Alternative definitions of potential?

Question

I hope this question is simple and can be quickly cleared up.

In a 1D conservative dynamical system, I've always been taught that the potential function is the function $V(x)$ such that:

$$F=-\frac{dV}{dx}$$

That makes sense to me, simply derived from the definitions of work and conservation of energy.

However, just reading through the book 'Nonlinear dynamics and chaos' by Steven Strogatz, in the first chapter he defines the potential for the most basic 1D system:

$$\dot{x} = f(x)$$

To be

$$f(x)=-\frac{dV}{dx}$$

What is happening here??? Usually, the value $-\frac{dV}{dx}$ is proportional to $\ddot{x}$ (because, with constant mass, $F$ is proportional to $\ddot{x}$) but now this value is equal to $\dot{x}$. is this an alternative definition of potential? I don't think the 2 definitions are equivalent.

Thanks!

score 4 · Accepted Answer · answered Oct 25 '11 at 03:59

While Jonathan's answer is correct, it is very general. I would like to explain the reason for the terminology in this specific problem.

When dealing with second order equations, like Newton's laws, the notion of potential is designed to give a conserved energy. This definition is natural in the inertial domain--- in the case that reversible inertial motion is what is going on.

When the basic motion is irreversible, however, the energy generally goes down, not up. In the limit of strong damping, the motion is a velocity in the direction of the force, reduced by the damping coefficient--- the object is always travelling in the direction of the force with terminal velocity. This could be called the Aristotelian limit (although it is a terrible abuse of history to give Aristotle's vapid writing the credit for this precise notion), which works for pollen in water, for biological cells, for overdamped oscillators, and generally for anything small in a thick fluid. The equations of motion in this case just drop the $m\ddot{x}$ term, and leave the friction and the force only.

The motion then becomes a gradient flow. This is motion down a gradient of a function, and this function is generally called a "potential", because it is the potential in the Aristotelian limit. Potential flow is mathematically important, because you can prove that the end result is that you get sucked to a local minimum. Proving that a flow is a potential flow is one of the most general ways of establishing global convergence results.

Thanks, I think that makes sense. If I gather correctly, because we have a first order equation, we can never come back to the same point twice (this is the irreversibility???). And the gradient of V as defined above would indeed give the velocity, rather than the acceleration. Very cool. I guess I really just wanted to confirm that the 2 definitions were indeed mathematically different - I was getting rather worried for a while :) — tom, Oct 25 '11 at 12:14
Irreversibility doesn't quite mean that you don't go back to where you started (but if you say come back arbitrarily close, then Poincare proved this in the case where the motion is reversible, Hamiltonian, and bounded). Reversible means that when x(t) solves the equation, so does x(-t), the motion going backwards in time. — Ron Maimon, Oct 25 '11 at 21:10

score 3 · Answer 2 · answered Oct 25 '11 at 01:22

Keep in mind that the definition of potential is dependent on context. There is nothing deep going on here. If he wishes to define the potential of the system $\dot{x}=f(x)$ to be a function $V$ such that $-\frac{dV}{dx}=f(x)$, he is free to do so. Now of course, because the system itself is different, you can't expect this potential to satisfy all of the properties you're used to. Nevertheless, this definition of potential is the most natural for the system at hand.

If you're familiar with electrodynamics, then you've seen something similar with the vector potential. $\vec{\nabla}\cdot \mathbf{B}=0$ implies that there is some vector field $\mathbf{A}$ such that $\vec{\nabla}\times \mathbf{A}=\mathbf{B}$. It's a different system, so of course this is not going to be the same as the scalar potential, nevertheless, it is the most natural, analogous definition for the potential in this context.

Alternative definitions of potential?

2 Answers2