Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself.
Original post:
The magnetic force is a strange animal. The F = q(E+v×B) = qE+qv×B formula implies that both its direction as well as its magnitude depend on the direction and the magnitude of the motion of the charge. The magnetic force is, just like the electric force, still proportional to the amount of charge (q), but then we have not one but two vectors co-determining its direction and magnitude, as expressed by the vector product v×B = |v|·|B|·sinθ = v·B·sinθ.
The presence of the velocity vector in the F = q(E+v×B) formula implies both the magnetic as well as the electric field are relative, as we wonder: “What velocity? With respect to which reference frame?” The (a) and (b) below illustrate the same interaction between some current-carrying wire and some negative charge q from two perspectives:
Diagram (a) below represents frame S, in which the wire is at rest, and the charge moves along the wire with velocity v0, while
Diagram (b) below represents frame S’, which coincides with the reference frame of the charge, so now it’s the wire that’s moving past the particle, instead of the other way around.
Because of relativity, all of our variables transform: we have time dilation, length contraction, and relativistic mass, as I explained in my posts on special relativity. So we cannot take any of the variables for granted and so we prime all of them: in S’, we have I’, v’, etcetera, and so we need to calculate their values using the Lorentz transformation rules.
Now, we know that the absolute speed of light connects both pictures, but that’s not enough to explain what’s going on. We need some other anchoring principle as well. We have such anchor: charges are always the same, moving or not. They are indestructible. They are never lost or created: they move from place to place but never appear from nowhere. In short, charge is conserved. So we also need to look at charge densities and see what happens to them.
The illustration above shows the current I going in the conventional direction, so that’s opposite to the actual direction of travel of the free drifting electrons. It’s a convention that makes sense because of all our other conventions, such as the right-hand rule for our vector cross-product v×B above, so we won’t touch it. Having said that, the illustration shows what’s going on in S: the positive charges in the wire don’t move, so we have some charge density ρ+ and a velocity v+ = 0. The electrons, on the other hand, do move, and so we have some charge density ρ− and a velocity v− = v. Now, we’re looking at an uncharged wire, so ρ+ must be equal to −ρ−. So the situation is rather simple: we have a current causing a magnetic field, and the force on our moving charge q(−) is F = v0×B.
However, the same situation looks very different from the S’ perspective: our q(−) charge is not moving and, therefore, there can be no magnetic force. Hence, if there’s any force on the particle, it must come from an electric field. But what electric field? If the wire is neutral, there can be no electric flux from it.
You’ll say: why should there be a force on it? Forces also look different in different reference frames, don’t they? They do: they’re subject to the same Lorentz transformation rules: F’ = γF with γ = (1−v2/c2)−1/2. So, yes, the force looks different, but they surely do not disappear! Especially not because the typical drift velocity of electrons in a conductor is exceedingly slow. In fact, it’s usually measured in centimeter per hour and, hence, the Lorentz factor γ is extremely close to 1. 🙂 So the forces in the two reference frames should be nearly identical. Hence, the conclusion must be that the electromagnetic force in the S’ reference frame appears as some electric force, which implies that… Well… The bold conclusion is that our wire must be charged in S’ and, therefore, causes an electric field, rather than a magnetic field!
Huh? How is that possible?
To simplify the calculations involved, Feynman analyzes a special case: he equates v with v0. So that gives us the variables in diagram (b) above: in reference frame S’, we have some charge density ρ’+ and a velocity v’+ = –v0 = −v, while the electrons don’t seem to move: we have some charge density ρ’− but the velocity v’− = 0. As mentioned above, we cannot assume that ρ’+ = ρ+ or that ρ’− = ρ− and, therefore, we cannot assume that I = I’.
[…] OK. Now that we’ve explained all the variables involved, we’re ready to actually do the calculation. The crux of the matter is that a charge density is some number expressed per unit volume, and that the volume changes because of the relativistic contraction of distances. That’s what’s shown below.
As I mentioned in my posts on relativity, of all of the effects of relativity, length contraction is probably the most difficult to grasp. How comes the same amount of charge is suddenly spread over a smaller volume? Well… It is what it is, and I cannot say more about it than what I already said in the mentioned posts, so let’s get on with it. The (a) and (b) situations above describe the same piece of wire: its length and area, as measured in the stationary reference frame S, is L0 and A0 respectively, so its volume is L0·A0. If we denote the total charge in this volume as Q, then the charge density ρ0 will be measured as ρ0 = Q/(L0·A0).
Now what changes if we change the reference frame, so we look at this piece of wire moving past at velocity v? The dimensions that are transverse to the direction of motion don’t change, so the area A0 remains what it is. What about Q? Well… Q doesn’t change either. As mentioned above, there’s no such thing as relativistic charge, so there’s no equivalent for the mv = γm0 (or, multiplied with c2, Ev = γE0) formula when charges are involved. How do we know that? Feynman answers that question appealing to common sense:
“Suppose that we take a block of material, say a conductor, which is initially uncharged. Now we heat it up. Because the electrons have a different mass than the protons, the velocities of the electrons and of the protons will change by different amounts. If the charge of a particle depended on the speed of the particle carrying it, in the heated block the charge of the electrons and protons would no longer balance. A block would become charged when heated. As we have seen earlier, a very small fractional change in the charge of all the electrons in a block would give rise to enormous electric fields. No such effect has ever been observed. Also, we can point out that the mean speed of the electrons in matter depends on its chemical composition. If the charge on an electron changed with speed, the net charge in a piece of material would be changed in a chemical reaction. Again, a straightforward calculation shows that even a very small dependence of charge on speed would give enormous fields from the simplest chemical reactions. No such effect is observed, and we conclude that the electric charge of a single particle is independent of its state of motion. So the charge q on a particle is an invariant scalar quantity, independent of the frame of reference. That means that in any frame the charge density of a distribution of electrons is just proportional to the number of electrons per unit volume. We need only worry about the fact that the volume can change because of the relativistic contraction of distances.”
OK. That’s clear enough. Let’s get back to the lesson. The upshot here is that we don’t need to worry about the charge but about the charge density. To be specific, the charge density, as measured in the reference frame S’, will be equal to:
Why? If the total charge Q is the same in both S and S’, then Q = ρ0·L0·A0 must be equal to ρ·L·A0, with L the measured length in the S’ reference frame. Now, because of the relativistic length contraction effect, we know that L = L0·(1−v2/c2)1/2 and, therefore, ρ must be equal to ρ = ρ0·(1−v2/c2)−1/2. Capito?
We’re almost there. Now we need to apply this more general result to the ρ’−/ρ− and ρ+/ρ’+ density ‘pairs’ that we mentioned at the start. Let me copy the illustration once again so you can see what we are talking about:
The analysis is straightforward but a bit tricky. For the positive charges, you should note that they are at rest in (a), so that’s in reference frame S and, therefore, we can just write:
However, for the negative charges, we see they’re at rest in (b), and so that’s in reference frame S’, so the ρ0 in our general formula is not ρ− but ρ’−! So you should be careful when applying the same formula. However, if you are careful, you’ll agree we can write:
Now, the total charge density ρ’ in reference frame S’ is, of course, the sum of ρ’− and ρ’+. Now, also noting that we were looking at an uncharged wire in reference frame S, so ρ+ = − ρ−, we get the following grand result:
So our wire appears to be positively charged in the S’ frame, with a charge that’s equal to the product of the positive charge density and a β2/(1−β2)1/2 factor. So that’s our Lorentz factor γ multiplied by β2 = (v/c)2. The graph below compares how that factor increases as β = v/c goes from 0 to 1. We’ve also inserted the graph of the Lorentz factor itself, so you can compare both. Interesting, isn’t it? 🙂
Now, because the wire is electrically charged in reference frame S’, we have an electric field E’ which, using the formula for the field of a uniformly charged cylinder, can be calculated as:
Now, as far as I am concerned, that’s it. But… Well… Of course, we should generalize the analysis for v ≠ v0. However, I’ll refer you to Feynman for that. He also takes care of the remainder of the calculations you’d probably want to see, like a formula which show that the force on the charge in S’ is indeed what we would expect it to be. Feynman also shows that all other variables we can possibly calculate in the S’ reference frame, such as the momentum of the charged particle after the force has acted on it for some time all turn out be what we’d expect them to be according to special relativity.
However, I have to limit this post and, hence, I’ll just copy Feynman’s grand conclusion:
“We have found that we get the same physical result whether we analyze the motion of a particle moving along a wire in a coordinate system at rest with respect to the wire, or in a system at rest with respect to the particle. In the first instance, the force was purely “magnetic,” in the second, it was purely “electric. If we had chosen still another coordinate system, we would have found a different mixture of E and B fields. Electric and magnetic forces are part of one physical phenomenon—the electromagnetic interactions of particles. The separation of this interaction into electric and magnetic parts depends very much on the reference frame chosen for the description. But a complete electromagnetic description is invariant; electricity and magnetism taken together are consistent with Einstein’s relativity.”
So… That’s basically it for today’s lesson. 🙂 I should just add one more thing so as to be as complete as I should be in regard to the issue on hand here. You know the Lorentz transformation rules for the space and time coordinates, and you may or may not remember we had similar relativistic four-vectors for energy and momentum. Now, it turns out that we also have similar equations to relate charges and currents in one reference frame to those in another. More in particular, to transform ρ and j to a coordinate system moving with velocity in the x-direction, you should use the following rules:
But that’s really it for today. Have fun reflecting upon it all! 🙂
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Reading Feynman 
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