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Take the positive constant ratio $\lambda =p_r/p_g$. The equation of state impliesThis is a polytropic equation of state of index three. Differentiating the pressure and using vertical hydrostatic equilibrium gives $4(1+\lambda )\mathcal{R}\rho T_z=-\rho {\Omega }^{2}z$. HenceDefine the semi-thickness $H$ by $T(H)=0$ and put $s=1-z^2/H^2$. Then the constant gas-radiation ratio vertical disc structure iswithThe surface density is the integral across both faces. The supplied polynomial integral gives ${\int }_0^1(1-x^2{)}^{3}dx=16/35$, so $\Sigma =(32/35){\rho }_{0}H$. Substitution yieldsAll coefficients here are local in radial position; the derivation fixes the vertical profile rather than a global accretion rate. The zero-temperature surface is a formal boundary of this bulk model; optically thick diffusion ultimately fails in a thin photospheric layer.

The first relation is vertical hydrostatic equilibrium. In a thin disc, the central object's gravity near the midplane is $g_z=-{\Omega }^{2}z$; the pressure gradient balances it. The second is an equation of state combining ideal gas pressure with isotropic radiation pressure. Write $\mathcal{R}=k/({\mu }_m{m}_p)$, so $p_g=\mathcal{R}\rho T$ and $p_r=4\sigma T^4/(3c)$; $\sigma$ is the Stefan-Boltzmann constant and ${\mu }_m$ the mean molecular weight in proton-mass units.

The third is local energy balance: the increase of upward radiative flux with height equals viscous dissipation. Here $\mu$ is dynamic viscosity, not mean molecular weight. A Keplerian disc has shear $rd\Omega /dr=-3\Omega /2$, so its heating rate is $\mu (r{\Omega }^{\prime }{)}^2=9\mu {\Omega }^2/4$. The fourth is optically thick radiative diffusion, with opacity per unit mass $\kappa$ and diffusion directed down the temperature gradient. Midplane symmetry gives $F(0)=0$ and oppositely directed fluxes from the two faces. The four relations express vertical force balance, gas–radiation thermodynamics, local heating, and diffusive cooling.

Assume positive fluid density, ${\rho }_0>0$, and write $m=1+3w>0$. A perfect fluid with constant equation of state obeys the cosmological perfect-fluid continuity equation, givingIn the stated units the Friedmann equation is $H^2=\rho /3-k/a^2+\Lambda /3$. Multiplication by $a^2/2$ yields the Friedmann effective potential for a constant-equation-of-state fluid,The allowed region has $V\le 0$. The Friedmann acceleration equation is equivalently $\ddot{a}=-V^{\prime }(a)$, including turning points by continuity. Thus a zero-energy mechanical trajectory reproduces the cosmological evolution, with $a>0$.

For $k=0$, $\Lambda <0$, the potential increases strictly from negative infinity to positive infinity. Its unique zero givesExpansion from the Big Bang stops there, with negative acceleration, then reverses into a Big Crunch. Both the turning point and the final singularity occur in finite proper time: $dt=da/\sqrt{-2V}$ is integrable near a simple turning point and behaves as a constant times $a^{m/2}da$ near zero.

For $\Lambda =0$, $k=+1$, $V$ increases from negative infinity to $1/2$ and crosses zero atThis is again expansion followed by finite-time recollapse. For $k=-1$, the same increasing curve has asymptote $-1/2$ and never meets zero. Expansion continues without a finite maximum; asymptotically $\dot{a}\to 1$ and $a\sim t$, as spatial curvature dominates the diluted fluid.

For $k=0$, $\Lambda >0$, $V$ tends to negative infinity at both ends. It has a maximum atThere is no turning point. Expansion initially decelerates, then accelerates once $a>a_{\ast }$, and approaches de Sitter spacetime expansion $a\propto e^{\sqrt{\Lambda /3}t}$. Thus positive flat dark-energy expansion is unbounded; negative flat dark energy and positive curvature without dark energy recollapse.

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.

Critical exponents describe leading power-law singularities as the reduced temperature $t=(T-T_c)/T_c$ and the conjugate field tend to zero. Their amplitudes depend on microscopic details, while the exponents are characteristic of a universality class. Define $m_0\sim (-t{)}^{\beta }$ on a selected ordered branch, $\chi \sim |t{|}^{-\gamma }$ at zero field, $C_{V,s}\sim |t{|}^{-\alpha }$ for the singular heat capacity, and $m(t=0,h)\sim \mathrm{sgn}(h)|h{|}^{1/\delta }$. Also define $\xi \sim |t{|}^{-\nu }$ and $G(r,t=0)\sim r^{-(D-2+\eta )}$. In these statements $\beta$ is the order-parameter critical exponent, not inverse temperature.

Assume $r=r_{t}t$ with $r_t>0$ and a nonzero positive quartic coefficient $u$. Minimizing the quartic potential gives $m_0=\sqrt{-r/u}$, hence $\beta =1/2$. Differentiating its equation of state givesso $\gamma =1$ with different amplitudes. At $r=0$, $h=u{m}^3$, giving $\delta =3$. The minimized potential is zero above the transition and $-r^2/(4u)$ below, so its second temperature derivative has a finite jump: $\alpha =0$.

The quadratic fluctuation kernel around the stable uniform minimum is $\kappa p^2+V^{\prime \prime }(m_0)$, so ${\xi }^2=\kappa /V^{\prime \prime }(m_0)$. Since $V^{\prime \prime }$ is $r$ above and $-2r$ below, $\nu =1/2$. Its critical momentum dependence is $p^{-2}$, giving $\eta =0$. Thus the ordinary mean-field critical exponents areThese are predictions of the Landau approximation, whose validity is constrained by the Ginzburg criterion; they need not be the exponents of the fluctuating theory below its upper critical dimension.

The one-particle distribution function is a scalar function on the future mass shell. Choose its normalization so that, in a local orthonormal tetrad, $f{d}^{3}x{d}^{3}p$ is the expected photon number in a phase-space cell on a constant-time slice. Spin degeneracy and any phase-space normalization factors are included in $f$; with an occupation-number convention they would instead appear explicitly in the measure.

More covariantly, the photon number crossing a spacelike element is $f{p}^{\mu }d{\Sigma }_{\mu }{d}^{3}p/E$. The factor $p^{\mu }d{\Sigma }_{\mu }$ reduces to $E{d}^{3}x$ on a local constant-time slice, recovering the cell interpretation. The measure $d^{3}p/E$ is Lorentz invariant on the mass shell.

A photon carries energy $E$ and momentum $p^i$, while its velocity is $p^i/E$. Energy density, momentum density and momentum flux are thereforeThey assemble into the stress-energy tensorThe tetrad momenta are related to coordinate components by the tetrad basis. For an isotropic distribution, angular averaging gives $T^{\hat{i}\hat{j}}={\rho }_{\gamma }{\delta }^{ij}/3$ and $T^{\hat{0}\hat{i}}=0$. Thus the photon gas has pressure ${\rho }_{\gamma }/3$, consistent with its massless equation of state.

Both gas and radiation pressures are nonnegative, so $0\le \beta \le 1$. Rearranging the constant-fraction result givesThis is the Eddington luminosity for the effective opacity constant of this model. Positive gas support gives a strict inequality, with equality only in the radiation-only limiting case. Exceeding the limit would require a negative gas-pressure fraction, which is incompatible with the assumed equation of state.