Proca Lagrangian

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  • #1
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Could someone explain how can one go from

$$ \int dx\ \frac{-1}{4}F^{\mu \nu}F_{\mu \nu}$$

where $$F_{\mu \nu} = \partial_{\mu} \phi_{\nu}-\partial_{\nu} \phi_{\mu}$$

to

$$\int dx\ \frac{-1}{2}(\partial_{\mu} \phi^{\nu})^2 + \frac{1}{2}(\partial_{\mu} \phi^{\mu})^2 $$

I assume it has something to do with integration by parts but I can't see it
 

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  • #2
Orodruin
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It follows directly from the form of ##F_{\mu\nu}##.
 
  • #3
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It follows directly from the form of ##F_{\mu\nu}##.
No it doesn't, the second term is slightly different. They are supposed to be the same up to total divergence according to these notes https://cds.cern.ch/record/292286/files/B00008237.pdf on page 4 it says you need to apply the divergence theorem but I don't see how.
 
  • #4
Orodruin
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No it doesn't, the second term is slightly different. They are supposed to be the same up to total divergence according to these notes https://cds.cern.ch/record/292286/files/B00008237.pdf on page 4 it says you need to apply the divergence theorem but I don't see how.
Yes, it is a trivial application of integration by parts to move the derivatives from one ##\phi## to the other. The assumption is that the boundary terms vanish.
 
  • #5
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Yes, it is a trivial application of integration by parts to move the derivatives from one ##\phi## to the other. The assumption is that the boundary terms vanish.
Thanks for explaining that it is trivial that made it easier to understand.
 
  • #6
Orodruin
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Thanks for explaining that it is trivial that made it easier to understand.
Well, it is difficult to see why you have a problem with it. You are familiar with paritial integration I assume? For the derivative ##\partial_\mu##, just perform the partial integration in the ##x^\mu## direction. There really is nothing else to it.
 
  • #7
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## \frac{-1}{4} \int dx\ F^{\mu \nu}F_{\mu \nu} =\frac{-1}{4} \int dx\ (\partial_{\mu} \phi_{\nu}-\partial_{\nu} \phi_{\mu})( \partial^{\mu} \phi^{\nu}-\partial^{\nu} \phi^{\mu})##

## = \frac{-1}{2} \int dx\ (\partial_{\mu} \phi^{\nu})^2 - \partial_{\nu} \phi_{\mu} \partial^{\mu}\phi^{\nu} ##

## = \frac{-1}{2} \int dx\ (\partial_{\mu} \phi^{\nu})^2 - \partial_{\nu}(\phi_{\mu} \partial^{\mu} \phi^{\nu}) + \phi_{\mu} \partial_{\nu} \partial^{\mu} \phi^{\nu} ##

## = \frac{-1}{2} \int dx\ (\partial_{\mu} \phi^{\nu})^2 - \partial_{\nu}(\phi_{\mu} \partial^{\mu} \phi_{\nu}) + \phi_{\mu} \partial^{\mu} \partial_{\nu} \phi^{\nu} ##

## = \frac{-1}{2} \int dx\ (\partial_{\mu} \phi^{\nu})^2 - \partial_{\nu}(\phi_{\mu} \partial^{\mu} \phi_{\nu}) + \partial^{\mu}(\phi_{\mu} \partial_{\nu} \phi^{\nu}) - \partial^{\mu}\phi_{\mu} \partial_{\nu} \phi^{\nu} ##

And the two total divergences go to zero.

Posting for anyone else who didn't find it trivial.
 
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  • #8
vanhees71
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Well, that' reminds me of one of the many stories about Pauli. When giving a lecture, he told the students something's trivial. The student asked, whether it is really trivial. Pauli answered, he had to think about, left the room for 10 minutes and then declared that the point made was really trivial. A clear answer, one must admit ;-)).
 
  • #9
dextercioby
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Incidentally, the thread's title is a little misleading when compared to the contents of the 1st post, as Proca's field is massive, hence the Lagrangian density contains a mass term proportional with the field.
 
  • #10
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Incidentally, the thread's title is a little misleading when compared to the contents of the 1st post, as Proca's field is massive, hence the Lagrangian density contains a mass term proportional with the field.
Sorry the actual question I had was about the gauge invariance of the massive version and was entitled 'Procca Lagrangian'. But in the question the Lagrangian was written as what I wanted to derive + a mass term so I wanted to show the non mass terms were equivalent to the obviously gauge invariant form of (Fuv)^2 + total divergences so I could say that the issue with gauge invariance lies in the mass term only.

Annoyingly it didn't come up in the exam.
 
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  • #13
vanhees71
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Thanks, that was the second part of the question, here it is.
Well, if this is a problem to be solved by YOU, please post in the homework section of this forum. There you get help to solve the problem YOURSELF instead of giving simply the answer. That's much better for learning the subject than just reading the final answer!
 
  • #14
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Well, if this is a problem to be solved by YOU, please post in the homework section of this forum. There you get help to solve the problem YOURSELF instead of giving simply the answer. That's much better for learning the subject than just reading the final answer!
It was an exam question from like 6 years ago, I was revising, that is why I wrote "Annoyingly it didn't come up in the exam" and regardless I did solve it myself.
 

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