Let $M^n$ be a smooth manifold equipped with a nondegenerate Lagrangian $L:TM\to\mathbb R$, $L=L(x,y)$, $x\in M$, $y\in T_xM$. The stationary points of the corresponding integral functional on curves are the solutions of the Euler-Lagrange equation, which in coordinates reads
$$
\frac d{dt} \frac{dL}{dy_i}(x(t),\dot x(t)) - \frac{\partial L}{\partial x_i}(x(t),\dot x(t)) = 0, \qquad i=1,\dots,n .
$$
Consider a smooth curve $t\mapsto x(t)$ which is **not** stationary. Plugging it into the l.h.s. of the equation yields coordinates of a co-vector (from $T^*_{x(t)}M$) which depends on the curve but not on the coordinate system. The invariance of this co-vector can be seen e.g. from the first variation formula for the functional.

**Question:** Is there a coordinate-free definition of this co-vector?

Actually I am interested only in the case when $L$ is the Lagrangian associated to a Finsler metric (i.e. $L$ is quadratically homogeneous).

**Notes**

The equation itself (i.e. the property that the co-vector is zero) has an invariant expression e.g. with Hamiltonian formalism.

In the Riemannian case, the co-vector in question (for a unit-speed curve) corresponds to the geodesic curvature vector under the isomorphism between $TM$ and $T^*M$ defined by the metric. This can be defined invariantly via the Levi-Civita connection.