Deriving equations from the Lagrangian density

Question

I was working on problem 10.6 of the book "Problem Book in Relativity and Gravitation by A. Lightman, R. H. Price" where we derive the following equation for killing vectors:

$$\xi^{\nu;\lambda}_{\space\space\space\space\space\space;\lambda}+R^{\nu}_{\space\space\sigma}\xi^{\sigma}=0 \space\space\space\space\space\space\space\space\space\space\space(1)$$

They say that the following Lagrangian density reproduces equation (1) by varying the action or using the Euler Lagrange equations:

$$\quad\mathfrak{L}=\xi^{}_{\mu;\nu}\xi^{\mu;\nu}-\frac{1}{2}R_{\mu\nu}\xi^{\mu}\xi^{\nu}$$

but I'm having trouble getting there. I tried using the Euler Lagrange equations but I get confused when trying to compute the derivative of the Lagrangian with respect to the covariant derivative of $\xi^{\mu}$. I think for the derivative with respect to just $\xi^{\mu}$ I just have to use the chain rule on the second term and add them up to cancel the factor of 1/2 but then one of the indices of the Ricci tensor are out of place. Any help would be appreciated

Answer 1

You can do it in two ways.

1. The commutation of second derivatives for any vector satisfies: $$ \xi_{\mu;\nu\lambda}-\xi_{\mu;\lambda\nu} = R_{\mu\sigma\lambda\nu}\xi^\sigma. $$ If you know it's a Killing vector, then $\xi_{\mu;\nu} = -\xi_{\nu;\mu}$. Contracting $\mu$ and $\lambda$ so as to go from (a variant of) the Riemann tensor to (a variant of) the Ricci tensor: $$\xi^{\nu;\lambda}_{;\lambda}+R^\nu_\sigma\xi^\sigma = -(\xi_{;\mu}^\mu)^{;\nu}, $$ where the last term is $0$ because the Killing condition results in $\xi^\mu_{;\mu}=0$.

2. With an action $S$ of the type: $$ S = \int\mathrm{d}^4\mathbf{x}\sqrt{-g}\,\mathcal{L}(\psi, \nabla_a\psi, g^{ab}), $$ you can vary both $\delta\psi$ and $\delta(\nabla_a\psi)$ to get the same E-L equations as usual, but with the covariant derivative.
   But I end up needing a factor of $1/2$ on the 'kinetic' part of the lagrangian...