1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Lagrangian invariant but Action is gauge invariant

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    So I'm having some difficulty with my QFT assignment. I have to solve the following problem.

    In three spacetime dimensions (two space plus one time) an antisymmetric Lorentz tensor
    F[itex]^{\mu\nu}[/itex] = -F[itex]^{\nu\mu}[/itex] is equivalent to an axial Lorentz vector, F[itex]^{\mu\nu}[/itex] = e[itex]^{\mu\nu\lambda}[/itex]F[itex]_{\lambda}[/itex]. Consequently, in 3D
    one can have a massive photon despite unbroken gauge invariance of the electromagnetic
    field A[itex]_{\mu}[/itex]. Indeed, consider the following Lagrangian:

    L = -(1/2)*F[itex]_{\lambda}[/itex]F[itex]^{\lambda}[/itex] + (m/2)*F[itex]_{\lambda}[/itex]A[itex]^{\lambda}[/itex] (6)


    F[itex]_{\lambda}[/itex](x) = (1/2)*[itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex]F[itex]^{\mu\nu}[/itex] = [itex]\epsilon[/itex][itex]_{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]^{\mu}[/itex]A[itex]^{\nu}[/itex],

    or in components, F[itex]_{0}[/itex] = -B, F1 = +E[itex]^{2}[/itex], F[itex]_{2}[/itex] = -E[itex]^{1}[/itex].

    (a) Show that the action S = [itex]\int[/itex]d[itex]^{3}[/itex]x*L is gauge invariant (although the Lagrangian (6) is not invariant).

    So I tried substituting A[itex]^{\lambda}[/itex] -> A[itex]^{\lambda'}[/itex] = A[itex]^{\lambda}[/itex] + [itex]\partial[/itex][itex]^{\lambda}[/itex][itex]\Lambda[/itex]
    and F[itex]^{\lambda}[/itex] -> F[itex]^{\lambda'}[/itex] = [itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex]A[itex]_{\nu}[/itex]'

    then I obtained L' = L + (1/2)*[ F[itex]_{\lambda}[/itex] [itex]\epsilon^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex] [itex]\partial[/itex][itex]_{\nu}[/itex] [itex]\Lambda[/itex] + some other terms]

    What I don't understand is how these leftover terms would vanish after being integrated (to obtain S'), but they don't all vanish if they are not integrated (since L is not invariant). Is there some kind of special mathematical trick I have to use? I just don't see how I can integrate terms like [itex]\int[/itex]d[itex]^{3}[/itex]x F[itex]_{\lambda}[/itex][itex]\epsilon[/itex][itex]^{\lambda\mu\nu}[/itex][itex]\partial[/itex][itex]_{\mu}[/itex][itex]\partial[/itex][itex]_{\nu}[/itex][itex]\Lambda[/itex]
  2. jcsd
  3. Dec 29, 2011 #2
    Well, I can see how Fλ ϵλμν∂μ ∂ν Λ would vanish. The \mu and \nu are symmetric wrt to exchange for the partial derivatives, but the indices of the \epsilon symbol are totally antisymmetric. Multiplying a symmetric tensor and an antisymmetry one gives zero always.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook