# The action of a particle point is $$\int ds = \int d\tau 1. Jun 7, 2010 ### alejandrito29 The action of a particle point is [tex]\int ds = \int d\tau \sqrt{g_{uv} \dot{x}^u \dot{x}^v}$$ (eq1)

the action of hilbert is $$\int d^x R \sqrt{g}$$

eq1 is generalization of einstein hilbert action????

2. Jun 7, 2010

### Stingray

Re: action

No, those two actions are unrelated. The Einstein-Hilbert action is effectively the action only for the metric. Extra terms must be added in the presence of matter (if present).

The action you gave for a point particle is one possibility. At least formally, GR with a point particle could be formulated using the sum of your two actions. In practice, though, there are no solutions to the resulting equations of motion. The point particle action is useful mainly in a test particle regime (where the particle's own effects on the metric are ignored).

3. Jun 8, 2010

### haushofer

Re: action

You have to be carefull what your action describes :)

The Hilbert action describes the gravitational field. But it doesn't describe how particles move in it. The same situation you have with electromagnetic fields; the action of the EM-field alone is given by the usual F^2 term, but if you want to describe how charged particles move in this field, you have to describe coupling.

Your first action describes the idea that particles follow geodesics in spacetime. Notice that the metric appears in it, and the solution of this metric is given by your second action: the Hilbert action.