Lagrangian in weak field

1. Oct 7, 2012

raopeng

I just read some basic concepts on General Relativity, and this idea pops up: I know we should use variations of metrics for gravitational field in the Lagrangian. But considering the resemblance of gravitational field(weak-field) to electromagnetic field, can we construct a 4-potential similar to that of the electromagnetic field, say $A_{G} = (ψ(gravitational potential),0, 0, 0)$. So the Action for the effect of gravitational field would be: $\int -\frac{m}{c} A_{G}dx^{i}$. Would that be a good approximation for weak field?

2. Oct 7, 2012

haushofer

No. What you wrote down is identically zero. You should consider the weak field limit directly for the point particle action. So, take this action, write the metric as a
perturbation of e.g. Minkowski spacetime, and perform a Taylor expansion on the square root. Taking static gravity and slowly moving particles should then on its turn give newtonian gravity, as in e.g. Eqn 2.10 of arxiv:1206.5176.

In the weak field limit GR becomes Fierz-Pauli theory, massless spin 2. A vector potential would mean that gravity is represented by spin-1. In the Newtonian approx. gravity is effectively reduced to a Gallilean scalar field, which could be interpreted as spin 0 (but then with spin defined wrt to the Galilei group).

Last edited: Oct 7, 2012
3. Oct 7, 2012

raopeng

Eh I was trying to make a parallel comparison to $S_{mf}$, so the integral is actually $\int -\frac{m}{c}ψ d(ct)$ and when v is relatively small the whole expression does degrade into the classical Lagrangian. But it is a very imprudent thought as I only start to scratch the surface of Relativity Theory. Thank you for your reply that I know where my problem is.