# What is varing in the variational principle of GR

1. Jan 10, 2012

### spocchio

Consider the variational principle used to obtain that in the vacuum the Einstein tensor vanish.
So we set the lagrangian density as $L(g,\partial g)=R$
$0 = \delta S =\delta\int{d^4 x \sqrt{-g}L}$

proceeding with the calculus I finally have to vary R such that
$\delta R = \delta{g^{\mu\nu}R_{\mu\nu}}$
but what is $\delta g$??

i think (but not sure) $\delta{g^{\mu\nu}}=g^{\mu\nu}-{g'}^{\mu\nu}$
since g transform as a tensor
$\delta{g^{\mu\nu}}=g^{\mu\nu}-J_{\rho}^{\mu} J_{\sigma}^{\nu} {g}^{\rho\sigma}$

where J is the jacobian of the transformation..am I right?

2. Jan 10, 2012

### Bill_K

Usually the independent variables to be varied are taken as gμν. The equations you need are:

δgμν = - gμσ gντ δgστ

δ√-g = ½√-g gμν δgμν

δR = gμν δRμν - Rμν δgμν

δRμν = ½ gστ(gμν;στ + gστ;μν - gμσ;ντ - gντ;μσ)