Interaction between 2 charges moving at the same velocity

[adelmakram]

How does a moving point charge affect another point charge moving at the same velocity and separated by some distance in one direction relative to an observer whom the charge are moving? What formula describes that interaction?

 

[dauto]

If they have the same velocity than one is not moving with respect to the other. Thair interaction is described by Coulomb's law.

 

[Philip Wood]

Edit: what follows in this post was based on at the misunderstanding that the charges were separated by a distance transverse to the direction in which they are moving relative to the observer. I leave the post here, in case anyone is interested in this case as well.

The force in the frame of reference in which the charges are moving is \frac{1}{\gamma} times the force in the frame of reference in which the charges are both at rest, that is \frac{1}{\gamma} times the ordinary Coulomb's law force.

\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} in which v is the velocity of the charges, that is the relative velocity of the two frames of reference.

\frac{1}{\gamma} is less than 1, so the force is reduced.

The easy way to establish this result if by transforming the (transverse) force i.e. rate of change of transverse momentum between the frames. This is very easy because transverse momentum is the same in both frames. The \gamma enters through time dilation in computing the rate of change.

The result can also be reached – less easily – by electromagnetic theory. In the frame in which the charges are stationary there is a purely electric force between them. In the frame in which they are moving in parallel paths, there is both an electric (repulsive) and magnetic (attractive) force between them. The electric force is larger in this frame than in the other. But the additional magnetic force is greater in magnitude than the increase in electric force, so the overall repulsive force is reduced – by a factor of \frac{1}{\gamma}, it turns out.

 

[adelmakram]

The force in the frame of reference in which the charges are moving as you describe is \frac{1}{\gamma} times the force in the frame of reference in which the charges are both at rest, that is \frac{1}{\gamma} times the ordinary Coulomb's law force.

\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} in which v is the velocity of the charges, that is the relative velocity of the two frames of reference.

\frac{1}{\gamma} is less than 1, so the force is reduced.

But my example considers 2 charges moving one ahead of the other not in parallel. And according to the transformation of electromagnetic field, the electric field in the direction of the motion does not change, so does the force between them in this direction. E_x= E`<sub>x</sub> & Also B <sub>x</sub>= B`_x , specifically, B_x=0. This suppose to result into the electromagnetic force between 2 charges are the same for both frames in this direction ! Agreed?

 

[adelmakram]

The same would also apply if our scenario consists of 2 magnets moving one ahead of the other in one direction. B x= B`x , and there would be no change of the force between them in both frames! So if you agree on the 2 scenarios I would take you to more important result.

 

[Philip Wood]

Sorry that I misunderstood the set-up with the charges. Agree that if there is no transverse separation (\Delta y), the force between them will be unaltered if they are both moving x-wise at the same velocity.

Not sure about your logic for the case of the two magnets. If we model the magnets by current-carrying solenoids, it's clear that the (x-wise) force on magnet 2 doesn't arise from the B_x due to magnet 1, but from the B_y and B_z from magnet 1.

Nonetheless, I think your conclusion about the magnets may be right, because F'_x = F_x for forces between bodies which are stationary in the S frame.

 

[adelmakram]

Agree that if there is no transverse separation (\Delta y), the force between them will be unaltered if they are both moving x-wise at the same velocity.

Now consider, a train carries the 2 charges where one of them is attached to its near end and the other one is attached to the rear end. Again consider that the train comes to a sudden stop relative to a platform observer so as both near and rear ends stops immediately at the same time. As it does so, its length which marks the distance between the 2 charges is not altered relative to that observer. But for the train observer, where the train is originally at rest, the near end starts to move first toward the rear end which bring the near charge close to the rear one. This is suppose to increase the force between them contrarily to what is seen by the platform observer. So one event causes the 2 observers to record 2 different observations!

 

[Philip Wood]

Have you read up on Bell's spaceship paradox?

 

[A.T.]

This is suppose to increase the force between them contrarily to what is seen by the platform observer.

Both observes see an increase in force. The train observer, because the charges come closer together. The platform observer, because the length contracted E-fields expand to spherical symmetry again.

 

[adelmakram]

Both observes see an increase in force. The train observer, because the charges come closer together. The platform observer, because the length contracted E-fields expand to spherical symmetry again.

If both ends come to a simultaneous stop relative to the platform, then the distance between them should not altered before and after the stop. That may only happen when the rear end stops before the near end which is not my scenario.

 

[Meir Achuz]

How does a moving point charge affect another point charge moving at the same velocity and separated by some distance in one direction relative to an observer whom the charge are moving? What formula describes that interaction?

That is the configuration of the Trouton – Noble experiment.
The interaction between the two moving charges is given by equation (3) of
http://arxiv.org/pdf/physics/0603110.pdf,
but that leads to a paradox, which is resolved in the rest of the paper.

 

[A.T.]

But for the train observer, where the train is originally at rest, the near end starts to move first toward the rear end which bring the near charge close to the rear one.

If both ends come to a simultaneous stop relative to the platform, then the distance between them should not altered before and after the stop.

You just explained yourself that the distance is reduced in the train frame.

 

[Dale]

This topic is already being discussed in a separate thread by the OP in the relativity section. This duplicate thread is closed.