# The minimizers of energy and length of a curve

Mathematics Asked by W. mu on August 4, 2020

Assume S is a regular surface with a Riemannian metric $$g$$.

The energy of a curve $$c: [0,a] to S$$ is defined as :

$$E(c) = frac{1}{2} int_0^a g_{c(t)}(dot c(t),dot c(t))mathsf{dt}$$

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

$$L(c) = int_0^a sqrt{g_{c(t)} (dot c(t), dot c(t) )}mathsf{dt}$$

The Lemma 2.3 in Chapter 9 of Riemannian Geometry by Do Carmo is:

About the formula
$$aE(gamma)=(L(gamma))^2leq (L(c))^2leq aE(c),$$
the author has proved the inequality: $$(L(c))^2leq aE(c)$$
using the Schwarz inequality.

The question is: why $$aE(gamma)=(L(gamma))^2$$?

## One Answer

For any smooth curve $$c : [0,a] to M$$, choosing $$f equiv 1$$ and $$g=|c'(t)|$$ the Cauchy-Schwarz inequality $$Big( int_0^a fg ,, text{d}tBig)^2 leq int_0^a f^2 , text{d}t int_0^a g^2 , text{d}t,$$ become $$L(c)^2 leq aE(c)$$. But Cauchy-Schwarz inequality become an equality if and only if $$g$$ is constant, which means that $$L(c)^2 = aE(c)$$ if and only if $$|c'(t)| = text{const}$$.

Apply this inequality to the curve $$gamma$$ in the lemma, we certainly have $$L(gamma)^2 leq aE(gamma)$$. Since by hypotheses, $$gamma$$ is a geodesic ( i.e. $$Dgamma'/dt = 0$$ ), a simple computation $$frac{d}{dt} langle gamma',gamma' rangle = Biglangle frac{Dgamma'}{dt}, gamma'Bigrangle + Biglangle gamma', frac{Dgamma'}{dt} Bigrangle = 0,$$ implies the speed $$|gamma'(t)|$$ is constant. Therefore $$L(gamma)^2 = aE(gamma)$$.

Correct answer by Si Kucing on August 4, 2020

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