As I hinted a couple of posts back, I am interested in discussing the application of the calculus of variations to differential geometry. So fix a Riemannian manifold with metric denoted either by ${g}$ or ${\left \langle \cdot, \cdot \right \rangle}$, and let ${c: I \rightarrow M}$ be a smooth path in ${M}$. Define the energy as $\displaystyle E(c) := \frac{1}{2} \int g(c',c').$

The energy integral is closely related to the length function, though it is easier to deal with. Now we are interested in studying a variation ${H}$ of the curve ${c}$ and how the energy integral behaves with respect to ${H}$. Recall that ${H: I \times(-\epsilon,\epsilon) \rightarrow M}$ is a smooth map with ${H(t,0)=c(t), H(a,u)=c(a), H(b,u)=c(b)}$ for all ${t,u}$. The last two conditions mean that ${H}$ is a family of curves that keep the endpoints fixed.

Define ${H_u}$ as the curve ${J(\cdot, u)}$, and consider the function of ${u}$, $\displaystyle E(u) := E(H_u) = \frac{1}{2} \int_I g\left( \frac{\partial}{\partial t} H(t,u) , \frac{\partial}{\partial t} H(t,u) \right).$

Ultimately, we are interested in curves that minimize the energy integral, at least locally. This means that for any variation ${H}$ as above, ${E(u)}$ should have a local minimum at ${u=0}$. So we will compute ${\frac{d}{du} H(u)|_{u=0}}$, and, eventually, the second derivative too. The evaluations involve nothing more than a rehash of many standard tricks we have repeated already.

If we differentiate under the integral sign, legal because of all the smoothness, we get $\displaystyle \frac{d}{du} E(u) = \frac{1}{2} \int_I \frac{\partial}{\partial u} g\left( \frac{\partial}{\partial t} H(t,u) , \frac{\partial}{\partial t} H(t,u) \right)$

By the formula for differentiating inner products and the symmetry of the Levi-Civita connection, this becomes $\displaystyle \int_I g\left( \frac{D}{du} \frac{\partial}{\partial t} H(t,u), \frac{\partial}{\partial t} H(t,u) \right) = \int_I g\left( \frac{D}{dt} \frac{\partial}{\partial u} H(t,u), \frac{\partial}{\partial t} H(t,u) \right)$

If we let ${V(t)}$ be the variation vector field and set ${u=0}$, we find that $\displaystyle \frac{d}{du} E(u) |_{u=0} = \int_I g\left( \frac{D}{dt} V(t), \dot{c}(t) \right),$

which using similar identities becomes $\displaystyle \int_I \frac{d}{d t} g\left( V(t), \dot{c}(t) \right) - \int_I g\left( V(t), \ddot{c}(t) \right)$

(Note that ${\ddot{c}(t) := \frac{D}{dt} \dot{c}(t)}$ by definition.) The left integral is ${g(V(b), \dot{c}(b)) - g(V(a), \dot{c}(a)) = 0}$, since ${V(b)=V(a)=0}$ for a variation keeping endpoints fixed. Hence we have the following:

Theorem 1 (First Variation Formula)

For variations fixing the endpoints $\displaystyle \boxed{ \frac{d}{du}E(u)|_{u=0} = -\int_I g\left( V(t), \ddot{c}(t) \right) .}$

In particular, since ${V(t)}$ can be really chosen arbitrary with ${V(a)=V(b)=0}$, we see that a curve locally minimizes the energy ${E}$ only if it is a geodesic. There is a reason for this. First, by Cauchy-Schwarz, we have $\displaystyle l(c) \leq \sqrt{E(c)} \sqrt{length(I)}.$

There is equality precisely when ${c}$ moves at constant speed, i.e. ${|c'|}$ is constant. In particular, $\displaystyle \left(\frac{l(c)}{\sqrt{length(I)}}\right)^2 \leq E(c)$

with equality holding for a geodesic. If ${c}$ is not a geodesic, we can always find a shorter geodesic path between ${c(a),c(b)}$ that necessarily makes ${E}$ smaller by the above inequality (and equality).