How do you correctly account for a force in some imaginary world?
You are standing on a hill, but this particular hill changes as you move.
Typically you simply look at the local slope (this signifies no physical motion, you are just looking around), however, as the hill changes you will find a different force if you were to actually move.
I am under the impression that if you have the ability to move, you should travel a path as predicted by the force you would observe when accounting for this varying hill; however if you are completely rigid and can never move, you really have no path to travel; so what force do you feel?
In the end, does the force one feels depend on ones degrees of freedom?
(Situation is imaginary, but it is related to a real world physical situation)
Update:(zenophyle)
As far as I know, if you can define a potential landscape, a conservative force may be defined from the vector gradient of this potential.
I too prefer the concept of a charged particle in some electric potential, but rather than considering the time dependent potential, consider a potential that interacts with the charge. Consider a coupled system where the charged particle changes the potential landscape as it moves.
So the paradox is that the true potential, the one that defines the force determining the motion of this charge, requires that the charges physical motion in order to calculate the potential after interaction. If the charge is unable to move, it will never see the changing potential, so is the force it feels merely the gradient of the potential calculated without moving (just keeping separation position fixed and looking at gradient)? In other words does the force the charge feels at one position depend on its degrees of freedom?
Update 3:(zenophyle)
You seem to have grasped the essence of the problem fairly well.
Perhaps you can expand upon the concept of a charge in the vicinity of an infinite conducting plane. However, consider instead of a charge that is fixed, a dipole with some induced dipole moment (the problem I am thinking would require the charge to change with position, which is difficult to grasp, so an induced dipole moment should suffice). Now, for every location you not only have a different potential, but also a different dipole moment (or charge) that responds. How would the force be properly accounted for if the dipole was just placed in some location (0 initial velocity)? Is it based on the local gradient of the potential, or does it depend upon some infinitesimal perturbation that reveals how the potential and dipole moment (charge) changes in space?
Answers & Comments
Verified answer
This is an interesting question. And it sounds like charged particle in a changing electrical field or something like that.
To answer your question, let me first ask a question. Your trouble seems to come about because of your assumption that the force is the gradient of the potential. But is this correct? Isn't the potential the derivative of the force? Which one is the physically meaningful quantity, and which one is the derived from that quantity. I think you will find it easier if you think of the force as the 'real' quantity, and the potential as just a convenient artifact derived from the force.
This is somehow easier for me to see in the picture of a charge particle in an electric potential than it is in your picture of a hill. In the electric potential, the charged particle will feel the force caused by the Coloumb repulsion of the charged on the electrodes making that potential. If the potential is changing in time, the force at time t will correspond to force due to the electrodes at time t. Which will happen to also be the slope of the potential at that point. This is irrespective of whether the particle responds to that force by moving, or not.
Now, let's further examine the case where the particle is not moving, though what is said above still works. In that case, you aren't describing the whole system. Why is the particle not moving? There must be some other force and some other potential that causes it not to move. And the sum of the electric force and this second force equals zero. Or the sum of their potentials has no gradient. You need to keep track of what keeps the particle from not moving, otherwise not moving can't happen.
And now, back to your original hill analogy. What's the problem with that analogy? Why is it harder to see what is going on there? Well, because the height of a hill has no meaning without gravity to go along with it. We think of a gravitational potential as being the height of the hill, but that just isn't correct. So you do have to be careful making that analogy. The actual force the particle feels on a hill is the normal force, which is related directly to the slope of the hill and is there if the particle is stationary or not. The particle does not need to know what the level of the hill is in places the particle is not. The slope enters the equation simply because at the point the particle is, the slope happens to be perpendicular to the direction from that the normal force pushes.
EDIT
Hmmm, I see what you are saying. That is an interesting question.
Perhaps I am not picturing your problem correctly. But here are a couple other systems that popped to mind after reading your post.
If you have a charge next to an infinite conducting plane, the force the charge feels is related to the induced charge in the conductor. So the force depends on where the charge is. However, the potential surface you would construct in that case is constructed by figuring out what the force would be if the charge was at any given location. So it is some how independent of the motion of the particle.
If you have something like two charged particles orbiting each other (or two masses on a spring on frictionless surface... or an atom with a Vander Waals potential... etc), then certainly the motion will effect the force, and thus the potential the particle feels. However, in these cases, you can greatly simplify the problem by finding a useful coordinate system (the center of mass coordinate system) that breaks the motion into some motion that is relevant to the potential and motion that is not. In this coordinate system, the potential is no longer motion dependent.
However, there are things like a charged particle moving in a magnetic field. There the force is certainly motion dependent. So you can find a force that is inherently motion dependent, but there should still be something like motion (analogous) in this system that cancels out, if you look at it the right way.
But to answer your question, it seems to me like you should look for a proper coordinate system, and if you find it the motion dependence drops out (or something like motion dependence, as in the case of the B-field). However, in an incorrectly chosen coordinate system, the motion would not factor out. Basically, you need to find the 'eigen'-coordinate system. And this is only going to be possible in linear systems where the degrees of freedom are not coupled and an eigen-coordinate system exists. In non-linear systems, there can be chaotic systems where I would think you very much need to know the motion to know the potential.
EDIT
I'm not sure I understand your example. My guess is you are talking about an object that can have a dipole moment, let's say a straight conductive wire. And this wire is floating in space, with a spatially-varying electric potential applied to that space. The electric potential will induce a dipole in the wire, that will then be acted on by the potential, which will then alter induced dipole, and the process repeats. This reminds me of paramagnetism. Or the way a parallel plate capacitor will pull a dielectric piece between the plates if the piece is off to one side. I would then say that on the macroscale, one would need to perform some infinitesimal perturbation around the current location to find the potential and the force that macro-object will react with.
On the microscale, however, I still think you may be able to break it down far enough to find a potential that is not dependent on the exact position. It's just these microscale dynamics may not superpose in a nice way.
I also think you could look at the macroscale object and break it down into proper coordinates where there is a force analog that does not depend on a motion analog. I know that general graduate level mechanics texts, such as Goldstein, have some discussion of this sort of thing, usually in or near the Lagrangian section. For example, if you have a marble rolling on table, and the table has a hole in the center, and the marble is attached to a string, and that string is threaded through the hole and attached to a free hanging weight. The dynamics of that problem can be solved if the proper set of coordinates is picked (and those coordinates will not be Cartesian) and a force-like analog (a moment times a second time derivative of a coordinate), along with a motion like analog (the first derivative of the coordinate) can be found and those two things would be separable (not dependent on infinitesimal perturbation). So I would be inclined to believe that a proper coordinate system could be found for the induced dipole problem, if the electric potential was of a nice nature (linear or quadratic). But if the wire were not straight or if the electric potential was not nice, then I would guess you would get into a chaotic system where such a separable coordinate system would be difficult or impossible to find.
I suspect this is an analogy so I'll caution you that one has to really think hard about just how analogous to the real problem the "imaginary" problem is.
In the case you describe, the force felt by the person is the force of gravity. It always pulls down on the person with a froce equal to his or hers weight. The hill does not alter that. The shape of the hill alters the acceleration the person experiences if they are allowed to move under the influence of gravity - a steep slope will cause the person to have a greater acceleration than a shallow slope. But the person's weight will not change because the force due to gravity is unaffected by the hill.
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