Figure C.1: Solutions of eq.(C.6) are plotted for indices of $n=3$ or $\Gamma =4/3$ (the most extended one), $n=5/2$ or $\Gamma =7/5$ (the middle), and $n=3/2$ or $\Gamma =5/3$(the most compact one). For $n\ge 5$ gas extends infinitely and the solution has no zero-point.
\epsfxsize =.45\columnwidth \epsfbox{eps/Emden.ps}\end{figure}

If we choose the polytropic equation of stateC.1,

\end{displaymath} (C.1)

the hydrostatic balance is expressed by
\frac{1}{r^2}\frac{\partial }{\partial r}\left(\frac{r^2}{\rho}\frac{d p}{d r}\right)=-4\pi G\rho,
\end{displaymath} (C.2)

where we used equations (4.1) and (4.2). A hydrostatic gaeous star composed with a polytropic gas is called polytrope. Normalizing the density, pressure and radius as
$\displaystyle \rho$ $\textstyle \equiv$ $\displaystyle \rho_c \theta,$ (C.3)
$\displaystyle p$ $\textstyle \equiv$ $\displaystyle K\rho_c^{1+1/n}\theta^{n+1},$ (C.4)
$\displaystyle r$ $\textstyle \equiv$ $\displaystyle H\xi=\left(\frac{(n+1)K\rho_c^{(1-n)/n}}{4\pi G}\right)^{1/2}\xi$ (C.5)

we obtain a normalized equation as
\frac{1}{\xi^2}\frac{d }{d \xi}\left(\xi^2\frac{d \theta}{d \xi}\right)=-\theta^n,
\end{displaymath} (C.6)

which is called Lane-Emden equation of index $n$. The boundary condition at the center of polytrope should be
$\displaystyle \theta$ $\textstyle =$ $\displaystyle 1,$ (C.7)
$\displaystyle \frac{d \theta}{d \xi}$ $\textstyle =$ $\displaystyle 0,$ (C.8)

at $\xi=0$. Solution of this equation is plotted for several $n$ in Figure C.1.

Mass of the polytrope is written down as

$\displaystyle M$ $\textstyle =$ $\displaystyle \int_0^r 4\pi \rho r^2 dr,$  
  $\textstyle =$ $\displaystyle 4 \pi H^3 \int_0^\xi \theta^n \xi^2 d\xi,$  
  $\textstyle =$ $\displaystyle 4 \pi H^3 \int_0^\xi \frac{d }{d \xi} \xi^2 \frac{d \theta}{d \xi} d\xi,$  
  $\textstyle =$ $\displaystyle -4 \pi H^3 \xi^2 \left.\frac{d \theta}{d \xi}\right\vert _{\xi_1}$  
  $\textstyle =$ $\displaystyle -4 \pi \left[\frac{(n+1)K}{4\pi G}\right]^{3/2}
\rho_c^{(3-n)/2n} \xi^2 \left.\frac{d \theta}{d \xi}\right\vert _{\xi_1},$ (C.9)

where $\xi_1$ represents the zero point of $\theta$ or the surface radius normalized by $H$. For $n=3$ or $\Gamma\equiv 1+1/n=4/3$, equation (C.9) reduces to
M=-4\pi \left(\frac{K}{\pi G}\right)^{3/2}\xi^2 \left.\frac{d \theta}{d \xi}\right\vert _{\xi_1}.
\end{displaymath} (C.10)

Thus, the mass does not depend on the central density for $\Gamma =4/3$ polytrope. For $n=5/2$ or $\Gamma=1+1/n=7/5$, $M$ is written down as
M=-4\pi \left(\frac{7K}{8\pi G}\right)^{3/2}\rho_c^{1/10}
\xi^2 \left.\frac{d \theta}{d \xi}\right\vert _{\xi_1},
\end{displaymath} (C.11)

where $\xi_1=5.35528$ and $-\xi_1^2 \left.d \theta/d \xi \right\vert _{\xi_1}=2.18720$. Polytrope with $\Gamma =7/5$ gas, the mass-density relation becomes $\rho_c\propto M^{10}$. While, for $n=3/2$ of $\Gamma=1+1/n=5/3$, $M$ is written down as
M=-4\pi \left(\frac{5K}{8\pi G}\right)^{3/2}\rho_c^{1/2}
\xi^2 \left.\frac{d \theta}{d \xi}\right\vert _{\xi_1},
\end{displaymath} (C.12)

where $\xi_1=3.65375$ and $-\xi_1^2 \left.d \theta/d \xi \right\vert _{\xi_1}=2.71406$.

Kohji Tomisaka 2012-10-03