# What are the mathematical models for force, acceleration and velocity?

1. In mechanics, the space can be described as a Riemann manifold. Forces, then, can be defined as vector fields of this manifold. Accelerations are linear functions of forces, so they are covector fields. But what about velocities and many other kinds of vectors?

2. Of course velocities are not forces, so I don't think it is right to reuse vector fields of this manifold. But does this mean that this manifold has many different tangent spaces at each point?

3. This sounds very strange to me. I think the problem is that math models have no physical units, maybe somehow we can create a many-sorted manifold to accommodate units?

• "In mechanics, the space can be described as a Riemann manifold."...well, that depends. Hamiltonian mechanics usually describes physical system as symplectic manifolds. Also, related question: Is force a co- or contravariant vector? – ACuriousMind Mar 31 '15 at 15:58
• Answering this cleanly will require a bit more than what your'e providing. Are we talking about a Newtonian mechanics in a not-necessarily Euclidean space? Are we allowing for a relativistic dynamics here? Are we doing relativistic dynamics, but assuming that the metric can be decomposed into some sort of $-dt^{2} + f(t)g_{ij}dx^{i}dx^{j}$ ? Is this even a metric space? This sounds like a bunch of technical complaining, but the answer is actually different in all of these cases. – Jerry Schirmer Mar 31 '15 at 21:48

Velocities and Spatial Accelerations are twists and Forces and Momenta are wrenches. Both are screws (two-vectors) with one vector free and the other a spatial field. All of them transform with the same laws and their interactions have many dual properties.

NOTE: See "A treatise on the theory of screws", Stawell R Ball, https://archive.org/details/theoryscrews00ballrich  The proportionality tensor transforming twists to wrenches is the 6×6 spatial mass matrix converting motion into momentum and acceleration into forces.

For example below I am composing a velocity twist and a momentum wrench. Do you spot the similarities?

\begin{aligned} {\hat v} &= \begin{pmatrix} {\bf \omega} \\{\bf r} \times {\bf \omega} \end{pmatrix} & {\hat p} &= \begin{pmatrix} {\bf p} \\{\bf r} \times {\bf p} \end{pmatrix}\end{aligned}

In classical mechanics a system is described by a Lagrangian $\mathscr{L}\colon TQ\to \mathbb{R}$, with $Q$ being the configuration space and $TQ$ its tangent bundle, namely the union over $q\in Q$ of all tangent spaces $T_qQ$: $TQ = \cup_q T_qQ$. A local chart on $Q$ looks like $(q_1, \ldots, q_n)$, the $q_k$ being the degrees of freedom of the system. The Lagrangian is then $\mathscr{L}\equiv\mathscr{L}\big(q(t), v(t)\big)$ and the equations of motion are: $$\frac{d}{dt}\frac{\partial \mathscr{L}}{\partial v^{\mu}} - \frac{\partial \mathscr{L}}{\partial q^{\mu}}=0.$$ The solution is a collection of $\big(q^{\mu}(t), v^{\mu}(t)\big)$ that live on $TQ$; if we make the further requirement that, on those solutions, $v=\dot{q}$, then the path on $TQ$ projects uniquely onto a path on $Q$, whose flow is given by the velocity fields.

Wrong. Positions and velocities are coordinates of local charts $\phi$ from the tangent bundle $\phi\colon U\subset TQ\to\mathbb{R}$: as such, they transform contra-variantly. Forces, in the above formalism, are related to the conjugate momenta $p_{\mu}=\partial\mathscr{L}/\partial{v^{\mu}}$ and hence transform co-variantly, with the inverse matrix.