物理

【广义相对论速成版】2. Einstein引力场方程 2.5 附录：矩阵与行列式

1. 矩阵与逆矩阵

以$a_{\mu\nu}~(\mu,\nu=1,2,\cdots,n)$为元素的$n\times n$矩阵记为

$$\hat{a}=\begin{pmatrix}a_{11} & \cdots & a_{1n}\\\vdots & \ddots & \vdots\\a_{n1} & \cdots & a_{nn}\end{pmatrix}$$

即

$$\hat{a}=\left[a_{\mu\nu}\right]$$

$\hat{a}$的逆矩阵$\hat{a}^{-1}$定义为

$$\boxed{\hat{a}^{-1} \hat{a}=\hat{a} \hat{a}^{-1}=I}$$

$\hat{a}^{-1}$的元素记为$a^{\mu\nu}$，即 :

$$\hat{a}^{-1}=\left[a^{\mu\nu}\right]$$

则有

$$\boxed{a^{\mu \lambda} a_{\lambda\nu}=\delta_{\nu}^{\mu}}\quad\text{且}\quad\boxed{a_{\mu\lambda}a^{\lambda\nu}=\delta_{\mu}^{\nu}}$$

2. 矩阵的行列式

矩阵$\hat{a}=\left[a_{\mu\nu}\right]$的行列式记为

$$a=\det\left[a_{\mu\nu}\right]$$

其定义为

$$\boxed{\epsilon_{\nu_{1}\nu_{2}\cdots\nu_{n}}a=\epsilon^{\mu_{1} \mu_{2} \cdots \mu_{n}}a_{\mu_{1}\nu_{1}}a_{\mu_{2}\nu_{2}}\cdots a_{\mu_{n}\nu_{n}}}$$

将上式乘以$\epsilon^{\nu_{1}\nu_{2}\cdots\nu_{n}}$，利用

$$\epsilon^{\nu_{1}\nu_{2}\cdots\nu_{n}}\epsilon_{\nu_{1}\nu_{2}\cdots\nu_{n}}=n!$$

可得

$$\boxed{a=\frac{1}{n!}\epsilon^{\mu_{1}\mu_{2}\cdots\mu_{n}}\epsilon^{\nu_{1}\nu_{2}\cdots\nu_{n}}a_{\mu_{1}\nu_{1}}a_{\mu_{2}\nu_{2}}\cdots a_{\mu_{n}\nu_{n}}}$$

令$\hat{a}$的逆矩阵$\hat{a}^{-1}$的行列式记为

$$\overline{a}=\det\left[a^{\mu\nu}\right]$$

$$\epsilon^{\mu_{1}\mu_{2}\cdots\mu_{n}}\overline{a}=\epsilon_{\lambda_{1}\lambda_{2}\cdots\lambda_{n}}a^{\lambda_{1}\mu_{1}}a^{\lambda_{2}\mu_{2}}\cdots a^{\lambda_{n}\mu_{n}}$$

则

$$\begin{aligned}\epsilon^{\mu_{1} \mu_{2} \cdots \mu_{n}} a_{\mu_{1} \nu_{1}} a_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} \nu_{n}} \overline{a}=&\epsilon_{\lambda_{1} \lambda_{2} \cdots \lambda_{n}}\left(a^{\lambda_{1} \mu_{1}} a_{\mu_{1} \nu_{1}}\right)\left(a^{\lambda_{2} \mu_{2}} \alpha_{\mu_{2} \nu_{2}}\right) \cdots\left(a^{\lambda_{n} \mu_{n}} a_{\mu_{n} \nu_{n}} \right)\\=&\epsilon_{\lambda_{1} \lambda_{2} \cdots \lambda_{n}} \delta_{\nu_{1}}^{\lambda_{1}} \delta_{\nu_{2}}^{\lambda_{2}} \cdots \delta_{\nu_{n}}^{\lambda_{n}}\\=&\epsilon_{\nu_{1} \nu_{2} \cdots \nu_{n}}\end{aligned}$$

上式乘以$\epsilon^{\nu_{1}\nu_{2}\cdots\nu_{n}}$得到

$$\begin{aligned}\epsilon^{\mu_{1} \mu_{2} \cdots \mu_{n}}\epsilon^{\nu_{1}\nu_{2}\cdots\nu_{n}}a_{\mu_{1} \nu_{1}} a_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} \nu_{n}} \overline{a}&=\epsilon_{\nu_{1} \nu_{2} \cdots \nu_{n}}\epsilon^{\nu_{1}\nu_{2}\cdots\nu_{n}}\\\frac{1}{n!}\epsilon^{\mu_{1} \mu_{2} \cdots \mu_{n}}\epsilon^{\nu_{1}\nu_{2}\cdots\nu_{n}}a_{\mu_{1} \nu_{1}} a_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} \nu_{n}} \overline{a}&=\frac{1}{n!}\epsilon_{\nu_{1} \nu_{2} \cdots \nu_{n}}\epsilon^{\nu_{1}\nu_{2}\cdots\nu_{n}}\end{aligned}$$

故

$$a \overline{a}=1$$

即

$$\overline{a}=\frac{1}{a}=a^{-1}$$

故

$$\boxed{\det\left[a^{\mu\nu}\right]=a^{-1}=\frac{1}{a}}$$

其中，$a=\det\left[a_{\mu \nu}\right]$。

3. $a_{\mu\nu}$的余因子$A^{\mu\nu}$

定义 

$$A^{\mu \nu}=\frac{1}{n !} \epsilon^{\mu \mu_{2} \cdots \mu_{n}} \epsilon^{\nu \nu_{2} \cdots \nu_{n}} a_{\mu_{2} \nu_{2}} \cdot \cdots a_{\mu_{n} v_{n}}$$

可证明

$$\begin{aligned}A^{\mu \nu}a_{\mu\lambda}&=\frac{1}{n !} \epsilon^{\mu \mu_{2} \cdots \mu_{n}} \epsilon^{\nu \nu_{2} \cdots \nu_{n}} a_{\mu\lambda}a_{\mu_{2} \nu_{2}} \cdot \cdots a_{\mu_{n} v_{n}}\\&=\left(\frac{1}{n!}\epsilon_{\lambda\nu_{2}\cdots\nu_{n}}\epsilon^{\lambda\nu_{2}\cdots\nu_{n}}\right)\cdot\left(\frac{1}{n !} \epsilon^{\mu \mu_{2} \cdots \mu_{n}} \epsilon^{\nu \nu_{2} \cdots \nu_{n}} a_{\mu\lambda}a_{\mu_{2} \nu_{2}} \cdot \cdots a_{\mu_{n} v_{n}}\right)\\&=\left(\frac{1}{n!}\epsilon_{\lambda\nu_{2}\cdots\nu_{n}}\epsilon^{\nu \nu_{2} \cdots \nu_{n}}\right)\cdot\left(\frac{1}{n !} \epsilon^{\mu \mu_{2} \cdots \mu_{n}}\epsilon^{\lambda\nu_{2}\cdots\nu_{n}}  a_{\mu\lambda}a_{\mu_{2} \nu_{2}} \cdot \cdots a_{\mu_{n} v_{n}}\right)\\&=\frac{1}{n !}\epsilon_{\lambda\nu_{2}\cdots\nu_{n}} \epsilon^{\nu\nu_{2}\cdots\nu_{n}}a\end{aligned}$$

利用

$$\epsilon_{\lambda\nu_{2}\cdots\nu_{n}}\epsilon^{\nu\nu_{2}\cdots\nu_{n}}=\epsilon_{\lambda\nu_{2}\cdots\nu_{n}}\frac{1}{n}\epsilon^{\lambda\nu_{2}\cdots\nu_{n}}\delta_{\lambda}^{\nu}=\epsilon_{\lambda\nu_{2}\cdots\nu_{n}}\epsilon^{\lambda\nu_{2}\cdots\nu_{n}}\frac{1}{n}\delta_{\lambda}^{\nu}=n!\frac{1}{n}\delta_{\lambda}^{\nu}=(n-1)!\delta_{\lambda}^{\nu}$$

故

$$A^{\mu\nu}a_{\mu\lambda}=\frac{1}{n}\delta_{\lambda}^{\nu}a$$

由$\delta_{\nu}^{\nu}=n$，对上式取$\lambda=\nu$得到

$$\boxed{A^{\mu\nu}a_{\mu\nu}=a}$$

$A^{\mu\nu}$称为$a_{\mu\nu}$的余因子(Cofactor)。

4. $a_{\mu\nu}=a_{\nu\mu}$情况的余因子$A^{\mu\nu}$

当$a_{\mu\nu}=a_{\nu\mu}$ (即$a_{\mu\nu}$为对称时)，由

$$A^{\mu \nu} a_{\mu \lambda}=\frac{1}{n} \delta_{\lambda}^{\nu} a$$

可知

$$A^{\mu \nu} a_{\lambda \mu}=\frac{1}{n} \delta_{\lambda}^{\nu} a$$

即

$$\boxed{a_{\lambda\mu}A^{\mu\nu}=\frac{1}{n} \delta_{\lambda}^{\nu} a\qquad \text{当}a_{\lambda\mu}=a_{\mu\lambda}}$$

由于

$$a_{\lambda \mu} a^{\mu \nu}=\delta_{\lambda}^{\nu}$$

故

$$\boxed{A^{\mu \nu}=\frac{1}{n} a^{\mu \nu} a}\qquad a_{\mu\nu}\text{对称}$$

因

$$a_{\nu \mu} A^{\mu \nu}=\frac{1}{n} \delta_{\nu}^{\nu} a=a$$

即

$$\boxed{A^{\mu \nu}a_{\mu\nu}=a}\qquad a_{\mu \nu}\text{对称}$$

只有对称的情况才是矩阵乘法的意义。

5. $\partial_{\mu}a$公式

$$a=\frac{1}{n !} \epsilon^{\mu_{1} \mu_{2} \cdots \mu_{n}} \epsilon^{\nu_{1} \nu_{2} \cdots \nu_{n}} a_{\mu_{1} \nu_{1}} a_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} \nu_{n}}$$

当$a_{\mu \nu}=a_{\mu \nu}(x),~x=\left(x^{1}, x^{2} \cdots x^{n}\right)$，可证明

$$\partial_{\lambda} a=\frac{1}{(n-1) !} \epsilon^{\mu_{1} \mu_{2} \cdots \mu_{n}} \epsilon^{\nu_{1} \nu_{2} \cdots \nu_{n}}\left(\partial_{\lambda} a_{\mu_{1} \nu_{1}}\right) a_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} \nu_{n}}$$

因

$$A^{\mu \nu}=\frac{1}{n !} \epsilon^{\mu \mu_{2} \cdots \mu_{n}} \epsilon^{\nu \nu_{2} \cdots \nu_{n}} a_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} \nu_{n}}$$

故

$$\partial_{\lambda} a=n A^{\mu \nu}\left(\partial_{\lambda} a_{\mu \nu}\right)$$

当$a_{\mu \nu}$对称时

$$A^{\mu \nu}=\frac{1}{n} a^{\mu \nu} a$$

故

$$\boxed{\partial_{\lambda} a=a^{\mu \nu}a\left(\partial_{\lambda} a_{\mu \nu}\right)\qquad \text{当}a_{\mu \nu}=a_{\nu \mu}}$$

6. $a$的变分$\delta a$

$$\delta a=\frac{n}{n !} \epsilon^{\mu \mu_{2} \cdots \mu_{n}} \epsilon^{\nu \nu_{2} \cdots \nu_{n}} \delta \left(a_{\mu \nu}\right) a_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} u_{n}}$$

故

$$\boxed{\delta a=n A^{\mu \nu} \delta a_{\mu \nu}}$$

当$a_{\mu \nu}$对称时

$$A^{\mu \nu}=\frac{1}{n} a^{\mu \nu} a$$

故

$$\boxed{\delta a=\left(a^{\mu \nu} \delta a_{\mu \nu}\right) a\qquad \text{当}a_{\mu \nu}=a_{\nu \mu}}$$

7. $\frac{\delta a}{\delta a_{\mu\nu}}$

$$a=\frac{1}{n !} \epsilon^{\mu \mu_{2} \cdots \mu_{n}} \epsilon^{\nu \nu_{2} \cdots \nu_{n}}a_{\mu \nu} a_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} \nu_{n}}$$

看出$a$中不含$a_{\mu\nu}$的偏微商项

$$\frac{\delta a}{\delta a_{\mu\nu}}=\frac{\partial a}{\partial a_{\mu \nu}}$$

故

$$\frac{\delta a}{\delta a_{\mu \nu}}=\frac{n}{n !} \epsilon^{\mu \mu_{2} \cdots \mu_{n}} \epsilon^{\nu \nu_{2} \cdots \nu_{n}} \alpha_{\mu_{2} \nu_{2}} \cdots a_{\mu_{n} \nu_{n}}$$

因此

$$\boxed{\frac{\delta a}{\delta a_{\mu \nu}}=n A^{\mu \nu}}$$

当$a_{\mu \nu}=a_{\nu \mu}$时

$$A^{\mu \nu}=\frac{1}{n} a^{\mu \nu} a$$

则

$$\boxed{\frac{\delta a}{\delta a_{\mu \nu}}=a^{\mu \nu} a\qquad \text{当}a_{\mu \nu}=a_{\nu \mu}}$$