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Weinberg 5.9.34

  1. Jul 28, 2015 #1
    Weinberg 5.9.34

    "[...] Using this together with Eq. 5.9.23 gives the general antisymmetric tensor field for massless particles of helicity ##\pm 1## in the form ##f_{\mu\nu} = \partial_{ [ \mu } a_{ \nu ] }##. Note that this is a tensor even though ##a_{\mu}## is not a 4-vector."​

    Not a four vector? So the vector potential in the development that follows is not a vector, not Lorentz invariant, and most significantly, not generally covariant in this universe.

    If ##a## is not a vector in the construction of a Lagrangian, either the action is not a scalar or the charge-current density is not a tensor, or both.

    If we brush this under the carpet, a Lagrangian constructed to conserve charge is either not the Lagrangian of a conserved quantity (##dj \neq 0##) or the Lagrangian density is frame dependent, or both.

    Is this later resolved?
     
    Last edited: Jul 28, 2015
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  3. Jul 28, 2015 #2

    fzero

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    You should read the commentary below (5.9.31). The reason that ##a^\mu## here is not a 4-vector is that Weinberg has already reduced it to the physical transverse degrees of freedom. An arbitrary Lorentz transformation will introduce longitudinal and timelike components that were not included in the definition of ##a^\mu##. The solution is to not make the transverse restriction by hand but to use gauge invariance to do it. Then the unfixed ##a^\mu## is a 4-vector, but the gauge-fixing can be used to reduce the physical states to the transverse degrees of freedom.
     
  4. Jul 29, 2015 #3

    vanhees71

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    The point is that ##a^{\mu}## is a gauge field, which transforms according to a vector field modulo gauge transformations, i.e., you can always do the Lorentz transformation, introducing unphysical degrees of freedom and then gauge them away. Compared to the standard treatment, where you only do the Lorentz transformation, with an additional gauge transformation, you get the correct symmetric and gauge invariant energy-momentum-stress and center-momentum-angular-momentum tensors right away, without redefining it afterwards via an extra gauge transformation. At the end, both are the same, and a unique derivation of the energy-momentum-stress tensor is finally given by varying the metric, as we know from General Relativity.
     
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