Suppose that you want to find the geodesics of some metric $ds^2$. In order to do so, one must solve the geodesic equation $$\frac{d^2x^\mu}{d \tau^2}+\Gamma^\mu_{\rho\sigma}\frac{dx^\rho}{d \tau}\frac{dx^\sigma}{d \tau}=0$$. To solve this we can apply the EL equations $$\frac{\partial L}{\partial x^\mu}-\frac{d}{d \tau}\left(\frac{\partial L}{\partial x'^\mu}\right)=0$$ where $\forall x'=\frac{dx}{d \tau}$. My issue arises when trying to find the Lagrangian for an arbitrary metric. I have seen the Lagrangian for some metric $ds^2$ defined as $$L=\frac{1}{2}g_{\mu\nu}x'^{\mu}x'^{\nu}$$ $$L=g_{\mu\nu}x'^{\mu}x'^{\nu}$$ $$L=(-g_{\mu\nu}x'^{\mu}x'^{\nu})^{1/2}$$, etc. My question is; for any arbitrary metric $ds^2$ is there a general form that all Lagrangians will take? Or is there different formulas for different metrics; if so how can I deduce what formula to work with ?


All of these Lagrangians are classically equivalent if we take the geodesics to be affinely parametrized (see the edit below), meaning they lead to the exact same equations of motion. In other words, all of them lead to the geodesic equation. The equivalence between the last two expressions you gave is shown, for example, on Bob Wald's General Relativity, pp. 43–45. The equivalence between the first and second Lagrangian can be seen by noticing that if $\delta S = 0$ (where $S$ is the action), then $\delta(\alpha S) = 0$ for any constant $\alpha$. Think that multiplying a function by a constant will not change the points at which it achieves its extrema, even though it changes the values of those extrema. For example, both $g(x) = 2f(x) = 2 x^2 - 2$ and $f(x) = x^2 - 1$ have a minimum at $x = 0$, though the value $g(0) = 2$ differs from $f(0) = 1$. This is briefly discussed on the first chapter of Landau and Lifshitz's Mechanics (specially sections 2–4, if I recall correctly).

Edit: as pointed out by Bence Racskó in the comments, these Lagrangians need not be equivalent when one allows for non-affine parametrizations, for example. If you want to consider arbitrary parametrizations, the third option is the one to go with, and it will not always be equivalent to the first two. In Wald's discussion of these Lagrangians, he points out where one assumes the parametrization to be affine when proving them to be equivalent and why the third Lagrangian is the correct one.

Hence, you can use any of these three Lagrangians you wrote down. The first two expressions are often easier to do computations with, so we'll usually prefer them. I'm particularly fond of the first one, since it already leads to the standard $\frac{1}{2}$ factors we often use when writing the kinetic term for non-relativistic particles — we write $\frac{m \dot{x}^2}{2}$, but we could as well multiply the entire Lagrangian by $2$ (potential included) and use $m\dot{x}^2$.

In summary, the three formulas you have quoted are all correct and will lead to the correct geodesic equation. More details can be found on the pages I mentioned of Wald's book.

  • $\begingroup$ Thanks for the great explanation! $\endgroup$
    – aygx
    Nov 23 '21 at 2:51
  • $\begingroup$ @aygx You're welcome! If it fits your expectations, please consider accepting it $\endgroup$ Nov 23 '21 at 2:54
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    $\begingroup$ Just a quick nitpicking comment that the first/second and the third Lagrangians are equivalent only if the space of the variational problem is the set of all paths (as in curves modulo parametrizations). If one varies in the set of (appropriate) curves, then the Lagrangians define different systems, for example non-affinely parametrized geodesics ("pregeodesics") extremize the third Lagrangian, but not the first two. $\endgroup$ Nov 25 '21 at 11:35
  • $\begingroup$ @BenceRacskó Oops! Thanks for pointing that out! I've updated my answer accordingly $\endgroup$ Nov 25 '21 at 18:59

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