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For $p=K{\rho }^{1+1/n}$, the Lane-Emden mass formula gives $M\propto {\rho }_c^{(3-n)/(2n)}$ at fixed $K$, while $M/R\propto {\rho }_c^{1/n}$. For $3<n<5$, the Lane-Emden equation has a finite-radius solution, yet $M\to \infty$ as ${\rho }_c\to 0$. The relativistic Tolman–Oppenheimer–Volkoff equation approaches this limit since $p_c/{\rho }_c\to 0$. Monotone pressure and density alone therefore do not impose a finite stellar maximum mass.

A cold nonrelativistic white dwarf supported by electron degeneracy pressure has a polytropic equation of state $P=K_{\mathrm{NR}}{\rho }^{5/3}$, so its polytropic index is $3/2$. The Lane-Emden mass formula and polytropic mass-radius relation give $R\propto K_{\mathrm{NR}}{M}^{-1/3}$ at fixed composition. As density rises, relativistic corrections soften the exponent toward $4/3$ and lead toward the Chandrasekhar mass regime.

A stellar polytrope approximates a spherical self-gravitating star by the structural relationwhere $K$ is constant and $n$ is the polytropic index. Combining hydrostatic equilibrium with mass conservation givesThe Lane-Emden variables for a stellar polytrope separate the dimensional scales from the profile:For $n>0$ these substitutions give the Lane-Emden equationThe central conditions express finite central mass density and spherical symmetry. If ${\xi }_1$ is the first zero, a zero-pressure surface hasthe Lane-Emden mass formula. For $0<n<5$ this surface is at finite radius. For example, the polytrope of index one has $\theta =\sin \xi /\xi$ and ${\xi }_1=\pi$, giving the density profile shown in the overview figure.

An isentropic ideal monatomic gas has $p\propto {\rho }^{5/3}$ and hence $n=3/2$, a useful approximation for a fully convective ideal-gas star. Nonrelativistic electron degeneracy pressure has the same density exponent. The exponent $4/3$, giving $n=3$, occurs for ultrarelativistic electron degeneracy and in appropriate radiation-dominated stellar models. The structural index should not be confused with the adiabatic index controlling a perturbation unless the model is actually isentropic.

Polytropes give useful mass-radius scalings and central concentration with very few parameters. They do not by themselves specify nuclear fusion, energy transport, or the star's thermal evolution. Nor does a structural profile alone establish dynamical stability; that requires the response of pressure to the perturbation.

A stellar polytrope has a spatially constant coefficient $K$ in the relation $P=K{\rho }^{1+1/n}$; $n$ is its polytropic index. This relates pressure to mass density and need not describe an adiabatic gas. Combining hydrostatic equilibrium with mass conservation eliminates the enclosed mass:For $\rho ={\rho }_c{\theta }^n$, the polytropic equation of state gives $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Choose the Lane-Emden variables for a stellar polytrope withSubstitution then gives the dimensionless Lane-Emden equation:A regular stellar centre has $\theta (0)=1$, ${\theta }^{\prime }(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta =1-{\xi }^2/6+O({\xi }^4)$. For an isolated finite-radius stellar polytrope with negligible external pressure, the surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$. A core embedded in an envelope can instead end at a positive matching pressure, before this zero.

The integrated Lane-Emden equation is ${\xi }^2{\theta }^{\prime }=-{\int }_0^{\xi }{s}^2\theta (s{)}^{n}ds$. Thus mass conservation gives the Lane-Emden mass formula directly:For the proposed functional form, differentiation givesChoosing $\gamma =-1/2$ removes the extra polynomial factor; equality to $-{\theta }^5$ then requires $3\beta =1$. The regular polytrope of index five is thereforeIts enclosed mass and enclosed mean mass density areThere is no finite zero: its isolated radius is infinite. Nevertheless its total mass converges, since the outer mass density falls as $r^{-5}$:The zero mean mass density is a limiting volume average, not a statement that the material has zero mass density at finite radius.

To turn this density model into a CNO cycle burning model, use gas-dominated pressure and uniform mean molecular weight. The ideal gas law and the polytropic equation of state then imply $T/T_c=\theta$. Without this thermal assumption a polytropic equation of state for total pressure does not alone fix the temperature profile. Since the stellar energy-generation rate is power per unit mass, its contribution to luminosity is $ϵdm$, giving the luminosity integral of an index-five polytrope:Put $u=\xi /\sqrt{3+{\xi }^2}$. Then $\xi =\sqrt{3}u/\sqrt{1-u^2}$ and the dimensionless integral becomes $3\sqrt{3}{\int }_0^1{u}^2(1-u^2{)}^{8}du$. Evaluation givesThe PDF's power of $\alpha$ in the proposed luminosity scaling is a genuine dimensional error: it must be three, rather than two. The volume element supplies ${\alpha }^3$; replacing it by ${\alpha }^2$ would give power divided by length, and could not be repaired by a dimensionless numerical constant.

The polytrope of index five is unlikely to model a real burning stellar core well. An embedded core has a finite boundary, and a complete isolated star cannot have this infinite radius. More importantly, strongly temperature-sensitive CNO cycle heating concentrates the luminosity centrally and tends to produce a convection zone there. A well-mixed gas-pressure-dominated convective core is closer to an adiabatic stellar polytrope of index $3/2$ than to index five. A finite truncation would also change the integral defining $A$; its value above pertains to the full mathematical model.

For a spherical stellar polytrope in hydrostatic equilibrium, combine $d{m}_r/dr=4\pi r^2\rho$ with $dP/dr=-G{m}_r\rho /r^2$ to eliminate the enclosed mass:For $n>0$, put $P=K{\rho }^{1+1/n}$, $\rho ={\rho }_c{\theta }^n$, $P=P_c{\theta }^{n+1}$ and $r=a\xi$. Since ${\rho }^{-1}dP/dr=(n+1)P_c{\theta }^{\prime }/(a{\rho }_c)$, chooseThe Lane-Emden equation is thenA regular center requires $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$, with positive chosen central mass density and pressure. Locally $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. The stellar surface is the first positive zero ${\xi }_1$, where the idealized external pressure is zero; retain the positive solution before it. Thus $R=a{\xi }_1$ and the Lane-Emden mass formula isThe surface condition selects where to stop a centrally regular solution, rather than replacing its central regularity conditions.

For a polytrope of index zero, the density is constant and $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2$. Regularity givesThe pressure is $P_c{\theta }_0$ and $a^2=P_c/(4\pi G{\rho }_c^2)$, so $R^2=3P_c/(2\pi G{\rho }_c^2)$. Index zero is the structural incompressible limit: the expression $K{\rho }^{1+1/n}$ is not itself defined at $n=0$.

For a polytrope of index one, set $u=\xi \theta$. The equation becomes $u^{\prime \prime }+u=0$, while central regularity requires $u(0)=0$, $u^{\prime }(0)=1$. HenceThe mass is $4{\pi }^2{a}^3{\rho }_c$, so it can change with central mass density while the radius stays fixed.

The moment of inertia of a polytropic star about any axis through its center follows by integrating $r^2{\sin }^2\vartheta$ over spherical shells:For constant mass density this gives $I_0=2M{R}^2/5$. For index one, integration by parts gives ${\int }_0^{\pi }{\xi }^3\sin \xi d\xi =\pi ({\pi }^2-6)$, and thereforeThese are axial moments of inertia, not the scalar second mass moment $\int r^{2}dm$.

For a finite-radius centrally regular stellar polytrope with $0<n<5$ and $n\ne 1$, eliminate ${\rho }_c$ from $R=a{\xi }_1$ and $M=4\pi a^3{\rho }_c{\omega }_n$. The resulting polytropic mass-radius relation isHere ${\xi }_1,{\omega }_n$ are dimensionless functions of $n$. At $n=1$ no such single-valued mass-as-a-power-of-radius relation exists at fixed $K$: the radius is fixed instead. At $n=3$ the exponent is zero and $M=4\pi {\omega }_3(K/(\pi G){)}^{3/2}$ is independent of central mass density. For the incompressible case, $M=(4\pi {\rho }_c/3)R^3$ at fixed mass density. Regular $n\ge 5$ solutions have no finite zero-pressure surface, so the finite-radius formula does not apply to them.

For polytropic index three, the Lane-Emden mass formula cancels the central mass density:Substituting the preceding constant $K$ givesFor fixed composition this is $M\propto \sqrt{1-\beta }/{\beta }^2$, as required. Squaring and rearranging yields the Eddington quartic relation, $(1-\beta )/{\beta }^4\propto {\mu }^4{M}^2$. The larger masses in this model have smaller gas fractions and relatively more radiation support.

Assume a Newtonian, spherically symmetric, nonrotating body in hydrostatic equilibrium, supported by a polytropic equation of state $P=K{\rho }^{1+1/n}$ with constant $K>0$ and $n>0$. The enclosed mass and hydrostatic pressure support equation areEliminate $m$ by first writing $r^2{\rho }^{-1}dP/dr=-Gm$ and then differentiating:Let ${\rho }_c$ be central mass density, and introduce the Lane-Emden variables for a stellar polytropewhereThen $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Substitution cancels the dimensional factors and yields the Lane-Emden equationThe central conditions enforce the chosen central mass density and regular spherical symmetry. The local regular expansion is $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. If a finite first zero ${\xi }_1$ exists and surface pressure is negligible, the physical radius is $R=a{\xi }_1$, and the Lane-Emden mass formula is $M=4\pi a^3{\rho }_c[-{\xi }_1^2{\theta }^{\prime }({\xi }_1)]$. For $0<n<5$ the standard isolated solution has a finite surface; $n=5$ has infinite extent, so a finite surface must not be assumed for all indices. Irradiation, composition stratification and non-polytropic equations of state require more general structure equations.

Using the enclosed mass integral, $R=a{\xi }_1$, and the Lane-Emden equation,Substituting the value of $a$ from part (b) gives the Lane-Emden mass formula

Let $\omega =-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$ be the Lane-Emden surface mass constant. The Lane-Emden mass formula givesThus $R\propto {\rho }_c^{-1/6}$ while $M\propto {\rho }_c^{1/2}$ at fixed composition. Eliminating central mass density gives the polytropic mass-radius relationMore explicitly,Increasing the mass of a nonrelativistic white dwarf decreases its radius. This scaling ceases to apply when relativistic electron degeneracy pressure becomes important.

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 $1-\beta =\nu L/(4\pi cGM)$.

The regular $n=5$ Lane-Emden equation solution has no finite first zero. Its Lane-Emden mass formula gives $m(\xi )=4\pi {\rho }_c{\alpha }^3{\xi }^3/[3(1+{\xi }^2/3{)}^{3/2}]$, so its infinite-radius total mass is $4\pi \sqrt{3}{\rho }_c{\alpha }^3$. Its enclosed mean mass density is ${\rho }_c(1+{\xi }^2/3{)}^{-3/2}$ and tends to zero. A finite externally confined core instead requires a specified truncation pressure.