---
# System prepended metadata

title: Bonnor-Ebert sphereの構造(2)

---

# Bonnor-Ebert sphereの構造(2)
KrumHoltz p.143 Problem Set 2 (b)(c )
小問ごとのリンク[(a)](https://hackmd.io/IdrIWDqPQpOodQxyYvXnhw)[(b)(c )](https://hackmd.io/Ij09i_P-RH-WlOyI1Rda4g)[(d)](https://hackmd.io/ebZUoLp7StGS25T0oHaWmA)[(e)](https://hackmd.io/NxfRWzPnTiKauQMTGgRKZA)[(f)](https://hackmd.io/EkWtW1DTSVaUu140MZ3A2g)[(g)](https://hackmd.io/aEX4mHAHSuqE7AAc3k5voA)

## (b)Integrate the equation of hydrostatic balance to obtain an expression relating ρ, ρc, and $\phi$.

$-\frac{1}{\rho}\frac{dP}{dr} = \frac{d\phi}{dr}$： hydrostatic equilibrium
$\frac{c_s^2}{\rho}\frac{d\rho}{dr}=-\frac{d\phi}{dr}$
$ln \rho=-\frac{\phi}{c_s^2}+C$
$\rho=expC exp(-\frac{\phi}{c_s^2})$
$r=0$で$\phi=0$より$expC=\rho(r=0)=\rho_c$
$\therefore \frac{\rho}{\rho_c}=exp(-\frac{\phi}{c_s^2})$ 

## (c )Non-dimensionalized isothermal Lane-Emden equation
球対称でのポアソン方程式は
$\frac{1}{r^2}\frac{d}{dr}(r^2 \frac{d\phi}{dr})=4\pi G\rho$：　(c1)
$\psi=\phi/c_s^2$、$\xi=r/r_0$とおくと、$c_s^2d\psi=d\phi$、$r_0d\xi=dr$
$\frac{d\phi}{dr}=\frac{c_s^2}{r_0}\frac{d\psi}{d\xi}$
(c1)の左辺は
$\frac{1}{r^2}\frac{d}{dr}(r^2 \frac{d\phi}{dr})=\frac{1}{r_0^3 \xi^2}\frac{d}{d\xi}(r_0 c_s^2\xi^2 \frac{d\psi}{d\xi})=\frac{c_s^2}{r_0^2 \xi^2}\frac{d}{d\xi}(\xi^2 \frac{d\psi}{d\xi})$
(c1)の右辺は(b)の結果より
$4\pi G\rho=4\pi G\rho_c exp(-\psi)$
従って
$\frac{c_s^2}{r_0^2 \xi^2}\frac{d}{d\xi}(\xi^2 \frac{d\psi}{d\xi})=4\pi G\rho_c exp(-\psi)$
$\frac{1}{\xi^2}\frac{d}{d\xi}(\xi^2 \frac{d\psi}{d\xi})=\frac{4\pi G\rho_c r_0^2}{c_s^2} exp(-\psi)$：(c2)
ここで
$r_0=\frac{c_s}{\sqrt{4\pi G\rho_c}}$
となるように$r_0$を選べば、(c2)式は
$\frac{1}{\xi^2}\frac{d}{d\xi}(\xi^2 \frac{d\psi}{d\xi})=exp(-\psi)$：(c3)
これは無次元化された等温Lane-Emden方程式である。
