Lagrangian invariant but Action is gauge invariant

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SUMMARY

The discussion focuses on the gauge invariance of the action in a three-dimensional spacetime context, specifically regarding the Lagrangian L = -(1/2)*F_{\lambda}F^{\lambda} + (m/2)*F_{\lambda}A^{\lambda}. It establishes that while the Lagrangian is not gauge invariant, the action S = ∫d^{3}x*L remains gauge invariant under the transformation A^{\lambda} -> A^{\lambda'} = A^{\lambda} + ∂^{\lambda}Λ. The key insight is that the leftover terms from the transformation vanish upon integration due to the properties of symmetric and antisymmetric tensors.

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  • Understanding of gauge invariance in quantum field theory (QFT)
  • Familiarity with antisymmetric Lorentz tensors and axial vectors
  • Knowledge of Lagrangian mechanics and action principles
  • Proficiency in tensor calculus and integration in three-dimensional spacetime
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creepypasta13
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Homework Statement



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^{\mu\nu} = -F^{\nu\mu} is equivalent to an axial Lorentz vector, F^{\mu\nu} = e^{\mu\nu\lambda}F_{\lambda}. Consequently, in 3D
one can have a massive photon despite unbroken gauge invariance of the electromagnetic
field A_{\mu}. Indeed, consider the following Lagrangian:

L = -(1/2)*F_{\lambda}F^{\lambda} + (m/2)*F_{\lambda}A^{\lambda} (6)

where

F_{\lambda}(x) = (1/2)*\epsilon_{\lambda\mu\nu}F^{\mu\nu} = \epsilon_{\lambda\mu\nu}\partial^{\mu}A^{\nu},

or in components, F_{0} = -B, F1 = +E^{2}, F_{2} = -E^{1}.

(a) Show that the action S = \intd^{3}x*L is gauge invariant (although the Lagrangian (6) is not invariant).
So I tried substituting A^{\lambda} -> A^{\lambda'} = A^{\lambda} + \partial^{\lambda}\Lambda
and F^{\lambda} -> F^{\lambda'} = \epsilon^{\lambda\mu\nu}\partial_{\mu}A_{\nu}'

then I obtained L' = L + (1/2)*[ F_{\lambda} \epsilon^{\lambda\mu\nu}\partial_{\mu} \partial_{\nu} \Lambda + 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 \intd^{3}x F_{\lambda}\epsilon^{\lambda\mu\nu}\partial_{\mu}\partial_{\nu}\Lambda
 
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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.
 

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