# 莱恩－埃姆登方程

n = 0, 1, 2, 3, 4, 5, 6 時的莱恩－埃姆登方程解。
${\displaystyle {\frac {1}{\xi ^{2}}}{\frac {d}{d\xi }}\left({\xi ^{2}{\frac {d\theta }{d\xi }}}\right)+\theta ^{n}=0}$

${\displaystyle \xi =r\left({\frac {4\pi G\rho _{c}^{2}}{(n+1)P_{c}}}\right)^{\frac {1}{2}}}$

${\displaystyle \rho =\rho _{c}\theta ^{n}\,}$

${\displaystyle P=K\rho ^{1+{\frac {1}{n}}}\,}$

${\displaystyle P}$ 是代表壓力，${\displaystyle \rho }$ 則是密度，而 ${\displaystyle K}$ 則是比例常數。標準的邊界條件則是 ${\displaystyle \theta (0)=1}$${\displaystyle \theta '(0)=0}$。因此該方程式的解是描述恆星壓力和密度與半徑的關係，並且給定的多方指數 ${\displaystyle n}$ 也是多方球的多方指數 ${\displaystyle n}$。流體靜力平衡與位能、密度、壓力梯度有關；泊松方程與位能、密度有關。

## 推導

### 以流體靜力平衡推導

${\displaystyle {\frac {dm}{dr}}=4\pi r^{2}\rho }$

${\displaystyle {\frac {1}{\rho }}{\frac {dP}{dr}}=-{\frac {Gm}{r^{2}}}}$

${\displaystyle m}$  也是 ${\displaystyle r}$  的公式。再一次求導數可得：

{\displaystyle {\begin{aligned}{\frac {d}{dr}}\left({\frac {1}{\rho }}{\frac {dP}{dr}}\right)&={\frac {2Gm}{r^{3}}}-{\frac {G}{r^{2}}}{\frac {dm}{dr}}\\&=-{\frac {2}{\rho r}}{\frac {dP}{dr}}-4\pi G\rho \end{aligned}}}

${\displaystyle r^{2}{\frac {d}{dr}}\left({\frac {1}{\rho }}{\frac {dP}{dr}}\right)+{\frac {2r}{\rho }}{\frac {dP}{dr}}={\frac {d}{dr}}\left({\frac {r^{2}}{\rho }}{\frac {dP}{dr}}\right)=-4\pi Gr^{2}\rho }$

${\displaystyle {\frac {1}{r^{2}}}{\frac {d}{dr}}\left(r^{2}K\rho _{c}^{\frac {1}{n}}(n+1){\frac {d\theta }{dr}}\right)=-4\pi G\rho _{c}\theta ^{n}}$

${\displaystyle \alpha ^{2}=(n+1)K\rho _{c}^{{\frac {1}{n}}-1}/4\pi G}$ ,

${\displaystyle {\frac {1}{\xi ^{2}}}{\frac {d}{d\xi }}\left({\xi ^{2}{\frac {d\theta }{d\xi }}}\right)+\theta ^{n}=0}$

### 以泊松方程推導

${\displaystyle \nabla ^{2}\Phi ={\frac {1}{r^{2}}}{\frac {d}{dr}}\left({r^{2}{\frac {d\Phi }{dr}}}\right)=-4\pi G\rho }$

${\displaystyle {\frac {d\Phi }{dr}}={\frac {1}{\rho }}{\frac {dP}{dr}}}$

## 方程式解

### 解析解

${\displaystyle n}$  只在3個值時有解析解

#### ${\displaystyle n=0}$

${\displaystyle {\frac {1}{\xi ^{2}}}{\frac {d}{d\xi }}\left({\xi ^{2}{\frac {d\theta }{d\xi }}}\right)+1=0}$

${\displaystyle \xi ^{2}{\frac {d\theta }{d\xi }}=C_{1}-{\frac {1}{3}}\xi ^{3}}$

${\displaystyle \theta (\xi )=C_{0}-{\frac {C_{1}}{\xi }}-{\frac {1}{6}}\xi ^{2}}$

#### ${\displaystyle n=1}$

${\displaystyle n=1}$ ，方程式可展開如下：

${\displaystyle {\frac {d^{2}\theta }{d\xi ^{2}}}+{\frac {2}{\xi }}{\frac {d\theta }{d\xi }}+\theta =0}$

${\displaystyle \theta (\xi )={\frac {\sin \xi }{\xi }}}$

#### ${\displaystyle n=5}$

${\displaystyle \theta (\xi )={\frac {1}{\sqrt {1+\xi ^{2}/3}}}}$

${\displaystyle n=5}$ ，方程式的解將是循著徑向的無限大值。

### 數值解

${\displaystyle {\frac {d\theta }{d\xi }}=-{\frac {\phi }{\xi ^{2}}}}$
${\displaystyle {\frac {d\phi }{d\xi }}=\theta ^{n}\xi ^{2}}$

## 同調變數

### 同調不變方程

${\displaystyle U={\frac {d\log m}{d\log r}}={\frac {\xi ^{3}\theta ^{n}}{\phi }}}$

${\displaystyle V={\frac {d\log P}{d\log r}}=(n+1){\frac {\phi }{\xi \theta }}}$

${\displaystyle {\frac {1}{U}}{\frac {dU}{d\xi }}={\frac {1}{\xi }}(3-n(n+1)^{-1}V-U)}$

${\displaystyle {\frac {1}{V}}{\frac {dV}{d\xi }}={\frac {1}{\xi }}(-1+U+(n+1)^{-1}V)}$ .

${\displaystyle {\frac {dV}{dU}}=-{\frac {V}{U}}\left({\frac {U+(n+1)^{-1}V-1}{U+n(n+1)^{-1}V-3}}\right)}$

### 拓撲結構不變的同調方程

${\displaystyle {\frac {dU}{d\log \xi }}=-U(U+n(n+1)^{-1}V-3)}$

${\displaystyle {\frac {dV}{d\log \xi }}=V(U+(n+1)^{-1}V-1)}$

${\displaystyle (0,0)}$  ${\displaystyle 3}$  ${\displaystyle -1}$  ${\displaystyle (1,0)}$  ${\displaystyle (0,1)}$
${\displaystyle (3,0)}$  ${\displaystyle -3}$  ${\displaystyle 2}$  ${\displaystyle (1,0)}$  ${\displaystyle (-3n,5+5n)}$
${\displaystyle (0,n+1)}$  ${\displaystyle 1}$  ${\displaystyle 3-n}$  ${\displaystyle (0,1)}$  ${\displaystyle (2-n,1+n)}$
${\displaystyle \left({\frac {n-3}{n-1}},2{\frac {n+1}{n-1}}\right)}$  ${\displaystyle {\frac {n-5\pm \Delta _{n}}{2-2n}}}$  ${\displaystyle \left(1-n\mp \Delta _{n},4+4n\right)}$

## 延伸閱讀

• Lane, Jonathan Homer, On the Theoretical Temperature of the Sun under the Hypothesis of a Gaseous Mass Maintaining its Volume by its Internal Heat and Depending on the Laws of Gases Known to Terrestrial Experiment, The American Journal of Science and Arts, 2nd series, 1870, 50: 57–74.

## 參考資料

1. ^ Chandrasekhar, Subrahmanyan. An introduction to the study of stellar structure. Chicago, Ill.: University of Chicago Press. 1939.
2. ^ Horedt, Georg P. Topology of the Lane-Emden equation. A&A. 1987, 117 (1-2): 117–130 [2012-06-27].

## 外部連結

• Horedt, George Paul ( 1986 ) PDF ( 5.9MB ), Astrophysics and Space Science vol. 126, no. 2, Oct. 1986, p. 357–408. ( ISSN 0004-640X ). Collected at the Smithsonian/NASA Astrophysical Data System.