Are all critical points of energy geodesics?

Question

Let $\gamma$ be a smooth curve in a Riemannian manifold and consider the arclength functional $L(\gamma) = \int_a^b |\gamma'(t)|\, dt$ and the energy functional $E(\gamma) = \frac{1}{2}\int_a^b |\gamma'(t)|^2\, dt$. A standard Cauchy-Scwharz argument gives $$L(\gamma)^2 \leq 2(b-a) E(\gamma)$$ with equality if and only if $\gamma$ has constant speed. As a consequence, all constant speed critical points of energy are geodesics.

I am wondering: are there any critical points of energy which do not have constant speed? Do they have any interesting geometric properties?

Answer 1

If you use the Euler-Lagrange equations, you will see that any critical point will obey the differential equation at the end of this section: http://en.wikipedia.org/wiki/Geodesic#Riemannian_geometry

If you mess around with it, you should be able to show that $$ \frac{d}{dt} \left(g_{\lambda,\kappa} \frac{dx^\lambda}{dt} \frac{dx^\kappa}{dt} \right) = 0 .$$ You can get this by applying the product rule to the outer derivative, and then substituting in the formula for the Christoffel symbol.

Answer 2

Maybe this is a bit sloppy (though interpreting all symbols in the right way, it certainly isn't), but here we go.

Since for any variation of $\gamma$ with fixed endpoints, we have $$0=\delta E(\gamma) = \int_a^b \langle \dot \gamma, \delta \dot \gamma \rangle = -\int_a^b \langle \ddot \gamma, \delta \gamma \rangle,$$ and since $\delta \gamma$ can be any vector field along $\gamma$ vanishing at the endpoints, it follows that $\ddot \gamma = 0$. So any critical point of the energy functional is necessarily a constant speed geodesic.