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Gauge symmetry in EM by inspection

  1. Oct 22, 2011 #1

    I was under the impression that gauge symmetry was a property of the Lagrange density. Here is the Lagrangian for EM written out in its components:

    \mathcal{L}_{EM} &= J\cdot A +\frac{1}{2}\left(B^2-E^2\right) \quad eq.~1\\
    &=\rho \phi - Jx Ax - Jy Ay - Jz Az \\
    &+ \frac{1}{2}((\frac{\partial Az}{\partial y})^2 -2 \frac{\partial Az}{\partial y} \frac{\partial Ay}{\partial z} + (\frac{\partial Ay}{\partial z})^2 - (\frac{\partial Ax}{\partial t})^2 - 2 \frac{\partial Ax}{\partial t} \frac{\partial \phi}{\partial x} - (\frac{\partial \phi}{\partial x})^2 \\
    &+ (\frac{\partial Ax}{\partial z})^2 -2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x} + (\frac{\partial Az}{\partial x})^2 - (\frac{\partial Ay}{\partial t})^2 - 2 \frac{\partial Ay}{\partial t} \frac{\partial \phi}{\partial y} - (\frac{\partial \phi}{\partial y})^2 \\
    &+(\frac{\partial Ay}{\partial x})^2 -2 \frac{\partial Ay}{\partial x} \frac{\partial Ax}{\partial y} + (\frac{\partial Ax}{\partial y})^2 - (\frac{\partial Az}{\partial t})^2 - 2 \frac{\partial Az}{\partial t} \frac{\partial \phi}{\partial z} - (\frac{\partial \phi}{\partial z})^2 )\\

    Here is the Coulomb gauge:

    \frac{\partial Ax}{\partial x} + \frac{\partial Ax}{\partial x} + \frac{\partial Ax}{\partial x} = 0 \quad eq.~2

    Since there are no such derivatives in the Lagrange density of EM, one is free to pick the Coulomb gauge. To me, that is all that is needed - gaze into the Lagrange density, if it isn't there, pick the gauge if you want.

    Someone else claimed that the reason the Lagrange density of EM is invariant under a gauge transformation is due to the following well-known transformation:

    [tex]\vec{A} \rightarrow \vec{A}' = \vec{A} + \vec{\nabla} f \quad eq.~3[/tex]
    [tex]\phi \rightarrow \phi' = \phi - \frac{\partial f}{\partial t} \quad eq.~4[/tex]

    where f is a Lorentz invariant scalar field. Drop this into the definition of a B field, and the B field is not changed due to the no monopoles identity of EM. Drop this transformation into the definition of the E field, and the E field does not change because of the cancellation of the mixed derivative terms. That is almost the same as Faraday's law which has the additional curl of a gradient of a scalar function.

    To my eye, there are several problems with this common claim to a proof of gauge symmetry. The first is that as of October, 2011, we have never seen a fundamental field that is a Lorentz invariant scalar field f. Perhaps the Higgs will be the first, but until then, we haven't seen one. As a math exercise, it is flawless, but physics should be constrained to work with fields we see in nature. The (virtual) scalar field f would be spin 0, not spin 1 as needed for like charges to repel in EM. Finally, I don't see a direct connection to the Coulomb gauge. Sure, Del.A is a scalar field, but it is not a Lorentz invariant scalar field. The Lorentz invariance is a requirement for equations 3 and 4 to make sense.

    I find the "just look at the Lagrangian" approach appealing due to its simplicity. I also work with an algebra that does not have either the identities that lead to the no monopoles rule or Faraday's law. The gentleman who recites the gauge transformations above (eq. 3 and 4) views me with utmost contempt. Is that position justified?

  2. jcsd
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