テンソルの共変微分

ベクトルの共変微分の次はテンソルの共変微分の計算も行いましょう。

計算

\[\nabla_\mu A^\alpha = \partial_\mu A^\alpha + A^\beta \Gamma^\alpha_{\beta \mu}\] \[\nabla_\mu B_\alpha = \partial_\mu B_\alpha - B_\beta \Gamma^\beta_{\mu \alpha}\]

より

\[\begin{aligned} \nabla_\mu T_{\alpha \beta} &= \nabla_\mu (A_\alpha B_\beta) = (\nabla_\mu A_\alpha) B_\beta + A_\alpha (\nabla_\mu B_\beta) = (\partial_\mu A_\alpha)B_\beta - A_\nu \Gamma^\nu_{\mu \alpha} B_\beta + A_\alpha (\partial_\mu B_\beta) - A_\alpha B_\nu \Gamma^\nu_{\mu \beta} \\ &= \partial_\mu (A_\alpha B_\beta) - \Gamma^\nu_{\mu \alpha} A_\nu B_\beta - \Gamma^\nu_{\mu \beta} A_\alpha B_\nu \ \Longrightarrow \ \therefore \ \nabla_\mu T_{\alpha \beta} = \partial_\mu T_{\alpha \beta} - \Gamma^\nu_{\mu \alpha} T_{\nu \beta} - \Gamma^\nu_{\mu \beta} T_{\alpha \nu} \end{aligned}\]

計算例: 計量テンソルの共変微分

\[\begin{aligned} \nabla_\mu g_{\alpha \beta} &= \partial_\mu g_{\alpha \beta} - \Gamma^\gamma_{\mu \alpha} g_{\gamma \beta} - \Gamma^\gamma_{\mu \beta} g_{\alpha \gamma} \\ &= \partial_\mu g_{\alpha \beta} - \frac{1}{2} g^{\nu \gamma} g_{\gamma\beta} (\partial_\mu g_{\alpha \nu} + \partial_\alpha g_{\nu \mu} - \partial_\nu g_{\mu \alpha}) - \frac{1}{2} g^{\nu \gamma} g_{\alpha \gamma} (\partial_\mu g_{\beta \nu} + \partial_\beta g_{\nu \mu} - \partial_{\nu} g_{\mu \beta})\\ &= \partial_\mu g_{\alpha \beta} - \frac{1}{2} (\partial_\mu g_{\alpha \beta} + \partial_\alpha g_{\beta \mu} - \partial_\beta g_{\mu \alpha}) - \frac{1}{2} (\partial_\mu g_{\beta \alpha} + \partial_\beta g_{\alpha \mu} - \partial_{\alpha} g_{\mu \beta}) = \partial_\mu g_{\alpha \beta} - \frac{1}{2} 2 \partial_\mu g_{\alpha \beta} = 0 \end{aligned}\]

となり、計量テンソルの形に関係なく恒等的に0となります。