# 引力坍缩

## 恒星衰亡中的引力坍缩

### 恒星的相对论模型

${\displaystyle ds^{2}=-e^{2\alpha (r)}dt^{2}+e^{2\beta (r)}dr^{2}+r^{2}d\Omega ^{2}\,}$

${\displaystyle G_{\mu \nu }=R_{\mu \nu }-{\frac {1}{2}}Rg_{\mu \nu }=8\pi GT_{\mu \nu }\,}$

${\displaystyle G_{tt}={\frac {1}{r^{2}}}e^{2(\alpha -\beta )}\left(2r\partial _{r}\beta -1+e^{2\beta }\right)\,}$
${\displaystyle G_{rr}={\frac {1}{r^{2}}}\left(2r\partial _{r}\alpha +1-e^{2\beta }\right)\,}$
${\displaystyle G_{\theta \theta }=r^{2}e^{-2\beta }\left[\partial _{r}^{2}\alpha +\left(\partial _{r}\alpha \right)^{2}-\partial _{r}\alpha \partial _{r}\beta +{\frac {1}{r}}\left(\partial _{r}\alpha -\partial _{r}\beta \right)\right]\,}$
${\displaystyle G_{\phi \phi }=\sin ^{2}\theta G_{\theta \theta }\,}$

${\displaystyle T_{\mu \nu }=\left(\rho +p\right)U_{\mu }U_{\nu }+pg_{\mu \nu }\,}$

${\displaystyle U_{\mu }=\left(e^{\alpha },0,0,0\right)\,}$

${\displaystyle T_{\mu \nu }={\begin{pmatrix}e^{2\alpha }\rho &&&\\&e^{2\beta }p&&\\&&r^{2}p&\\&&&r^{2}\left(sin^{2}\theta \right)p\end{pmatrix}}}$

${\displaystyle {\frac {1}{r^{2}}}e^{-2\beta }\left(2r\partial _{r}\beta -1+e^{2\beta }\right)=8\pi G\rho \,}$

${\displaystyle rr\,}$分量为

${\displaystyle {\frac {1}{r^{2}}}e^{-2\beta }\left(2r\partial _{r}\alpha +1-e^{2\beta }\right)=8\pi Gp\,}$

${\displaystyle \theta \theta \,}$分量为

${\displaystyle e^{-2\beta }\left[\partial _{r}^{2}\alpha +\left(\partial _{r}\alpha \right)^{2}-\partial _{r}\alpha \partial _{r}\beta +{\frac {1}{r}}\left(\partial _{r}\alpha -\partial _{r}\beta \right)\right]=8\pi Gp\,}$

${\displaystyle m(r)={\frac {1}{2G}}\left(r-re^{2\beta }\right)\,}$

${\displaystyle e^{2\beta }=\left[1-{\frac {2Gm(r)}{r}}\right]^{-1}\,}$

${\displaystyle ds^{2}=-e^{2\alpha (r)}dt^{2}+\left[1-{\frac {2Gm(r)}{r}}\right]^{-1}dr^{2}+r^{2}d\Omega ^{2}\,}$

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

${\displaystyle m(r)=4\pi \int _{0}^{r}\rho \left(r^{\prime }\right)r^{\prime 2}dr^{\prime }\,}$

${\displaystyle M=m(R)=4\pi \int _{0}^{R}\rho (r)r^{2}dr\,}$

${\displaystyle m(r)\,}$的物理意义似乎就是对星体内部的能量密度在半径${\displaystyle r\,}$的范围内积分，亦即这一范围内的星体质量。不过，如果我们考虑在度规定义下的空间积分，积分的体元应该由下式给出

${\displaystyle {\sqrt {\gamma }}d^{3}x=e^{\beta }r^{2}\sin \theta drd\theta d\phi \,}$

${\displaystyle \gamma _{ij}dx^{i}dx^{j}=e^{2\beta }dr^{2}+r^{2}d\theta ^{2}+r^{2}\sin ^{2}\theta d\phi ^{2}\,}$

{\displaystyle {\begin{aligned}{\bar {M}}&=4\pi \int _{0}^{R}\rho (r)r^{2}e^{\beta (r)}dr\\&=4\pi \int _{0}^{R}{\frac {\rho (r)r^{2}}{\left[1-{\frac {2Gm(r)}{r}}\right]^{1/2}}}dr\end{aligned}}}

${\displaystyle {\frac {d\alpha }{dr}}={\frac {Gm(r)+4\pi Gr^{3}\rho }{r[r-2Gm(r)]}}\,}$

${\displaystyle (p+\rho ){\frac {d\alpha }{dr}}=-{\frac {dp}{dr}}\,}$

${\displaystyle {\frac {dp}{dr}}=-{\frac {(\rho +p)\left[Gm(r)+4\pi Gr^{3}p\right]}{r[r-2Gm(r)]}}\,}$

${\displaystyle p=K\rho ^{\gamma }\,}$

${\displaystyle \rho (r)={\begin{cases}\rho ,&rR\end{cases}}}$

${\displaystyle m(r)={\begin{cases}{\frac {4}{3}}\pi r^{3}\rho ,&rR\end{cases}}}$

${\displaystyle p(r)=\rho \left[{\frac {R{\sqrt {R-2GM}}-{\sqrt {R^{3}-2GMr^{2}}}}{{\sqrt {R^{3}-2GMr^{2}}}-3R{\sqrt {R-2GM}}}}\right]\,}$

${\displaystyle e^{\alpha (r)}={\frac {3}{2}}\left(1-{\frac {2GM}{R}}\right)^{1/2}-{\frac {1}{2}}\left(1-{\frac {2GMr^{2}}{R^{3}}}\right)^{1/2},\qquad r

${\displaystyle p(0)=\rho \left[{\frac {R{\sqrt {R-2GM}}-R{\sqrt {R}}}{R{\sqrt {R}}-3R{\sqrt {R-2GM}}}}\right]\,}$

${\displaystyle M={\frac {4}{9G}}R\,}$时这个表达式的值为无穷大，而任何大于这个值的质量${\displaystyle M\,}$在广义相对论中都没有对应的定态解。也就是说，当我们将一颗超过这个质量的恒星压缩到给定的半径${\displaystyle R\,}$之内后，这颗恒星会不断地坍缩直到形成一个恆星黑洞。实际上，任何定态的球对称星体的质量都受到${\displaystyle M<{\frac {4}{9G}}R\,}$这个关系的制约。

### II型超新星的引力坍缩

II型超新星是大质量恒星引力坍缩的结果。尽管相关的理论研究已经长达三十余年，以及对超新星SN 1987A的观测取得了相当宝贵的成果，在超新星引力坍缩的理论研究中仍有很多部分和细节完全没有弄清楚，它们坍缩的细节有可能彼此之间存在很大差异[3]。一般认为质量在9倍太阳质量以上大质量恒星在核聚变反应的最后阶段会产生铁元素的内核，其内核的坍缩速度可以达到每秒七万千米（约合0.23倍光速[6]，这个过程会导致恒星的温度和密度发生急剧增长。内核的这一能量损失过程终止于向外简并压力与向内引力的彼此平衡。在光致蜕变的作用下，γ射线将铁原子分解为氦原子核并释放中子，同时吸收能量；而质子和电子则通过电子俘获过程（不可逆β衰变）合并，产生中子和逃逸的中微子

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