Let $M^n$ be a smooth manifold and $g$ be a Riemannian metric on $M$.
Is there $\omega \in \Omega^1(M)$ such that $\int_c \omega = \ell(c)$ for every curve in $M$?
In general the answer seems to be no. In Spivak's A Comprehensive Introduction to Differential Geometry (volume $1$) he gives this argument for $\Bbb R^2$ with the usual metric: if $\omega$ is such a non-zero form, then $\ker \omega$ is a one dimensional distribution in $\Bbb R^2$ - and every curve contained in a integral submanifold of $\ker \omega$ will have length zero.
If in the general case we consider $\omega$ a non-zero form, then $\ker \omega$ will be a $(n-1)-$dimensional distribution in $M$, which, by Frobenius' theorem, will be integrable if and only if $\omega \wedge {\rm d}\omega = 0$ (see for example here). If such condition holds we proceed like in the $\Bbb R^2$ case (in which the condition holds trivially).
But I see no reason for that condition to necessarily hold. Is there an example of manifold with a one-form like this?
This also leads me to wonder (from here on it's mostly food for thought and I'll be happy with the previous question answered):
what if $g$ is pseudo-Riemannian? The possibility of lightlike curves in the integral submanifold seems to make the reasoning above fail.
what if we looked at energy instead of length? The situation seems the same, but energy depends on the parametrization, which ends up depending on a coordinate system (ok, maybe that second question is not so interesting after all)
Let $x\in M$, let $\gamma : [0, t] \to M$ be $\gamma(s) = \exp_x (sX)$ for some $X \in T_xM$, with $\|X\| =1$. Then $\ell (\gamma) = t$. Thus
$$ \int_0^t \omega_{ \gamma(s)} (\gamma'(s)) ds = t$$
for all $t$. Differentiating gives
$$1 = \omega_{\gamma(t)} (\gamma'(t))$$
for all $t$. Putting $t=0$ gives $1 = \omega_x (X)$. But $X \in T_pM$ is arbitrary and this is nonsense.