# How the inverse square law in electrodynamics is related to photon mass?

I have read somewhere that one of the tests of the inverse square law is to assume nonzero mass for photon and then, by finding a maximum limit for it , determine a maximum possible error in $\frac{1}{r^{2+\epsilon} }$ for $\epsilon$ . My question is: (in the context of classical physics)

• How are these two related and what is the formula relating them?

• Why measuring photon mass is easier than testing the inverse square law directly? (Indeed it seems more challenging!)

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Related answers: physics.stackexchange.com/q/4700/2451 Equivalent question for Newton's gravitational force instead of Coulomb force: physics.stackexchange.com/q/22010/2451 – Qmechanic Apr 27 '13 at 16:18
– Ben Crowell May 15 '13 at 19:27

However, if you accept that electromagnetic field is described by some local quadratic Lagrangian, then I believe that the fact that photon posseses some mass $m$ would imply for the potential something like: $$\phi\propto \frac{e^{-mr}}{r},$$ in units where $\hbar=c=1$. It is the Yukawa potential. Well, may be for EM it is a liitle bit different, but the point is that there is some exponential decay. There is an intuitive explanation for this fact: a virtual photon of mass $m$ can exist only for a time about $1/m$ so that it can propagate to a distance $1/m$. This means that the range of the EM intraction will be about $1/m$ ($\hbar/mc$ in usual units).