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Making Lagrangian gauge invariant

  1. Apr 3, 2015 #1
    1. The problem statement, all variables and given/known data

    The Lagrangian ##\mathcal{L}\frac{1}{2}(\partial_\mu\phi^\nu)^2+\frac{1}{2}(\partial_\mu\phi^\mu)^2+\frac{m^2}{2}(\phi_\mu\phi^\mu)^2## for the vector field ##\phi^\mu## is not invariant with respect to the gauge transformation ##\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha##. Introduce a new real scalar field ##\sigma## and find a new interacting Lagrangian ##\mathcal{L}'(\phi^\mu,\sigma)=\mathcal{L}(\phi^\mu)+\tilde{\mathcal{L}}(\phi^\mu,\sigma)## which is gauge invariant under the given transformation and satisfies ##\mathcal{L}'(\phi^\mu,0)=\mathcal{L}(\phi^\mu)##.

    Can we solve this with a ##\sigma## that has a canonical kinetic term ##-\frac{1}{2}(\partial_\mu\sigma)^2##?


    2. Relevant equations


    3. The attempt at a solution

    It's the part about the canonical kinetic term I don't understand. On the requirement that ##\sigma \rightarrow \sigma + \alpha## I found a ##\tilde{\mathcal{L}}(\phi^\mu,\sigma)## that satisfies the required properties. I won't write it because it's rather long. I was told the answer to whether or not there can be a canonical kinetic term is "no", but I don't know why.

    Is there a particular reason why this should be? Something about having unbounded negative energy perhaps?
     
  2. jcsd
  3. Apr 8, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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