# Why is the Lagrangian for space-like geodesics equal to 1?

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?

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 $$s$$.