Schwarzschild metric circular orbits and kepler's 3rd law

I have been looking at the Schwarzschild metric presented to me as the following within lectures:

$$ds^2=-\frac{\textrm{d}r^2}{1+\frac{\gamma}{r}}-r^2\textrm{d}\theta^2-r^2\sin^2\theta\textrm{d}\phi^2+c^2\left(1+\frac{\gamma}{r}\right)\textrm{d}t^2,$$

where $$\gamma=-\frac{2GM}{c^2}$$.

For circular motion the radius can be taken as constant and $$\theta=\frac{\pi}{2}$$ can be set without a loss of generality.

From this metric and taking into consideration the constant variables,the Lagrangian is

$$L= -\frac{\dot{r}^2}{1+\gamma/r}-r^2\dot{\phi}^2+c^2\left(1+\gamma/r\right)\dot{t}^2.$$

The Euler-Lagrange equations are

$$\frac{\textrm{d}}{\textrm{d}s}\left[\frac{\partial L}{\partial \dot{x}^{\mu}}\right]-\frac{\partial L}{\partial x^{\mu}}=0.$$

My question arises here, up until this point whenever a variable has been given as a constant, in the corresponding part of the metric, I have been able to effectively eliminate this part of the metric as the derivative of a constant is equal to 0. From inspecting the Lagrangian I can see that this has been done in the case of $$\theta$$, but even though r has been stated to be a constant it is kept in the statement of the Lagrangian, why is this?

Through further manipulations considering the case of the Lagrangian for $$\mu=1$$, it is possible to get the following

$$r\dot{\phi}^2=\frac{GM}{r^2}\dot{t}^2,$$

whereby a statement of Kepler's law is obtained

$$\Omega^2=\left(\frac{\textrm{d}\phi}{\textrm{d}t}\right)^2=\frac{GM}{r^3}.$$

I am having trouble understanding how this is valid given that it seems to rely on $$\dot{r}$$ being kept in the Lagrangian even though $$r$$ is stated as a constant and therefore $$\dot{r}$$ should be 0.

• +1; my interest in this question is especially to effect another appraisal of answers such as this: PSE/a/174711. Apr 21, 2015 at 21:32
• I don't know if the OP is still around, but what is the notation $\mu$?
– user4552
Feb 4, 2019 at 22:22
• It's been awhile since I posted this question and I can't remember off the top of my head but I am happy to check my notes when I get home. Although re-reading the question, it looks like it just represents the order of the derivative? Feb 5, 2019 at 6:20

If we get rid of the $$\dot{r}^2$$ term in the Lagrangian and apply the Euler-Lagrange equations, we get $$\dot{\phi}=\text{const.}$$ and $$\dot{t}=\text{const.}$$ This is certainly true for a circular orbit, but as the OP notes, it doesn't give the desired connection between $$\dot{\phi}$$ and $$\dot{t}$$.

If we could obtain the correct dynamics just by deleting the $$\dot{r}^2$$ term in the Lagrangian before variation, then we would be finding that the spherical shell at fixed $$r$$ was its own little universe, and that once we were inserted into that little universe, like Mr. A. Square from Flatland, we would never be able to detect the fact that the $$r$$ dimension existed. But this is clearly false, because by blasting a rocket engine in a tangential direction, we can get out of the spherical sub-universe.

As a simpler example from nonrelativistic mechanics, I don't think you can obtain the correct dynamics for a free particle in the plane if you adopt polar coordinates and then throw out the $$\dot{r}^2$$ term in the Lagrangian.

Lagrangian mechanics works because the trajectories are stationary with respect to small variations. This example shows that you do need to have these small variations available to you in some neighborhood of the correct trajectory, otherwise you can't get the dynamics. The variation has to happen before you eliminate the unwanted degree of freedom.

This was a nice question. Thanks, Aesir, for the opportunity to think about it.

• This seems wrong. A test particle doesn't radiate because its mass is infinitesimal. Circular orbits are stable down to a given $r$ (of the innermost stable circular orbit).