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:
$$xi^{nu;lambda}_{spacespacespacespacespacespace;lambda}+R^{nu}_{spacespacesigma}xi^{sigma}=0 spacespacespacespacespacespacespacespacespacespacespace(1)$$
They say that the following Lagrangian density reproduces equation (1) by varying the action or using the Euler Lagrange equations:
$$quadmathfrak{L}=xi^{}_{mu;nu}xi^{mu;nu}-frac{1}{2}R_{munu}xi^{mu}xi^{nu}$$
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

SuperCiocia · Accepted Answer

You can do it in two ways.

The commutation of second derivatives for any vector satisfies:
$$ xi_{mu;nulambda}-xi_{mu;lambdanu} = R_{musigmalambdanu}xi^sigma. $$
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:
$$xi^{nu;lambda}_{;lambda}+R^nu_sigmaxi^sigma = -(xi_{;mu}^mu)^{;nu}, $$ where the last term is $0$ because the Killing condition results in $xi^mu_{;mu}=0$.

With an action $S$ of the type:
$$ S = intmathrm{d}^4mathbf{x}sqrt{-g},mathcal{L}(psi, nabla_apsi, g^{ab}), $$ you can vary both $deltapsi$ and $delta(nabla_apsi)$ 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...

Deriving equations from the Lagrangian density

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