Euler-Lagrange equation of energy of length function on Riemann manifold

Question

$(M,g)$ is a Riemann manifold. $\gamma:[0,1]\rightarrow M$ is a curve.The length of $~\gamma $ is $$ L(\gamma)=\int^1_0 ||\dot\gamma (t)||_g ~dt $$ The energy is $$ E(\gamma)=\frac{1}{2}\int^1_0 ||\dot\gamma (t)||^2_g ~dt $$ Now ,how to compute the Euler-Lagrange equation of energy of geodesics? In fact, I confuse with it.If I compute is as below $$ \frac{d}{ds}|_{s=0}E(\gamma+s\xi)=0 $$ What is the mean of $\gamma+s\xi$ ? If $\xi$ is a curve, what is the adding of two curve ?

Answer 1

You need to look at one-parameter families $$V(s,t):=(-\varepsilon, \varepsilon)\times [0,r]\rightarrow M$$ of curves, with $V(0,.)=c(.)$, say, being the solution the existence of which you assume.

In case you need to write this down explicilty: if $M$ is a submanifold of some Euclidean space you can just add a small vectorfield which is orthogonal to the curve, (i.e. you look at something like $c(t) + s V(t)$ with $V\in T_{c(t)}M$) and project it orthogonally back to $TM$. In the general case you need other means like the so called exponential map. In all cases the variation $V$ needs to cover all possible directions in which the curve may be varied, i.e. $X(t)=\frac{\partial V}{\partial s}V(.,t)|_{s=0}$ is some arbitrary vector field along $c$ (usually orthogonal to $c$ can be assumed). The case of fixed endpoints is covered by requiring $X(0)=0, X(r)=0$.