Where the angular momentum has gone?

I came across an interesting problem when I prepared for the preliminary exam on electromagnetism. Below is the problem in its original words:

A metallic sphere of mass, $m$, and radius, $a$, carries a net charge, $Q$, and is magnetized with uniform magnetization, $M$, in the $z$-direction.

(a) Determine the total angular momentum associated with the electromagnetic field.

(b) The magnetization of the sphere is now reduced to zero. Assuming that no external mechanical forces act on it (PS: this is possible. For instance, just heat up the sphere and the magnetization will be eliminated ), determine the angular velocity and the sense of rotation of the sphere when $M=0$.

(c) Describe the origin of the torque that causes the sphere to rotate.

For part (a), it is straight forward to calculate electric and magnetic field E, H inside and outside of the sphere. Generated by the net charge $Q$ uniformly distributed on the surface, the electric field $E$ is isotropic outside the sphere and vanishes inside. The magnetic field H is generated by magnetization $M$ and can be calculated using the "magnetic scalar potential" formalism since there is no free current. The result is that H is also uniform inside the sphere and takes the dipole form outside the sphere.

The total angular momentum $J$ is obtained by integrating over the region where a local angular momentum density of the EM field exists, i.e. outside of the sphere. It is in the $z$-direction and I find it to be

$$J= \frac{2 Q M a^{2}}{9 \epsilon_0 c^{2}}$$

For part (b) and (c), it is interesting to ask why the sphere would rotate after magnetization $M$ is erased. What I think up is that as $M$ decreases to zero, a electric field with non-zero curl is generated, as implied by Faraday's law. This electric field is in the azimuthal direction, and will act on the surface charge $Q$ to provide a torque. However, a straightforward calculation can show that the angular momentum this torque can transfer is

$$J'= \frac{Q M a^{2}}{9 \epsilon_0 c^{2}}$$

Obviously a factor of 2 is missing, so this only accounts for half of the angular momentum stored originally in the EM field. So I am perplexed by where the other half has gone?

I come up with 2 possibilities:

1. since the charged sphere begins to rotate, there is also non-zero angular momentum stored in EM field after magnetization is removed. But it could hardly account for the missing amount, since in that case the electric field $E$ is proportion to $Q$, the magnetic field $H$ to $Q^2$ ( one $Q$ from the charge and a second $Q$ from the angular velocity). Hence one expects the angular momentum stored is proportional to $Q^3$, which is very different from $J$'s linear dependence on $Q$.

2. maybe in the process of magnetization removal, EM radiation carries away some angular momentum. But this possibility has to be validated quantitatively before it can convince people. What I feel is that if the process of removal is infinitely slow, then the effect of radiation can be neglected.

Then what is the correct answer? I provide quantitative result for $J$ and $J'$ in the hope that someone can check them and point out that I have made a wrong calculation lol. But I am really eager to hear from any explanation that sounds real ... Electromagnetism is as interesting as it was 3 years ago, but I just don't remember as much as I did then ...

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Did you took into account that during the process of the magnetization removal the whole EM field rotates as a whole?(Consider an infinitely slow removal) –  Martin Gales Aug 3 '11 at 10:03
This was asked by one of my friends, so I do not really know.. –  Kerry Aug 4 '11 at 5:25
Can you add the calculations of $J$ and $J'$. –  Martin Gales Aug 4 '11 at 6:34

http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect

Edit:

Quote:

Calculations based on a model of electron spin as a circulating electric charge underestimate this magnetic moment by a factor of approximately 2, the Landé g-factor. A correct description of this magnetic moment requires a treatment based on quantum electrodynamics.

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@whoplisp: Of course, this describes the phenomenon but does not answer the question. –  Martin Gales Aug 4 '11 at 6:35
@Martin Gales I urge anyone with the experience to give an answer with calculation. I'm not up to the task. –  whoplisp Aug 4 '11 at 7:22
The charge on the sphere maybe is some "catch question" ? –  Georg Aug 4 '11 at 10:04

So when calculating the magnetic flux through a surface, did you choose a surface inside the sphere to get your azimuthally generated electric field? And if so, did you make sure to note the the $\mathbf{B}$ field inside the sphere is $\frac{2}{3}\mu_0\mathbf{M}$ where $\mathbf{M}$ is the magnetization vector? (before the magnetization is changed, of course) Because when I did that, and and got an azimuthal electric field, and then found the differential force on the surface of the sphere, and used that to get the differential torque, I found that when integrating the torque over the whole sphere, and then integrated the torque $dt$ from $t_0$ to $\infty$, that all of the angular momentum is accounted for.

P.S. It is convenient the the legendre G-factor is 2, but we are not dealing with a quantum description of reality when were are dealing with a macroscopic sphere. When doing a similar calculation to compute the magnetic moment of an electron, you get an answer that is half of what it should be. But that is when a classical model is employed.

Also: the electric and magnetic fields cannot just carry off momentum, because they extend to infinity so are part of the system. And since the electromagnetic angular momentum is zero after the sphere is done magnetizing, all of that angular momentum must be imparted to the sphere (no external forces). Yes, the spinning sphere will have a magnetic field, but the electric field will not be that of a static configuration. Thinking about it, it is not so surprising that the magnetic field that the charges on the sphere produce when moving cannot produce and electric field that can act on them, because it would be rather absurd especially classically if charges could act on themselves.

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you can always edit your own posts to include additional information. –  Neuneck Jul 15 '13 at 6:56
Not only can, but should: comments are subject to deletion without notice and without cause. They are not first class objects ans should never be relied upon to provide important information. –  dmckee Jul 15 '13 at 16:57

The angular momentum is in the energy that was added into the object. After the object has cooled, the angular momentum is in the heat radiation.

But: If the object radiates mostly from the center parts, the radiation has less angular momentum than it has when it is the outer parts that radiate the most. Can you solve this?

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