# Understanding Newtonian mechanics using concepts from differential geometry

In a book I'm reading (Friedrich and Agricola), I encountered the following definition of a "Newtonian system":

An autonomous Newtonian system is a triple ($$M^m$$, $$g$$, $$X$$) consiting of a manifold $$M^m$$, a Riemanian metric $$g$$, and a vector field on the space $$TM$$ such that $$d\pi \circ X = Id_{TM^m}$$, where $$\pi$$ is the canonical projection from $$TM$$ to $$M$$.

Rationale for definition is that "not all vector fields are allowed, since a force can act only in space".

I'm having trouble understanding the definition, because the composition of $$d\pi$$ and $$X$$ being identity is equivalent to $$X$$ having only components of the form $$\frac{\partial}{\partial x^i}$$ (and no components $$\frac{\partial}{\partial v^i}$$).

The book then provides the following example on $$TM \cong \mathbb{R^2}$$ where $$M=\mathbb{R}$$:

$$X = \dot{x} \frac{\partial}{\partial x} + \frac{1}{m}(-k^2x-\rho\dot{x})\frac{\partial}{\partial \dot{x}}$$ which does not seem to fit!

I am going to consider only the one-dimensional case to simplify the demonstration. If we view a point $$(p, v) \in TM$$ as a position $$p$$ and velocity $$v$$, then the canonical projection $$\pi: TM \to M, (p,v)\mapsto p$$ projects away the velocity. Then $$d\pi: T(TM) \to TM$$, which is a map from the tangent tangent bundle to the tangent bundle, is going to be such that it projects away both $$v$$ and $$\partial_v.$$ That is, if $$X = a(p,v)\,\partial_p + b(p,v)\,\partial_v$$ then,

$$(d\pi\circ X)(p,v) = a(p,v)\,\partial_p = (p, a(p,v)),$$

where the last equality uses the position-velocity notation for elements of $$TM$$.

Now, consider the requirement that $$d\pi \circ X = \mathrm{Id}_{TM}.$$ We need this to be true for all possible points on $$TM$$. That is, if you feed in a $$(p, v)$$ into $$d\pi \circ X$$ you must get $$(p, v)$$ back out. Notice immediately that the only way this is true is if it is the case that $$a(p,v)=v$$ everywhere. This determines that the rate of change of position is velocity, i.e. the coefficient of the component $$\partial_p$$ of your vector field $$X$$ on $$TM$$ must be $$v$$. Equivalently, $$X$$ takes the form

$$X(p,v) = v\,\partial_p + b(p,v)\,\partial_v.$$

When $$m > 1$$, you will find that instead that $$X$$ must have $$v^k\,\partial_{p^k}$$ terms. The only term feeding into the dynamics of $$p^k$$ is the k-th velocity component $$v^k$$.

By strictly enforcing that the dynamics of position are velocity, you are saying the only variation allowed in the dynamics are in the dynamics of velocity, i.e. acceleration. So in your particular example, it is the fact that $$X$$'s position dynamics are precisely $$\dot x\,\partial_x$$ that make it compatible with the definition.