Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

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).

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.