# Deriving equations from the Lagrangian density

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

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...
• Some examples for GR specifically are shown here, but again I can't get your equation - I would need a factor of $1/2$ on the kinetic part of the Lagrangian... Oct 15, 2020 at 17:36
• Sorry for the super late response, but I also need a factor of $\frac{1}{2}$ in the kinetic term. Dec 6, 2020 at 20:26