# Does the electromagnetic field “spin”?

Due to electron "spin", a small magnetic field is produced. Maxwell's equations imply that magnetic fields are due to changes in electric fields. Is the magnetic field produced then because the electric field is "spinning" with the "spinning" electron, in the quantum sense of "spinning" and this change in electric field is generating the magnetic field?

Can one generalize to say that the magnetic field thus would spin when a magnet is spun?

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–  Qmechanic Jan 31 '13 at 14:48
Maxwell equations state that a varying electric field produces a magnetic field, but not necessarily every magnetic field is produced by a varying electric field. The spin of a particle produces (for not straightforward reasons) a magnetic dipole, which produces a magnetic field independent on charge motions. –  Bzazz Jan 31 '13 at 16:01
Thanks. Do you know how to calculate the magnetic field produced then without Maxwell's equations. –  Mew Jan 31 '13 at 23:14

If by "spin", you mean rotate around its axis, like the earth does every 24h, then it is incorrect that the electric field of a point particle spins. A point particle doesn't have dimensions so it doesn't have an axis to rotate around and thus no magnetic field is produced.

The property of "spin" of elementary particles is not caused because of their rotation.

Now, the magnetic fields are different because there are no magnetic monopoles, so a magnetic field does rotate when the magnet is spun.

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How do you explain the magnetic field produced by "electron spin" if not by a change in electric field? –  Mew Jan 31 '13 at 14:59
Spin is an intrinsic property, as far as we know, just as it's mass. It is associated with an angular momentum, but it is quantized, and the group of symmetries is not the same as the group of symmetries of spinning macroscopic objects. The latter is SO(3) and the former SU(2). The difference,roughly, is that in the second one, after a rotation of 360 degrees, you get to the initial state with a minus sign. –  Barefeg Jan 31 '13 at 15:06
I understand that. I am wondering how the magnetic field results from this "spin". –  Mew Jan 31 '13 at 15:06
You cannot obtain the magnetic field as you would in the classical case. Because charge zero particle also have spin so you cannot suppose that the charge is distributed in a tiny sphere and work out the magnetic field. –  Barefeg Jan 31 '13 at 15:46
Well, the quantum dynamical description of the electron (spin 1/2 particles) is given by Dirac's equation. When you couple this with an external electromagnetic field (vector potential) you get a gauge theory. The equations of motion in that theory include a term which is the interaction energy of the magnetic field with the intrinsic magnetic moment of the particle. That therm predicts the correct magnetic moment of an electron. I'm not going to repeat the calculation here because is rather lengthy, but you can check a quantum mechanics book I guess. –  Barefeg Feb 1 '13 at 5:34

Spin corresponds to quantized angular momentum. However a substantial fraction of the spin angular momentum of an electron is included in its surrounding electromagnetic field where a nonzero Poynting vector does exist everywhere outside of its spin axis. This electron-bound Poynting vector corresponds to electromagnetic energy-momentum density circulating around the electron. The local magnetic field at a given point is given by the electron’s dipole field while the electrostatic field results from the Coulomb-field of a point-like charge [1].

Please also note that neither an electrostatic field nor a magnetostatic field can rotate like a rigid body. This misconception would contradict Maxwell's and relativistic electrodynamics. See Spinning magnets and Jehle’s model of the electron.

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Your statement "Maxwell's equations imply that magnetic fields are due to changes in electric fields." is not complete.

A corrected statement is that Maxwell's equations imply that magnetic fields are due to changes in electric fields AND due to currents (which can be stationary):

$$\nabla\times\mathbf{H} = \mathbf{J}+\partial\mathbf{D}/\partial t$$

As you can see, the magnetic field $\mathbf{H}$ has two "sources": the $\partial\mathbf{D}/\partial t$ part is due to varying electric fields as you said (where $\mathbf{D}$ is the electric displacement), but the $\mathbf{J}$ is the part due to the free currents. This is why a coiled wire with a constant current running through it creates a magnetic field (without need of changing electric fields).

In the case of the electron spin, this goes beyond my areas of knowledge, but according to my limited understanding of quantum mechanics, the magnetic field comes from the stationary particle current associated with the wave-function of the electron. So it is similar to the magnetic field arising from a coil of wire with a current.

As a further related note: interestingly, Maxwell's equations apply to any inertial frame, so you could argue that an observer moving with respect to the electron will see a changing electric field (because the electron is moving), and this will create a magnetic field which apparently does not exist for a stationary observer. This is because different observers will not agree on the electric and magnetic fields separately, however they will agree on the existence of an electromagnetic tensor (which includes the electric and magnetic fields as its "parts"), and they will agree on the physical effects produced by it.

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