Does the angular momentum density $\mathbf r\times(\mathbf E\times\mathbf B)$ account for the spin? 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?
 A: 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.
