $G(r)=r^{-(D-2+\eta )}{f}_G(r/\xi )$ in the continuum long-distance regime. For $r\ll \xi$, $f_G$ tends to a finite nonzero constant. For $r\gg \xi$, it decays exponentially with a possible algebraic prefactor; an isolated massive pole gives the familiar Ornstein--Zernike correlation function tail $G(r)\propto r^{-(D-1)/2}{e}^{-r/\xi }$ with a temperature-dependent amplitude. Exponential decay is the essential feature; the same critical algebraic power need not remain the exact large-distance prefactor.

Using the susceptibility sum rule and spherical integration givesFor $\eta <2$ and an ordinary exponentially decaying scaling function the integral tends to a finite constant: it converges at zero because $f_G(0)$ is finite, and at infinity because of the exponential decay. Microscopic distances add a regular background. Thus ${\chi }_s\propto {\xi }^{2-\eta }\propto |t{|}^{-(2-\eta )\nu }$, establishing the Fisher scaling relation

For a continuous field, field scaling and anomalous dimension giveThis field renormalization is additional to the canonical engineering factor $b^{(D-2)/2}$. It makes $G(r)=b^{-2x_{\sigma }}{G}^{\prime }(r/b)$. Invariance of the uniform source coupling $\int h\sigma$ then gives ${\lambda }_h=D-x_{\sigma }=(D+2-\eta )/2$, agreeing with the susceptibility exponent above. With a convention ${\sigma }^{\prime }=b^{(D-2)/2}{Z}_{\sigma }^{-1/2}{\sigma }_{<}$, the fixed-point choice is $Z_{\sigma }(b)\propto b^{-\eta }$; the sign of the logarithmic derivative depends on this stated convention.

An acoustic analogy is an exact rearrangement of the fluid equations into a chosen linear propagation operator acting on an acoustic variable, with everything left over placed on the right as effective forcing. The rearrangement becomes a sound-prediction method only after a reference medium, boundary conditions and approximations to the forcing are specified. In particular, a right-hand-side term need not represent independently generated sound.

Use Einstein summation convention and write ${\partial }_i=\partial /\partial x_i$. Differentiating the continuity equation in time and taking the divergence of the momentum equation eliminates ${\partial }_t(\rho u_i)$:Set $q=p-{\hat{p}}_0$ and $W_{ij}=\rho u_i{u}_j-{\sigma }_{ij}$. Since $p+\chi =q+P$, the constant $P$ has zero Laplacian. The prescribed reference mass density and reference speed of sound are independent of time, soCombining the two identities givesOnly conservation of mass and conservation of momentum have been used; no equation of state or energy equation was needed. Spatial variation of ${\hat{c}}_0$ creates no omitted derivative in this identity, because it multiplies a time derivative. The double divergence of the momentum flux tensor has the structure of an acoustic quadrupole.

For a localized flow, choose the reference fields to match the stationary surrounding medium: ${\hat{\rho }}_0={\rho }_0$ and ${\hat{c}}_0^2=c_0^2$ outside the flow, with $P$ chosen so ${\hat{p}}_0=p_0$ there. A uniform surrounding fluid permits ambient constant mass density and adiabatic sound speed, giving the familiar homogeneous wave equation. A nonuniform surrounding fluid calls for its actual stationary reference profiles, extended sensibly through the flow region. This makes the acoustic variable vanish in the unperturbed exterior and minimizes artificial contrast terms. In a uniform isentropic exterior the leading acoustic relation $p^{\prime }=c_0^2{\rho }^{\prime }$ also makes $Q$ vanish to first order. In a stratified exterior, propagation and entropy-advection effects can remain in $Q$, as the next part demonstrates.

At rest, all time derivatives and advective terms vanish. The continuity equation and the stated energy equation are then identities, while the momentum equation reduces to $\boxed{\nabla (p_0+\chi )=0}$. The reference state need not have uniform mass density or entropy. Assume smooth positive reference pressure and mass density, and retain only first-order perturbations.

The linearized continuity equation, momentum balance and adiabatic energy relation areThe perturbation of $c^2$ multiplies a vanishing reference material derivative, so it does not enter at first order. Differentiate the last equation in time, then substitute the linearized momentum equation:Expanding the divergence yields the stratified acoustic pressure equationFor a perfect gas, $c_0^2{\rho }_0=\gamma p_0$. Put $a(\mathbf{x})=p_0^{1/\gamma }/{\rho }_0$. ThenThus the divergence-form stratified acoustic pressure equation isThis is homogeneous linear propagation through the static medium, including its inhomogeneity. In the acoustic analogy of part (a), $W$ has no first-order contribution when viscosity is neglected and the reference velocity is zero. Nevertheless the chosen left-hand operator lacks the gradient terms in the genuine propagation equation. Those effects must consequently appear in $Q_{tt}$. They describe propagation of an existing disturbance, rather than an independent source. Since reference-field choices also change the division between the operator and forcing, $Q_{tt}$ cannot be identified unambiguously with newly generated noise. Its presence in the right-hand side is a consequence of the chosen analogy, not a source-classification theorem.

Let $h=\log \psi$, with $\psi >0$, and set $f=2\alpha h_{\theta }$. Direct differentiation givesThe heat equation ${\psi }_Z=\alpha {\psi }_{\theta \theta }$ implies $h_Z=\alpha (h_{\theta \theta }+h_{\theta }^2)$, so the bracket vanishes. Conversely its being independent of $\theta$ can be absorbed into a $Z$-dependent multiplicative normalization of $\psi$, leaving $f$ unchanged. This proves the Cole-Hopf transformation with the positive sign appropriate to the negative-flux Burgers convention. Although the algebra works for nonzero $\alpha$, the given forward Gaussian function diffusion kernel and a physical vanishing-viscosity limit require $\alpha >0$.

Take $U,L>0$ for the stated Burgers N-wave. Integrating ${\partial }_{\theta }\log \psi (0,\theta )=f_0(\theta )/(2\alpha )$ and normalizing the exterior value to one givesThis function is continuous at $\pm L$; its logarithmic derivative has the specified jumps. Convolution with the heat kernel is positive and solves the heat equation for $Z>0$. Splitting the integral into the exterior baseline and the interior correction yieldsPut $a=1+UZ$. Completing the square givesChanging the interior integration variable to $\eta =a\phi$ changes its limits to $\pm La$ and the Gaussian function width to $Za$. The Jacobian and normalization leave the factor $a^{-1/2}$. Thus the Cole-Hopf solution for a Burgers N-wave isHere $I_{\alpha }$ is the normalized Gaussian function mass of its indicated interval. Its explicit error function representation isFor fixed $w>0$, the Gaussian function concentrates at $\phi =\theta$ as $\alpha \downarrow 0$. Consequentlyaway from the endpoints; at $\theta =\pm L$ the limit is $1/2$. The transition layer has width $O(\sqrt{\alpha w})$. This is an approximate identity argument, not a uniform step approximation across the endpoints.

For fixed $Z>0$ and $|\theta |\ll L$, both interval masses tend to one, and the exponentially large positive $\hat{\psi }$ dominates the exterior correction. Its logarithmic derivative therefore givesFor $|\theta |\gg La$, both interval masses are exponentially small and the weighted interior integral is also negligible, while the exterior contribution tends to one. Therefore $\boxed{f(Z,\theta )\simeq 0}$ in the specified far exterior.

An exponentially weighted Gaussian function tail should not be discarded solely because its unweighted interval mass tends to zero. In fact, comparing the order-one exterior term with $\hat{\psi }$ gives the sharper inviscid Burgers N-wave fronts $|\theta |=L\sqrt{1+UZ}$, not $La$. Away from these fronts, the vanishing-viscosity limit isInside $|\theta |<La$, this follows from the sign of $L^2-{\theta }^2/a$ in the exponential. Outside $La$, the constrained Gaussian function maximum lies at an interval endpoint and has negative exponent. The right shock wave's speed is $UL/(2\sqrt{a})$, equal both to the derivative of $L\sqrt{a}$ and to minus half the sum of its two limiting states; the left shock wave is its reflection. This checks consistency with the Rankine-Hugoniot condition and shows that the requested near-center and far-exterior approximations are compatible with the full entropy limit.

For a uniform-density stellar model, $\rho =3M/(4\pi R^3)$ and $m(r)=4\pi \rho r^3/3$. With zero surface pressure, integrate the stellar hydrostatic equation:Using the ideal gas equation of state, $T(r)=P(r)/(\mathcal{R}\rho )$. ThereforeEquivalently $P_c=3G{M}^2/(8\pi R^4)$ and $T_c=GM/(2\mathcal{R}R)$. The constant mass density is imposed as a structural approximation; a perfect gas can realize it hydrostatically by allowing its temperature and entropy to vary with radius.

The stellar adiabatic exponents are fixed-composition, constant-specific entropy derivatives:The chain rule gives ${\Gamma }_1({\Gamma }_2-1)/{\Gamma }_2={\Gamma }_3-1$. Define the pressure derivatives ${\chi }_{\rho }=(\partial \log P/\partial \log \rho {)}_T$ and ${\chi }_T=(\partial \log P/\partial \log T{)}_{\rho }$. The first law of thermodynamics, with $du=Tds+Pd\rho /{\rho }^2$, givesCombining this with $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$ on an adiabat givesThus the specific-heat ratio is not generally equal to the three stellar adiabatic exponents.

For the mixture, ${\chi }_{\rho }=\beta$ and ${\chi }_T=4-3\beta$. The specific heat capacity at constant volume, obtained by differentiating $u$ at fixed mass density, isTherefore the adiabatic exponents of a monatomic gas-radiation mixture and its specific-heat ratio areIf a relation involving only the exponents is wanted, eliminate $\beta$ from ${\Gamma }_1=\beta +(4-3\beta )({\Gamma }_3-1)$:For a pure monatomic perfect gas, $\beta =1$ and $\gamma ={\Gamma }_1={\Gamma }_2={\Gamma }_3=5/3$. For any calorically perfect gas with constant heat capacities, the same equality holds with its own ${\gamma }_g$. In a genuine gas-radiation mixture, $0<\beta <1$, $\gamma ={\Gamma }_1/\beta$ while the exponents are given separately above.

In the radiation limit, ${\Gamma }_1={\Gamma }_2={\Gamma }_3=4/3$ and the adiabatic temperature gradient is $1/4$. This follows independently from photon entropy: a comoving volume $V$ has $S\propto V{T}^3$, so an adiabat obeys $T\propto V^{-1/3}$ and $P\propto V^{-4/3}$. However, $c_P/c_V$ is singular in the pure-radiation limit, not $4/3$. This is the pure-radiation constant-pressure heat-capacity singularity. The equation $P=a{T}^4/3$ fixes temperature whenever pressure is fixed, so an ordinary constant-pressure temperature derivative is not available; along the mixture limit $\gamma \to \infty$ and $\beta \gamma \to 4/3$. For photons alone, mass-specific quantities additionally require a material mass label. The often quoted radiation index $4/3$ is its pressure-density adiabatic exponent, not a finite constant-pressure/constant-volume heat-capacity ratio.

For a fixed polytropic index $n>0$ and fixed equation-of-state constant $K$, let ${\xi }_1$ be the first zero of the regular Lane-Emden equation solution. A finite-radius model requires such a zero; for the usual nonnegative indices this holds for $n<5$. The surface radius and mass follow from $r=\alpha \xi$ and $\rho ={\rho }_c{\theta }^n$:The last equality integrates the Lane-Emden equation; define the positive Lane-Emden surface mass constant ${\omega }_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. Substituting $\alpha =C_1{\rho }_c^{(1-n)/(2n)}$ givesThe central-density exponents cancel when the required powers are taken. Thus the polytropic mass-radius relation isIt is independent of ${\rho }_c$, with $n,K$ and composition held fixed. For $n=1$ the radius is fixed; for $n=3$ the mass is fixed. These limiting powers need no division by a vanishing exponent.

For $n=0$, the literal pressure-density power and the supplied expression for $\alpha$ are singular. Interpret this case separately as an incompressible planetary interior with constant density. Then $M=4\pi \rho R^3/3$, or $R\propto M^{1/3}$ at fixed density. A weakly compressed rocky body, or a rough uniform-density approximation to Earth, is an example; realistic terrestrial planets are stratified and compressible.

For $n=3/2$, $P\propto {\rho }^{5/3}$. This describes a cold nonrelativistic degenerate electron gas of fixed composition, or a fully convective monatomic ideal gas at fixed entropy. Examples are a nonrelativistic white dwarf, a sufficiently cooled partly degenerate brown dwarf as an approximation, and an approximately fully convective low-mass star for the ideal-gas version. The fixed-$K$ scaling is $R\propto M^{-1/3}$. It must not be applied to an entire main-sequence stellar sequence with different entropies; the entropy dependence of a polytropic mass-radius relation explains the distinction.

The mass-radius curve of solar-composition substellar objects reflects the transition from weak compression to pressure ionization and electron degeneracy pressure, followed by sustained hydrogen burning. A schematic joining representative object classes is:

The illustration is not an age-specific numerical evolutionary model. In particular, ice giants contain much more heavy material than a solar-composition giant, so a single uniform-composition equation of state does not describe the entire joined curve.

Planet/brown-dwarf naming conventions and deuterium burning do not define a universal discontinuity in the mass-radius relation. Composition, age and irradiation move the curves; the hydrogen-burning transition changes the long-term energy source more fundamentally.

The hot-Jupiter radius inflation problem is that some strongly irradiated giant planets have radii much larger than standard age-, mass- and composition-dependent cooling models predict. Greater internal entropy generally means a larger radius at fixed mass. The two broad classes of explanation are retaining existing heat by delaying cooling and depositing additional energy into the deep planet.

For delayed cooling of an inflated giant planet:

For heating of an inflated giant planet:

Surface or upper-atmosphere heating that is promptly reradiated need not raise deep entropy. The depth and long-term power budget distinguish an effective inflation mechanism from one that merely changes a photospheric temperature.

Use inverse temperature ${\beta }_T=(k_{B}T{)}^{-1}$, reserving $\beta$ without a subscript for the critical exponent. Write $c=qJ$; each undirected bond occurs once, so the number of bonds is $Nq/2$. In the mean-field theory of the Ising model, write each Ising spin as ${\sigma }_i=m+\delta {\sigma }_i$ and neglect the product of fluctuations:The resulting independent-spin statistical Hamiltonian isThe constant corrects the double counting of interaction energy. The Ising spin sums now factorize and can all be evaluated:This is the approximate partition function at an assumed mean field; equilibrium fixes $m$ self-consistently. The corresponding Ising auxiliary mean-field free energy isIts equilibrium value gives the mean-field Helmholtz free energy in the imposed field. Away from a stationary point its parameter $m$ is an assumed field variable, not necessarily the actual mean Ising spin of that independent-spin distribution.

For the full small-$m$ expansion at fixed $h$, set $t_h=\tanh ({\beta }_{T}h)$ and $s_h={\mathrm{sech}}^2({\beta }_{T}h)$. Differentiating $\log \cosh z$ gives successive derivatives $t_h$, $s_h$, $-2t_h{s}_h$, and $2s_h(2-3s_h)$ at $z={\beta }_{T}h$. ThereforeAt zero field, spin inversion symmetry eliminates odd powers and this simplifies toThe quadratic coefficient changes sign and the quartic coefficient is positive atThus the zero-field mean-field prediction is a continuous, continuous phase transition: the stable zero spin magnetization develops two symmetry-related nonzero minima continuously below $T_c$.

Both requested routes give the same mean-field self-consistency equation. First, the one-spin expectation in the effective field isSecond, differentiating the Ising auxiliary mean-field free energy giveswhose stationary condition is precisely that mean-field self-consistency equation. Choose its stable, lowest-free-energy branch rather than every algebraic solution.

An equally useful mean-field approximation parametrizes the trial distribution by its actual mean Ising spin. Its probabilities are $(1\pm m)/2$, giving the Bragg-Williams free energy of the Ising modelThis energy-minus-entropy function has expansionIts stationary equation is $h=k_{B}T\mathrm{arctanh}m-cm$, again equivalent to the mean-field self-consistency equation. The two functions differ away from equilibrium, but agree on stationary branches: for $z={\beta }_T(h+cm)$ and $m=\tanh z$, the entropy bracket equals $mz-\log (2\cosh z)$. Their small-$m$ coefficients therefore need not agree at arbitrary temperature; at $T_c$ they have the same leading critical quartic coefficient $c/12$. This distinction prevents confusing the auxiliary-field expansion with the physical-magnetization variational expansion.

For the order-parameter critical exponent, expand the equation of state at $h=0$:On a nonzero stable branch,Hence $\beta =1/2$. The spontaneous magnetization is understood by selecting a branch with an infinitesimal field after the thermodynamic limit; a finite symmetric sample has zero exact zero-field mean Ising spin.

For the magnetic susceptibility per site, define $\chi =\partial m/\partial h$ with $h$ in energy units. Implicit differentiation givesAbove $T_c$, $m=0$ and $\chi =1/[k_B(T-T_c)]$. Below $T_c$, evaluated on a selected ordered branch, the expansion of $m^2$ gives $\chi \sim 1/[2k_B(T_c-T)]$. Thus the magnetic-susceptibility critical exponent is $\gamma =1$ on both sides, with different amplitudes. At $T=T_c$, the equation of state becomesso the critical-isotherm exponent is $\delta =3$. These are mean-field critical exponents, not a claim that neglecting fluctuations gives the exact Ising transition in every dimension.

Take a homogeneous comoving cell of physical volume $V_c=a^3{V}_0$ and total entropy $S=s{V}_c$. At vanishing chemical potential, the reversible first law of thermodynamics readsFor an extensive equilibrium system, the thermodynamic Euler relation is $E=TS-P{V}_c+\mu N$. At $\mu =0$ this givesThe supplied pressure identity verifies consistency directly: $dP=(\rho +P)dT/T$ impliesThe thermodynamic Euler relation fixes the additive normalization that a differential identity alone would leave undetermined.