# Classical Limit of Schwarzschild Metric

The orbit of a test particle orbiting a black hole can be described by the Lagrangian $$\mathcal{L} = -\frac{1}{2}\left(-\left(1-\frac{2 G m}{c^2 r}\right) \dot{t}^2 + \left(1-\frac{2 G m}{c^2 r}\right)^{-1}\dot{r}^2 + r^2 \dot{\varphi^2}\right),$$ where the dot denotes the derivative with aspect to a parameter $s$, and $m$ is the mass of the black hole.

In the classical limit $c \rightarrow \infty$, so that the terms $\propto G$ vanish. So if the Equations of Motion, from Eulers Equations, are written down, which also show that $t = s$, there are no terms containing $G$, since they all vanish in the considered limit.

But since we are talking about the classical limit, and about a test particle, the Lagrangian is of the form $$\tilde{\mathcal{L}} = \frac{m_{Testparticle}v^2}{2} + \frac{G m m_{Testparticle}}{r},$$ so there are terms containing $G$ and therefor the potential energy.

To me, this seems to be contradictory.

I see a couple problems here. First off, there's a mistake in your Lagrangian. If you're going to use non-geometrized units, you should probably be consistent about it. $$\mathcal{L} = -\frac{1}{2}\left[-\left(1-\frac{2 G m}{c^2 r}\right) c^2 \dot{t}^2 + \left(1-\frac{2 G m}{c^2 r}\right)^{-1}\dot{r}^2 + r^2 \dot{\varphi^2}\right].$$ Note the extra $c$ in the time term.
Now, the $c \to \infty$ limit looks dubious. This should be your clue that it's not such a meaningful limit; it's more of a hand-wavy mnemonic. In fact, the Newtonian limit is $v/c \to 0$, which is different as you've just discovered. If you expand in powers of $v/c$, this is called a "post-Newtonian" expansion.
In fact, in your version you've effectively taken the limit $2Gm/c^2 r \to 0$, which is the classical limit of flat spacetime. What you've done is taken the lowest-order term in a "post-Minkowski" expansion. You've simply found that removing curvature from GR removes gravity. If you want to find Keplerian gravity, just do the expansion, and keep another term around. Of course, you always need to be careful when interchanging the orders of limits (which includes taking derivatives here).