Conservative overdamped harmonic oscillator?

• June 12th, 2013, 04:48 PM
inkliing
Conservative overdamped harmonic oscillator?
This isn't homework. I'm reviewing calculus and basic physics after many years of neglect.
I want to show that a damped harmonic oscillator in one dimension is nonconservative. Given F = -kx - v, if F were conservative then there would exist P(x) such that . I want to show that no such function, P(x), exists.

The easy way would be to find a closed curve around which the integral of Fdx would be zero, but since Fdx is a 1-dimensional 1-form, this doesn't seem to be a meaningful way to do it.

So I think brute force has to prevail. It should be true that: So let For underdamped  Therefore x(t) is not 1-1 is multivalued implies W is not a function implies p(x) doesn't exist (since W=- P) implies F is not conservative. Similarly for .

But in the overdamped case, >1, x(t) is a non-oscillating decaying exponential which never crosses equilibrium, implying x(t) is 1-1, implying W is a function, implying F is conservative. But how can this be? How can a frictional damping force, which dissipates energy as heat, ever be conservative?
• June 12th, 2013, 05:56 PM
tk421
Quote:

But in the overdamped case, >1, x(t) is a non-oscillating decaying exponential which never crosses equilibrium, implying x(t) is 1-1, implying W is a function, implying F is conservative. But how can this be? How can a frictional damping force, which dissipates energy as heat, ever be conservative?

In short, it can't, just as your math shows. A conservative force would yield path-independent answers. Friction, however, acts continuously all along the direction of motion, so that the energy dissipated depends on the total distance traversed, not just on the endpoints. So, friction isn't conservative.
• June 13th, 2013, 09:54 PM
PhyMan
Quote:

This isn't homework. I'm reviewing calculus and basic physics after many years of neglect.

Id like to compliment you on your efforts to stay sharp. A place like this can be very rewarding in such endevors. I myself utilize them when I get stuck myself as in my recent example of a bullet fired straight up into the air. So Bravo! Good for you sir! :)

I think its great that you want to keep these things fresh in you mind. I myself am doing the same thing. I imagine that we'll probably be helping each other on this topic.

A very simply way to show that the force is non-conservative is to evaluate around a closed path. Since the work is not zeo around a closed path then the force is not conservative. Clearly the force you gave is not conservative as applying it to the case for going through from x = 0 to x = L and then back to x = 0 will give a non-zero value. Notice how the sign of the velocity changes depending on the direction of integration therefore there is no way for the integration to evaluate to zero,
• June 15th, 2013, 01:14 AM
inkliing
I guess you're right. I had thought that since Fdx is a 1-dimensional 1-form on a 1-dim manifold, that an integral around a closed loop would be meaningless. But I guess there's nothing wrong with evaluating such an integral, which of course will imply that the force is nonconservative.

I'm starting to understand that what I've written as the work function is a functional (I vaguely remember such things, calculus of variations, I suppose), not a function, even when x(t) is 1-1. I intuitively knew this, but I could not state it clearly. I still can't clearly and straightforwardly state exactly why it's not always possible to find a function, f(x), such that . Suggestions?