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According to the answer of @PhysicsDave here, there can be energy stored in an electromagnetic field even when the $E$ and $M$ fields are zero everywhere.

However, according to Griffiths, page 380, equation 9.53, the energy per unit volume in an electromagnetic field is given by $$\frac12\left(\epsilon_0\mathbf{E}^2+\frac{1}{\mu_0}\mathbf{B}^2\right).$$ So when $\mathbf{E}=\mathbf{B}=0$, this would equate to zero.

How can this contradiction be explained?

Riemann
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    In my mind, the "resolution to this contradiction" is that PhysicsDave's statement in that answer is just wrong. I'm honestly not sure what he could be referring to. Perhaps this question would have been better as a comment on that answer. – Michael Seifert Jun 06 '22 at 13:09
  • Good point, I have now placed a comment there – Riemann Jun 08 '22 at 15:13
  • @MichaelSeifert I am merely pointing to the fact that 2 photons crossing opposing paths in space will lead to zero E and B for a very brief time ... but the energy can NOT be destroyed... therefore the energy is still in the field. A similar analogy to water waves, they can pass thru each other and travel on their respective opposing paths .... but for a brief time the water is flat the energy having being stored in the medium. – PhysicsDave Jun 08 '22 at 21:47
  • @PhysicsDave, what do you think of the formula $\frac12(\epsilon_0\mathbf{E}^2+\frac{1}{\mu_0}\mathbf{B}^2)$? If it is temporarily zero everywhere, then where is the energy? – Riemann Jun 09 '22 at 09:00
  • That is exactly the point .... where is the energy! We are taking about photons passing each other either perfectly or even at slight angles .... which we know is possible. We are also taking about things on a very small scale ... in the real world every photon created by an electron/atom is eventually absorbed by another electron/atom .... energy is transferred but never lost. – PhysicsDave Jun 09 '22 at 10:43
  • Because energy is conserved the formula is correct on the larger scale .... but if we look at the quantum scale obviously there are exceptions .... at the small quantum scale things are very interesting. – PhysicsDave Jun 09 '22 at 10:49
  • Could you cite a source for two photons passing each other (almost) perfectly? – Riemann Jun 10 '22 at 15:45

2 Answers2

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If standard Poynting expressions for EM energy are adopted, then vanishing fields $\mathbf E,\mathbf B$ in vacuum indeed imply zero EM energy density there. So in this context, PhysicsDave is wrong.

Although PhysicsDave did not explain what he means, his statement could be true if non-standard expressions for EM energy density and EM energy flux vector (other than the Poynting expressions) were adopted (which is entirely consistent with EM theory).

Density of EM energy $\rho$ and vector of EM energy flow $\mathbf S$ are ambiguous, because they are defined based on the desired equation form

$$ \partial_t \rho + \nabla \cdot \mathbf S = -\mathbf j\cdot \mathbf E $$ which does not fix exact expressions for $\rho, \mathbf S$.

One possible set is the Poynting set, but given any pair of functions $\rho, \mathbf S$ which obey the above condition, we can define another pair via

$$ \rho' = \rho - \nabla \cdot \mathbf u $$ $$ \mathbf S' = \mathbf S + \partial_t \mathbf u $$ where $\mathbf u$ is any vector field differentiable in both space and time coordinates. With a convenient choice of $\mathbf u$, and using Poynting's expressions for $\rho,\mathbf S$, we can get non-zero $\rho'$ even in places where $\mathbf E = \mathbf B = \mathbf 0$.

Ján Lalinský
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  • For OP: this choice of transformation, although does effect field energy, and the expression for the poynting vector. It does not, effect the work done on charges. So in the sense that although there is "field energy" in regions of E=B=0, this energy is useless and cannot be used for anything. ( as far as I am aware) and is (probably*) a mathematical artifact, than anything physic – jensen paull Jun 06 '22 at 15:16
  • As such, Including this energy In poynting theorem does not mean extra energy is needed to create the fields either, it exists independant to the location of charges and currents – jensen paull Jun 06 '22 at 15:20
  • However: $\partial_{t}u = 0$, has some very similar properties to dark energy. time invariant to the universes expansion, provided this energy still interacts with gravity. – jensen paull Jun 06 '22 at 15:21
  • @jensenpaull I am merely pointing to the fact that 2 photons crossing opposing paths in space will lead to zero E and B for a very brief time ... but the energy can NOT be destroyed... therefore the energy is still in the field. A similar analogy to water waves, they can pass thru each other and travel on their respective opposing paths .... but for a brief time the water is flat the energy having being stored in the medium. – PhysicsDave Jun 08 '22 at 21:45
  • @jensenpaull I agree with your assessment that at a certain point this energy is "useless" nontheless the EM field is able to recover this energy as the "interfering" photons continue along their merry way a brief time later. – PhysicsDave Jun 08 '22 at 22:25
  • @PhysicsDave > "2 photons crossing opposing paths in space will lead to zero E and B for a very brief time" Why do you think this? Two oppositely travelling EM waves do not cancel EM field everywhere, not even for an instant. – Ján Lalinský Jun 09 '22 at 01:01
  • https://readingpenrose.files.wordpress.com/2015/01/photon-wave.gif For example in this image if red and blue are 2 waves representing E field they will net to zero when they superimpose in opposite directions. – PhysicsDave Jun 09 '22 at 01:33
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    @PhysicsDave and what happens to magnetic fields? They superimpose in the same direction! – Ján Lalinský Jun 09 '22 at 02:01
  • @JánLalinský So now we have a 2 photons for a brief period that is all B and no E?! And we also must consider photon polarization/spin .... this leads to more permutations. And also if if we take 4 photons then it is possible to make 2 pairs with opposite B fields and now we have zero E and B? Superimposed photons present many interesting possibilities in theory .... but in reality these quanta all eventually become absorbed. – PhysicsDave Jun 09 '22 at 02:16
  • @PhysicsDave not sure if you understand my comment. I was commenting on Jan's answer and the mathematical freedoms of poyntings theorem. There is redundant field energy that cannot interact electromagnetically. This phenomena is completely seperate to the issue of destructive and constructive interference for EM waves , which I might add, in the context of maxwells equations, I dissagree with you. When E destructively interferes, B constructively interferes. In the context of photons, I would not be able to comment. – jensen paull Jun 09 '22 at 13:30
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If you are letting (M) represent the magnetization within a ferromagnetic material, it might be possible for a superconductor to maintain a (B) without an (E).

R.W. Bird
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