The action of a particle point is [tex]\int ds = \int d\tau

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The action of a point particle is defined by the integral \(\int ds = \int d\tau \sqrt{g_{uv} \dot{x}^u \dot{x}^v}\), which is distinct from the Einstein-Hilbert action \(\int d^x R \sqrt{g}\). The Einstein-Hilbert action pertains solely to the metric, requiring additional terms to account for matter. While the point particle action can be combined with the Hilbert action in theory, practical solutions to the equations of motion are non-existent. The point particle action is primarily applicable in a test particle regime, where the particle's influence on the metric is negligible.

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The action of a particle point is \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?
 
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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).
 


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.
 

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