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Electromagnetic energy is not Gauge invariant?

  1. Feb 8, 2009 #1
    I assume I am making a mistake here. Can you please help me learn how to fix them?

    In electrodynamics, the gauge transformations are:
    [tex]\vec{A} \rightarrow \vec{A} + \vec{\nabla}\lambda[/tex]
    [tex]V \rightarrow V - \frac{\partial}{\partial t}\lambda[/tex]

    These leave the electric and magnetic fields unchanged. The electromagnetic energy density is proportional to:
    [tex] u \propto (E^2 + B^2)[/tex]
    So the energy density should be gauge invariant. A guage transformation shouldn't even shift it by a constant.


    However, the energy density can also be written in terms of the potentials and charge distributions as:
    [tex] u = \frac{1}{2}(\rho V + \vec{j} \cdot \vec{A})[/tex]

    If I let [itex]\lambda[/itex] = e^(-r^2), then V is unchanged, and A is just changed by a radial field which vanishes at infinity. Here, the energy density IS changed, and not by a constant amount either ... it changes by an amount depending on j. What gives?

    Even weirder, is some textbooks start with the potentials form to derive the fields form. I don't see any place they fix any gauge in doing such derivation. Please help.
     
  2. jcsd
  3. Feb 8, 2009 #2

    samalkhaiat

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  4. Feb 8, 2009 #3
    I wanted to talk about the electromagnetic energy density (proportional to [itex](E^2+B^2)[/itex]), but you have instead talked about a relativistic scalar density (proportional to [itex](E^2-B^2)[/itex]). Therefore I am not sure how to relate your response back to the original question. Are you saying [itex]A_{\mu}J^{\mu}[/itex] is gauge invariant but the electromagnetic energy density is NOT?


    I understand that
    [tex]\partial_{\mu} J^{\mu}=0[/tex]
    because it is a statement of conservation of charge. But I don't understand why the following is zero for every possible J or lambda
    [tex]\int d^{3} x \ \delta (A_{\mu}J^{\mu}) = \int d^{3} x \ J^{\mu} \partial_{\mu} \lambda = 0[/tex].


    Let's try just a simple circulating current, and a lambda polynomial in x and y:
    [tex]\vec{j} = y\hat{x} + x\hat{y}[/tex]
    [tex]\lambda = xy[/tex]
    noting that for this choice
    [tex]\partial_{\mu} \lambda = y\hat{x} + x\hat{y}[/tex]

    Looking at the result:
    [tex]\int d^{3} x \ J^{\mu} \partial_{\mu} \lambda =
    \int d^{3} x \ (y\hat{x} + x\hat{y}) \cdot (y\hat{x} + x\hat{y}) \neq 0 [/tex]

    So that doesn't seem to be gauge invariant either!?
     
  5. Feb 11, 2009 #4

    samalkhaiat

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    Last edited: Feb 11, 2009
  6. Feb 14, 2009 #5
    Ah, okay. That is why my "counter-example" fails.
    The rest of what you wrote makes sense as well.

    Thanks for your help.
     
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