Physics:Palatini identity

From HandWiki
Short description: Variation of the Ricci tensor with respect to the metric.

In general relativity and tensor calculus, the Palatini identity is

[math]\displaystyle{ \delta R_{\sigma\nu} = \nabla_\rho \delta \Gamma^\rho_{\nu\sigma} - \nabla_\nu \delta \Gamma^\rho_{\rho\sigma}, }[/math]

where [math]\displaystyle{ \delta \Gamma^\rho_{\nu\sigma} }[/math] denotes the variation of Christoffel symbols and [math]\displaystyle{ \nabla_\rho }[/math] indicates covariant differentiation.[1]

The "same" identity holds for the Lie derivative [math]\displaystyle{ \mathcal{L}_{\xi} R_{\sigma\nu} }[/math]. In fact, one has

[math]\displaystyle{ \mathcal{L}_{\xi} R_{\sigma\nu} = \nabla_\rho (\mathcal{L}_{\xi} \Gamma^\rho_{\nu\sigma}) - \nabla_\nu (\mathcal{L}_{\xi} \Gamma^\rho_{\rho\sigma}), }[/math]

where [math]\displaystyle{ \xi = \xi^{\rho}\partial_{\rho} }[/math] denotes any vector field on the spacetime manifold [math]\displaystyle{ M }[/math].

Proof

The Riemann curvature tensor is defined in terms of the Levi-Civita connection [math]\displaystyle{ \Gamma^\lambda_{\mu\nu} }[/math] as

[math]\displaystyle{ {R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda} \Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma} }[/math].

Its variation is

[math]\displaystyle{ \delta{R^\rho}_{\sigma\mu\nu} = \partial_\mu \delta\Gamma^\rho_{\nu\sigma} - \partial_\nu \delta\Gamma^\rho_{\mu\sigma} + \delta\Gamma^\rho_{\mu\lambda} \Gamma^\lambda_{\nu\sigma} + \Gamma^\rho_{\mu\lambda} \delta\Gamma^\lambda_{\nu\sigma} - \delta\Gamma^\rho_{\nu\lambda} \Gamma^\lambda_{\mu\sigma} - \Gamma^\rho_{\nu\lambda} \delta\Gamma^\lambda_{\mu\sigma} }[/math].

While the connection [math]\displaystyle{ \Gamma^\rho_{\nu\sigma} }[/math] is not a tensor, the difference [math]\displaystyle{ \delta\Gamma^\rho_{\nu\sigma} }[/math] between two connections is, so we can take its covariant derivative

[math]\displaystyle{ \nabla_\mu \delta \Gamma^\rho_{\nu\sigma} = \partial_\mu \delta \Gamma^\rho_{\nu\sigma} + \Gamma^\rho_{\mu\lambda} \delta \Gamma^\lambda_{\nu\sigma} - \Gamma^\lambda_{\mu\nu} \delta \Gamma^\rho_{\lambda\sigma} - \Gamma^\lambda_{\mu\sigma} \delta \Gamma^\rho_{\nu\lambda} }[/math].

Solving this equation for [math]\displaystyle{ \partial_\mu \delta \Gamma^\rho_{\nu\sigma} }[/math] and substituting the result in [math]\displaystyle{ \delta{R^\rho}_{\sigma\mu\nu} }[/math], all the [math]\displaystyle{ \Gamma \delta \Gamma }[/math]-like terms cancel, leaving only

[math]\displaystyle{ \delta{R^\rho}_{\sigma\mu\nu} = \nabla_\mu \delta\Gamma^\rho_{\nu\sigma} - \nabla_\nu \delta\Gamma^\rho_{\mu\sigma} }[/math].

Finally, the variation of the Ricci curvature tensor follows by contracting two indices, proving the identity

[math]\displaystyle{ \delta R_{\sigma\nu} = \delta {R^\rho}_{\sigma\rho\nu} = \nabla_\rho \delta \Gamma^\rho_{\nu\sigma} - \nabla_\nu \delta \Gamma^\rho_{\rho\sigma} }[/math].

See also

Notes

References