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A free particle Lagrangian in a 3D curvilinear coordinate system can be written as an inner product with the metric $g$: $$ L = \frac{1}{2}m\sum_{i,j=1}^3v^ig_{ij}v^j. $$

This equation was taught to me in the context of curved coordinate systems that can be transformed to be flat, for example polar coordinates. However, it seems awfully tempting to take a "fundamentally curved" metric from GR, like the Schwarzschild metric, to use in this equation. I know that even in flat space, this will give me the wrong answer if the velocity gets near $c$. However, so long as $|v| \ll c$, will there be any problems in treating it this way? What must be true about the coordinate system and the dynamics of the system in order for this scheme to work?

Qmechanic
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Retracted
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1 Answers1

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Your proposed Lagrangian will sadly give wrong equations no matter the limit. In the Schwarzschild weak field limit, it is

$$L = \frac12 m (\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2\theta\, \dot{\varphi}^2) + \frac{GmM}{r} \frac{\dot{r}^2}{c^2}$$

instead of

$$ L = \frac12 m (\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2\theta\, \dot{\varphi}^2) + \frac{GmM}{r}.$$

You can see that it's essential that the relativistic Lagrangian contain something involving the time component of the four-velocity, which will, in the non-relativistic limit, turn into the velocity independent potential.

The conceptual point behind this is that the "time" part in "curvature of spacetime" is essential. It's not just a relativistic correction; it's where the whole physics lies. Newtonian gravitation is not just GR restricted to just space curvature.

Javier
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