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It seems like the identities of curl of gradient, divergence of curl, and the simple derivations of electromagnetic waves from Maxwell equations all rely on the symmetry (interchangeability of their order) of partial derivatives.

But mathematically speaking, this tacitly assumes that the electric/magnetic fields in question are continuously double differentiable, something which needs justification.

Why do we assume that in the real world, the fields are smooth enough for the identities to work, atleast almost everywhere?

Sidd
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    Possible duplicate of: Electromagnetic field and continuous and differentiable vector fields – John Rennie Jul 22 '16 at 15:45
  • @JohnRennie: The answer to the question you linked talks about the electric and magnetic fields possibly being discontinuous distributions in $\mathscr{S}'(\mathbb{R}^3)$. It doesn't give a reason why we assume the fields to be nicely behaved, and more importantly, only talks about them naturally belonging to spaces like $L^2$, but that doesn't imply continuous second derivatives. – Sidd Jul 22 '16 at 15:55
  • Related: http://physics.stackexchange.com/q/1324/2451 and links therein. – Qmechanic Jul 22 '16 at 16:19
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    The question is indeed thought provoking. – hsinghal Jul 22 '16 at 17:51
  • Have you ever seen a non-continuous, non-differentiable physically measurable function? I have not. – CuriousOne Jul 22 '16 at 21:42
  • @CuriousOne electric field of a point charge has singularity at 0 and thats where it become non continuous. – hsinghal Jul 23 '16 at 02:38
  • @hsinghal: I would challenge you to try and find that singularity on a real charge. :-) This is just another case of mistaken identity: the picture of the thief is not the thief. – CuriousOne Jul 23 '16 at 04:53
  • The question JohnRennie linked is a sufficient duplicate. What is perhaps not so clear to you in the answer there is that many $L^2$ functions do have a notion of "second derivative" even if they are not continuous or differentiable in the strong sense you are thinking of, cf. weak derivative. – ACuriousMind Jul 23 '16 at 11:39

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