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# The photon mass

A common question asked in physics is whether we can truly know if the photon has no mass.

In this connection I would like to relate an anecdote, something from a conversation after a cocktail party in Paris some years ago. There was a time at which all the ladies mysteriously disappeared, and I was left facing a famous professor, solemnly seated in an armchair, surrounded by his students. He asked, "Tell me, Professor Feynman, how sure are you that the photon has no rest mass?" I answered "Well, it depends on the mass; evidently if the mass is infinitesimally small, so that it would have no effect whatsoever, I could not disprove its existence, but I would be glad to discuss the possibility that the mass is not of a certain definite size. The condition is that after I give you arguments against such mass, it should be against the rules to change the mass." The professor then chose a mass of $10^{-6}$ of an electron mass.
My answer was that, if we agreed that the mass of the photon was related to the frequency as $\omega = \sqrt{k^2 + m^2}$, photons of different wavelengths would travel with different velocities. Then in observing an eclipsing double star, which was sufficiently far away, we would observe the eclipse in blue light and red light at different times. Since nothing like this is observed, we can put an upper limit on the mass, which, if you do the numbers, turns out to be of the order of $10^{-9}$ electron masses. The answer was translated to the professor. Then he wanted to know what I would have said if he had said $10^{-12}$ electron masses. The translating student was embarrassed by the question, and I protested that this was against the rules, but I agreed to try again.
If the photons have a small mass, equal for all photons, larger fractional differences from the massless behavior are expected as the wavelength gets longer. So that from the sharpness of the known reflection of pulses in radar, we can put an upper limit to the photon mass which is somewhat better than from an eclipsing double star argument. It turns out that the mass had to be smaller than $10^{-15}$ electron masses.
After this, the professor wanted to change the mass again, and make it $10^{-18}$ electron masses. The students all became rather uneasy at this question, and I protested that, if he kept breaking the rules, and making the mass smaller and smaller, evidently I would be unable to make an argument at some point. Nevertheless, I tried again. I asked him whether he agreed that if the photon had a small mass, then from field theory arguments the potential should go as $\exp(—mr)/r$. He agreed. Then, the earth has a static magnetic field, which is known to extend out into space for some distance, from the behavior of the cosmic rays, a distance at least of the order of a few earth radii. But this means that the photon mass must be of a size smaller than that corresponding to a decay length of the order of 8000 miles, or some $10^{-20}$ electron masses. At this point, the conversation ended, to my great relief.

## The massive Maxwell Lagrangian

By default, the Lagrangian of electromagnetism is the Yang-Mill gauge Lagrangian for $\mathrm{U}(1)$, with action

$$S[A] = \int_M \left( \mathrm{Tr}(F \wedge \ast F) + L_{\mathrm{Source}} \right)$$

with $F = dA$. In terms of coordinates, we get the classic action

$$S[A] = \int_M \left( F_{\mu\nu} F^{\mu\nu} + L_{\mathrm{Source}} \right)$$

with $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. The electromagnetic tensor has simply the variation

$$\frac{\delta F_{\mu\nu}}{\delta A_\sigma} =$$

The usual massive version of this is the Proca Lagrangian,

$$S[A] = \int_M \left( \mathrm{Tr}(F \wedge \ast F) + m \langle A, A \rangle + L_{\mathrm{Source}} \right)$$
Last updated : 2021-05-26 10:37:22
Tags : physics , electromagnetism