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Consider two electrons moving parallel to each other in the same direction with same constant velocity. Will they experience any force due to either of them?

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4 Answers 4

Yes, they will feel both electric and magnetic force. If you apply a Lorentz boost and get into the frame where they are at rest, they will simply feel the electric field of each other.

However if we stay in the lab frame we will notice that the force they experience goes down with $1/\beta \gamma^2$ (relativistic factors), approaching zero as they approach the speed of light. You can picture this also noticing that the time is flowing slower in their fast moving frame of reference and so you kinda see their radial movement in slow-motion.

This is a key point in accelerator physics where you want to get the beam of particles travel as fast as possible in the minimum amount of space thus reducing the radial defocusing effect due the so called "space charge".

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Yes. In their rest frame there is only the electric Coulomb force. If the E field is Lorentz transformed to their moving frame, there will be a magnetic force and the electric force will be modified.

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No. Each electron has zero velocity relative to the other. Their movement relative to something else contributes nothing to the magnetic force between them.

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This is wrong. The magnetic force will be zero in their rest frame, but nonzero in the moving frame. –  David H Nov 3 '13 at 7:37
    
I'm often wrong. Are you saying the trajectories remain parallel in one inertial frame and not in another? Please explain. –  DaveV Nov 3 '13 at 11:26
    
No, they repel in every frame. The repulsive force is a frame dependent combination of a repulsive coulomb force minus a smaller attractive magnetic force, both of which increase in magnitude with velocity. But difference is the same in all inertial frames –  David H Nov 3 '13 at 11:46
    
Right, I phrased that carelessly, but I think I'm beginning to understand. Also, I have assumed that the electrons travel at the same velocity, which was not actually stated in the original question. The question becomes interesting, though, if their velocities are equal. I think you are saying that the trajectories appear to diverge at different rates in different reference frames, due to relativistic contraction of measured distances in the direction of their travel. That makes sense to me now. –  DaveV Nov 3 '13 at 13:23

If you consider two charges moving in the same direction, with same velocity, parallel to each other, due to the motion of charges, magnetic field will be produced around them along with their initial electric field.

Now, force between the charges depends on whether the charges are identical or opposite.

If charges are identical, they repel each other.

If you consider opposite charges, there is two possibilities, one possibility is, they may attract each other, as they are oppositely charged, and another possibility is, they may not experience any force, as the net charge becomes zero, vector fields get cancelled.[frame of reference is chosen with respect to observer]

By the way, when would they attract and when would they feel no force, is another matter of question, which is currently not associated with the present one.

All the things even depend on the frame of reference. If you consider two trains ,moving with the same velocity in the same direction, and if you assume, you are in one of the train, and I am in one of the train.It seems as if, both of our trains were at rest, considering that, there was no reference for us, to notice the change in position of either of the trains (like tree moving backwards).In the frame of reference of the man standing outside, both the trains are moving.This clarifies you, how frame of reference matters.

If you choose frame of reference with respect to the observer above cases will be applicable, if you consider with respect to either of the charges as frame of reference, they are at rest, they would not produce any magnetic field. But, they do have electric field, again force acts or not, depends on type of charges, as explained above.

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