Rotating Electromagnetic Field Transformation

Question

Lets say we have the following problem:

We are given some charge configuration in 3D, $\rho (s,\theta)$ (cylindrical coordinates), which doesn't depend on z. Let's say that the total charge is zero.

This charge distribution creates some electric field $E(s,\theta)$. The magnetic field is, obviously, zero everywhere.

Now we start rotating this charge configuration around the z-axis. What happens to the EM-field? I think that the electric field will remain the same (except that it rotates), and there will be some magnetic field induced. Is there some universal transformation formula for EM-fields under rotation? The rotation velocity is non relativistic!

Hint: I am looking for a formula like $\vec{E'} = \vec E+\vec v \times \vec B$, which transforms the magnetic field.

Thank you!

Actually, I don't have a specific example. But if it makes life easier, let's take two parallel infinite lines of opposite charge, which are 1 m apart. Both at the same distance from the z-axis, at opposite sites. — Samuel, Mar 24 '17 at 17:23
Note that the concept of a rigidly rotating object in special relativity is deeply problematic; for more details see the Ehrenfest paradox and related issues. As such, it's entirely possible that there is no satisfactory answer because the situation you propose is either unphysical or insufficiently specified, but in any case these are troubled waters and one needs to pay special attention to the subtleties of the relativistic treatment. — Emilio Pisanty, Mar 24 '17 at 18:01
Yes, I am aware that this is quite complicated. But for now, let's just assume that $c=\infty$ :) — Samuel, Mar 24 '17 at 19:16
@EmilioPisanty OP says that "The rotation velocity is non relativistic". Am I missing something or your concern is not applicable here? — aaaaa says reinstate Monica, Mar 24 '17 at 19:24
@aaaaaa Any magnetic effects are ultimately caused by relativistic considerations (or can be seen to do so), so it's certainly not a trivial concern at the very least. — Emilio Pisanty, Mar 24 '17 at 19:40

score 0 · Answer 1 · edited Mar 24 '17 at 17:40

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I think since you have axial symmetry, $\rho=\rho(r,\theta)$, you can decompose your charge density into infinitely thin loops carrying current of

$$ I(r,\theta)drd\theta=r\omega\rho(r,\theta)rd\theta dr $$

$\omega$ here is your system's angular speed.

Now you just use equation for field of current loop and integrate over $r$ and $\theta$.

You can see how this is applied in case of the rotating charged disk here.

Finally, you will have to go and integrate over $z$ as well, of course.

edited Mar 24 '17 at 17:40

Eric Angle

2,926

answered Mar 24 '17 at 17:27

aaaaa says reinstate Monica

1,229

Hey, thanks for the quick response. Yes, I agree that this would be a very elegant solution. But unfortunately, there is no axial symmetry, as $\rho$ depends on $\theta$ :( – Samuel Mar 24 '17 at 17:42
@SamuelBosch ok, maybe i don't understand something here. What is your coordinate system? Cylindrical or spherical coordinates? In any way, idea is to present each small fraction of $]rho$ as single loop with current proportional to $\omega$ and integrate. $\rho$ can be anything, symmetries just make integration easier in practice – aaaaa says reinstate Monica Mar 24 '17 at 18:29
Sorry, I should have made it more clear that I was thinking about cylindrical coordinates. Therefore it is not a very good approximation to take small current loops. – Samuel Mar 24 '17 at 19:18
if coordinates are cylindrical and charge density does not depend on Z then thin loops of current should work OK. Just figure out $B(z,r)$ for current loop first. Also cylindrical coordinates are usually $(r, \theta, z)$, not $(s, \theta)$ – aaaaa says reinstate Monica Mar 24 '17 at 19:27

Rotating Electromagnetic Field Transformation

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