# I Experimental check of photon mass and Coulomb's law

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1. Oct 13, 2016

### Prometeus

Despite some effort I still dont understand how Coulombs law can be used for experimental search for mass of photon.

From wikipedia:
If a photon did have non-zero mass, there would be other effects as well. Coulomb's law would be modified and the electromagnetic field would have an extra physical degree of freedom. These effects yield more sensitive experimental probes of the photon mass than the frequency dependence of the speed of light. If Coulomb's law is not exactly valid, then that would allow the presence of an electric field to exist within a hollow conductor when it is subjected to an external electric field. This thus allows one to test Coulomb's law to very high precision.[25] A null result of such an experiment has set a limit of m ≲ 10−14 eV/c2.[26]

Why would Coulombs law be needed modified if photon had non-zero invariant mass?

2. Oct 13, 2016

### CrackerMcGinger

I want to answer this, but give me about......3-5 days to do some research.

3. Oct 13, 2016

### vanhees71

The photon propagator is the Green's function for the electromagnetic interaction. The static case (Coulomb field) is described by $q^0=0$, i.e.,
$$G=-\frac{1}{\vec{p}^2},$$
which is familiar from classical electrodynamics. It's just the Fourier transform for the coulomb field (potential), which is $\propto \frac{1}{|\vec{r}|}$.

Now with a photon mass the propagator reads
$$G=-\frac{1}{\vec{p}^2+m^2}.$$
This corresponds to an interaction law following a Yukawa potential $\propto \exp(-mr)/r$.

4. Oct 13, 2016

### Prometeus

Thanks for answer, but it seems that this topic is above my current knowledge.

5. Oct 13, 2016

### Ibix

The E field can carry energy and momentum. If I'm reading @vanhees71 correctly (and I may not be, so see what he says to this) then energy and momentum aren't conserved by an E field with a $1/r$ potential if the photon has mass. So a check that a static E field looks like a $1/r$ potential is equivalent to a check on the masslessness of the photon.

6. Oct 13, 2016

### robphy

7. Oct 15, 2016

### vanhees71

Energy and momentum are conserved for any kind of fields in relativstic QFT. The mass only enters in the dispersion relation. For a free particle it reads $E=\sqrt{\vec{p}^2+m^2}$. In any scattering process the four-momentum, i.e., energy and momentum, is conserved.

8. Oct 15, 2016

### Prometeus

Have tried to study the proposed topics, but despite my effort is seems to me, that all these arguments towards Proca equations, Yukawa potential and Coulombs law are kind of circular arguments. I will try to explain why. So here we have Coulombs law including vacuum permittivity:

https://wikimedia.org/api/rest_v1/media/math/render/svg/2c30719f1efcb507ce62de711b04c7f51f64983b

https://wikimedia.org/api/rest_v1/media/math/render/svg/2c30719f1efcb507ce62de711b04c7f51f64983b

But lets say, that theoretically photon has some tiny mass, the measured speed of light would include this mass and also measured vacuum permittivity would include that mass, because speed of light can be expressed by

https://wikimedia.org/api/rest_v1/media/math/render/svg/f83cb831ce59c5745ee4e7af60b4f68710405b6b

https://wikimedia.org/api/rest_v1/media/math/render/svg/f83cb831ce59c5745ee4e7af60b4f68710405b6b

So theoretically we could say that mass of photon is included in measured c and vacuum permittivity and you can use Yukawa potential to find mass of particles with mass above mass of photon, but not to find mass of photon itself.

I understand, that this line of reasoning is with very high probability wrong, but could somebody point out where is the mistake in my understanding?

Last edited: Oct 15, 2016
9. Oct 15, 2016

### robphy

10. Oct 15, 2016

### Prometeus

In theory yes, practically no. Just look on the measurements of speed of neutrino (quote from Wikipedia):
Since it is established that neutrinos possess mass, the speed of neutrinos should be slightly lower than the speed of light in accordance with special relativity. Existing measurements provided upper limits for deviations of approximately 10−9, or a few parts per billion. Within the margin of error this is consistent with no deviation at all.

11. Oct 15, 2016

### vanhees71

The only consequence concerning the "speed of light" if the electromagnetic field (photon) had a mass term would be that we had to rename that speed the "limiting speed of relvaitivity". There is no doubt that spacetime is better described relativistically than non-relativistically and this implies the existence of such a "limiting speed", which in fact within relativity is just a conversion parameter from the arbitrary choice of time and spatial distance units.

12. Oct 15, 2016

### Prometeus

Sure, you are completely right, but this is other topic. Im trying to get to the bottom of reasoning behind experimental checks of photon mass.

13. Oct 15, 2016

### robphy

14. Oct 16, 2016

### vanhees71

I don't know, how often I've written this already in this thread, but again. A pure state is NOT represented by a (normalized) vector $|\psi \rangle$ but by the corresponding ray. This is very important to keep in mind to understand very basic concepts of QT, e.g., the existence of half-integer spins etc. Pure states, however, are not enough to describe physics since they are rare and have to be usually carefully prepared in the lab. The general description of a state is through a statistical operator $\hat{\rho}$ which is a self-adjoint positve semidefinite operator of trace 1. A statistical operator $\hat{\rho}$ represents a pure state if and only if $\hat{\rho}^2=\hat{\rho}$. Then there exists one and only one eigenvector with eigenvalue $1$, $|\psi \rangle$, and all other eigenvectors are to the eigenvalue $0$. Thus a pure state is uniquely represented by the projector $\hat{\rho}=|\psi \rangle \langle \psi|$.

15. Oct 18, 2016

### Staff: Mentor

The speed of electromagnetic waves can be measured much more precisely than the speed of neutrinos, but the corresponding upper limits are still much worse than tests of the quasi-static electromagnetic field in astronomy.

16. Oct 18, 2016

### Prometeus

Yes, thats true, but nobody bothered to point out where is my logical flaw when Im assuming that electromagnetic tests are useless because of circular reasoning behind it. Because when photon has mass, that means speed of light is corresponding to its mass and this also means that value of vacuum permittivity is also including mass of photon and vacuum permittivity is directly affecting Couloubs field and also Yukawa potential, meaning that electromagnetic tests are showing no difference.

17. Oct 18, 2016

### Staff: Mentor

No. You cannot use a framework developed for massless light to study the case of massive light.

The speed limit of special relativity is independent of light. If light has a mass, it is slower than this speed limit, and the changed propagator also means you do not get a 1/r field any more (similar to the Z boson: it has mass, therefore it has a short range). The difference between Yukawa and Coulomb potential is testable: put the same charge at twice the distance and see if the force goes down by a factor 4 (very simplified description).

18. Oct 20, 2016

### Prometeus

It seems you dont understand what Im trying to say. Testing difference between Yukawa and Coulomb potential could be completely useless because measured value of vacuum permittivity could include (unmeasured) mass of photon. If my understanding of equations of Yukawa and Coulomb is right, it will still go down by factor 4 even when vacuum permittivity includes mass of photon. Measuring mass of Z boson is possible via Yukawa potential, because if vacuum permittivity includes mass of photon, the difference would show mass of Z boson greater as mass of photon, so it would be still in line of existing expectations.

19. Oct 20, 2016

### Staff: Mentor

As I said, you cannot apply frameworks for massless particles in the case of massive particles. As long as you try that you cannot come to the right anwer.

20. Oct 20, 2016

### Orodruin

Staff Emeritus
It seems to me that the fundamental flaw in the OP's reasoning is to assume that the speed of light referred to in SR is the speed at which light travels regardless of whether it has mass or not. This is not the case. If light has mass it will be able to travel at any speed below c. It is merely a historical coincidence that we call the invariant speed the speed of light, based on light being the first observed entity traveling at that speed (within experimental error).

Also, the discovery of neutrino masses had nothing to do with measuring their velocities.