Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!)

share|cite|improve this question
Related answers: Equivalent question for Newton's gravitational force instead of Coulomb force: – Qmechanic Apr 27 '13 at 16:18

Regarding your first question. When you are asking it, you should understand that it has an answer only in some model -- there is no universal relation that holds in every imaginable model of electromagnetic interactions. I personally do not know a model that would break the inverse square law in the way you want.

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).

Also, quantum corrections alter the inverse square law at small distances in a rather complicated way, but it is a different story. Also, in priciple EM field could develop an anomalous scaling dimension due to quantum effects, but I think that it is protected by gauge invariance. (Someone correct me if I'm wrong).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.