物理

【广义相对论速成版】1. Riemann几何 1.5 Riemann曲率张量(2)

3. Bianchi恒等式(Bianchi Identity)

由于Riemann几何是无挠率的，即：

$$T_{\mu\nu}^{\sigma}=0$$

设$\phi_{\sigma}$是协变矢量，则有

$$[\nabla_{\mu},\nabla_{\nu}]\phi_{\sigma}=R_{~\sigma\mu\nu}^{\rho}\phi_{\rho}$$

我们对上式左作用一个协变散度算符得到

$$\nabla_{\lambda}[\nabla_{\mu},\nabla_{\nu}]\phi_{\sigma}=\nabla_{\lambda}\left(R_{~\sigma\mu\nu}^{\rho}\phi^{\sigma}\right)=\left(\nabla_{\lambda}R_{~\sigma\mu\nu}^{\rho}\right)\phi_{\rho}+R_{~\sigma\mu\nu}^{\rho}\left(\nabla_{\lambda}\phi_{\rho}\right)$$

还有

$$[\nabla_{\mu},\nabla_{\nu}]\nabla_{\lambda}\phi_{\sigma}=R_{~\lambda\mu\nu}^{\rho}\nabla_{\rho}\phi_{\sigma}+R_{~\sigma\mu\nu}^{\rho}\nabla_{\lambda}\phi_{\rho}$$

上面两式相减得到

$$[\nabla_{\lambda},[\nabla_{\mu},\nabla_{\nu}]]\phi_{\sigma}=\left(\nabla_{\lambda}R_{~\sigma\mu\nu}^{\rho}\right)\phi_{\rho}-R_{~\lambda\mu\nu}^{\rho}\nabla_{\rho}\phi_{\sigma}$$

轮换指标($\lambda\leftarrow\mu$,$\mu\leftarrow\nu$,$\nu\leftarrow\lambda$)得到

$$[\nabla_{\mu},[\nabla_{\nu},\nabla_{\lambda}]]\phi_{\sigma}=\left(\nabla_{\mu}R_{~\sigma\nu\lambda}^{\rho}\right)\phi_{\rho}-R_{~\mu\nu\lambda}^{\rho}\nabla_{\rho}\phi_{\sigma}$$

轮换指标($\mu\leftarrow\nu$,$\nu\leftarrow\lambda$,$\lambda\leftarrow\mu$)得到

$$[\nabla_{\nu},[\nabla_{\lambda},\nabla_{\mu}]]\phi_{\sigma}=\left(\nabla_{\nu}R_{~\sigma\lambda\mu}^{\rho}\right)\phi_{\rho}-R_{~\nu\lambda\mu}^{\rho}\nabla_{\rho}\phi_{\sigma}$$

根据$[a,[b,c]]+[b,[c,a]]+[c,[a,b]]=0$，我们将上面三式相加

$$\begin{aligned}&\left\{[\nabla_{\lambda},[\nabla_{\mu},\nabla_{\nu}]]+[\nabla_{\mu},[\nabla_{\nu},\nabla_{\lambda}]]+[\nabla_{\nu},[\nabla_{\lambda},\nabla_{\mu}]]\right\}\phi_{\sigma}\\=&\left(\nabla_{\lambda}R_{~\sigma\mu\nu}^{\rho}+\nabla_{\mu}R_{~\sigma\nu\lambda}^{\rho}+\nabla_{\nu}R_{~\sigma\lambda\mu}^{\rho}\right)\phi_{\rho}-\left(R_{~\lambda\mu\nu}^{\rho}+R_{~\mu\nu\lambda}^{\rho}+R_{~\nu\lambda\mu}^{\rho}\right)\nabla_{\rho}\phi_{\sigma}=0\end{aligned}$$

根据逆变曲率张量循环等式

$$R_{~\lambda\mu\nu}^{\rho}+R_{~\mu\nu\lambda}^{\rho}+R_{~\nu\lambda\mu}^{\rho}=0$$

而协变矢量$\phi_{\sigma}$取值是任意的，因此可证明Bianchi恒等式

$$\nabla_{\lambda}R_{~\sigma\mu\nu}^{\rho}+\nabla_{\mu}R_{~\sigma\nu\lambda}^{\rho}+\nabla_{\nu}R_{~\sigma\lambda\mu}^{\rho}=0$$

(上式的证明用$SO(N)$规范场张量$F_{\mu\nu}^{ab}$证明十分简单。)