yoshitake-hのブログ

Einstein 方程式

phys

Lorentz 多様体 $(M,\, g,\, \nabla)$ 上のベクトル場 $X=X^\mu \partial_\mu$ に対し，

$\mathrm{tr}(\nabla X)\omega_g = L_X\omega_g.$

証明

$\mathrm{tr}(\nabla X)\omega_g = (\nabla_\mu X)^\mu \sqrt{-\det g}\; \omega_0.$

$(\nabla_\mu X)^\nu = \partial_\mu X^\nu + {\Gamma^\nu}_{\mu \rho}X^\rho = \partial_\mu X^\nu + \frac{1}{2}g^{\nu \sigma}(\partial_\mu g_{\sigma \rho} + \partial_\rho g_{\mu \sigma} - \partial_\sigma g_{\mu \rho})X^\rho.$

$(\nabla_\mu X)^\mu = \partial_\mu X^\mu + \frac{1}{2} g^{\mu \sigma}(\partial_\rho g_{\mu \sigma})X^\rho.$

$L_X\omega_g = \partial_\mu (X^\mu \sqrt{-\det g})\omega_0 = (\partial_\mu X^\mu + \frac{1}{2} g^{\rho \nu}(\partial_\mu g_{\nu \rho})X^\mu ) \sqrt{-\det g}\; \omega_0.\quad \square$

Hilbert 作用

$S[g] = \int_M R\omega_g = \int_M R_{\mu \nu}g^{\nu \mu} \sqrt{-\det g}\; \omega_0$

変分

$(\delta R_{\mu \nu})g^{\nu \mu} = (\delta {R^\rho}_{\mu \rho \nu})g^{\nu \mu} = (\nabla_\mu \delta {\Gamma^\rho}_{\rho \nu})g^{\nu \mu} = \nabla_\mu (\delta {\Gamma^\rho}_{\rho \nu} g^{\nu \mu})$

より， $(\delta R_{\mu \nu})g^{\nu \mu}\omega_g$ は完全形式．

$\delta \sqrt{-\det g} = \frac{1}{2} g^{\nu\mu} (\delta g_{\mu \nu})\sqrt{-\det g} = -\frac{1}{2} (\delta g^{\nu\mu}) g_{\mu \nu}\sqrt{-\det g}.$

したがって，

$\delta (R_{\mu \nu}g^{\nu \mu} \sqrt{-\det g} )=\partial_\mu ( (\delta {\Gamma^\rho}_{\rho \nu}) g^{\nu \mu} \sqrt{-\det g} )+ (R_{\mu \nu}-\frac{1}{2}Rg_{\mu \nu})(\delta g^{\nu \mu}) \sqrt{-\det g}.$

Einstein 方程式

$R_{\mu \nu}-\frac{1}{2}Rg_{\mu \nu} =0.$