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In Schwarzschild spacetime, the Lagrangian can be defined as

$$ L = -\left( 1 - \frac{2M}{r} \right) \dot{t}^2 + \left( 1- \frac{2M}{r} \right)^{-1} \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2\theta \dot{\phi}^2 $$

where the dot denotes a derivative $\frac{\mathrm d}{\mathrm d\lambda}$ for an affine parameter $\lambda$.

From Noether's theorem, one can easily prove that the Lagrangian itself if an integral of motion. One can also prove that $\theta = \text{const} = \pi/2$ without loss of generality. Hence, the quantity $$Q = -\left( 1 - \frac{2M}{r} \right) \dot{t}^2 + \left( 1- \frac{2M}{r} \right)^{-1} \dot{r}^2 + r^2 \dot{\phi}^2$$ is a constant of motion.

In my lecture notes, it is claimed without proof that $Q = 1$ for space-like geodesics. It seems like this is rather obvious from the way it is stated but I can't seem to find any proof of this anywhere in my lecture notes nor from Google (after a little bit of searching). Hence, my question is why is this the case and how can one prove this?

Qmechanic
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alfred
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1 Answers1

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  1. The solutions to the Euler-Lagrange (EL) equations for OP's non-square root Lagrangian $L$ are affinely parametrized geodesics.

  2. Noether's theorem yields that $L$ is a constant (of motion) along a geodesic, but it is not necessarily $1$. For more information, see e.g. my Phys.SE answer here.

  3. The constant is $1$ if one picks the parameter $\lambda$ to be arclength $s$.

Qmechanic
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