# What if $F\neq \frac{dp}{dt}$?

I was thinking of this idea that maybe there are esoteric cases where the force is not given in classical mechanics as $F=dp/dt$ but as some function of $F=F(p,q,\dot{p},\dot{q})$

E.g, something like: $k\cdot \frac{dp}{dq}$ with a suitable constant $k$, or any other sort of function of p,q and its time derivatives.

Are there any toy models that theorists suggest on this idea?

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If you're using $q$ and $p$ then you're implicitly using Hamiltonian formalism. Then $\dot{p}=u(q,p)$ and $\dot{q}=v(q,p)$ so $F=F(p,q,\dot{p},\dot{q})\equiv F(q,p)$. In your particular proposal $k\displaystyle\frac{dq}{dp}=0$ because $q$ and $p$ are treated as independent variables. In classical mechanics the choice of the independent variables is of capital importance.

Let's take point particle electromagnetism. The Hamiltonian is:

$$H=\frac{1}{2}\left(\vec{p}-e\vec{A} \right)^2+e\phi$$

and Hamilton's equation for the momentum reads:

$$\dot{p}_{i}=-\frac{\partial H}{\partial q_{i}}=0$$

but $\vec{F}\neq 0$. I think the problem is mixing terms about Newtonian Mechanics (i.e. Force) and terms about Hamiltonian Mechanics (momentum).

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I don't think choosing $q$ and $p$ as independent variable means Hamiltonian formalism. It is only Hamiltonian formalism when $-\partial H/\partial q=\dot{p}$ and $\partial H/\partial p=\dot{q}$. – Sankaran Feb 2 '13 at 22:37
Isn't $F = dp/dt$ the definition of force? – Paul J. Gans Feb 3 '13 at 1:05
@PaulJ.Gans: In Newtonian mechanics, yes. In Hamiltionian/Lagrangian, no: en.wikipedia.org/wiki/Generalized_forces – Manishearth Feb 3 '13 at 14:16