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For a stellar polytrope of index $n>0$, write $P=K{\rho }^{1+1/n}$ and combine hydrostatic equilibrium with mass conservation to obtainIntroduce Lane-Emden variables for a stellar polytrope, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=\alpha \xi$, whereSince ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/{\rho }_c\alpha$, the mechanical equation reduces to the Lane-Emden equationA constant-density interior corresponds to the formal polytrope of index zero, with $\rho ={\rho }_c{\theta }^0={\rho }_c$ where $\theta >0$. At $n=0$, regularity first gives ${\xi }^2{\theta }^{\prime }=-{\xi }^3/3$, and a second integration givesHere $P=P_c\theta =P_c(1-r^2/R^2)$ and ${\alpha }^2=P_c/(4\pi G{\rho }_c^2)$, reproducing $P_c=2\pi G{\rho }_c^2{R}^2/3$. The mass density jumps from its constant interior value to zero at the surface, while pressure vanishes continuously. The relation $P=K{\rho }^{1+1/n}$ is singular at $n=0$; the regular dimensionless pressure/mass density formulation defines this incompressible structural limit. It does not mean that the gas's perturbative stellar adiabatic exponent is infinite: the hydrostatic ideal-gas toy model and its adiabatic response are distinct choices.

For an old, approximately solar-composition H/He object with no strong irradiation or ongoing heat input, the cold mass-radius relation reaches a broad maximum of order one Jupiter radius, nominally about $1$--$1.2R_J$, at a few Jupiter masses. This is a property of that composition and low-entropy sequence, not a composition-independent theorem from hydrostatic equilibrium alone. Young isolated substellar objects can be larger while cooling; strict thermal steady state of a non-burning isolated body would require its internal luminosity to vanish.

At low mass, an incompressible fixed-composition approximation gives $M\simeq 4\pi \bar{\rho }{R}^3/3$, so $R\propto M^{1/3}$. Ice giants contain dense heavy material and modest H/He envelopes, so real radii depend strongly on composition and are not obtained by extending a pure-hydrogen curve through them.

Gas giants approach a regime in which Coulomb interactions, compression and partial electron degeneracy pressure compete. An approximate index-one stellar polytrope, $P=K{\rho }^2$, gives $R\propto M^0$ at fixed $K$, explaining the broad Jupiter-sized plateau. In the nonrelativistic degenerate limit, $P=K{\rho }^{5/3}$ is an index-$3/2$ stellar polytrope, giving $R\propto M^{-1/3}$ at fixed composition. This is a limiting slope; finite-temperature brown dwarfs do not follow that exact power law over their entire observed mass range.

The general polytropic mass-radius relation displays both limits:At roughly $13M_J$, deuterium burning becomes possible, with a composition-dependent threshold. This is a conventional planet/brown-dwarf division rather than a unique formation definition. At roughly $0.075$--$0.08M_{\odot }$, about $75$--$85M_J$, sustained hydrogen burning can support a low-mass star. Its nuclear energy supply maintains a specific entropy that prevents indefinite cooling onto the cold substellar sequence; its approximate main-sequence relation is $R\propto M$ over a limited low-mass interval.

Ice giants chiefly lose formation heat and some gravitational/radiogenic energy, with convection in fluid regions and thermal conduction or inhibited convection where composition stratifies the interior. Gas giants and non-burning brown dwarfs radiate stored heat and contraction energy; their deep interiors are largely convective, with radiative photospheres. Helium separation can supply extra gravitational energy in mature giant planets, while brown dwarfs above the deuterium threshold have a temporary nuclear contribution. Low-mass stars are powered primarily by hydrogen fusion and are largely or fully convective at the lowest stellar masses. The transition to sustained fusion is the essential change from a cooling body to a long-lived thermal balance.

A mass-radius curve of solar-composition substellar objects must specify age, irradiation and composition. For a mature, weakly irradiated hydrogen–helium sequence, the qualitative mass-radius relation has three main trends:

The conventional deuterium-burning mass near $13$ Jupiter masses marks a commonly used giant-planet/brown dwarf classification scale, not a sharp discontinuity in radius or a universal formation boundary. Young brown dwarfs are larger and cool by contraction; irradiation, a heavy-element core and hot-Jupiter radius inflation shift planets away from a single solar-composition curve.

The drawing is a qualitative mature sequence, not an interpolation of a specified evolutionary model grid. Three relevant observations, using representative historical measurements, are:

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.

At the centre, spherical symmetry and regularity require a finite central density $\rho (0)={\rho }_c$ andAt the stellar surface $r=R$, the pressure vanishes:For a stellar polytrope these become the dimensionless conditions $\theta (0)=1$, ${\theta }^{\prime }(0)=0$, and the surface is the first zero $\theta ({\xi }_1)=0$; specifying ${\rho }_c$ fixes the overall scale.

Radiative diffusion in a star can be writtenDividing by hydrostatic equilibrium giveswhich is constant by assumption. Since both $P_{\mathrm{rad}}$ and $P$ vanish at the surface, integration gives $P_{\mathrm{rad}}=(1-\beta )P$ with constant $\beta$.

Eliminating $T$ between the gas and radiation equations of state now givesThus the star is an $n=3$ stellar polytrope. PutThe structure equations become the Lane-Emden equationThe surface is the first zero ${\xi }_1\simeq 6.89685$. Writingthe Lane-Emden mass formula givesThereforewhereThe additional radiative-equilibrium relation is $1-\beta =\nu L/(4\pi cGM)$.

For a spherical stellar polytrope with $P=K{\rho }^{1+1/n}$, the Lane-Emden equation givesEliminating the central density at fixed composition and entropy gives the polytropic mass-radius relationAn incompressible rocky body has $n=0$ and $\beta =1/3$. A moderately massive gas giant is approximately an $n=1$ polytrope and has $\beta =0$, explaining its weak radius dependence on mass. In a more strongly degenerate nonrelativistic regime, $n=3/2$ gives $\beta =-1/3$.

Write $q=32/31$. The supplied approximation $x=qy(qy+1)$ and $x\propto T^{1/2}$ implyThereforeand direct logarithmic differentiation givesThus the model is locally a stellar polytrope withSince $y=\beta /(1-\beta )$,The radiation-pressure and gas-pressure limits are

Suppose first that the gas-pressure fraction $\beta =P_g/P$ is spatially constant, with $0<\beta <1$. This is a sufficient condition for a stellar polytrope. Sincethe ratio $P_{\mathrm{rad}}/P_g=(1-\beta )/\beta$ givesThereforewithFor an ordinary positive polytropic index one also assumes $0\le s<3$.

The radiative diffusion in a star equation can be written asBecause $P_{\mathrm{rad}}=(1-\beta )P$ and $\beta$ is constant, hydrostatic equilibrium givesEquating them and using $\eta =(L_r/L)/(m_r/M)$ yields

For $P=K{\rho }^2$, the specific enthalpy isHydrostatic equilibrium says $\nabla (H+\Phi )=0$. Taking a Laplacian and using the gravitational Poisson equation gives the Helmholtz equation

For a spherical star, regularity at the centre selects the $n=1$ stellar polytropeIts first zero is $kR=\pi$, henceDirect integration gives $M=4{\rho }_c{R}^3/\pi$, and therefore

On the cube, the separated positive solutionvanishes on all six faces. It solves the same Helmholtz equation whenThe mean of each sine over $[0,L]$ is $2/\pi$, so

The interior fields formally solve the local structure equations, but an isolated fluid surface must be an equipotential and its interior gravitational field must match a decaying exterior solution with continuous normal derivative. A cube does not satisfy the global free-boundary conditions for a nonrotating self-gravitating barotrope. Such sharp planar faces and edges are also not observed in stars; ordinary pressure and gravity smooth the body toward a sphere.

A stellar polytrope of index $n$ obeysCombining hydrostatic equilibrium, $dP/dr=-Gm\rho /r^2$, with mass conservation, $dm/dr=4\pi r^2\rho$, gives the Lane-Emden equationRegularity and normalization at the stellar centre requireIntegrating the Lane-Emden equation from the centre then gives the enclosed mass

For $n=5$, direct substitution into the equation findsso $\beta =1/3$ and $\gamma =-1/2$. This profile has no finite first zero and hence has infinite radius, but its total mass converges:Its mean density over the full, infinite configuration is consequently zero.

For a perfect gas, $T\propto P/\rho \propto {\rho }^{1/5}$, so $T/T_c=\theta$. The luminosity from CNO cycle burning with $ϵ={ϵ}_0\rho T^{11}$ is thereforewhereThus $A$ is of order unity. The model is physically poor because the $n=5$ polytrope has infinite radius and zero mean density, while strongly temperature-sensitive CNO burning changes the thermal gradient and commonly creates convection. Real cores also have evolving composition, non-polytropic opacity and energy transport, and boundaries supplied by the surrounding star.

A stellar polytrope with polytropic index one has $\nabla \Phi =-2K\nabla \rho$ in hydrostatic equilibrium. Combining this with the Poisson equation for Newtonian gravity gives the Helmholtz equationThe spherical solution regular at the origin is $\rho (r)={\rho }_c\sin (kr)/(kr)$. Its first zero gives $R=\pi /k=\sqrt{\pi K/(2G)}$, independent of central mass density, and $\bar{\rho }/{\rho }_c=3/{\pi }^2$.