The equation for conservation of angular momentum in electrodynamics is

\begin{equation} \frac{\mathrm d\vec{L}_\mathrm{mec}}{\mathrm dt} + \frac{\mathrm d\vec{L}_\mathrm{EM}}{\mathrm dt} = \int M \cdot \mathrm d \vec{S} \end{equation}

where $\vec{L}_\mathrm{EM}=\int \vec{r} \times (\frac{\vec{E} \times \vec{B}}{4\pi c})\mathrm dV$ is the electromagnetic angular momentum and $M_{il}=\epsilon_{ijk}r_j T_{kl}$ with $T_{ij}$ the Maxwell tensor.

The question is: does $\vec{L}_\mathrm{EM}$ include both orbital and intrinsic angular momentum? Looking at the definition it just seems like an orbital angular momentum $\vec{r}\times\vec{p_{EM}}$ but I think that intrinsic angular momentum should also be included somewhere... is it?

  • $\begingroup$ Perhaps there is a QED-analogon for this expression, however in the classical way you wrote it down, there is no reason why intrinsic angular momentum -- usually considered as a purely quantum mechanical quantity -- should be incorperated. (Maybe it's wrong to say that intrinsic ang.mom. is purely q.m. as, as far as I know, there are ways to include spin semiclassically. However usual Electro-Dynamics doesn't make any effort towards this) $\endgroup$ – Peter Wildemann Jul 30 '17 at 16:58
  • $\begingroup$ isnt polarization the intrinsic angular momentum in classical electromagnetism? $\endgroup$ – P. C. Spaniel Jul 30 '17 at 17:00

Yes, the integral you've specified is the total angular momentum and it includes both intrinsic and extrinsic angular momentum. This is a tricky subject because the decomposition of angular momentum into intrinsic and extrinsic components is normally not unique, but generally speaking the literature tends to recognize two contributions to that angular momentum, an orbital angular momentum and a spin contribution, which read $$ \mathbf J = \frac{1}{\mu_0c^2} \int\mathrm d\mathbf r \: \mathbf r \times (\mathbf E\times\mathbf B) = \frac{1}{\mu_0c^2} \int\mathrm d\mathbf r \: (E_i (\mathbf r \times\nabla) A_i ) + \frac{1}{\mu_0c^2} \int\mathrm d\mathbf r \: \mathbf E\times \mathbf A =\mathbf L + \mathbf S , $$ and which can be obtained via an integration-by-parts argument as in this previous answer of mine. For more details see the references in that thread or this PhD thesis and its references.

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  • $\begingroup$ Thanks! does intrinsic angular momentum have to do with polarization? is there a way to see this in the equation? $\endgroup$ – P. C. Spaniel Jul 30 '17 at 19:35
  • $\begingroup$ @P.C.Spaniel Yes, the intrinsic component (in this decomposition) is directly tied to the polarization properties, which is evident in the absence of spatial derivatives. $\endgroup$ – Emilio Pisanty Jul 30 '17 at 20:23

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