# Proca Lagrangian (Math troubles with four vectors)

1. Sep 12, 2011

### Elwin.Martin

I'm reading Griffith's Elementary particles and I'm stuck on the math for one of the examples, could anyone show me what I'm missing or point me in the right direction?

I attached a pdf (of the word doc I was using) that shows what I did so far since I'm really bad with LaTeX and it would've taken me an hour to write it for this post.

Thanks in advanced for any and all help/direction; I feel so useless for having to ask, but I need to get through this.

Elwin

**any review material on the math would be nice if you have recommendations!**

#### Attached Files:

• ###### Lagrangians.pdf
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2. Sep 12, 2011

### jfy4

I'll write something up for you to help, let me compose it.

So we know the last part doesn't do anything, so lets not even put it in for now till we need it.
Lets multiply out the Lagrangian to see explicitly whats going on with the indices.
$$\mathcal{L}=-\frac{1}{16\pi}(\partial^a A^b\partial_a A_b -\partial^a A^b \partial_b A_a -\partial^b A^a \partial_a A_b +\partial^b A^a \partial_b A_b)$$
Now anytime you sum over repeated indices, it doesn't matter what the letter used is, because you are summing over it. So change them so that you can simplify. If you look closer at this, there are really only two different expressions with the partial derivative and the vector potential: $\partial^c A^d \partial_c A_d$ and $\partial^d A^c \partial_c A_d$. So you have two of these, that is,
$$\mathcal{L}=-\frac{1}{8\pi}(\partial^c A^d \partial_c A_d-\partial^d A^c \partial_c A_d)$$
With this in mind, try to do what you did before with this new information in your mind.

Last edited: Sep 12, 2011
3. Sep 12, 2011

### Elwin.Martin

Thank you very much!

4. Sep 13, 2011

### vanhees71

Why shouldn't the last term contribute anything? Of course it does contribute something, namely the mass of the vector particle.

5. Sep 13, 2011

### jfy4

Yes, of course. I meant for
$$\frac{\partial\mathcal{L}}{\partial (\partial_a A_b)}$$
we can ignore it. It's definitely important, as the distinguishing factor between E&M.

6. Sep 13, 2011

### vanhees71

I see. BTW, it's definitely simpler to directly use Hamilton's principle to derive the eoms. I'll write the Proca action in modern Heaviside-Lorentz units, leading to

$$S[A_{\mu}]=\int \mathrm{d}^4 x \left (-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{1}{2} A_{\mu} A^{\mu} \right ).$$

I use the west-coast metric. That's why I have a + sign in front of the mass term. Variation with respect to $A_{\mu}$ and using the anti-symmetry of the Faraday tensor, $F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$ leads to

$$\delta S=\int \mathrm{d}^4 x \left ( -F^{\mu \nu} \partial_{\mu} \delta A_{\nu} + m^2 A_{\nu} \delta A^{\nu} \right )= \int \mathrm{d}^4 x \left (\partial_{\mu} F^{\mu \nu} + m^2 A^{\nu} \right ) \delta A_{\nu} \stackrel{!}{=}0.$$

From this we have the equation of motion

$$\partial_{\mu} F^{\mu \nu} + m^2 A^{\nu}=0.$$

First from this you get by taking the four-divergence of this equation, using $\partial_{\mu} \partial_{\nu} F^{\mu \nu}=0$

$$m^2 \partial_{\nu} A^{\nu}=0.$$

The Proca field with non-vanishing mass is thus necessarily transversal (contrary to the massless photon field).

From this we can rewrite the eom.
$$\partial_{\mu} F^{\mu \nu}=\partial_{\mu} (\partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu})=\Box A^{\nu}=-m^2 A^{\nu}.$$

This shows that $m$ is indeed the mass of the vector particle.

One should also mention that a massive vector field is not necessarily an (at least abelian) gauge field as turns out to be the case for massless vector fields. However to build renormalizable interacting field theories one can make the Proca Lagrangian U(1)-gauge invariant by adding an additional scalar field (the Stueckelberg ghost) and then use the usual minimal-coupling approach to couple the massive vector field to other matter fields (e.g., scalar and/or Dirac fields) to get a gauge-invariant theory with massive gauge fields. It turns out that after gauge fixing the Stueckelberg ghost doesn't couple to the other fields as don't the Feynman-Faddeev-Popov ghosts in this abelian-gauge model.

This socalled Stueckelberg realization of massive vector fields doesn't work for non-abelian gauge theories, i.e., then the only way to make the non-abelian gauge fields massive without destroying local gauge invariance is via the Higgs mechanism, where you absorb the would-be Goldstone bosons into the gauge field but always keep at least one additional massive Higgs boson left in the physical particle spectrum.

7. Sep 13, 2011

### jfy4

Thanks vanhees71,

That's nice to see