# Potentials which depend on positions of two particles

1. Sep 4, 2011

### BrunoIdeas

Hello. I have been asked if it is correct to say that the force of a spring is conservative.

So, for a mass attached to a spring on the roof it's OK.
Now consider two masses linked by a spring. Now the potential will depend on both coordinates. It is now not so clear to me any of the properties of conservative forces.
Closed loop integral = 0 and so forth. Neither the fact rot F = 0 => consrevative

2. Sep 4, 2011

### Ken G

It's conservative because any force that depends only on configuration is always conservative. That's because if the force depends only on configuration, then the work done by that force will always be expressible in terms of a potential energy function, via the line integral of that force over deformations of the configuration. The simplest way to see this, especially for 1D deformations like that of a spring, is to simply reverse the path-- if the force depends only on the 1D deformation, then you can get all the work back just by reversing the path of the deformation. If you can always get the work back, it's conservative. Note this doesn't require that either end of the spring be locked down, the work done on the endpoints depends only on the force and the change in the stretch of the spring, because that change is the difference in work on the two ends (given that the force on the two ends is equal and opposite, as per Newton's third law).

3. Sep 4, 2011

### BrunoIdeas

I am so grateful for your answer. I particularly apreciate you've given me the intuiton about understanding conservative forces as position dependent and the notion of getting work back. Using this one evidently sees why friction is not, and probably why it conflicts with time invariance or reversal of Newton's eqs.

I however do not fully understand what is after the bold text.

Once again thanks.

4. Sep 5, 2011

### Ken G

I think you understand completely, the stuff after the bolded text was in there just because it sounded like you could understand this better if one end of the spring was locked down, but if both ends were free to move around it wasn't so clear. I was trying to show that it doesn't matter if either end is locked down, what matters is that if the configuration of the spring is all you need to know the force it is applying at both ends, then the net work done only depends on the configuration (i.e., the stretch) of the spring, and re-extracting that work is also only dependent on returning to the original configuration, even if the spring has waffled around in the mean time.

5. Sep 5, 2011

### BrunoIdeas

Fine. Thanks, I get that. Now I would like to ask what about energy conservation in such cases.
1) Consider a mass attached to a spring hooked from the ceiling with no gravity (or with it, it does not matter). We say that energy is conserved because forces are conservative.
Question: Does the potential energy belong to the spring, to the mass, or to the system.

I understand this question may be misleading but answer me as if I knew nothing.

2) Consider the two masses linked by a spring. Is energy conserved? I understand/guess that for the system it it, not for the masses, but I cannot give good arguments. Could you help me?

Thanks Ken G, and everyone.

6. Sep 5, 2011

### Ken G

The potential energy belongs to the spring. It is stored in interparticle forces within the spring. If you remove the mass completely, all the potential energy is still in that spring, as anyone has discovered to their chagrin if they did remove the mass suddenly!
Energy is conserved if there are two free masses linked by a spring. The sum of the kinetic energies of the two masses, plus kx2/2 for the spring, will always be the same. I'm not sure what aspect of this is where your question is focused.