As we know, if we use the Lagrangian of electrodynamics we can find that the photon has no mass. If the photon had mass, it would even have 3 polarizations, which is a consequence of having mass. My question is how can we see or approximate the range of interaction of the bosons? In the case of the photon, if it had mass, the range of interaction would decrease, since the $W$'s and $Z$ bosons have mass, their range is short, so as can be seen from the equations, such interaction will be short or longer.


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Using the basic formulae $$\Delta \frac{1}{r}=-4\pi \delta^{(3)}(\vec{x}) \quad \text{with} \quad r=|\vec{x}| \tag{1}$$and $$\Delta \frac{f(r)}{r} =\frac{f^{\prime \prime}(r)}{r} \quad \text{for}\quad r\ne 0,\tag{2}$$ the one-line calculation $$\Delta \frac{e^{-m r}}{r}=\Delta \frac{e^{-mr}-1}{r}+\Delta \frac{1}{r}=m^2 \,\frac{e^{-mr}}{r}-4\pi \delta^{(3)}(\vec{x}) \tag{3}$$ reveals that the Yukawa potential $$\frac{e^{-mr}}{r} \tag{4} \label{4}$$ is a solution of the inhomogeneous Klein-Gordon equation with a static source term, $$(m^2+\square)\frac{e^{-mr}}{r}=4\pi \delta^{(3)}(\vec{x}). \tag{5}$$ The solution given in \eqref{4} falls off rapidly for $r \gt 1/m$ showing that $\lambda=1/m$ is measure for the range of the corresponding interaction. Converting from natural units used here to "engineering" units, $\lambda =\hbar/Mc$ is nothing else than the Compton length of a particle with mass $M$. The limit $M\to 0$ corresponds to the range $\lambda \to \infty$, where \eqref{4} becomes the (scalar) potential $1/r$ of the electro(magnetic) field generated by a point charge sitting at the origin.

By the way, the relation $M=\hbar/\lambda c$ was used by Hideki Yukawa in his famous 1935 paper to estimate the mass of the exchange particle mediating the (strong) nuclear force between proton and neutron from the known size of the nucleus.


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