Given a Lagrangian $L:TM \rightarrow \mathbb{R}$ defined on a tangent bundle, Hamilton's principle states that a curve $\gamma:[a,b] \rightarrow M$ is a stationary point of the action functional $S[\gamma]:=\int_a^b L(\gamma(t),\gamma'(t)) dt$ among variations with fixed endpoints iff the Euler-Lagrange equations $\frac{d}{dt}\left(\frac{\partial L}{\partial q'} \right) – \frac{\partial L}{\partial q} = 0$ are satisfied.
Now we consider the Lagrangian given by $L(q,q')= \sqrt{g(q',q')}$, that is the length of a tangent vector. Then the action functional becomes arc length. The Euler-Lagrange equations then become the geodesic equations.
My question is the following:
Is a stationary point of the length functional $S_l[\gamma]:=\int_a^b\sqrt{(g(\gamma'(t), \gamma'(t)))} dt$ automatically a local minimum for this length functional $S_l$ among all variations with fixed endpoints?
In other words, if $\gamma$ is a stationary point of $S_l$ (equivalently, if it satisfies the Euler-Lagrange equation, that is in this case the geodesic equations) is it then true that for any variation $f:(-\epsilon, \epsilon) \times [a,b] \rightarrow M$ of $\gamma$ with fixed endpoints (that is $f(0,t)=\gamma(t)$ for all $t \in [a,b]$ and $f(s,a)=\gamma(a)$ and $f(s,b) = \gamma(b)$ for all $s \in ( -\epsilon, \epsilon)$) we have that for all small enough $s$ $$S_l(f(s, \cdot))= \int_a^b \sqrt{g \left(\frac{\partial f}{\partial t}(s,t)),\frac{\partial f}{\partial t}(s,t) \right)} dt \geq S_l(f(0, \cdot))=S_l[\gamma] \ \ \ ?$$
(This is different from the fact that a geodesic is not necessarily a global minimum of the length functional among all admissible path connecting two given points. And it is different from the property of geodesics to be length-minimizing within a geodesic ball.)
Best Answer
No, consider e.g. a great circle $\Gamma$ on a 2-sphere $\mathbb{S}^2$ that starts and ends at the same fixed point $p\in\mathbb{S}^2$. Now $\Gamma$ satisfies the geodesic equation and is hence a stationary curve. Next deform infinitesimally $\Gamma$ to a smaller circle $\gamma$ that also starts and ends at the same fixed point $p$.