= Solution
For a spherical <stellar polytrope> in <hydrostatic equilibrium>, combine $dm_r/dr=4\pi r^2\rho$ with $dP/dr=-Gm_r\rho/r^2$ to eliminate the <enclosed mass>:
$$
\frac1{r^2}\frac d{dr}\left(\frac{r^2}{\rho}\frac{dP}{dr}\right)=-4\pi G\rho.
$$
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 $\rho^{-1}dP/dr=(n+1)P_c\theta'/(a\rho_c)$, choose
$$
a^2=\frac{(n+1)P_c}{4\pi G\rho_c^2}
=\frac{(n+1)K}{4\pi G}\rho_c^{(1-n)/n}.
$$
The <Lane-Emden equation> is then
$$
\boxed{\frac1{\xi^2}\frac d{d\xi}(\xi^2\theta')=-\theta^n.}
$$
A regular center requires \b[$\theta(0)=1$ and $\theta'(0)=0$], 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> is
$$
M=4\pi a^3\rho_c\omega_n,\qquad
\omega_n=-\xi_1^2\theta'(\xi_1)
=\int_0^{\xi_1}\xi^2\theta^n\,d\xi.
$$
The 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 $(\xi^2\theta')'=-\xi^2$. Regularity gives
$$
\boxed{\theta_0=1-\xi^2/6,\quad\xi_1=\sqrt6,\quad R=\sqrt6\,a.}
$$
The <pressure> is $P_c\theta_0$ and $a^2=P_c/(4\pi G\rho_c^2)$, so $R^2=3P_c/(2\pi G\rho_c^2)$. \b[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 $u''+u=0$, while central regularity requires $u(0)=0$, $u'(0)=1$. Hence
$$
\boxed{\theta_1=\frac{\sin\xi}{\xi},\quad\xi_1=\pi,\quad
R=\pi a=\sqrt{\frac{\pi K}{2G}},\quad\omega_1=\pi.}
$$
The mass is $4\pi^2a^3\rho_c$, 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:
$$
I=\frac{8\pi}{3}\int_0^R\rho(r)r^4\,dr
=\frac{8\pi}{3}\rho_ca^5\int_0^{\xi_1}\xi^4\theta^n\,d\xi.
$$
For constant <mass density> this gives \b[$I_0=2MR^2/5$]. For index one, <integration by parts> gives $\int_0^\pi\xi^3\sin\xi\,d\xi=\pi(\pi^2-6)$, and therefore
$$
\boxed{I_1=\frac23\left(1-\frac6{\pi^2}\right)MR^2\simeq0.26138MR^2.}
$$
These are axial <moments of inertia>, not the scalar second mass moment $\int r^2dm$.
For a finite-radius centrally regular <stellar polytrope> with $0<n<5$ and $n\ne1$, eliminate $\rho_c$ from $R=a\xi_1$ and $M=4\pi a^3\rho_c\omega_n$. The resulting <polytropic mass-radius relation> is
$$
\boxed{M=AR^{p(n)},\quad p(n)=\frac{n-3}{n-1},\quad
A=4\pi\omega_n\,\xi_1^{-p(n)}
\left(\frac{(n+1)K}{4\pi G}\right)^{n/(n-1)}.}
$$
Here $\xi_1,\omega_n$ are dimensionless functions of $n$. \b[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 $n\ge5$ solutions have no finite zero-pressure surface, so the finite-radius formula does not apply to them.
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