- #1
etotheipi
- Homework Statement
- Derive the Penrose equation for a vacuum spacetime$$\nabla^{\lambda} \nabla_{\lambda} R_{\mu \nu \rho \sigma} = 2 {R^{\kappa}}_{\mu \lambda \sigma} {R^{\lambda}}_{\rho \kappa \nu} - 2{R^{\kappa}}_{\nu \lambda \sigma} {R^{\lambda}}_{\rho \kappa \mu} - {R^{\kappa}}_{\lambda \sigma \rho} {R^{\lambda}}_{\kappa \mu \nu}$$
- Relevant Equations
- N/A
Okay so for this one we can consider the Bianchi identity again$$\begin{align*}
\nabla_{[\lambda}R_{\sigma \rho] \mu \nu} = 2 \nabla_{\lambda} R_{\sigma \rho \mu \nu} + 2\nabla_{\rho} R_{\lambda \sigma \mu \nu} + 2\nabla_{\sigma} R_{\rho \lambda \mu \nu} &= 0 \\
\nabla^{\lambda} \nabla_{\lambda} R_{\sigma \rho \mu \nu} &= \nabla^{\lambda} \nabla_{\rho} R_{\sigma \lambda \mu \nu} + \nabla^{\lambda} \nabla_{\sigma} R_{\lambda \rho \mu \nu} \\
\nabla^{\lambda} \nabla_{\lambda} R_{\sigma \rho \mu \nu} &= g^{\epsilon \lambda} \nabla_{\epsilon} \nabla_{\rho} R_{\sigma \lambda \mu \nu} + g^{\epsilon \lambda} \nabla_{\epsilon} \nabla_{\sigma} R_{\lambda \rho \mu \nu} \end{align*}$$Here I wrote it in a form that's suggestive of another identity$$(\nabla_a \nabla_b - \nabla_b \nabla_a){T^{c_1, \dots, c_k}}_{d_1,\dots,d_l} = \sum_{j=1}^{l} {R_{abd_j}}^e {T^{c_1, \dots, c_k}}_{d_1, \dots, e, \dots d_l}
- \sum_{i=1}^{k} {R_{abe}}^{c_i} {T^{c_1, \dots, e, \dots, c_k}}_{d_1, \dots, d_l}
$$Let's say we consider ##T = R##, i.e.$$\begin{align*}(\nabla_{\epsilon} \nabla_{\rho} - \nabla_{\rho} \nabla_{\epsilon}) R_{\sigma \lambda \mu \nu} &= {R_{\epsilon \rho \sigma}}^{e} R_{e\lambda \mu \nu} + {R_{\epsilon \rho \lambda}}^{e} R_{\sigma e \mu \nu} + {R_{\epsilon \rho \mu}}^{e} R_{\sigma \lambda e \nu}+ {R_{\epsilon \rho \nu}}^{e} R_{\sigma \lambda \mu e}\end{align*}$$Is there a way to tidy this up using some other symmetries of the Riemann tensor and obtain the desired result? Thanks
\nabla_{[\lambda}R_{\sigma \rho] \mu \nu} = 2 \nabla_{\lambda} R_{\sigma \rho \mu \nu} + 2\nabla_{\rho} R_{\lambda \sigma \mu \nu} + 2\nabla_{\sigma} R_{\rho \lambda \mu \nu} &= 0 \\
\nabla^{\lambda} \nabla_{\lambda} R_{\sigma \rho \mu \nu} &= \nabla^{\lambda} \nabla_{\rho} R_{\sigma \lambda \mu \nu} + \nabla^{\lambda} \nabla_{\sigma} R_{\lambda \rho \mu \nu} \\
\nabla^{\lambda} \nabla_{\lambda} R_{\sigma \rho \mu \nu} &= g^{\epsilon \lambda} \nabla_{\epsilon} \nabla_{\rho} R_{\sigma \lambda \mu \nu} + g^{\epsilon \lambda} \nabla_{\epsilon} \nabla_{\sigma} R_{\lambda \rho \mu \nu} \end{align*}$$Here I wrote it in a form that's suggestive of another identity$$(\nabla_a \nabla_b - \nabla_b \nabla_a){T^{c_1, \dots, c_k}}_{d_1,\dots,d_l} = \sum_{j=1}^{l} {R_{abd_j}}^e {T^{c_1, \dots, c_k}}_{d_1, \dots, e, \dots d_l}
- \sum_{i=1}^{k} {R_{abe}}^{c_i} {T^{c_1, \dots, e, \dots, c_k}}_{d_1, \dots, d_l}
$$Let's say we consider ##T = R##, i.e.$$\begin{align*}(\nabla_{\epsilon} \nabla_{\rho} - \nabla_{\rho} \nabla_{\epsilon}) R_{\sigma \lambda \mu \nu} &= {R_{\epsilon \rho \sigma}}^{e} R_{e\lambda \mu \nu} + {R_{\epsilon \rho \lambda}}^{e} R_{\sigma e \mu \nu} + {R_{\epsilon \rho \mu}}^{e} R_{\sigma \lambda e \nu}+ {R_{\epsilon \rho \nu}}^{e} R_{\sigma \lambda \mu e}\end{align*}$$Is there a way to tidy this up using some other symmetries of the Riemann tensor and obtain the desired result? Thanks