It is often mentioned that the density of angular momentum of an electromagnetic field is given (up to constants) by the expression (e.g. in the Wikipedia page): $$\mathbf M\equiv \mathbf r\times(\mathbf E\times\mathbf B).\tag A$$ We also know that there are two "types" of angular momentum that an EM field can carry: spin and orbital angular momentum. At least from a quantum mechanical perspective, these are quite different beasts: the spin being an intrinsically two-dimensional degree of freedom, while the orbital one being related to the spatial distribution of photons/light.

I wonder, from a classical perspective, does (A) account for both spin and orbital angular momenta? Is there any easy way to see this?


Yes it does, paradoxically. It goes like this. To reach explicitly gauge independent conservation laws the well known GI Lagrangian is adopted. Unfortunately, the Noether currents that come out are not GI. The energy-momentum tensor is asymmetric and not GI. The angular moment consists of an orbital term of the form ${\bf r} \times {\bf P}$, involving the em tensor, and an intrinsic, spin term independent of $\bf r$. This is corrected by making an ad hoc modification of the energy-momentum tensor density making it symmetric and GI. Substitution of this for the Noether EM tensor in the angular momentum leads to a GI total angular momentum that is of the orbital form. There is no spin term. Indeed, it is impossible to explicitly account for spin with a GI expression.

Not everybody is happy with this, notably one person and the is me. That is why I published a paper with a completely valid alternative approach in which spin and orbital momentum are separately conserved mechanical variables, at the expense of gauge invariance.

  • $\begingroup$ by "GI lagrangian" do you mean $-F^2$? I don't quite understand what you are saying though. I don't see this Lagrangian used anywhere in the wiki page you link. Also, you say that $r\times P$ accounts for spin at the beginning, but then you seem to be saying that the GI lagrangian also gives an additional spin term. Or do you mean that the spin is included in term when the "corrected" GI Lagrangian is used instead? $\endgroup$ – glS Jan 14 '20 at 0:31
  • $\begingroup$ also, is there a direct connection between this and the calculations in the linked answer? Are the two components they derive the same that are obtained using the unmodified GI Lagrangian? $\endgroup$ – glS Jan 14 '20 at 0:33
  • $\begingroup$ 1. Indeed, $-\frac{\epsilon_0}{4} F_{\mu\nu} F^{\mu\nu} $. 2. The GI expression accounts for the total, volume integrated angular momentum. $\endgroup$ – my2cts Jan 14 '20 at 8:46
  • $\begingroup$ As to the linked answer, the first three paragraphs deal with the paradox that a plane wave cannot have angular momentum along its propagation direction. The solution is that the volume integrated AM is correct as the integral picks up AM in regions where the wave is no longer plane. Correct, but it does not alter the fact that all AM is in the form.of orbital momentum as I state in my answer. $\endgroup$ – my2cts Jan 14 '20 at 8:52
  • $\begingroup$ The linked answer then goes on to back engineer a non GI expression for spin very similar to the original Noether expression. Why do complicated. The conclusion still the same: either you have GI and no spin or spin and no GI. $\endgroup$ – my2cts Jan 14 '20 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.