# 引力坍縮

## 恆星衰亡中的引力坍縮

### 恆星的相對論模型

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

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

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

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

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

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

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

$rr\,$分量為

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

$\theta\theta\,$分量為

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

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

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

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

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

$m(r) = 4\pi \int^r_0 \rho \left( r^\prime \right) r^{\prime 2}dr^{\prime}\,$

$M = m(R) = 4\pi \int^R_0 \rho(r)r^2dr\,$

$m(r)\,$的物理意義似乎就是對星體內部的能量密度在半徑$r\,$的範圍內積分，亦即這一範圍內的星體質量。不過，如果我們考慮在度規定義下的空間積分，積分的體元應該由下式給出

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

$\gamma_{ij} dx^idx^j= e^{2\beta}dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\,$

\begin{align} \bar{M} & = 4\pi \int^R_0 \rho(r)r^2e^{\beta(r)}dr\\ & = 4\pi \int^R_0\frac{\rho(r)r^2}{\left[ 1 - \frac{2Gm(r)}{r} \right]^{1/2}}dr \end{align}

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

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

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

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

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

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

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

$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

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

$M = \frac{4}{9G}R\,$時這個表達式的值為無窮大，而任何大於這個值的質量$M\,$在廣義相對論中都沒有對應的定態解。也就是說，當我們將一顆超過這個質量的恆星壓縮到給定的半徑$R\,$之內後，這顆恆星會不斷地坍縮直到形成一個黑洞。實際上，任何定態的球對稱星體的質量都受到$M <\frac{4}{9G}R\,$這個關係的制約。

### II型超新星的引力坍縮

II型超新星是大質量恆星引力坍縮的結果。儘管相關的理論研究已經長達三十餘年，以及對超新星SN 1987A的觀測取得了相當寶貴的成果，在超新星引力坍縮的理論研究中仍有很多部分和細節完全沒有弄清楚，它們坍縮的細節有可能彼此之間存在很大差異[3]。一般認為質量在9倍太陽質量以上大質量恆星在核聚變反應的最後階段會產生鐵元素的內核，其內核的坍縮速度可以達到每秒七萬千米（約合0.23倍光速[6]，這個過程會導致恆星的溫度和密度發生急劇增長。內核的這一能量損失過程終止於向外簡併壓力與向內引力的彼此平衡。在光致蛻變的作用下，γ射線將鐵原子分解為氦原子核並釋放中子，同時吸收能量；而質子和電子則通過電子俘獲過程（不可逆β衰變）合併，產生中子和逃逸的中微子

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