To find the geodesics of a space(time), what we need to do is extremizing the functional ##\displaystyle \int_{\lambda_1}^{\lambda_2}\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}} d\lambda ##. But sometimes the presence of the square root makes the equation of motion too complicated and it would be good if we could use the functional ## \displaystyle\int_{\lambda_1}^{\lambda_2}g_{\mu \nu} \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda} d\lambda ## instead.(adsbygoogle = window.adsbygoogle || []).push({});

It turns out that if we use an affine parameter (s) to parameterize the geodesic, then the two Lagrangians become equivalent:

## \frac{d}{ds}\frac{\partial L^2}{\partial \dot x^\mu}=\frac{\partial L^2}{\partial x^\mu} \rightarrow \frac{d}{ds}\left(2L\frac{\partial L}{\partial \dot x^\mu}\right)=2L\frac{\partial L}{\partial x^\mu} \rightarrow \frac{dL}{ds}\frac{\partial L}{\partial \dot x^\mu}+L \frac {d}{ds}\frac{\partial L}{\partial \dot x^\mu}=L\frac{\partial L}{\partial x^\mu} \xrightarrow{\frac{dL}{ds}=0} \frac {d}{ds}\frac{\partial L}{\partial \dot x^\mu}=\frac{\partial L}{\partial x^\mu}##.

My problem is, ##\frac{dL}{ds}=0## means that the Lagrangian is a constant and its possible to use a parametrization in which ## L=1 ##. In this case, what does it mean to extremize the functional ## \int L^2 ds ##? Also a constant Lagrangian gives 0=0 as the EL equations of motion. What's wrong here?

Thanks

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# A Finding geodesics using the squared Lagrangian

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