# Proca Lagrangian

• I
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

Orodruin
Staff Emeritus
Homework Helper
Gold Member
It follows directly from the form of ##F_{\mu\nu}##.

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.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

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.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

## \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.

vanhees71
vanhees71
Gold Member
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 ;-)).

dextercioby
Homework Helper
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.

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.

vanhees71
Gold Member
In fact you can formulate the massive vector particle as a U(1) gauge symmetry. The mass term is not disturbing gauge invariance in the abelian case. That's known as the Stueckelberg formalism. See the very nice review

Ruegg, Henri, Ruiz-Altaba, Marti: The Stueckelberg field, Int. J. Mod. Phys. A 19, 3265–3348, 2004
http://dx.doi.org/10.1142/S0217751X04019755
https://arxiv.org/abs/hep-th/0304245

In fact you can formulate the massive vector particle as a U(1) gauge symmetry. The mass term is not disturbing gauge invariance in the abelian case. That's known as the Stueckelberg formalism. See the very nice review

Ruegg, Henri, Ruiz-Altaba, Marti: The Stueckelberg field, Int. J. Mod. Phys. A 19, 3265–3348, 2004
http://dx.doi.org/10.1142/S0217751X04019755
https://arxiv.org/abs/hep-th/0304245
Thanks, that was the second part of the question, here it is.

vanhees71