# Does the Einstein -Hilbert Lagrangian has a potential term?

1. Nov 7, 2006

If you can derive the Eisntein equations from:

$$L=\int_{V} d^{4}x\sqrt (-g)R$$ but does L has a potential term so we can do Qm with it?..

2. Nov 8, 2006

### dextercioby

Nope, the lagrangian is purely kinetic at first sight. However, perturbative expansion reveals an infinite series in the (self)coupling constant for the graviton, $\kappa$. So you can think of the Pauli-Fierz lagrangian as the purely kinetic term and the rest of the lagrangian (depending on powers of $\kappa$ from 1 to infinity) as the potential one.

Daniel.

Last edited: Nov 8, 2006
3. Nov 8, 2006

### Epicurus

Yes you can to QM with it, Feyman derived the rules for the expansion. The problem is that it is not renormalisable

This does not make sense. The 00 component of the stress tensor will give the energy density, integrted to give the energy. The expansion will give corrections to the interactions, however as said, these are not renormalisable. The first thing one is required to do is construct an interacting theory from gauge symmetry arguements, as these are renormalisable and at present is the only theoretically consistent way of non-empirically including interactions. This has not been done for gravitation.

4. Nov 8, 2006

### dextercioby

Who said anything about the stress tensor of the field? I was merely talking about a series expansion of the $\sqrt{|g|} \ R$ aroud a flat spacetime metric, the usual Minkowski metric.

The cubic term, the quartic term, etc. can all be viewed as self-interaction terms, hence can be considered potential energy, just like the $\frac{\lambda}{4!}\varphi^{4}$ can be considered that way for the scalar field theory.

Daniel.

5. Jan 12, 2012

### bilt7am

Hi, I am not sure, but your Lagrangian would give "free field", where there is no generating source for the gravitational field (you would have "R-gR/2=0") - like for gravitation waves. Actually, any field can be added as "second" term to you Lagrangian (even QM Lagrangian), which would end up on right side of Einstein equations as the "field source".