# Why does massive carrier implies finite range?

1. Jun 28, 2008

### tomkeus

Back on lecture about Standard model, professor said that original Weinberg-Salam-Glashow theory had to be augmented with Higgs mechanism because it didnt give rise to massive gauge fields, yet since weak interaction had finite range they knew it had to have a massive carrier, well three of them due to three SU(2) generators. When I asked him why massive carrier implies finite range, and massless infinite, I didnt get satisfactory (any) explanation.

So, can anyone please shed any light on this subject?

2. Jun 28, 2008

### humanino

This is a very important fact. You are right to search for a rigorous answer.

The "bad reason" why this is important is technical. You should be able to calculate the associated potential using Cauchy's theorem.

The "good reason" is conceptual. Gauge invariance imposes the photon to be massless, but this is quite annoying to fix gauge invariance. A useful trick is to set a finite mass, that physically you claim is to small to be observable. Do your calculations, and set m=0 at the very end. It is a quite non-trivial property that this procedure will still respect gauge invariance at the end of the day. But many people use it.

Do you have Zee's "QFT in a nutshell" ? He discusses about all this at the conceptual level. This is a very good introductory book.

So the thing is something like : calculation of the massive scalar potential with a point source at the origin $$\rho=g\delta(\vec{x})$$
Klein-Gordon : $$\partial_{\mu}\partial^{\mu}U+m^{2}U=\rho$$
Time independence : $$(-\Delta^{2}+m^2)U=g\delta(\vec{x})$$
Fourier transform : $$U(\vec{x})=\frac{g}{2\pi^3}\int d^3k\frac{e^{i\vec{k}\vec{x}}}{\vec{k}^2+m^2}$$
Go to polar coordinate, trivial angular integration, radial integration done be closing the contour (say) in the positive imaginary part you pick up the pole at $$k=im$$. This is the important technical part. Details upon request (try it before !).
Finally
$$U(\vec{x})=\frac{g}{4\pi}\frac{e^{-mr}}{r}$$
This is what we mean by finite range, because exponential is decreasing fast. Setting a vanishing mass, you recover the Coulomb potential

3. Jun 28, 2008

### isospin

I guess what you meant is that if the intermedia boson is massive, such as W-boson, then the related interaction should be short-ranged. For W,Z+,Z-, it is weak interaction.
Well~~why~~I am not sure. We can calculate the propogators in coordinate-space, and use the C-S equation to get the details.
May the reason is that these massive bosons decay during the propogating process due to their lower velocities than that of Light. As we know, "All three particles(W,Z+,Z-) are very short-lived with a mean life of about 3×10^-25 s".

4. Jun 29, 2008

### ismaili

A conceptual argument is from uncertainty principle. $$\Delta E \Delta t \sim \hbar$$
Like what we often said, the higher the energy of the accelerator, the shorter the distance one can probe.
So, if an intermediate boson has exactly zero energy, then the range of the interaction should be infinite, this is the case of QED. However, if the intermediate boson has finite mass, we take the deviation $$\Delta E$$ to be the mass of the boson, then we get a finite range of the interaction.
However, although the intermediate bosons, gluons, are massless, the range of QCD is still short-ranged due to confinement.
I tried to look at the range of interaction from the propagator in coordinate space. For the massless case, I can work out the Fourier transform and the result is 1/x^2. But for the case of massive boson, the propagator of real space is kinda complicated, I can't work out the integral.