Einstein-Hilbert action.

[Klaus_Hoffmann]

if we have the Einstein Hilbert action

I= \int_{V} dV (-g)^{1/2}R or I= \int_{V} dV \mathcal L (g_{ab}, \Gamma_{kl}^{i}

then my question is if we can obtain Einstein equations by varying the metric and Christofell symbols independently i mean you get Einstein Field equations by

\frac{\delta I}{\delta g_{ab} =0 \frac{\delta I}{\delta \Gamma_{kl}^{i} =0

i mean , you consider the metric g_{ab} and \Gamma_{kl}^{i} as independent variables for your theory.

 

[nrqed]

if we have the Einstein Hilbert action

I= \int_{V} dV (-g)^{1/2}R or I= \int_{V} dV \mathcal L (g_{ab}, \Gamma_{kl}^{i}

then my question is if we can obtain Einstein equations by varying the metric and Christofell symbols independently i mean you get Einstein Field equations by

\frac{\delta I}{\delta g_{ab} =0 \frac{\delta I}{\delta \Gamma_{kl}^{i} =0

i mean , you consider the metric g_{ab} and \Gamma_{kl}^{i} as independent variables for your theory.

Yes, this is done in the so-called Palatini fromulation of GR. Google it to find more information. (if the action is simply the EH action, the Palatini formulation gives the same result as the usual metric approach. If the action is generalized to included other terms (liek a 1/R term as is done in some modedl of the so-called f(R) gravity, the Palatini formulation gives a different result than th emetric formulation.)

Patrick

 

[kharranger]

Yes, if you vary the metric+connection EH action with respect to the connection, you find the algebraic equation: connection=levi-civita connection. Thus you can insert these equations of motion inside the action, recovering the metric-only EH action, and the two are classically equivalent.