11
Nov
【外微分浅谈】7. 有力的计算
By 苏剑林 | 2016-11-11 | 1470位读者这里我们将展示上面一节的方法对于计算黎曼曲率张量的计算是多少的有力！我们再次列出我们得到的所有公式。首先是概念式的
$$\begin{aligned}&\omega^{\mu}=h_{\alpha}^{\mu}dx^{\alpha}\\
&d\boldsymbol{r}=\hat{\boldsymbol{e}}_{\mu} \omega^{\mu}\\
&ds^2 = \eta_{\mu\nu} \omega^{\mu}\omega^{\nu}\\
&\langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle = \eta_{\mu\nu}\end{aligned} \tag{65} $$
然后是
$$\begin{aligned}&d\eta_{\mu\nu}=\omega_{\nu\mu}+\omega_{\mu\nu}=\eta_{\nu\alpha}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha}\\
&d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0\end{aligned} \tag{66} $$
这两个可以帮助我们确定$\omega_{\nu}^{\mu}$；接着就是
$$\mathscr{R}_{\nu}^{\mu} = d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha} \tag{67} $$
最后你要正交标架下的$\hat{R}^{\mu}_{\nu\beta\gamma}$，就要写出：
$$\mathscr{R}_{\nu}^{\mu}=\sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma} \tag{68} $$
如果你要原始标架下的$R^{\mu}_{\nu\beta\gamma}$，就要写出
$$(h^{-1})_{\mu'}^{\mu}\mathscr{R}^{\mu'}_{\nu'}h_{\nu}^{\nu'} = \sum_{\beta < \gamma} R^{\mu}_{\nu\beta\gamma}dx^{\beta}\land dx^{\gamma} \tag{69} $$
然后依次读出$R^{\mu}_{\nu\beta\gamma}$，就像制表一样。
29
Aug
计算夏至的精确时刻2——提高精确度
By 苏剑林 | 2010-08-29 | 784位读者
8
Jul
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