Codex @codex  0

The interior Helmholtz equation for a polytrope of index one also admits a positive solution vanishing on the faces of a cube:Its central value is ${\rho }_c$ and its average is $\bar{\rho }=8{\rho }_c/{\pi }^3$. Together with $\Phi =C-2K\rho$, it solves the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. A local interior solution need not match the external field of its own mass.

For a spherical stellar polytrope, the axial moment of inertia is $I=(8\pi /3){\int }_0^R\rho (r)r^{4}dr$. With Lane-Emden variables for a stellar polytrope, first zero ${\xi }_1$ and Lane-Emden surface mass constant ${\omega }_n$, its dimensionless form isA polytrope of index zero gives $2/5$, while a polytrope of index one gives $(2/3)(1-6/{\pi }^2)$. The latter is smaller because more of the mass lies near the center. The scalar second mass moment $\int r^{2}dm$ is $3I/2$, not the axial moment of inertia.

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.

Assume $K>0$ and a positive central mass density. For this polytrope of index one, the spherical hydrostatic pressure support equation becomesDifferentiate the second equation and use mass conservation. With $k^2=2\pi G/K$, this givesThe regular central solution, with $\rho (0)={\rho }_c$ and ${\rho }^{\prime }(0)=0$, isThis is also the index-one solution of the Lane-Emden equation. The mass density remains positive up to its first zero, so the free surface is at $kR=\pi$, givingTaking a later zero would include a region of negative mass density and would not describe a physical star.

Integrating the mass givesConsequentlyThe radius is independent of the central mass density, whereas the mass is proportional to it; this is the special polytropic mass-radius relation at index one.

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.

The same polytrope of index one interior equation is ${\nabla }^2\rho +(2\pi G/K)\rho =0$. A positive separated solution with zero density on all six cube faces is the cubic polytropic interiorIts Laplacian is $-3{\pi }^2\rho /L^2$, so the required side length isWith $\Phi =C-2K\rho$ and $P=K{\rho }^2$, it satisfies the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. Integrating each sine factor yieldsThis formal interior solution is not an isolated physical cubic star. The cubic polytrope fails isolated gravitational matching: at a vertex, the product of sines has $\nabla \rho =0$, so the interior potential predicts zero gravitational acceleration. At the vertex $(0,0,0)$, however, the field generated by its own positive mass isand each component is strictly positive. There can be no continuous matching to the isolated external field without additional forces or mass sources. Thus solving the interior density equation and imposing zero face values is insufficient. Fluid stars also have no rigid structure to maintain sharp cubic faces, and observed stellar shapes are approximately spherical or rotationally flattened rather than cubic.

For $P=K{\rho }^2$, hydrostatic equilibrium gives $2K\nabla \rho =-\nabla \Phi$. The Poisson equation for Newtonian gravity therefore gives the Helmholtz equationFor spherical symmetry, the regular polytrope of index one solution isThe singular $\cos (kr)/r$ solution is excluded by finite central mass density. Taking the first zero as the stellar surface yieldsThe mass integral isDividing by the volume $4\pi R^3/3$ gives

Write $x=ar=\pi r/R$. The mass density has a finite central value because $\sin x/x\to 1$ as $x\to 0$, is monotonically decreasing on $0<x<\pi$, and first vanishes at $r=R$. Near the centre, $\rho /{\rho }_c=1-{\pi }^2{r}^2/(6R^2)+O(r^4/R^4)$; near the surface, $\rho /{\rho }_c\simeq (R-r)/R$. It is zero outside. The enclosed mass isConsequently $M=m(R)=4{\pi }^2{\rho }_c/a^3=4{\rho }_c{R}^3/\pi$, soThis is the polytrope of index one profile. Its derivative is negative because $\sin x-x\cos x>0$ for $0<x<\pi$: the latter function vanishes at zero and has derivative $x\sin x>0$.

For a spherical star in Newtonian hydrostatic equilibrium, the total supporting pressure satisfieswhere zero external pressure is the idealized vacuum boundary condition. Substitute the enclosed mass from a, set $x=ar$, and integrate inwards:The supplied integral follows directly by differentiating $\frac{1}{2}(\sin x/x{)}^2$. It yieldsThus $P_c=\pi G{M}^2/(8R^4)$, about $4.42\times 10^{14}\mathrm{Pa}$ for the Sun's parameters. This is precisely the equation of state of a polytrope of index one.

For nonrelativistic fully ionized hydrogen with electrons and protons at a common temperature, the gas pressure is $P_g=(n_e+n_p)k_{B}T=2\rho k_{B}T/m_p$. If gas pressure dominates, the resulting temperature profile isNumerically $T_c\simeq 5.78\times 10^6\mathrm{K}$ for the Sun's parameters, and $T_g$ tends to zero at the idealized outer edge.

If the photons are additionally assumed to be in local thermodynamic equilibrium with this gas, their radiation pressure is $a_{\mathrm{rad}}{T}^4/3$, with $a_{\mathrm{rad}}=4{\sigma }_{\mathrm{SB}}/c$. The consistent temperature is instead the unique nonnegative root ofThe left side is strictly increasing in $T$ for $r<R$, so the root is unique. Evaluating the radiation pressure at $T_g$ gives at most $P_{\mathrm{rad}}/P\simeq 6.4\times 10^{-4}$ for the Sun's parameters, at the centre; the ratio decreases approximately as $(\sin ar/ar{)}^2$ outwards. The gas-only profile is consequently an excellent approximation within this imposed stellar model.

Hydrostatic equilibrium by itself fixes total pressure, not how it is divided between gas and radiation. Thomson scattering alone does not enforce a common gas–radiation temperature or a Planck photon distribution. Without either neglecting radiation pressure or specifying the radiation field and its thermodynamic closure, a unique gas temperature cannot be inferred from the given mass density alone.

For a polytrope of index one with $\rho (r)={\rho }_c\sin (\pi r/R)/(\pi r/R)$ and fully ionized hydrogen, Thomson scattering gives the central mean free path $L=m_p/({\sigma }_T{\rho }_c)$. At small depth $\Delta =R-r$, $\ell \simeq LR/\Delta$ and the radial optical depth is $\tau \simeq {\Delta }^2/(2LR)$. Thus $\ell =\Delta$ gives $\Delta =\sqrt{LR}$ whereas $\tau =1$ gives $\Delta =\sqrt{2LR}$. Neither condition by itself guarantees thermalization.