大家知道的最长（复杂）的公式是什么？

杂然赋流形丶，物理学本科在读，MHD萌新

居然没人提到爱因斯坦场方程。

物理学家都是懒货，总是喜欢发明一些符号简化方程，让你误以为这很简单

（从此你便走上一条不归之路）

一、爱因斯坦场方程

你以为的爱因斯坦场方程：

$\boxed{1.\ \ R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = 8\pi G T_{\mu\nu}}$

（这不是弱爆了嘛？一行就能写完的式子哪里复杂了？）

这是一个二阶张量方程，重复指标代表求和

其中标曲率

$R$

是由里奇张量

$R_{\mu\nu}$

求迹而得：

$2.\ \ R=g^{\mu\nu}R_{\mu\nu}$

而里奇张量

$R_{\mu\nu}$

是由黎曼曲率张量

$R^{\lambda}_{\ \ \mu\nu\rho}$

两个指标缩并而得

$3. \ \ R_{\mu\nu}=R^{\lambda}_{\ \ \mu \lambda \nu}$

所以你不得不计算黎曼曲率张量，而黎曼曲率张量是由克里斯托弗联络计算而来：

$4.\ \ R^{\rho}_{\ \ \sigma\mu\nu}= \partial_{\mu}\Gamma_{\nu\sigma}^{\rho} - \partial_{\nu}\Gamma_{\mu\sigma}^{\rho} + \Gamma_{\mu\lambda}^{\rho}\Gamma_{\nu\sigma}^{\lambda}- \Gamma_{\nu\lambda}^{\rho} \Gamma_{\mu\sigma}^{\lambda}$

为了得到黎曼曲率张量你不得不计算克里斯托弗联络系数：

$5. \ \ \Gamma_{\mu\nu}^{\sigma} = \frac{1}{2}g^{\sigma\rho}(\partial_{\mu}g_{\nu\rho} + \partial_{\nu}g_{\mu\rho} - \partial_{\rho}g_{\mu\nu})$

全部代入爱因斯坦场方程中，你便把场方程的左边全部换到度规系数

$g_{\mu\nu}$

上了。

二、哑指标展开

这么说，可能还是有点抽象，接下来让我们丧心病狂地把哑指标全部展开

（当然它们的分量并不完全独立，不过我们旨在展示它的复杂程度）

（仅以一个分量为例）：

一个克里斯托弗联络

$\Gamma_{\mu\nu}^{\sigma}$

，共

$12$

项：

$\begin{align} \Gamma_{\mu\nu}^{\sigma} &= \frac{1}{2}g^{\sigma\rho}(\partial_{\mu}g_{\nu\rho} + \partial_{\nu}g_{\mu\rho} - \partial_{\rho}g_{\mu\nu})\\[.2cm] &=\frac{1}{2}g^{\sigma0}(\partial_{\mu}g_{\nu0} + \partial_{\nu}g_{\mu0}- \partial_{0}g_{\mu\nu})+\frac{1}{2}g^{\sigma1}(\partial_{\mu}g_{\nu1} + \partial_{\nu}g_{\mu1}- \partial_{1}g_{\mu\nu})\\[.2cm] &\quad +\frac{1}{2}g^{\sigma2}(\partial_{\mu}g_{\nu2} + \partial_{\nu}g_{\mu2}- \partial_{2}g_{\mu\nu})+\frac{1}{2}g^{\sigma3}(\partial_{\mu}g_{\nu3} + \partial_{\nu}g_{\mu3}- \partial_{3}g_{\mu\nu}) \end{align}$

共计

$12$

项

一个黎曼曲率张量

$R^{\rho}_{\ \ \sigma\mu\nu}$

，共

$120$

项：

$\begin{align} R^{\rho}_{\ \ \sigma\mu\nu}={}& \partial_{\mu}\Gamma_{\nu\sigma}^{\rho} - \partial_{\nu}\Gamma_{\mu\sigma}^{\rho} + \Gamma_{\mu\lambda}^{\rho}\Gamma_{\nu\sigma}^{\lambda}- \Gamma_{\nu\lambda}^{\rho} \Gamma_{\mu\sigma}^{\lambda}\\[.2cm] ={}&\partial_{\mu}\Gamma_{\nu\sigma}^{\rho} - \partial_{\nu}\Gamma_{\mu\sigma}^{\rho} \\[.2cm] {}&+\left(\Gamma_{\mu0}^{\rho}\Gamma_{\nu\sigma}^{0}+\Gamma_{\mu1}^{\rho}\Gamma_{\nu\sigma}^{1}+\Gamma_{\mu2}^{\rho}\Gamma_{\nu\sigma}^{2}+\Gamma_{\mu3}^{\rho}\Gamma_{\nu\sigma}^{3}\right)\\[.2cm] {}&-\left(\Gamma_{\nu0}^{\rho} \Gamma_{\mu\sigma}^{0}+\Gamma_{\nu1}^{\rho} \Gamma_{\mu\sigma}^{1}+\Gamma_{\nu2}^{\rho} \Gamma_{\mu\sigma}^{2}+\Gamma_{\nu3}^{\rho} \Gamma_{\mu\sigma}^{3}\right) \end{align}$

按理来说我应该把每个

$\Gamma_{\mu\nu}^{\sigma}$

代进去的，奈何实在敲不动

$\LaTeX$

，见谅见谅

共计

$10$

项，代入每个

$\Gamma_{\mu\nu}^{\sigma}$

的

$12$

项，对于一个

$R^{\rho}_{\ \ \sigma\mu\nu}$

你需要计算

$120$

项，

里奇张量

$R_{\mu\nu}$

，共

$480$

项：

$\begin{align} R^{\mu\nu} &= R^{\lambda}_{\ \ \mu \lambda \nu}\\[.2cm] &= R^{0}_{\ \ \mu0\nu}+ R^{1}_{\ \ \mu1\nu}+R^{2}_{\ \ \mu2\nu}+R^{3}_{\ \ \mu3\nu} \end{align}$

共计

$4$

项，代入每个

$R^{\rho}_{\ \ \sigma\mu\nu}$

的

$120$

项，对于一个

$R_{\mu\nu}$

你需要计算

$480$

项，

标曲率

$R$

，共

$7680$

项：

$\begin{align} R={}&g^{\mu\nu}R_{\mu\nu}\\[.2cm] ={}&g^{00}R_{00} + g^{01}R_{01}+g^{02}R_{02}+g^{03}R_{03} \\[.2cm] {}&+g^{10}R_{10} + g^{11}R_{11}+g^{12}R_{12}+g^{13}R_{13} \\[.2cm] {}&+g^{20}R_{20} + g^{21}R_{21}+g^{22}R_{22}+g^{23}R_{23} \\[.2cm] {}&+g^{30}R_{30} + g^{31}R_{31}+g^{32}R_{32}+g^{33}R_{33} \\[.2cm] \end{align}$

共计

$16$

项，代入每个

$R_{\mu\nu}$

的

$480$

项，对于标曲率

$R$

你需要计算

$7680$

项。

算到这里，你才仅仅算出来爱因斯坦场方程中的一个方程（这还没考虑物质模型和坐标条件）。

要是想完全计算出爱因斯坦场方程（不包括能量动量张量

$T_{\mu\nu}$

），

克里斯托弗联络

$\Gamma_{\mu\nu}^{\sigma}$

共计

$4^3=64$

个分量，故需要计算

$768$

项，

黎曼曲率张量

$R^{\rho}_{\ \ \sigma\mu\nu}$

共计

$4^4=256$

个分量，故需要计算

$30720$

项，

里奇张量

$R_{\mu\nu}$

共计

$4^2=16$

个分量，故需要计算

$7680$

项，

标曲率

$R$

，共

$7680$

项。

三、代入场方程

稍微运算一下，你可以得到下面这玩意儿^[1]：

$\begin{align} &\frac{1}{2}g^{\alpha\beta} \partial_{\alpha}\partial_{\mu} g_{\beta\nu} + \frac{1}{2}g^{\alpha\beta} \partial_{\alpha}\partial_{\nu}g_{\mu\beta} \\[.2cm] -{}& \frac{1}{2}g^{\alpha\beta}\partial_{\alpha}\partial_{\beta}g_{\mu\nu} -\frac{3}{2}g^{\alpha\beta}\partial_{\mu}\partial_{\nu}g_{\alpha\beta}\\[.2cm] -&\frac{1}{2}g^{\beta\lambda}g^{\alpha\rho}\partial_{\alpha}g_{\rho\lambda}\partial_{\mu}g_{\beta\nu}-\frac{1}{2}g^{\beta\lambda}g^{\alpha\rho}\partial_{\alpha}g_{\rho\lambda}\partial_{\nu}g_{\mu\beta}\\[.2cm] +&\frac{1}{4}g^{\beta\lambda}g^{\alpha\rho}\partial_{\nu}g_{\alpha\lambda}\partial_{\mu}g_{\rho\beta}+\frac{1}{4|g|}g^{\alpha\beta}\partial_{\beta}|g|\partial_{\nu}g_{\mu\alpha}\\[.2cm] -&\frac{1}{4|g|}g^{\alpha\beta}\partial_{\beta}|g|\partial_{\alpha}g_{\mu\nu} -\frac{1}{4|g|}g^{\alpha\beta}\partial_{\beta}|g|\partial_{\mu}g_{\alpha\nu} \\[.2cm] =&{} 8\pi GT_{\mu\nu} \end{align}$

如果你对爱因斯坦求和约定没有什么概念的话，可以把求和号写出来看看，它长下面这样：

$\begin{align} &\frac{1}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^{3}g^{\alpha\beta} \partial_{\alpha}\partial_{\mu} g_{\beta\nu} + \frac{1}{2}\sum_{\alpha=0}^3\sum_{\beta=0}^{3}g^{\alpha\beta} \partial_{\alpha}\partial_{\nu}g_{\mu\beta} \\[.2cm] -{}& \frac{1}{2}\sum_{\alpha=0}^3\sum_{\beta=0}^3g^{\alpha\beta}\partial_{\alpha}\partial_{\beta}g_{\mu\nu} -\frac{3}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^{3}g^{\alpha\beta}\partial_{\mu}\partial_{\nu}g_{\alpha\beta}\\[.2cm] -&\frac{1}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3\sum_{\rho=0}^3\sum_{\lambda=0}^3g^{\beta\lambda}g^{\alpha\rho}\partial_{\alpha}g_{\rho\lambda}\partial_{\mu}g_{\beta\nu}-\frac{1}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3\sum_{\rho=0}^3\sum_{\lambda=0}^3g^{\beta\lambda}g^{\alpha\rho}\partial_{\alpha}g_{\rho\lambda}\partial_{\nu}g_{\mu\beta}\\[.2cm] +&\frac{1}{4}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3\sum_{\rho=0}^3\sum_{\lambda=0}^3g^{\beta\lambda}g^{\alpha\rho}\partial_{\nu}g_{\alpha\lambda}\partial_{\mu}g_{\rho\beta}+\frac{1}{4|g|}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3g^{\alpha\beta}\partial_{\beta}|g|\partial_{\nu}g_{\mu\alpha}\\[.2cm] -&\frac{1}{4|g|}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3g^{\alpha\beta}\partial_{\beta}|g|\partial_{\alpha}g_{\mu\nu} -\frac{1}{4|g|}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3g^{\alpha\beta}\partial_{\beta}|g|\partial_{\mu}g_{\alpha\nu} \\[.2cm] =&{} 8\pi GT_{\mu\nu} \end{align}$

如果你还是没对这个方程没概念的话，以第一项求和展开为例，它长下面这样：

$\begin{align} &\frac{1}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3g^{\alpha\beta}\partial_{\alpha}\partial_{\nu} g_{\mu\beta}\\[.2cm] ={}&\frac{1}{2}g^{00}\partial_{0}\partial_{\nu}g_{\mu0} +\frac{1}{2}g^{01}\partial_{0}\partial_{\nu}g_{\mu1} +\frac{1}{2}g^{02}\partial_{0}\partial_{\nu}g_{\mu2} +\frac{1}{2}g^{03}\partial_{0}\partial_{\nu}g_{\mu3}\\[.2cm] &{}+\frac{1}{2}g^{10}\partial_{1}\partial_{\nu}g_{\mu0} +\frac{1}{2}g^{11}\partial_{1}\partial_{\nu}g_{\mu1}+ \frac{1}{2}g^{12}\partial_{1}\partial_{\nu}g_{\mu2}+ \frac{1}{2}g^{13}\partial_{1}\partial_{\nu}g_{\mu3}\\[.2cm]&{}+ \frac{1}{2}g^{20}\partial_{2}\partial_{\nu}g_{\mu0}+ \frac{1}{2}g^{21}\partial_{2}\partial_{\nu}g_{\mu1}+ \frac{1}{2}g^{22}\partial_{2}\partial_{\nu}g_{\mu2}+ \frac{1}{2}g^{23}\partial_{2}\partial_{\nu}g_{\mu3}\\[.2cm]&{}+ \frac{1}{2}g^{30}\partial_{3}\partial_{\nu}g_{\mu0}+ \frac{1}{2}g^{31}\partial_{3}\partial_{\nu}g_{\mu1}+ \frac{1}{2}g^{32}\partial_{3}\partial_{\nu}g_{\mu2}+ \frac{1}{2}g^{33}\partial_{3}\partial_{\nu}g_{\mu3} \end{align}$

你这才展开第一项的求和，后面还有好几项等着你展开。

第二项展开

$\begin{align} &\frac{1}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3g^{\alpha\beta}\partial_{\alpha}\partial_{\mu} g_{\beta\nu}\\[.2cm] ={}&\frac{1}{2}g^{00}\partial_{0}\partial_{\mu}g_{0\nu} +\frac{1}{2}g^{01}\partial_{0}\partial_{\mu}g_{1\nu} +\frac{1}{2}g^{02}\partial_{0}\partial_{\mu}g_{2\nu} +\frac{1}{2}g^{03}\partial_{0}\partial_{\mu}g_{3\nu}\\[.2cm] &{}+\frac{1}{2}g^{10}\partial_{1}\partial_{\mu}g_{0\nu} +\frac{1}{2}g^{11}\partial_{1}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{12}\partial_{1}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{13}\partial_{1}\partial_{\mu}g_{3\nu}\\[.2cm]&{}+ \frac{1}{2}g^{20}\partial_{2}\partial_{\mu}g_{0\nu}+ \frac{1}{2}g^{21}\partial_{2}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{22}\partial_{2}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{23}\partial_{2}\partial_{\mu}g_{3\nu}\\[.2cm]&{}+ \frac{1}{2}g^{30}\partial_{3}\partial_{\mu}g_{0\nu}+ \frac{1}{2}g^{31}\partial_{3}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{32}\partial_{3}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{33}\partial_{3}\partial_{\mu}g_{3\nu} \end{align}$

第三项展开

$\begin{align} &\frac{1}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3g^{\alpha\beta}\partial_{\alpha}\partial_{\beta} g_{\mu\nu}\\[.2cm] ={}&\frac{1}{2}g^{00}\partial_{0}\partial_{0}g_{\mu\nu} +\frac{1}{2}g^{01}\partial_{0}\partial_{1}g_{\mu\nu} +\frac{1}{2}g^{02}\partial_{0}\partial_{2}g_{\mu\nu} +\frac{1}{2}g^{03}\partial_{0}\partial_{3}g_{\mu\nu}\\[.2cm] &{}+\frac{1}{2}g^{10}\partial_{1}\partial_{0}g_{\mu\nu} +\frac{1}{2}g^{11}\partial_{1}\partial_{1}g_{\mu\nu}+ \frac{1}{2}g^{12}\partial_{1}\partial_{2}g_{\mu\nu}+ \frac{1}{2}g^{13}\partial_{1}\partial_{3}g_{\mu\nu}\\[.2cm]&{}+ \frac{1}{2}g^{20}\partial_{2}\partial_{0}g_{\mu\nu}+ \frac{1}{2}g^{21}\partial_{2}\partial_{1}g_{\mu\nu}+ \frac{1}{2}g^{22}\partial_{2}\partial_{2}g_{\mu\nu}+ \frac{1}{2}g^{23}\partial_{2}\partial_{3}g_{\mu\nu}\\[.2cm]&{}+ \frac{1}{2}g^{30}\partial_{3}\partial_{0}g_{\mu\nu}+ \frac{1}{2}g^{31}\partial_{3}\partial_{1}g_{\mu\nu}+ \frac{1}{2}g^{32}\partial_{3}\partial_{2}g_{\mu\nu}+ \frac{1}{2}g^{33}\partial_{3}\partial_{3}g_{\mu\nu} \end{align}$

第四项展开

$\begin{align} &\frac{3}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3g^{\alpha\beta}\partial_{\mu}\partial_{\nu} g_{\alpha\beta}\\[.2cm] ={}&\frac{3}{2}g^{00}\partial_{\mu}\partial_{\nu}g_{00} +\frac{3}{2}g^{01}\partial_{\mu}\partial_{\nu}g_{01} +\frac{3}{2}g^{02}\partial_{\mu}\partial_{\nu}g_{02} +\frac{3}{2}g^{03}\partial_{\mu}\partial_{\nu}g_{03}\\[.2cm] &{}+\frac{3}{2}g^{10}\partial_{\mu}\partial_{\nu}g_{10} +\frac{3}{2}g^{11}\partial_{\mu}\partial_{\nu}g_{11}+ \frac{3}{2}g^{12}\partial_{\mu}\partial_{\nu}g_{12}+ \frac{3}{2}g^{13}\partial_{\mu}\partial_{\nu}g_{13}\\[.2cm]&{}+ \frac{3}{2}g^{20}\partial_{\mu}\partial_{\nu}g_{20}+ \frac{3}{2}g^{21}\partial_{\mu}\partial_{\nu}g_{21}+ \frac{3}{2}g^{22}\partial_{\mu}\partial_{\nu}g_{22}+ \frac{3}{2}g^{23}\partial_{\mu}\partial_{\nu}g_{23}\\[.2cm]&{}+ \frac{3}{2}g^{30}\partial_{\mu}\partial_{\nu}g_{30}+ \frac{3}{2}g^{31}\partial_{\mu}\partial_{\nu}g_{31}+ \frac{3}{2}g^{32}\partial_{\mu}\partial_{\nu}g_{32}+ \frac{3}{2}g^{33}\partial_{\mu}\partial_{\nu}g_{33} \end{align}$

第五项展开

$\begin{align} &{}\frac{1}{2}\sum_{\alpha=0}^{3}\sum_{\beta=0}^3\sum_{\rho=0}^3\sum_{\lambda=0}^3g^{\beta\lambda}g^{\alpha\rho}\partial_{\alpha}g_{\rho\lambda}\partial_{\mu}g_{\beta\nu}\\[.2cm]={}& \frac{1}{2}g^{00}g^{00}\partial_{0}g_{00}\partial_{\mu}g_{0\nu}+ \frac{1}{2}g^{00}g^{10}\partial_{1}g_{00}\partial_{\mu}g_{0\nu}+ \frac{1}{2}g^{00}g^{20}\partial_{2}g_{00}\partial_{\mu}g_{0\nu}+ \frac{1}{2}g^{00}g^{30}\partial_{3}g_{00}\partial_{\mu}g_{0\nu}\\[.2cm]&{}+ \frac{1}{2}g^{10}g^{00}\partial_{0}g_{00}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{10}g^{10}\partial_{1}g_{00}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{10}g^{20}\partial_{2}g_{00}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{10}g^{30}\partial_{3}g_{00}\partial_{\mu}g_{1\nu}\\[.2cm]&{}+ \frac{1}{2}g^{20}g^{00}\partial_{0}g_{00}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{20}g^{10}\partial_{1}g_{00}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{20}g^{20}\partial_{2}g_{00}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{20}g^{30}\partial_{3}g_{00}\partial_{\mu}g_{2\nu}\\[.2cm]&{}+ \frac{1}{2}g^{30}g^{00}\partial_{0}g_{00}\partial_{\mu}g_{3\nu}+ \frac{1}{2}g^{30}g^{10}\partial_{1}g_{00}\partial_{\mu}g_{3\nu}+ \frac{1}{2}g^{30}g^{20}\partial_{2}g_{00}\partial_{\mu}g_{3\nu}+ \frac{1}{2}g^{30}g^{30}\partial_{3}g_{00}\partial_{\mu}g_{3\nu}\\[.2cm]&{}+ \frac{1}{2}g^{00}g^{01}\partial_{0}g_{10}\partial_{\mu}g_{0\nu}+ \frac{1}{2}g^{00}g^{11}\partial_{1}g_{10}\partial_{\mu}g_{0\nu}+ \frac{1}{2}g^{00}g^{21}\partial_{2}g_{10}\partial_{\mu}g_{0\nu}+ \frac{1}{2}g^{00}g^{31}\partial_{3}g_{10}\partial_{\mu}g_{0\nu}\\[.2cm]&{}+ \frac{1}{2}g^{10}g^{01}\partial_{0}g_{10}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{10}g^{11}\partial_{1}g_{10}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{10}g^{21}\partial_{2}g_{10}\partial_{\mu}g_{1\nu}+ \frac{1}{2}g^{10}g^{31}\partial_{3}g_{10}\partial_{\mu}g_{1\nu}\\[.2cm]&{}+ \frac{1}{2}g^{20}g^{01}\partial_{0}g_{10}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{20}g^{11}\partial_{1}g_{10}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{20}g^{21}\partial_{2}g_{10}\partial_{\mu}g_{2\nu}+ \frac{1}{2}g^{20}g^{31}\partial_{3}g_{10}\partial_{\mu}g_{2\nu}\\[.2cm]&{}+ \cdots\cdots\cdots\cdots\cdots \end{align}$

好家伙，这还成功测试到了知乎公式上限

（这一项总共是

$256$

项的和，上面也才仅仅展出了

$28$

项）

总之，后面还有第六项，第七项，第八项，第九项，第十项的展开

当你哼哧哼哧全部展开之后，你会发现左边总共有

$880$

项！

你要知道爱因斯坦场方程是个二阶张量方程，它足足有

$16$

个类似你刚才展出来的方程！！

$\begin{align} &R_{00} - \frac{1}{2}g_{00}R = 8\pi G T_{00}\\[.2cm] &R_{01} - \frac{1}{2}g_{01}R = 8\pi G T_{01}\\[.2cm] &R_{02} - \frac{1}{2}g_{02}R = 8\pi G T_{02}\\[.2cm] &R_{03} - \frac{1}{2}g_{03}R = 8\pi G T_{03}\\[.2cm] &R_{10} - \frac{1}{2}g_{10}R = 8\pi G T_{10}\\[.2cm] &R_{11} - \frac{1}{2}g_{11}R = 8\pi G T_{11}\\[.2cm] &R_{12} - \frac{1}{2}g_{12}R = 8\pi G T_{12}\\[.2cm] &R_{13} - \frac{1}{2}g_{13}R = 8\pi G T_{13}\\[.2cm] \end{align}$

$\begin{align} &R_{20} - \frac{1}{2}g_{20}R = 8\pi G T_{20}\\[.2cm] &R_{21} - \frac{1}{2}g_{21}R = 8\pi G T_{21}\\[.2cm] &R_{22} - \frac{1}{2}g_{22}R = 8\pi G T_{22}\\[.2cm] &R_{23} - \frac{1}{2}g_{23}R = 8\pi G T_{23}\\[.2cm] &R_{30} - \frac{1}{2}g_{30}R = 8\pi G T_{30}\\[.2cm] &R_{31} - \frac{1}{2}g_{31}R = 8\pi G T_{31}\\[.2cm] &R_{32} - \frac{1}{2}g_{32}R = 8\pi G T_{32}\\[.2cm] &R_{33} - \frac{1}{2}g_{33}R = 8\pi G T_{33}\\[.2cm] \end{align}$

每一个方程你都可以把左边展成

$880$

项，也就意味着你总共需要展成

$14080$

项！！！

你这才解决场方程左边，还有右边没有算呢！

四、物质模型和坐标条件

然而要完全求解爱因斯坦场方程你还需要以下两个东西：

物质模型，坐标条件

场方程的右侧的能量动量张量

$T_{\mu\nu}$

根具体的物质（或者说选取的物质模型）有关，

且爱因斯坦场方程每个方程并不独立（实质上仅有 6 条独立方程），为了完全求解

$g_{\mu\nu}$

还需引入谐和坐标条件。

由于爱因斯坦场方程的高度非线性，实际上找到解析解基本是不可能的事情，就算是数值解也颇为复杂。甚至为了求解爱因斯坦场方程还专门发展了一门“数值相对论”的学科。

不过在一些特殊情况下是可以求出解析解的，

比如史瓦西大佬在爱因斯坦发表广义相对论论文后的一个月就得到了真空、球对称（也是稳态）的解（那个时候他甚至还在参加一战的战场上）：

$\mathrm{d}s^2 = -\left(1- \frac{2GM}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2GM}{r}\right)^{-1}\mathrm{d}r^2 + r^2\ \mathrm{d}\theta^2 + r^2\sin^2\theta\ \mathrm{d}\phi^2$