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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.

In stellar homology, the dimensionless radial profiles are the same after scaling radius, enclosed mass, pressure, temperature and luminosity. At corresponding radii let $r^{\prime }=\xi r$, $m_r^{\prime }=m{m}_r$, $T^{\prime }=tT$, $P^{\prime }=pP$ and $L_r^{\prime }=l{L}_r$. The printed $R^{\prime }=\xi r$ is understood as this local radial scaling, with surface radius $R^{\prime }=\xi R$.

Let $q={\rho }^{\prime }/\rho$ and keep composition, opacity coefficient and nuclear coefficient fixed first. Mass conservation gives $q=m/{\xi }^3$, and hydrostatic equilibrium gives $p=mq/\xi =m^2/{\xi }^4$. The ideal gas law then gives $t=p/q=m/\xi$. Scaling the stellar energy-generation rate equation and the stellar radiative temperature gradient gives respectivelyThe second follows from $dT/dr\propto -\kappa \rho L_r/(r^2{T}^3)$, not from energy production. Equating the two luminosity scalings givesThis assumes the denominator is nonzero and a consistent homologous family exists. If the mean molecular weight differs by $u={\mu }^{\prime }/\mu$, then $t=um/\xi$ andRatios of the opacity and energy-generation coefficients multiply the right-hand side. Thus fixed composition is a real restriction, not an automatic property of every stellar sequence.

Use the Stefan–Boltzmann law $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$ to locate the family on a Hertzsprung-Russell diagram. It implies $T_{\mathrm{eff}}\propto M^{(Y-2X)/4}$ andFor proton–proton chain burning take the usual local approximation $\eta =4$, with Kramers opacity law $\lambda =1$, $\nu =-7/2$. ThenFor the CNO cycle with electron-scattering opacity, $\lambda =\nu =0$, soA conventional local choice $\eta =16$ gives $R\propto M^{15/19}$ and slope $76/9\simeq 8.44$. Choosing $\eta =17$ instead gives slope $60/7\simeq 8.57$. The source specifies no numerical nuclear exponents, and the effective exponent changes with temperature; the general expression is the unambiguous answer.

The Hertzsprung-Russell diagram places hotter stars to the left. Its branches rise toward higher luminosity and mass, and the CNO cycle/electron-scattering opacity branch has the steeper logarithmic slope. Their illustrative joining point and normalization are arbitrary because the proportional opacity and reaction laws do not specify absolute stellar scales. These are the fully radiative ideal-gas homology predictions, not exact observed main sequence relations: convection and increasing radiation support limit those assumptions in real stars.

The total pressure is the sum of ideal gas and radiation pressure contributions. The gas fraction and its complement will be obtained from the structure equations, rather than prescribed independently at each radius.

Put $\mathcal{R}=k_B/m_u$, with $m_u$ the atomic mass constant and $a_r$ the radiation energy-density constant. The ideal gas and radiation pressure contributions obey $P_g=\rho \mathcal{R}T/\mu$ and $P_r=a_r{T}^4/3$. With the constant stellar gas-pressure fraction, eliminate $T=\mu \beta P/(\mathcal{R}\rho )$ to getHenceSince $\mu$ and $\beta$ are spatially constant, so is $K$. Comparing with $P=K{\rho }^{1+1/n}$ gives polytropic index $n=3$. This structural stellar polytrope relation does not assert that every perturbed fluid parcel has stellar adiabatic exponent $4/3$.

Write $\mathcal{R}=k/(\mu H)$, where $H$ is the atomic mass unit used in the mean molecular weight convention, and let $\kappa$ be the Rosseland mean opacity per unit mass. For a static spherical star with negligible thermal storage, the stellar structure equations areHere $ϵ$ is the net stellar energy-generation rate per unit mass; $a$ is the radiation energy-density constant and $c$ the speed of light. The last differential equation follows from radiative diffusion, $F=-(c/3\kappa \rho )d(a{T}^4)/dr$, with $L=4\pi r^{2}F$. There is no gravothermal storage term in the specified stellar energy balance equation. Neglecting radiation pressure in force balance does not remove radiation as an energy carrier. Regular central conditions are $m(0)=L(0)=0$; take negligible external pressure, $P(R)=0$, for the analytic hydrostatic equilibrium model below.

Set $x=r/R$. Integrating the enclosed mass directly givesThusThe stellar hydrostatic equation then givesFactorizing makes the surface behaviour transparent:The ideal gas equation of state gives the temperature profileNo small-$x$ approximation was needed. In particular, close to the centre,These exact hydrostatic formulas reveal a limitation of the prescribed linear-density stellar model. Its mass density has a central cusp, and $T^{\prime }(0^{+})=T_c/R>0$. A positive opacity with positive outward luminosity would instead require $T^{\prime }<0$ in radiative diffusion. Moreover, regular positive nuclear heating gives $L=O(r^3)$ and therefore $T^{\prime }=O(r)$ at the centre. Thus the imposed profile is a hydrostatic toy model, not a complete positive-heating radiative-equilibrium solution of all the stellar structure equations. The requested algebraic profiles and nuclear-luminosity integral can still be computed consistently as properties of that toy model.

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.

Let $g=Gm/r^2>0$, $H_P=-dr/d\log P>0$, $\nabla =d\log T/d\log P$, and ${\nabla }_{\mu }=d\log \mu /d\log P$. Consider a small radial fluid displacement $\xi$ whose sound-crossing time is short enough to keep its pressure equal to its surroundings. Its composition is frozen and its heat exchange negligible. This is the local stellar convective stability test.

For gas-pressure-dominated ideal gas matter, $\rho \propto \mu P/T$. The ambient gradient is thereforeThe displaced element conserves specific entropy and mean molecular weight, giving $(d\log \rho /dr{)}_{\mathrm{parcel}}=-(1-{\nabla }_{\mathrm{ad}})/H_P$, where ${\nabla }_{\mathrm{ad}}=({\Gamma }_2-1)/{\Gamma }_2$. The parcel-minus-environment mass density difference isIts buoyancy acceleration is minus $g$ times this difference. ThusPositive stellar buoyancy frequency squared gives a restoring force. The gas-pressure-dominated Ledoux criterion is thereforeEquality is marginal in this ideal adiabatic test. For monatomic gas ${\nabla }_{\mathrm{ad}}=2/5$. An inward increase of mean molecular weight has ${\nabla }_{\mu }>0$ and stabilizes the layer; uniform composition recovers the Schwarzschild criterion. For a radiative layer one substitutes the stellar radiative temperature gradient, ${\nabla }_{\mathrm{rad}}=3\kappa LP/(16\pi acGm{T}^4)$.

More generally define $\delta =-(\partial \log \rho /\partial \log T{)}_{P,\mu }$ and $\phi =(\partial \log \rho /\partial \log \mu {)}_{P,T}$. The same displacement argument gives $N^2=(g/H_P)[\delta ({\nabla }_{\mathrm{ad}}-\nabla )+\phi {\nabla }_{\mu }]$ and stability for $\nabla <{\nabla }_{\mathrm{ad}}+(\phi /\delta ){\nabla }_{\mu }$. Both coefficients equal one in the gas-dominated ideal-gas limit used above. This is local dynamical stability against convection; it is distinct from global radial collapse and from instabilities requiring heat or composition diffusion.

Use a dry ideal gas of fixed composition with specific gas constant $\mathcal{R}=C_p-C_v$ and constant specific heat capacity at constant pressure $C_p$. For a fixed-mass parcel, $P{V}^{\gamma }=$ constant and $PV=m_{\mathrm{parcel}}\mathcal{R}{T}_g$ implyAlong a hydrostatic adiabat, $dP/dz=-{\rho }_{g}g$; hence the dry adiabatic lapse rate isFor an actual pressure-balanced parcel rising in an ambient atmosphere, $dP/dz=-{\rho }_{\mathrm{env}}g$ instead gives $d{T}_g/dz=-(g/C_p)(T_g/T_{\mathrm{env}})$. The usual lapse-rate expression is exact for a hydrostatic adiabatic column and is the local first-order result at the launch point where $T_g=T_{\mathrm{env}}$, as needed in a linear stability test. Treating an already much hotter parcel as an exact copy of the ambient hydrostatic column would be an extra approximation.

After a small upward displacement $\delta z$ from temperature equilibrium, its temperature excess isAt equal pressure, warmer gas is less dense and continues to rise. Thus the Schwarzschild criterion in altitude form isThe supplied non-strict inequality includes the marginal case; strict growth requires the strict inequality. A downward displacement gives the same stability conclusion. Efficient convection normally adjusts an initially superadiabatic gradient to a nearly adiabatic one.

Deep envelopes of gas giants and ice giants commonly transport intrinsic heat by convection, as do the planetary tropospheres of many weakly irradiated atmospheres. Earth's dry troposphere provides another approximate example, with moisture changing the lapse rate. Strongly irradiated hot Jupiters can still have deep convective interiors, while their upper radiative regions need not be convective. Composition gradients can modify the homogeneous-gas criterion and inhibit overturning even in an interior.

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 three basic modes are radiation, convection and thermal conduction. Radiative transfer dominates transparent atmospheric layers; convection carries heat through sufficiently unstable fluid regions; thermal conduction matters especially in solids or highly conducting dense material.

Consider a chemically homogeneous ideal gas atmosphere and a small, dry parcel displacement, with rapid pressure equilibration and negligible heat exchange during the displacement. The parcel's first law of thermodynamics per unit mass isFor an adiabatic process, $ds=0$. Using hydrostatic equilibrium, $dP/dz=-\rho g$, the parcel followsAfter an upward displacement $\delta z>0$, its temperature relative to its new surroundings isHere $C_p$ is the specific heat capacity at constant pressure, and $N$ is the buoyancy frequency. At the same pressure, a warmer parcel has lower density and accelerates upward. Thus strict instability requires $dT/dz<-g/C_p$. Equivalently, the parcel equation is $\ddot{\delta z}=-N^2\delta z$, withNegative $N^2$ gives growing displacements. The printed non-strict inequality includes the onset boundary: equality is neutrally stable, not a strictly growing instability. This dry parcel buoyancy in a homogeneous atmosphere is the dry Schwarzschild criterion; condensation and composition gradients require modified criteria.

The spherical stellar structure equations apply to a planet as well, with its appropriate equation of state:The last equation is the local first law, including gravitational contraction through the specific entropy change. For an isolated non-burning planet, ${ϵ}_{\mathrm{nuc}}\simeq 0$, soIf the convective interior has nearly uniform specific entropy $s(t)$, positive outgoing luminosity gives $\dot{s}=-L_{\mathrm{int}}/\int Tdm<0$. This entropy loss of an isolated convective planet describes its secular cooling.

During assembly, the gravitational energy scale $G{M}^2/R$ provides a source of heat. Integrating hydrostatic equilibrium gives the virial theorem, $\Omega +3\int PdV=0$, neglecting the surface-pressure term. For a monatomic ideal gas this becomes $2U+\Omega =0$, hence $E=U+\Omega =\Omega /2$. Contraction makes $\Omega$ more negative, releases radiation, and heats part of the gas. A characteristic formation temperature has scale $k_{B}T\sim \mu m_{H}GM/R$ up to structure factors. The retained fraction depends on the accretion shock, producing hot and cold starts of a giant planet.

Formation releases heat, and an isolated planet subsequently loses specific entropy and intrinsic luminosity as it approaches a cooler, more degenerate state. The structure equations alone do not fix one initial temperature, and the early contraction of an ideal gas can increase its central temperature while its total energy decreases. “Cools” therefore need not mean that every depth becomes colder at every instant. Once degeneracy limits contraction, the thermal reservoir declines more directly; the emergent effective temperature usually falls along a fixed-mass cooling track.

This makes young giant planets attractive for exoplanet direct imaging: their thermal near/mid-infrared emission can be far brighter than that of old planets of the same mass. Imaging and spectroscopy measure projected separation, astrometry, luminosity, an exoplanet emission spectrum, temperatures and atmospheric molecular and exoplanet cloud signatures. Radius follows from luminosity and fitted temperature; mass and initial specific entropy generally require age/evolutionary information, or an independent dynamical mass measurement. A detected luminosity alone is not a model-independent planet mass.