I found in the following reference (p. 10) an interesting decomposition for the electromagnetic angular momentum in terms of an orbital terms $\vec{L_{orb}}$ and an spin term $\vec{L_{spin}}$. However, I've been struggling for a few days on how they did the derivation, in particular what identities and integration by parts they used from (1) to (2) and from (2) to (3).

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I've looked for corresponding identities everywhere (Jackson, Wikipedia, Stewart) but I can't see to find how. I would only like if you could point me out the necessary vector identity or integration by parts, I'll work the rest by myself. One thing to note is that the fields vanish at infinity (hence why we can go from (4) to (5)).

PS: I tried asking in Math Exchange but it apparently it was downvoted because it's not related to the site, hence why I decided to ask here.

  • $\begingroup$ Electromagnetic spin is problematic as it is not gauge invariant. That may be why in Jackson Electrodynamics spin has a very low profile. It is only discussed in problem 7.19. This problem addresses your question. $\endgroup$ – my2cts Mar 12 '19 at 20:06
  • $\begingroup$ Please do not post images of mathematics. We have MathJax active on the site, and that is both editable and accessibly to screen readers. $\endgroup$ – dmckee --- ex-moderator kitten Mar 21 '19 at 3:10
  • $\begingroup$ @dmckee understood, I won't do it again, my apologies. $\endgroup$ – Charlie Mar 21 '19 at 4:50

(4) -> (5) seems to be about Gauss's law and integration by parts.

For simplicity, lets define vector $\mathbf{G}=\mathbf{r}\times\mathbf{A}$. Your second term in Eq. (4) is then:

$\mathbf{II}=\int_V d^3 r \left[-\frac{\left(\mathbf{E}.\boldsymbol{\nabla}\right)\mathbf{G}}{4\pi c}\right]$

I am not sure if you are familiar with abstract index notation, so I will avoid using it and apply arguments based on simple Cartesian basis. The quantity above is a vector. Let us project it along some other constant vector $\mathbf{\hat{u}}$. This vector could be $\mathbf{\hat{u}}=\mathbf{\hat{x}}$ or $\mathbf{\hat{y}}$ or $\mathbf{\hat{z}}$, or a combination of them, as long as it does NOT depend on position (so it is unaffected by $\boldsymbol{\nabla}$). So:

$\mathbf{\hat{u}}.\mathbf{II}=\int_V d^3 r \mathbf{\hat{u}}.\left[-\frac{\left(\mathbf{E}.\boldsymbol{\nabla}\right)\mathbf{G}}{4\pi c}\right]=\int_V d^3 r \left[-\frac{\left(\mathbf{E}.\boldsymbol{\nabla}\right)\mathbf{\hat{u}}.\mathbf{G}}{4\pi c}\right]=\int_V d^3 r \left[-\frac{\left(\mathbf{E}.\boldsymbol{\nabla}\right)G_u}{4\pi c}\right]$

Now we can integrate by parts:

$\mathbf{\hat{u}}.\mathbf{II}=\int_V d^3 r \left[-\frac{\left(\mathbf{E}.\boldsymbol{\nabla}\right)G_u}{4\pi c}\right]=\int_V d^3 r \boldsymbol{\nabla}.\left[-\frac{\left(\mathbf{E}G_u\right)}{4\pi c}\right]-\int_V d^3 r\left[-\frac{\left(\boldsymbol{\nabla}.\mathbf{E}\right)G_u}{4\pi c}\right]$

Apply Gauss's law to the first term.

$\mathbf{\hat{u}}.\mathbf{II}=\oint_{\partial V} d^2 r \mathbf{\hat{n}}.\mathbf{E}\left[-\frac{G_u}{4\pi c}\right]-\int_V d^3 r\left[-\frac{\left(\boldsymbol{\nabla}.\mathbf{E}\right)G_u}{4\pi c}\right]$

Now comes the hand-waving to make the first term (surface term) vanish (for large $V$). One option. The field in the far-field decays as $1/r^2$ or faster which is sufficient to make the surface term vanish. Second option. Say that your light is eminating from a localized source, then in the far-field the electric field will be transverse, and thus $\mathbf{\hat{n}}.\mathbf{E}=0$. Third option. Postulate some small absorption in the free-space (dust in the air for example), which will eventually supress the light, thus the surface term will be zero, if $V$ is large enough.

Then you are done

$\mathbf{\hat{u}}.\mathbf{II}=\int_V d^3 r\left[\frac{\left(\boldsymbol{\nabla}.\mathbf{E}\right)G_u}{4\pi c}\right]$

but since this applies for any $\mathbf{\hat{u}}$, the statement is general:

$\mathbf{II}=\int_V d^3 r\left[\frac{\left(\boldsymbol{\nabla}.\mathbf{E}\right)\mathbf{G}}{4\pi c}\right]$

Assuming trivial diagonal metric (i.e. Cartesian coordinates),

$\left(\mathbf{r}\times\mathbf{E}\times\boldsymbol{\nabla}\times\mathbf{A}\right)_\alpha=\epsilon_{\alpha\beta\gamma}r_\beta \epsilon_{\gamma\mu\nu}E_\mu\epsilon_{\nu\kappa\phi}\partial_\kappa A_\phi=\epsilon_{\alpha\beta\gamma}r_\beta \left(\delta_{\gamma\kappa}\delta_{\mu\phi}-\delta_{\gamma\phi}\delta_{\mu\kappa}\right)E_\mu\partial_\kappa A_\phi$ $\left(\mathbf{r}\times\mathbf{E}\times\boldsymbol{\nabla}\times\mathbf{A}\right)_\alpha=\epsilon_{\alpha\beta\gamma}r_\beta \left(E_\mu\partial_\gamma A_\mu-E_\mu\partial_\mu A_\gamma\right)=\left(E_\mu\left(\mathbf{r}\times\boldsymbol{\nabla}\right)A_\mu - \mathbf{r}\times \left(\mathbf{E}.\boldsymbol{\nabla}\right)\mathbf{A}\right)_\alpha$

  • $\begingroup$ Thanks a lot. Indeed I had suspected how to go from (4) to (5) through integration by parts but I was missing the use of Gauss theorem to fully prove it, so you made it very clear. On the other hand, do you know what could apply to (1) to (2)? I found an identity in Jackson's: $A\times (\nabla\times B)=\nabla (A\cdot B)-(A\cdot\nabla)B-(B\cdot\nabla)-B\times(\nabla\times A)$ which could potentially help if I make the last three terms zero through a similar procedure. $\endgroup$ – Charlie Mar 12 '19 at 23:18
  • $\begingroup$ I think (1)->(2) is a straight evaluation with no integration by parts or any such stuff. Simply try to apply the same $\mathbf{\hat{u}}$ trick and evaluate the integrands of (1) and (2) completely. You should then see that they are equal. $\endgroup$ – Cryo Mar 12 '19 at 23:32
  • $\begingroup$ Are you familiar with Levi-Civita tensors and index notation? $\endgroup$ – Cryo Mar 12 '19 at 23:35
  • $\begingroup$ Yes I am familiar with Levi-Civita tensors and index notation, I'll try working with it then. $\endgroup$ – Charlie Mar 13 '19 at 0:25
  • $\begingroup$ I added a proof based on index notation and contraction of Levi-Civita tensors $\endgroup$ – Cryo Mar 13 '19 at 0:47

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