2
$\begingroup$

On the solution of problem 10.6 of the book "Problem Book in Relativity and Gravitation by A. Lightman, R. H. Price" they mention using the Killing equation:

$\xi^{}_{\mu;\nu}=-\xi^{}_{\nu;\mu}$

and contracting both the $\mu$ and $\lambda$ indicies to arrive at the equation:

$\xi^{\nu;\lambda}_{\space\space\space\space\space\space;\lambda}+R^{\nu}_{\space\space\sigma}\xi^{\sigma}=-(\xi^{\mu}_{\space\space;\mu})^{;\nu}$

Here is my attempt:

Killing equation:

$\nabla^{}_{\nu}\xi^{}_{\mu}+\nabla{}_{\mu}\xi^{}_{\nu}=\nabla^{}_{\lambda}\nabla^{}_{\nu}\xi^{}_{\mu}+\nabla^{}_{\lambda}\nabla{}_{\mu}\xi^{}_{\nu}=0$

Definition of the Riemann curvature tensor that they give us:

$\nabla^{}_{\lambda}\nabla{}_{\nu}\xi^{}_{\mu}-\nabla^{}_{\nu}\nabla{}_{\lambda}\xi^{}_{\mu}= R^{}_{\mu\sigma\lambda\nu}\xi^{\sigma}$

I subtracted , in the Killing equation, the second term in the above equation and got:

$\nabla^{}_{\lambda}\nabla^{}_{\nu}\xi^{}_{\mu}-\nabla^{}_{\nu}\nabla{}_{\lambda}\xi^{}_{\mu}+\nabla^{}_{\lambda}\nabla{}_{\mu}\xi^{}_{\nu}=-\nabla^{}_{\nu}\nabla{}_{\lambda}\xi^{}_{\mu}$

Substituded the definition of the Riemann tensor for the first two terms and used $g^{\mu\lambda}$ for the contraction witch resulted in:

$R^{}_{\sigma\nu}\xi^{\sigma}+\nabla^{\mu}\nabla{}_{\mu}\xi^{}_{\nu}=-\nabla^{}_{\nu}\nabla^{\mu}\xi^{}_{\mu}$

I am almost at the final result but can´t quite get it.

$\endgroup$

1 Answer 1

4
$\begingroup$

You should use the definition of the curvature tensor $$[\nabla_\mu, \nabla_\nu]\xi_\kappa = R_{\mu\nu\kappa}^{\ \ \ \ \ \ \lambda}\xi_\lambda.$$ Then by differentiating the Killing condition we obtain $$\nabla_\kappa\nabla_\mu\xi_\nu + \nabla_\kappa\nabla_\nu\xi_\mu = 0.$$ One can then add a term to both sides to get a commutator and insert the above definition of the curvature tensor: $$R_{\kappa\mu\nu}^{\ \ \ \ \ \ \lambda}\xi_\lambda + \nabla_\kappa\nabla_\nu\xi_\mu = -\nabla_\mu\nabla_\kappa\xi_\nu.$$ Finally by contracting using the metric $g^{\nu\kappa}$ (and using the fact that the Levi-Civita connection is metric-compatible) you arrive at your formula.

EDIT: I messed up the indices in the definition of the curvature tensor and I made a sign error, this has been corrected.

$\endgroup$
2
  • $\begingroup$ I tried using your method (I edited the question above) and Im close but didn´t get the exact equation that they got, I have some indicies in the wrong places. $\endgroup$ Oct 6, 2020 at 9:21
  • $\begingroup$ I made some small mistakes (this has now been corrected but my apologies if it resulted in confusion). The result, however, remains the same. The only thing you miss is the symmetry of the Ricci curvature and raising the $\nu$-index. $\endgroup$
    – NDewolf
    Oct 6, 2020 at 10:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.