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If I consider equations of motion derived from the pinciple of least action for an explicilty time dependend Lagrangian

$$\delta S[L[q(\text{t}),q'(\text{t}),{\bf t}]]=0,$$

under what circumstances (i.e. which explicit functional $t$-dependence) is the force conservative?

By force I understand here the term on the right hand side of the equation, if I shove everything to the right except the expression $mq''(t)$.


As a sidenote, besides the technical answer I'd be interested here in some words about the physical motivations involved. I'm somewhat unhappy with a formal $\text{curl}[F]=0$ condition, since it seems to be to easy to fulfill (namely we have to consider closed circles only at single points in time, respecively). The physical motivation behind conservative forces is the conservation of energy on closed paths, where any parametrization $q(s)$ of curves can be considered. But practically, only loops tracked in finite time are physically realizable, i.e. we would move in a circle while t changes.

I guess as soon as one computes the r.h.s. for the equations of motion, one would also be able to define a more physical alternaltive to the above stated idea of conservative forces in this case. I.e. a ask-if-the-forces-integrate-to-zero-on-a-closed-loop functional for a rout between two points in time $t_1$ and $t_2$. This would be an integral where the momentarily force along the point in the path I'm taking would be taken into account. It wouldn't be path independend of course. (We could then even construct another optimization problem on its own, by asking for path with the smallest energy difference, which really would be a sensible question if friction is involved.)

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Rot? You mean curl? Zero curl of a field is hardly an easy condition to fulfill. –  Siyuan Ren May 7 '12 at 1:50
    
@KarsusRen: I mean both, but I should have written curl, yes. And it's not "easy", but still to easy, that's the point. –  NikolajK May 7 '12 at 7:11
    
Why do you identify the force with anything that equals $mq''(t)$? Even for "normal" conditions this fails - think of taking $q$ as an angular variable (so $mq''$ is dimensionally wrong). Shouldn't force be thought of then as anything that equals $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$? –  Emilio Pisanty May 7 '12 at 12:16
    
Other than that, what exactly do you mean by conservative forces? I'm guessing you mean a criterion on $\frac{\partial L}{\partial q}$ such that $H=p\dot{q}-L$ is an integral of the motion regardless of any time dependence, but that is just not realistic - think of adding a time-varying but spatially constant potential to an otherwise time-independent, conservative system. –  Emilio Pisanty May 7 '12 at 12:20
    
@episanty: Yes, could be. I'm not even sure if $\frac{d}{dt}\frac{\partial L}{\partial{q'}}$ is a better definition, I don't know. The Lagrangian doesn't display force as obvious as the Newtonian one, or energy as clear as the Hamiltonian approach. And what does not realistic mean here? The task is to pic out the cases in which it works I guess. –  NikolajK May 7 '12 at 12:21

2 Answers 2

Every time dependent force is non-conservative, because you can take the particle from point A to nearby point B at a time where the force is large, using a system of levers to transfer the work elsewhere, then hold the particle fixed and wait for the force to change, and bring the particle back to A.

The notion of conservative force is one which is conservative for every additional imposed constraint, for every system of walls and pullies you introduce, not one which conserves energy in the special case of solutions of the equation of motion only.

EDIT: What is conservative with constraints

The constraint business is trying to say that the conservation of energy in a system with forces is more than the statement that the energy is conserved for the equations of motion. For example, if you have a particle falling under gravity or a particle sitting on a platform which is slowly moving down under gravity, the work done on the particle is the same in both cases, but in the second case, all the energy is trasferred to the platform, and you need to consider the extra platform energy and whatever is holding up the platform.

When we say a force is conservative, we mean that that force is conservative when you add an arbitrariy platform and allow the force to push slowly. For example, if I imagine that I have a particle falling, and gravity only acts with a nonzero near the line of fall of the particle, the force conserves energy during the fall. But if you lower the particle on a platform, move it over a little on the platform, then raise the particle where gravity doesn't act, you get an energy non-conserving cycle. The statement of conservation of energy is over every concievable physical situation you can set up, and involves the platform constraint forces that keep the particle from falling into the platform.

The result is that when the force on the particle is the gradient of a potential, then any platforms and pullies you add can't be used to extract energy indefinitely from the force field, because if you take the particle on a closed loop, even in the presence of constraints, the total work is zero.

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I don't get the second paragraph. (v1) –  NikolajK May 7 '12 at 18:44
    
A conservative force is one which can be written as the gradient of a potential function as taught in standard physics books. It would be helpful if you could expand on your second paragraph because it doesn't make sense to me. –  John McVirgo May 7 '12 at 22:17
    
Ok, I'll do it. –  Ron Maimon May 8 '12 at 1:38
    
what does the second paragraph mean? –  Larry Harson Jun 7 '12 at 18:48

If the force is time dependent, you do not have energy conservation:

$m \frac{dV}{dt}= F(t)$

$d(\tfrac{1}{2}mV^2)=F V dt=F dx$

For a time dependant potential:

$dU=\mathbb{grad}(U(t))dx + \frac{∂U}{∂t}dt $

But

$F=-\mathbb{grad}(U(t)) \Rightarrow Fdx=-\mathbb{grad}(U(t))dx =-dU+\frac{∂U}{∂t}dt $

So you get:

$d(\tfrac{1}{2}mV^2+U(t))=\frac{∂U}{∂t}dt$

The energy is conserved if and only if U is time independant.

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Thank's Nick for editing my post. I am too busy for figuring out the tex math. –  Shaktyai Aug 6 '12 at 22:10
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The community norm is to use TeX. No one else is "too busy" for it, and StackExchange makes it very easy. –  Mark Eichenlaub Sep 6 '12 at 1:15

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