Definition of the Energy of a curve

Question

The energy of a curve $c: I \to S$ assuming S is a regular surface with a Riemannian metric $g$ is defined as :

$$ E[c] = \frac{1}{2} \int_I g_{c(t)}(\dot c(t),\dot c(t))\mathsf{dt} $$

This is quite similar to the definition of the length in terms of the defining functions i.e.:

$$L[c] = \int_I \sqrt{g_{c(t)} (\dot c(t), \dot c(t) )}\mathsf{dt}$$ I proved that the former is not an invariant under reparametrisation while the latter is. And moreover a parametrisation proportional to arc length is desirable to minimise energy. However I wish for a more physical connotation to this result.

Also rather vaguely, why is the energy functional defined thus? The definition of length is motivated by partitions and taking limits and gives rise to the distance metric as the infimum of the lengths of all possible curves connecting two points. Would it be feasible to incorporate the energy here?

Answer 1

To perhaps lend some physical insight lets examine this problem in a simpler context,

Physically the kinetic energy of a free particle is defined to be, $E = \frac{1}{2}m v^2$, where $v$ is the magnitude of the velocity vector $\vec{v} = \frac{d\vec{x}}{dt}$.

Thinking of this in $\mathbb{R}^3$ with the Euclidean metric we can say the following.

The average energy per unit time of a path is given by,

$$ \langle E \rangle = \frac{1}{2}m \int_C v^2 dt$$

The total length of a path is given by,

$$ L = \int_C v dt. $$

Physically, a reparametrization of the curve means changing the time at which you arrive at different points on the curve, i.e., $\vec{x}(t) \rightarrow \vec{x}(t')$. So we are now perhaps spending more time a some segments than we did before. This velocity distribution changes and since the average velocity ($L$) can't change we aren't surprised that the variance ($E$) does.

In this specific context the average energy per unit time is proportional to something called "The Action" in physics. This is usually denoted with an $S$ and it depends on the path taken in space between two points.

$$ S[x(t)] = \int_{t_1}^{t_2} \mathcal{L} dt = \int_{t_1}^{t_2} [\text{Kinetic Energy} - \text{Potential Energy}] dt $$

Hamilton's principle in classical mechanics states that for fixed end points ($\vec{x}_1,\vec{x}_2$) the path that minimizes the action is the one which is consistent with newtons laws of motion. In this case that means a free particle traveling in a straight line at a constant velocity (there are no forces acting on the particle in this case). For such a trajectory the distance traveled is proportional to the time spent traveling. Which would require you to parametrize in terms of arc length to minimize the functional.

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For more examples you could look to general relativity (which will have $g \neq 1 $). In that context your $E$ functional is related to the proper time. It is a principle in GR that the true trajectory of a particle is always the one which maximizes the total proper time.

Answer 2

Here is an explanation why people define energy in that way: For the length of a curve, we have $$L[c] = \int_I \sqrt{g_{c(t)} (\dot c(t), \dot c(t) )}\mathsf{dt}$$ as you said. However, dealing with the square root is difficult. For example, $\sqrt{t}$ is not $C^\infty$. So we would rather dealing with square of it, that is, the energy $$E[c] = \frac{1}{2} \int_I g_{c(t)}(\dot c(t),\dot c(t))\mathsf{dt}.$$ I think this is similar to the question of why people define variance as $\sum_{i=1}^n(x_i-\overline{x})^2$ where $\overline{x}$ is the average, instead of defining $\sum_{i=1}^n|x_i-\overline{x}|$.

Moreover, we don't lose anything when we consider $E$ instead of $L$. As you may have known, the minimizer of $L$ is the same as the minimizer of $E$. More precisely, the minimizer of $L$ must be minimizing geodesic, and the minimizer of $E$ is also minimizing geodesic. See Lemma 2.3 in Chapter 9 of Riemannian Geometry by Do Carmo.