Simplifying the functional derivative δg_μν/δg^κλ

[unchained1978]

Can anyone help me simplify the expression $\frac{\delta g_{\mu\nu}}{\delta g^{\kappa\lambda}}$? I haven't seen a term like this before and I don't know how to proceed. It seems like it might be the product of some kronecker delta's but I'm not sure.

 

[tom.stoer]

You get a pair of Kronecker deltas

 

[Bill_K]

Write the Kronecker delta as δ^μ_ν = g^μα g_να. take the variation: 0 = δg^μα g_να + g^μα δg_να. Now multiply by g_μβ to solve for δg_να.

 

[w4k4b4lool4]

Can anyone help me simplify the expression $\frac{\delta g_{\mu\nu}}{\delta g^{\kappa\lambda}}$? I haven't seen a term like this before and I don't know how to proceed. It seems like it might be the product of some kronecker delta's but I'm not sure.

This is a functional derivative, defined as follows:
$$
\frac{\delta F[g(x)]}{\delta g(y)} \equiv \lim_{\epsilon\rightarrow 0}\frac{F[g(x)+\epsilon \delta(x-y)]-F[g(x)]}{\epsilon}
$$
In your case this leads to,
$$
\frac{\delta g_{\mu\nu}(x)}{\delta g_{\rho\sigma}(y)} =\frac{1}{2}\Big(\delta_{\mu}^{\rho}\delta_{\nu}^{\sigma}+\delta_{\nu}^{\rho}\delta_{\mu}^{\sigma} \Big) \delta^D(x-y),
$$
with D the dimensionality of spacetime. Notice that both sides of the equation have the same symmetries on the indices.