Let $(M, g)$ be a Riemannian manifold. The length of an admissible curve $\gamma : [a, b] \to M$ is defined by $L(γ) = \int_a^b ∥\dot \gamma (t)∥dt$. 1)Compute the Euler–Langrange equations for the variation of the length functional. 2)I want to show that these equations are the geodesic equation if and only if the curve is parametrized by constant speed $∥\dot\gamma (t)∥$
The E-L equation:
My attempt:
- I just plugged $\scr L=\|\dot \gamma (t)\|$ in the E-L equation and got this:
$\frac{d}{dt}(\frac{1}{2\sqrt{g_{ij}\dot \gamma^i\dot \gamma^j}}2g_{ki}\dot \gamma^i\dot \gamma^j)-\partial_k g_{ij}\dot \gamma^i\dot \gamma^j=0\tag 1$ I didn't wanted to do the time derivative as the square root term makes it a pretty nasty equation Am I on the right track here?
- I think the Geodesic equation is (3.24) here:
I am not sure if this is indeed the equation that I should be obtaining, as as quoted in the picture (3.24 ) is the geodesic equation for the energy (3.23) and in my case I am working with the length, then the geodesic equation should be different, shouldn't it?
I am not sure how to do this part. Assuming 3.24 is indeed the geodesic equation that I want to obtain. There are two parts , since it is an iff:
i) If $∥\dot\gamma (t)∥=const \implies (*)$ symplifies to $(3.24)$
ii) If $(*)$ is the geodesic equation, that is $(*)=(3.24)$, (not sure how to use this.. does it means one equation is the other times a constant?) then $∥\dot\gamma (t)∥=$const.
How do I complete this?