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Assume we have an Lagrangian $\mathcal{L}$ of matter field $\psi$ including interactive part mediated by gauge bosons $A_\mu$, eg structured "similary" to usual QED Lagrangian like

$$ D_\mu \psi^{\dagger}D_\mu \psi -\frac{1}{4}F_{\mu \nu}F^{\mu \nu} -V(\psi) $$

where $F_{\mu \nu}=\partial_{\nu}A_\mu-\partial_{\mu}A_\nu$ and $ V(\psi) $ does not depend on derivative of $\psi$ .

Here there is no "mass term" $c \cdot A_{\mu}A^{\mu }$, and one often says that since these gauge bosons are massless in this theory, the " force/ interaction mediated by them is "long ranged", see eg in the the free available script "Global and Local symmetries" by Ling-Fong Li, but the're of course myriad scripts on QFT using this formulation.

Question: Can this phrase be expressed more precisely (in terms of mathematical notions $f(x) \sim g(x)$ for asymptotics ) how the mass of gauge bosons is related to the resulting characteristic range of carried interaction/ force?

user267839
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1 Answers1

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Of course. It's a bit complicated and model (the gauge group) dependent but ideally if the force carrying particle has mass $\mu$ then the potential it induces is:

$V(r) = \frac{-g^2}{4\pi} \frac{e^{-\mu r}}{r}$

The characteristic range than is $R \approx1/\mu$ as you can see by $e^{-r/R}$

The approach to obtain it is by evaluation of the gauge boson propagator, which is then fourier transformed.

LolloBoldo
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