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I am confused on one of the earlier sections of Feynman Lectures Vol 2. I've posted the relevant snippet below. Since it's in the public domain (http://www.feynmanlectures.caltech.edu/) I think I can paste it here. Anyhow, the question is about two charges moving in space at same speed and parallel, OR two currents. When I first read everything it made sense, but then I made the mistake of rereading.

My question is this. For the two currents in parallel Feynman goes to some lengths to point out that the electrical charges are balanced so that all the remains is a magnetic force, even though the current moves at .01 cm/sec. We also know (from the previous pages) that two wires will attract each other if current is going in the same direction. However, I'm having trouble understanding what happens if I travel at the same speed as the current and observe the two wires. Wouldn't there now be no current so no magnetic force? Do electrical forces somehow come back into play and exactly compensate for the lost magnetic forces? I saw some answers that seemed to refer to "bunching up" of charges when we change frames. I'm afraid I don't get this at all. I find it hard to believe the wires would not be attracted by me simply walking at the same pace as the current. I suppose that could be tested by arranging for the wires to be close enough and currents strong enough that they actually touch! Presumably something nasty (arcing, short circuit, burn the lab down, etc.) would happen in all reference frames. If the electrical force is reintroduced in other reference frames and makes up for the magnetic force, please explain how. Feynman does not seem to mention it and as I said, goes out of his way to mention, "near-perfect cancellations of electrical effects".

If you read this far, Thank you! I'm also (more) confused by his first example. If you have two charges travelling through space, I assume you have two forces between them: electrical and magnetic. If the charges are the same sign, the electrical forces should cause the charges to repel, right? But the magnetic forces would cause an attraction due to two current in same direction attracting? So what actually happens? If I'm watching them travel past me at some speed, do they attract or repel? I.e. which force is stronger? And then adding to my confusion, if I travel along with the charges at exactly the same speed, won't there be zero magnetic force and thus the charges will repel? Shouln't the total force between them be the same in all reference frames though?

Many thanks and if there is a better newsgroup ("Physics for Learners"?) please point me to it. If there is a particular good reference (or you've read ahead in Feynman and the answer is there!) please point me to it. -Dave

From Feynman:

In the case of the magnetic field we can make the following point: Suppose that you finally succeeded in making up a picture of the magnetic field in terms of some kind of lines or of gear wheels running through space. Then you try to explain what happens to two charges moving in space, both at the same speed and parallel to each other. Because they are moving, they will behave like two currents and will have a magnetic field associated with them (like the currents in the wires of Fig. 1–8). An observer who was riding along with the two charges, however, would see both charges as stationary, and would say that there is no magnetic field. The “gear wheels” or “lines” disappear when you ride along with the object! All we have done is to invent a new problem. How can the gear wheels disappear?! The people who draw field lines are in a similar difficulty. Not only is it not possible to say whether the field lines move or do not move with charges—they may disappear completely in certain coordinate frames.

What we are saying, then, is that magnetism is really a relativistic effect. In the case of the two charges we just considered, travelling parallel to each other, we would expect to have to make relativistic corrections to their motion, with terms of order v2/c2. These corrections must correspond to the magnetic force. But what about the force between the two wires in our experiment (Fig. 1–8). There the magnetic force is the whole force. It didn’t look like a “relativistic correction.” Also, if we estimate the velocities of the electrons in the wire (you can do this yourself), we find that their average speed along the wire is about 0.01 centimeter per second. So v2/c2 is about 10−25. Surely a negligible “correction.” But no! Although the magnetic force is, in this case, 10−25 of the “normal” electrical force between the moving electrons, remember that the “normal” electrical forces have disappeared because of the almost perfect balancing out—because the wires have the same number of protons as electrons. The balance is much more precise than one part in 1025, and the small relativistic term which we call the magnetic force is the only term left. It becomes the dominant term.

It is the near-perfect cancellation of electrical effects which allowed relativity effects (that is, magnetism) to be studied and the correct equations—to order v2/c2—to be discovered, even though physicists didn’t know that’s what was happening. And that is why, when relativity was discovered, the electromagnetic laws didn’t need to be changed. They—unlike mechanics—were already correct to a precision of v2/c2.

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  • $\begingroup$ Since writing this question I found a good description in Griffiths, Introduction to Electrodynamics 2nd edition. Section 10.3.1. "Magnetism as a Relativistic Phenomenon". Still studying it but I sort of get that because of Lorenz contraction, electrons and proton (which are travelling in different directions) will be spread out more or less, thus making the wire NOT neutral. However, for fixed length wires, I would have thought total charge should remain constant so charge should still be neutral? Also, all my questions about two charges travelling through space (not wires) still remain. $\endgroup$
    – Dave
    Commented Nov 24, 2019 at 15:51

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If you have two charges traveling through space, I assume you have two forces between them: electrical and magnetic. If the charges are the same sign, the electrical forces should cause the charges to repel, right?

Yes, that's correct.

But the magnetic forces would cause an attraction due to two current in the same direction attracting? So what actually happens?

Yes, the magnetic forces cause attraction but remember that the produced magnetic field, and its force consequentially, are not greater than the electric field. For instance, if the electric field is $E_y^\prime$ in the charge rest frame along with no magnetic field, the lab observer $-$ WRT whom the charge moves at $v$ $-$ detects two fields:

1- An electric field which is increased to $E_y=\gamma E_y^\prime$, and

2- A magnetic field of $B_z=\gamma \frac{v}{c^2}E_y^\prime$.

As it is clear, the relevant electric force $(\gamma qE_y^\prime)$ is always greater than the Lorentz force of $(\gamma q\frac{vw}{c^2}E_y^\prime)$, where $w$ is the speed of the other charge ($q$) WRT the lab observer.

If I'm watching them travel past me at some speed, do they attract or repel? I.e. which force is stronger?

They repel. The E-field is stronger.

And then adding to my confusion, if I travel along with the charges at exactly the same speed, won't there be zero magnetic force and thus the charges will repel?

YES. You are right: A zero magnetic field and the charges repel.

Shouldn't the total force between them be the same in all reference frames though?

Nope. The total force can remain unchanged only for when the forces are parallel to the motion direction, otherwise its magnitude would vary from one reference frame to another.

ADDED:

Indeed, three-force and three-acceleration have complicated transforms. To judge what really an observer measures for forces/accelerations in a moving frame of reference depends on many things esp. the direction of the force/acceleration, i.e., their angles with the motion direction. For example, if you exert a force $F^\prime$ against a wall in your room, the moving observer who moves parallel to the force direction would measure it the same $(F=F^\prime)$, however, the observer who moves perpendicular to the force direction measures it as $F=\frac{F^\prime}{\gamma}$. For any arbitrary given orientation, the calculations become slightly complicated.

The electromagnetic forces are not an exception. You need to consider their orientations and velocities in their rest frame.

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  • $\begingroup$ Thank you very much Mohammad! Can you say a little more about your last statement: "The total force can remain unchanged only for when the forces are parallel..." Just intuitively, I would have though the force between the two charges would remain the same. Otherwise you would get different results in different frames of reference. By your #2 above, the magnetic field would be exactly the same as the electric field if the charges were traveling at the speed of light so they would neither repel nor attract, But just by travelling with the charges, I would see them repel! $\endgroup$
    – Dave
    Commented Dec 2, 2019 at 17:26
  • $\begingroup$ Or perhaps you have a good tutorial or reference for your last comment? I definitely find this fascinating. Thanks! $\endgroup$
    – Dave
    Commented Dec 2, 2019 at 17:39
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    $\begingroup$ @Dave My pleasure. Can you say a little more about your last statement: "The total force can remain unchanged only for when the forces are parallel..." I've edited my answer in this regard. By your #2 above, the magnetic field would be exactly the same as the electric field if the charges were traveling at the speed of light so they would neither repel nor attract, But just by traveling with the charges, I would see them repel! Yes, they do repel. But remember that for the speeds very close to that of light, clocks stop running. That is, any transverse motion slows down to zero ... $\endgroup$ Commented Dec 2, 2019 at 18:43
  • $\begingroup$ ... the charges repel each other but they recede from each other tangibly at an infinite time. $\endgroup$ Commented Dec 2, 2019 at 18:45

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