# Action for a point particle in a curved spacetime

Is this action for a point particle in a curved spacetime correct? $$\mathcal S =-Mc \int ds = -Mc \int_{\xi_0}^{\xi_1}\sqrt{g_{\mu\nu}(x)\frac{dx^\mu(\xi)}{d\xi} \frac{dx^\nu(\xi)}{d\xi}} \ \ d\xi$$

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Yes, but if you're trying to derive the geodesic equation, you lose nothing but headaches by substituting this action with $S = \int ds (g_{ab}{\dot x}^{a}{\dot x}^{b})$, where ${\dot x}^{a} = \frac{\partial x^{a}}{\partial s}$, since a minimum of $\int f(x)$ is also going to be a minimum of $\int (f(x))^{2}$ –  Jerry Schirmer Dec 4 '12 at 21:32

It depends what you mean; that is the action of a test particle in a background gravitational field given by a metric $g_{\mu\nu}$. If you minimize it, you will get the geodesic equation. That is NOT the dynamical action for the gravitational field; your test particle does not change the curvature of the background spacetime. The action for the gravitational field the Einstein-Hilbert one,

$S=\frac{1}{\kappa}\int RdV$

where $R$ is the scalar curvature, $\kappa$ is the coupling constant.

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It is misleading to write $\mbox d^4 V$. That makes it seem like it is integrating over a $4(4)=16$ dimensional manifold. –  Dimensio1n0 Jul 17 '13 at 14:07
You are 100% correct. I will edit that. –  levitopher Jul 17 '13 at 20:29

The end points of your action must be events, therefore must be an d+1 dimensional object. Action must be extremized over all paths that start and end at the given space time points.

A path can be be parametrized by 4 functions of space and time, $x^\mu(\xi)$ of a one parameter object $\xi$. So it would be wrong to label the end points in terms of $\xi$. Instead $\xi$ must be treated as intermediate label to describe paths. Otherwise the functional form the integral is correct.

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Is this action for a point particle in a curved spacetime correct? $$\mathcal S =-Mc \int ds = -Mc \int_{\xi_0}^{\xi_1}\sqrt{g_{\mu\nu}(x)\frac{dx^\mu(\xi)}{d\xi} \frac{dx^\nu(\xi)}{d\xi}} \ \ d\xi$$

Yes, it is.

The second expression/term/whatever comes from trivially integrating the energy $\mathrm{d}t$.

You can also try deriving the geodesic equation from this, as a fun homework exercise.

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## protected by Qmechanic♦Jan 4 '13 at 11:07

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