In my GR lectures on the derivation of geodesic equations via extremal length, my lecturer wrote that the worldline action $S$ of a point particle with mass $m$ is given by $$S=-m\int\sqrt{-g_{\alpha\beta}\dot{x}^\alpha\dot x^\beta}d\tau,\tag{1}$$ where $$\dot x(\tau)=\frac{dx(\tau)}{d\tau}.\tag{2}$$

Where does this equation come from? I read from wikipedia that the worldline action of a relativistic point particle is $$S=-mc^2\int d\tau,\tag{3}$$ where $\tau$ is the proper time. Are theses two equations related in any way?

  • 1
    $\begingroup$ Do you know how the proper time relates to the metric? (I.e. write out the line element and compare the two integral expressions above). Also note the indices on your $x$ terms should be raised not lowered. $\endgroup$
    – Eletie
    Mar 2, 2021 at 11:54

1 Answer 1


Yes, the main point is that the arc length$^1$ $$ds~=~cd\tau~=~\sqrt{-g_{\mu\nu}(x) dx^{\mu}dx^{\nu}}\tag{A}$$ in spacetime is the speed of light $c$ times the proper time $d\tau$, so that $$ c\dot{\tau}~=~\dot{s}~=~\sqrt{-g_{\mu\nu}(x) \dot{x}^{\mu}\dot{x}^{\nu}},\tag{B}$$ where dot mean differentiation wrt. the worldline (WL) parameter $\lambda$ (which is not necessarily the proper time $\tau$). The action is therefore $$ S~=~ mc^2\Delta \tau~=~ mc\Delta s~=~mc\int_{\lambda_i}^{\lambda_f}\!d\lambda ~\sqrt{-g_{\mu\nu}(x) \dot{x}^{\mu}\dot{x}^{\nu}}, \tag{C}$$ cf e.g. this Phys.SE post and links therein.


$^1$ In this answer, we use the $(-,+,+,+)$ Minkowski sign convention.


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