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A Maxwell's equations and exterior algebra

  1. Feb 1, 2019 #1
    Maxwell's equations in differential form notation appeared as a motivating example in a mathematical physics book I'm reading. However, being a mathematical physics book it doesn't delve much into the physical aspects of the problem. It deduces the equations by setting dF equal to zero and d(*F) equal to J, but it doesn't explain why it is doing so. This article I found online does exactly the same thing. I'm left with the impression that this condition is just set in order to obtain the equations. Is that so? If not, is there any physical interpretation here to the exterior derivative? And what about the exterior derivative of the Hodge operator applied to the tensor being equal to the current density?
     
  2. jcsd
  3. Feb 1, 2019 #2
    I think I can actually make sense of the first condition, since the electromagnetic tensor is defined as the exterior derivative of the electromagnetic four-potential and that by Poincaré's lemma we must have d(dw) = 0, it follows that indeed dF must equal zero. But then again, that's just math, I'm wondering about the physics involved here.
     
  4. Feb 15, 2019 at 9:00 PM #3
    If anyone is interested at all, I believe I found an answer here and here. If I understood it correctly, the physics behind the first condition is just gauge invariance? I don't get the cohomology bit though. The second condition is easier to understand as it just translates to conservation of charge.
     
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