I know you can find geodesic equations with respect to proper time $\tau $ using the variational principle, i.e. using Euler-Lagrange equations $$ \frac{\partial}{\partial x^{\mu}}L-\frac{d}{d\tau}\frac{\partial}{\partial\dot{x}^{\mu}}L=0 \tag{1} $$ for Lagrangian $$ L=g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}.\tag{2} $$

Now can you use this same method to find geodesic equations with respect to coordinate time $t$ using Euler-Lagrange equations for Lagrangian $$ L=g_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}~?\tag{3} $$

More generally do you get equations that gives geodesics for solutions when you use Euler-Lagrange equations for a Lagrangian of the form $$ L=g_{\mu\nu}\frac{dx^{\mu}}{dx^{\sigma}}\frac{dx^{\nu}}{dx^{\sigma}},\tag{4} $$ where $x^{\sigma}$ is one of the coordinates?

  • $\begingroup$ To get the geodesic, you have to take the $\sqrt L$ which is Lorentz invariance $\endgroup$ – Eli May 20 '19 at 12:23

A geodesic can be parameterized by any parameter that increases monotonically along the geodesic. For timelike geodesics, which is what we have in mind when we derive them by minimising $\int d\tau$, we can use a coordinate $t$ as the parameter, provided the corresponding vector field $\partial/\partial t$ is everywhere timelike. This ensures that $t$ increases monotonically along any timelike worldline. The catch is that the geodesic equation comes from extremizing of the integral of $\sqrt{L}$, which doesn't necessarily give the usual Euler-Lagrange equation for $L$.

To see how this works, start with $$ L(\lambda) = g_{ab}\dot x^a \dot x^b $$ where $\dot x^a \equiv dx^a/d\lambda$, where $\lambda$ is any parameter that increases monotonically along timelike worldlines. The quantity $$ \int \sqrt{L(\lambda)}\,d\lambda $$ is reparameterization-invariant (intuitively, the $d\lambda$'s cancel), so it's the same no matter what parameter $\lambda$ we use. In particular, it's the same whether or not $\lambda$ ends up being the worldline's proper time; it doesn't even need to be an affine parameter. In particular, it can be one of the coordinates if that coordinate increases monotonically along all timelike worldlines.

The geodesic condition is $$ 0=\delta \int \sqrt{L(\lambda)}\,d\lambda. $$ Use \begin{align} \delta \int \sqrt{L(\lambda)}\,d\lambda &\propto \int \frac{1}{\sqrt{L}}\left( \frac{\partial L}{\partial x^a}\delta x^a + \frac{\partial L}{\partial \dot x^a}\delta \dot x^a \right)d\lambda \\ &= \int \left( \frac{1}{\sqrt{L}}\frac{\partial L}{\partial x^a} - \frac{d}{d\lambda}\left[\frac{1}{\sqrt{L}} \frac{\partial L}{\partial \dot x^a}\right] \right)\delta x^a\,d\lambda \end{align} to conclude that the geodesic equation is $$ \frac{\partial L}{\partial x^a} - \sqrt{L}\frac{d}{d\lambda}\left[\frac{1}{\sqrt{L}} \frac{\partial L}{\partial \dot x^a}\right] =0. $$ In the special case where the parameter $\lambda$ is set equal (in hindsight) to the wordline's proper time, the equation simplifies because $L=1$ in that case. More generally, for an affine parameter, we have (by definition) $L=$ constant, so the equation again simplifies in the same way, leaving the usual Euler-Lagrange equation. But for a general parameter, such as a timelike coordinate, the equation does not simplify in that way; the derivative with respect to $\lambda$ will act nontrivially on the factor $\sqrt{L}$, leading to an extra term compared to the usual form of the geodesic equation. Despite the extra term, the result is correct. The reparameterization-invariance of $\int \sqrt{L(\lambda)}\,d\lambda$ implies that the resulting equation selects the same set of worldlines (the ones we call geodesics) no matter what monotonic parameterization we used.


Comments to the post (v2):

  1. Note that one cannot use proper time $\tau$ (or arclength) as the independent parameter $\lambda$ before applying the principle of stationary action to find geodesics. This is explained in my Phys.SE answer here, which also explains the connection to the corresponding square root Lagrangian.

  2. For the non-square root Lagrangian, only after the variation is performed, a stationary solution is then affinely parametrized wrt. proper time $\tau$ (if the geodesic is timelike, i.e. if the point particle is massive).


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