# I Proca Lagrangian

1. May 22, 2016

### decerto

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

2. May 22, 2016

### Orodruin

Staff Emeritus
It follows directly from the form of $F_{\mu\nu}$.

3. May 22, 2016

### decerto

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. May 22, 2016

### Orodruin

Staff Emeritus
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. May 22, 2016

### decerto

Thanks for explaining that it is trivial that made it easier to understand.

6. May 22, 2016

### Orodruin

Staff Emeritus
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. May 22, 2016

### decerto

$\frac{-1}{4} \int dx\ F^{\mu \nu}F_{\mu \nu} =\frac{-1}{4} \int dx\ (\partial_{\mu} \phi_{\nu}-\partial_{\nu} \phi_{\mu})( \partial^{\mu} \phi^{\nu}-\partial^{\nu} \phi^{\mu})$

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

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

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

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

And the two total divergences go to zero.

Posting for anyone else who didn't find it trivial.

8. May 24, 2016

### vanhees71

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. May 24, 2016

### dextercioby

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. Jun 11, 2016

### decerto

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.

11. Jun 11, 2016

### vanhees71

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

12. Jun 11, 2016

### decerto

Thanks, that was the second part of the question, here it is.

13. Jun 12, 2016

### vanhees71

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. Jun 13, 2016

### decerto

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.