# Massive boson propagator - problem with step in derivation

1. Nov 28, 2007

### Ben1729

I've been going through a derivation of the massive boson propagator. The only trouble is I don't feel satisfied by one particular step. I've started getting the problem written up in tex so here goes:

$$\newcommand{\del}{\partial} \newcommand{\eqn}[1]{$$\begin{array}{rl}#1\end{array}$$} \newcommand{\where}[1]{\\\textrm{where}\quad #1} \newcommand{\dm}{\del_{\mu}} \newcommand{\dn}{\del_{\nu}} \newcommand{\dM}{\del^{\mu}} \newcommand{\dN}{\del^{\nu}} \newcommand{\vm}{V_{\mu}} \newcommand{\vn}{V_{\nu}} \newcommand{\vM}{V^{\mu}} \newcommand{\vN}{V^{\nu}} \newcommand{\dmvn}{\dm\vn} \newcommand{\dnvm}{\dn\vm} \newcommand{\dMvN}{\dM\vN} \newcommand{\dNvM}{\dN\vM} \newcommand{\vmn}{V_{\mu\nu}} \newcommand{\vMN}{V^{\mu\nu}} For a free (non-interacting) massive vector field V the lagrangian is: \eqn{ L&=-\frac{1}{4}V_{\mu\nu}V^{\mu\nu}+\frac{M^2}{2}V^{\mu}V_{\mu}\\ \where{V_{\mu\nu}&=\dmvn-\dnvm} } To extract the massive boson propagator, let us expand the first term: \eqn{ V_{\mu\nu}V^{\mu\nu}&=(\dmvn-\dnvm)(\dMvN-\dNvM)\\ &=\dmvn\dMvN-\dmvn\dNvM-\dnvm\dMvN+\dnvm\dNvM\\ &=2[\dmvn\dMvN-\dmvn\dNvM]\\ &=2[\dm(\vn\dMvN)-\vn\dm\dMvN-\dm(\vn\dNvM)+\vn\dm\dNvM] } Now in order to get a simple expression for the propagator, we'd quite like the sum of the 1st and 3rd terms to be zero: \eqn{ \dm(\vn\dMvN)-\dm(\vn\dNvM)=0\\ \dm(\vn(\dMvN-\dNvM))=0\\ \dm(\vn\vMN)=0 }$$

So my question now is, what is the sufficient condition to make this last equation true?

Some books just say 'integrate by parts' and voila, you mystically end up with only the 2nd and 4th terms of the last line of equation 2, and then you can go on to define the propagator.

But going through it myself, and iterating the steps explicitly as above, I'm not convinced by this. Have I done something really dumb? Or is there some condition I'm missing - does a 'free' field mean something here?

B.

2. Nov 28, 2007

### QuantumDevil

$$\int dx^{4}\partial_{\mu}[V_{\nu}V^{\nu\mu}]=\int dS_{\mu} V_{\nu}V^{\nu\mu}$$

As you know if you take boundry at infinity the letter integral gives zero. So this kind of expressions give no contribution to action integral and can be removed from lagrangian density.

PS: dS is a 3-dimentional hypersurface in Minkowski space

3. Nov 28, 2007

### Ben1729

Ah yes. Good old Stokes.
But I'm still not sure what zero divergence for such a strange quantity as:
$$V_\nu V^{\mu\nu}$$
implies for $$V_\nu$$ itself

4. Nov 28, 2007

### Avodyne

It does NOT have zero divergence. The INTEGRAL of the divergence is zero, because the integral of ANY divergence (with suitable boundary conditions at infinity) is zero.

5. Nov 28, 2007

### Ben1729

Er...yes. Sorry. Thanks!

I still can't quite visualise this quantity:
$$V_{\nu} V^{\mu\nu}$$