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For a homogeneous ideal gas of constant specific-heat ratio $\gamma >1$, the Schwarzschild criterion applied to the square-root exponential atmospheric profile givesFor $T_0>0$ and $\alpha >0$, put $s=\sqrt{T-T_0}$ and $A={\nabla }_{\mathrm{ad}}$. The quadratic inequality $A\alpha s^2-2s+A\alpha T_0<0$ has solutions precisely when $A\alpha \sqrt{T_0}<1$, between $s_{\pm }=[1\pm \sqrt{1-(A\alpha {)}^2{T}_0}]/(A\alpha )$. This is a local instability of the imposed temperature gradient; actual convection adjusts that gradient and the prescribed profile need not survive.

For a spherical stellar polytrope of finite radius and zero surface pressure, the specific enthalpy is $(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =-GM/R$, hence $(n+1)\int PdV=-G{M}^2/R-2\Omega$. Eliminating the pressure integral with the stellar virial theorem proves the displayed formula. For a gas adiabatic stellar polytrope, $n=1/(\gamma -1)$ and $U=G{M}^2/[(5\gamma -6)R]$; a general polytropic index need not have this relation to the specific-heat ratio.

The intrinsic planetary luminosity $L_{\mathrm{int}}$ is the power supplied by interior cooling, contraction, nuclear reactions if present, or internal heating, excluding reflected or passively reradiated stellar light. For fixed mass and no accretion or nuclear source, global conservation of energy gives $L_{\mathrm{int}}=Q_{\mathrm{deep}}-d(U+\Omega )/dt$, where $Q_{\mathrm{deep}}$ is any additional power deposited in the interior, such as tidal heating or Joule heating. For $Q_{\mathrm{deep}}=0$ and a gas-supported quasi-static sphere of fixed specific-heat ratio $\gamma$ and negligible surface pressure, the stellar virial theorem gives $\Omega =-3(\gamma -1)U$, henceFor a dynamically stable gas-supported model take $\gamma >4/3$. For $\gamma =5/3$ this is half the gravitational release. Degenerate giants require a different internal energy reservoir; the half-factor is not universal.

For a spherical hydrostatic star with negligible surface pressure, the stellar virial theorem isFor a monatomic nonrelativistic perfect gas, $U=\frac{3}{2}\int PdV$, so $2U+\Omega =0$ and the total energy is $E=U+\Omega =\Omega /2$. Half the released binding energy raises the internal energy; only the other half can be radiated. With fixed mass, no nuclear supply and no external work, conservation of energy givesFor the uniform-density stellar model, this becomes the contraction-luminosity evolution equationIt relates luminosity to the contraction rate; it does not by itself prescribe $L(t)$. Given $L(t)$,For constant luminosity, $R(t)=R_0/[1+t/t_0]$ with $t_0=3G{M}^2/(10R_{0}L)$. This describes quasi-static Kelvin-Helmholtz contraction, not dynamical free fall. If the gas has a different constant specific-heat ratio ${\gamma }_g$, the appropriate stellar virial theorem is $3({\gamma }_g-1)U+\Omega =0$, so the radiated fraction of binding release is $(3{\gamma }_g-4)/[3({\gamma }_g-1)]$ rather than universally one half.

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.

Let $e$ be specific internal energy, $Q$ the heat supplied per unit volume per unit time, and $E=\rho (e+|\mathbf{u}{|}^2/2)$. For an inviscid perfect gas in a prescribed Newtonian gravitational potential $\Phi$, the fluid total-energy equation isThere is no gravitational term if no external body is present. For a time-independent $\Phi$, adding potential energy gives the equivalent conservative form ${\partial }_t(E+\rho \Phi )+\nabla \cdot [(E+p+\rho \Phi )\mathbf{u}]=Q$. Subtracting the kinetic-energy equation and using mass conservation givesFor a perfect gas with constant specific-heat ratio $\gamma$, $e=p/[(\gamma -1)\rho ]$. The isothermal equation of state therefore makes $e=c_s^2/(\gamma -1)$ constant. Hence the required heat supply and heat loss areCompression requires cooling; expansion requires heating. The isothermal sound speed $c_s$ differs from the adiabatic sound speed $\sqrt{\gamma }{c}_s$.

Define the isothermal Mach numbers by ${\mathcal{M}}_i=u_i/c_s$. Orient the normal to a stationary isothermal shock along the flow, so the mass flux $j={\rho }_1{u}_1={\rho }_2{u}_2>0$. Tangential velocity is continuous for this planar hydrodynamic shock. Normal momentum conservation givesFor a genuine discontinuity $u_1\ne u_2$, factorization gives $u_1{u}_2=c_s^2$. Thus the isothermal shock jump relations areAcross a thin shock, gravitational potential and tangential kinetic energy are unchanged. The specific enthalpy $h=\gamma c_s^2/(\gamma -1)$ is the same on both sides, so the energy removed per unit area per unit time isPositive net cooling requires $u_1>u_2$, equivalently ${\rho }_2>{\rho }_1$ and ${\mathcal{M}}_1>1>{\mathcal{M}}_2$. Cooling across an isothermal shock permits only compression shocks; the formal expansion discontinuity requires energy supplied by the surroundings. The continuous state $u_1=u_2$ is not a shock.

For steady radial flow, mass conservation gives $r^2\rho u_r=\text{constant}$. Differentiating it and eliminating the mass density gradient from radial momentum balance givesA regular sonic point has $|u_r|=c_s$, so the right-hand side must vanish there as well. ThereforeThis is the Bondi sonic point for accretion; the same critical radius appears in an isothermal Parker wind. A general steady solution need not pass through a sonic point, but a smooth transonic one must obey both critical conditions.

Put $x=r/r_s$ and $\mathcal{M}=|u_r|/c_s$ on a branch whose flow direction is fixed. The radial equation becomesIntegrating yields the integrated isothermal Bondi flow relation:For a transonic branch, $x=\mathcal{M}=1$ gives $C=-3/2$. Expansion about the critical point gives $(\mathcal{M}-1{)}^2=(x-1{)}^2+O((x-1{)}^3)$, with slope $-1$ for inflow supplied by gas at rest at infinity and slope $+1$ for a transonic outflow.

For Isothermal Bondi accretion, integrate radial momentum balance once more, using $u_r\to 0$ and $\rho \to {\rho }_{\infty }$ at infinity:At the Bondi sonic point, $GM/r_s=2c_s^2$ and $u_r^2=c_s^2$, so ${\rho }_s=e^{3/2}{\rho }_{\infty }$. The positive inward isothermal Bondi accretion rate is consequentlyThe critical mass density and rate are fixed by the regular transonic solution and the specified reservoir at infinity; an arbitrary subsonic solution does not share this accretion rate.

Assume a homogeneous ideal gas, constant specific-heat ratio $\gamma >1$, and an upward displacement that is an adiabatic process and maintains pressure balance with its surroundings. If the ambient temperature falls faster with decreasing pressure than the parcel's adiabatic temperature gradient, the parcel becomes hotter and less dense than its surroundings; buoyancy amplifies the displacement. This is the Schwarzschild criterion:The square-root exponential atmospheric profile hasAn atmospheric thermal inversion with $\alpha <0$ has $\nabla <0$ and is stable to these ordinary adiabatic displacements. For $\alpha >0$, put $s=\sqrt{T-T_0}$. The condition becomes the quadratic inequalityintersected with $s\ge 0$ and $T_0+s^2>0$. In the usual case $T_0>0$, the maximum gradient occurs at $s=\sqrt{T_0}$, namely $T=2T_0$, and is $1/(\alpha \sqrt{T_0})$. Hence a convectively unstable interval exists precisely whenEquality in the existence condition gives a single neutrally stable point. A molecular-hydrogen-dominated hot Jupiter with rotational modes active but vibrational excitation and dissociation negligible has approximately $\gamma =7/5$, so $A=2/7$. The existence condition is then $\alpha \sqrt{T_0}<7/2$. For $T_0\le 0$ the general inequality above remains valid, but its roots must be intersected with the positive-temperature domain; the positive-$T_0$ maximum formula must not be reused. Composition gradients require the Ledoux criterion, and dissociation or variable heat capacities change $A$. Sustained convection normally adjusts a superadiabatic profile toward an adiabatic temperature gradient.

For a quasi-static self-gravitating object of fixed mass, negligible surface pressure, and no accretion, nuclear source or additional deep heating, conservation of energy gives the intrinsic planetary luminosityHere $U$ is internal energy, and $\Omega$ is negative gravitational potential energy. Gravitational contraction makes $\Omega$ more negative and releases power $L_{\mathrm{grav}}\equiv -d\Omega /dt>0$. Some of that release raises the internal energy; it cannot all be radiated while the object maintains hydrostatic equilibrium.

To quantify the split assume thermal gas pressure with constant specific-heat ratio $\gamma >4/3$ and negligible rotation or magnetic support. This excludes the dynamically unstable gas-supported regime. The stellar virial theorem givesIt follows thatFor a monatomic ideal gas, $\gamma =5/3$, so $L_{\mathrm{int}}=L_{\mathrm{grav}}/2$: half the released gravitational energy heats the gas and half escapes. If the density profile remains homologous, write $\Omega =-qG{M}^2/R$, with $q>0$ a fixed structure constant. Thenup to the structure and virial coefficients, where ${\tau }_{\mathrm{KH}}$ is the Kelvin-Helmholtz cooling time.

For a diatomic gas with $\gamma =7/5$ the idealized fraction is $1/6$, already demonstrating that the half-factor is not universal. Partial electron degeneracy pressure, dissociation, changing structure, accretion and deep heating require the full energy equation. The general link between luminosity and Kelvin-Helmholtz contraction is $L_{\mathrm{int}}=-d(U+\Omega )/dt$, with any genuine source terms added explicitly.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

Consider a spherical stellar polytrope with $P=K{\rho }^{1+1/n}$, constant $K$, $0<n<5$, zero surface pressure and finite radius $R$. Its specific enthalpy, measured from zero density, is $h={\int }_0^{P}d{P}^{\prime }/\rho =(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =\Phi (R)=-GM/R$. Integrating over mass therefore yieldsThe static stellar virial theorem gives $3\int PdV=-\Omega$, and eliminating the pressure integral proves the gravitational energy of a stellar polytrope:For an ideal gas with constant specific-heat ratio, the accompanying internal energy isTo express both energies using $\gamma$ alone one additionally identifies the model as an adiabatic stellar polytrope, so $n=1/(\gamma -1)$. ThenFor $\gamma =5/3$ these become $\Omega =-6G{M}^2/(7R)$ and $U=3G{M}^2/(7R)$. A general polytropic index need not be determined by the thermodynamic specific-heat ratio; without the adiabatic identification both $n$ and $\gamma$ remain.

For an ideal gas with constant specific-heat ratio $\gamma$, the specific internal energy is $u=c_{V}T$ and $P=\rho (c_P-c_V)T=(\gamma -1)\rho u$. HenceThe stellar virial theorem consequently reads $\ddot{I}/2=2\mathcal{T}+3(\gamma -1)U-3P_{S}V+\Omega$. For a static nonrotating star with negligible surface pressure, $\mathcal{T}=\ddot{I}=P_S=0$, so $\Omega =-3(\gamma -1)U$.

Fully ionized nonrelativistic ions and electrons form a monatomic ideal gas, with $\gamma =5/3$, provided electron degeneracy pressure and radiation pressure are negligible. ThusBeing steady alone does not eliminate rotation or other macroscopic kinetic energy; those are additional assumptions needed for this relation.

At constant volume, the First law of thermodynamics givesAt constant pressure,Differentiating $pV=N{k}_{B}T$ at fixed $p$ givesConsequently,and the Mayer relation isThe specific-heat ratio is therefore

Let $P=p_2/p_1$ and $D={\rho }_2/{\rho }_1$ be the pressure and mass density ratios across a normal shock wave in a perfect gas with specific-heat ratio $\gamma$. Eliminating the velocities from the Rankine-Hugoniot conditions for a perfect gas givesAs $P\to \infty$, the ratio tends to the shock compression ratio $(\gamma +1)/(\gamma -1)$.

For pure blackbody radiation, $P=a{T}^4/3$ depends only on temperature. Constant pressure therefore fixes temperature, so a usual finite constant-pressure temperature derivative is unavailable. In the material gas-radiation limit the specific-heat ratio diverges, although all three stellar adiabatic exponents approach $4/3$. The radiation index $4/3$ is an adiabatic pressure-volume response, not a finite heat-capacity ratio.

Specific enthalpy is enthalpy per unit mass. For a simple compressible fluid of fixed composition, the first law of thermodynamics giveswhere $T$ is temperature, $s$ is specific entropy, $p$ is pressure, and $\rho$ is mass density. Along an isentropic flow, $dh=dp/\rho$. For a perfect gas with constant specific-heat ratio $\gamma$, $h=\gamma p/[(\gamma -1)\rho ]$.

Specific entropy is entropy per unit mass. For a perfect gas with constant specific heat capacity $c_v$ at constant volume and specific-heat ratio $\gamma$, differences obeywhere $p$ is pressure and $\rho$ is mass density. Thus a homentropic flow has constant $p/{\rho }^{\gamma }$.