This is a pretty specific physics question and may not be appropriate for this board, but I haven’t been able to get the answer anywhere else – so what the hell.
If a localized system is not moving it’s total linear momentum has to be zero. This very obviously means that static electric and magnetic fields cannot store linear momentum.
On the other hand, I know for a fact that static E and B fields can store angular momentum. Why would these static fields be able to store angular momentum when they can’t store linear momentum. Is a system that is rotating considered to be stationary? Or does moving invariably mean translation?
My E&M is a little shaky, but I’ll throw out my thoughts:
When you say “E and B fields can store angular momentum”, you’re referring to spin angular momentum, correct? I think the answer is that the intrinsic spin of the photons does not have a linear equivalent. (Electric and magnetic field interactions are mediated by photons, which is why I’m talking about them.) However, photons can have both linear/angular momentum if you move them in a line/circle.
Moving does not invariably mean translation. A rotating system is not stationary, unless you’re in the rotating frame of the system.
Actually, I’m asking about strictly classical electromagnetism - nothing to do with QM at all. However I may have phrased the question wrong. I should probably have said, " a non moving localized system can have no net linear momentum."
The electromagnetic linear momentum density p[sub]em[/sub]= mu[sub]0[/sub] eps[sub]0[/sub]S where S is the Poynting vector (S = 1/ mu[sub]0[/sub] (E X B). But this is exactly offset by the hidden relativistic effects of the current flow.
Whereas the angular momentum = r X p[sub]em[/sub] and is not offset by anything. Hence my original question.
My E&M is also pretty rusty, but I’m not quite sure what you mean by saying that the field linear momentum is offset by relativistic effects.
As I understand it, the total linear momentum is conserved when the Maxwell stress tensor is taken into account (that is, the time derivative of the total linear momentum density is equal to the divergence of the stress tensor). Is this what you mean? If so, there’s certainly an equivalent for angular momentum.
I don’t see that it’s such a problem… I mean, if I spin a top on the table, it’s not going anywhere, either (zero net linear momentum), but it’s got angular momentum. Likewise, you can have a static arrangement of E and B fields such that the Poynting vector is not zero everywhere, but it’s pointing around in a circle, so the net momentum of the whole field is zero.
On second thought, Chronos, I don’t think your post answers the underlying question.
To wit:
If you can’t have linear momentum without linear motion how is it that you can have angular momentum without rotational motion. Your top example, for instance, did have rotational motion.
