- #1
spocchio
- 20
- 0
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
and asks for the condition
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?
So we set the lagrangian density as L(g,\partial g)=R
and asks for the condition
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?