How does one write Newtons 2nd Law using the language of forms? Newton's second law says that $F=ma$.
Supposing that the force is conservative and can thus be expressed in terms of a potential $V$ we have that $F=-dV$.
We have that $V$, being a function, can also be considered as a 0-form; so $dV$, and therefore $F$ is a 1-form.
So we ought to consider that $ma$ needs to be expressed as a 1-form; the natural suggestion is $a dx$; but $a$ is a second derivative.
How do I express $a$ as a 1-form naturally?
 A: To my mind, Newton's equation makes the most sense as an equation of vector fields. Let $(M,g)$ be a (Pseudo-)Riemannian manifold with Riemannian connection $\nabla$. Then the equations of motion for a position-dependent conservative force $F$ are given by $$m {}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}=F\circ \gamma=-(\nabla V) \circ \gamma,$$
where $\gamma: \mathbb{R} \supset I \to M$ is the wanted curve, ${}^{\gamma}\nabla_{\frac{d}{dt}}$ is the pullback of the Riemannian connection, and $\nabla V$ is the gradient vector field of $V$ defined through $dV(\cdot)=g(\nabla V, \cdot)$.
Now if you want to write this in terms of differential forms, you'd need to convert ${}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}$ into form language. Do you consider something like $$m (g\circ \gamma)({}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}, \cdot)=-(dV \circ \gamma)(\cdot)$$
as an equation for differential forms along $\gamma$ natural? I don't, and I cannot see a "natural" way of getting around this.
A: You can write
$$\text dV = - m v \text d v$$
where $v$ is the velocity.
