# Can physics describe a massive photon?

1. Dec 22, 2013

### ChrisVer

My question comes from seeing this:
http://pdg.lbl.gov/2009/tables/rpp2009-sum-gauge-higgs-bosons.pdf
From where we know that the photon has mass less than something. First of all it means that experimentalists have looked at it.
So, what would be the impact on the physics we already know, if the mass of the photon comes to be non-zero?

2. Dec 22, 2013

### ShayanJ

Take a look at here
Studying about virtual particles specially virtual photons may help too!

3. Dec 22, 2013

### Staff: Mentor

Don't misunderstand that table. It's not telling us that the photon has a mass less that $10^{-18}$ eV. It's telling us that the best experiments done have a measurement error no greater than that, so even if current theory is incomplete and the rest mass of the photon turns out to be non-zero, it still has to be less than $10^{-18}$ eV. In other words: To the limits of experimental accuracy, experimental tests agree with the theoretical prediction that the photon has no mass.
It would be amazing and exciting and a lot of fun, as we would be challenged to extend the physics we already know in new and interesting directions. It wouldn't "overthrow" physics - although the popular press would report it that way.

4. Dec 22, 2013

### bcrowell

Staff Emeritus
Here are some sources of information:

http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

https://www.physicsforums.com/showpost.php?p=2514054&postcount=19

http://physics.stackexchange.com/a/64713/4552

How stable is the photon?, Julian Heeck, http://arxiv.org/abs/1304.2821

Davis, PRL 35 (1975) 1402, http://prl.aps.org/abstract/PRL/v35/i21/p1402_1

R.S. Lakes, "Experimental limits on the photon mass and cosmic magnetic vector potential", Physical Review Letters , 1998, 80, 1826-1829, http://silver.neep.wisc.edu/~lakes/mu.html

Luo et al., â€œNew Experimental Limit on the Photon Rest Mass with a Rotating Torsion Balanceâ€, Phys. Rev. Lett, 90, no. 8, 081801 (2003)

Goldhaber and Nieto, Photon and Graviton Mass Limits, http://arxiv.org/abs/0809.1003

Goldhaber and Nieto, "Terrestrial and Extraterrestrial Limits on The Photon Mass," Rev. Mod. Phys. 43 (1971) 277â€“296

5. Dec 22, 2013

### Naty1

Hasn't that been experimentally confirmed?

And blechman posts #19:

so this seems to become rather complex mathematically!

Separately, Is it possible for the photon to have mass and still travel at 'c'.....I don't think so. Could we then define the wavefunction of a single photon? I think so.

Would gravitational waves would still be expected to travel at c..if photons don't ?

6. Dec 22, 2013

### bcrowell

Staff Emeritus
No, but since the mass is small, photons of ordinary energies would still travel at very close to c -- just like neutrinos.

Yes, but only in the case where the photon's energy was nonrelativistic -- which would not be in any practical situation.

Yes. The c in relativity isn't the speed of light.

7. Dec 22, 2013

### bcrowell

Staff Emeritus
There is a discussion of this in Zee, p. 141. He addresses thw question of why we can't detect an arbitrarily small mass thermodynamically, since a photon gas would have an extra d.f. The answer is that equilibration time goes to infinity as the mass goes to zero.

8. Dec 22, 2013

### ChrisVer

Well I already know how to interpret the table, I am not stating it has mass or that the table contradicts theory, but what might happen if it did indeed.
From wiki I liked the paper dealing with Proca fields.

9. Dec 22, 2013

### ChrisVer

I am not quiet sure about it. What is c in relativity then? the speed you get null vectors or zero ds2 (like a fundamental definition of it).
In the normal SR books, I guess they deal with c as the speed of light, which according to Einstein should be frame invariant.

10. Dec 22, 2013

### Staff: Mentor

in the modern understanding of SR, c is fundamentallly the "universal speed limit" or "universal invariant speed." We still often call c the "speed of light" for historical reasons and because as far as we can tell from experiment to date, light actually does travel at that speed.

We've progressed in our fundamental understanding of SR, in the 108 years since Einstein first published it. However, popular and introductory textbook treatments of SR still usually stick with the traditional approach which considers light as fundamentally special.

Last edited: Dec 22, 2013
11. Dec 23, 2013

### vanhees71

Indeed, that electromagnetic waves travel with the universal relativistic boundary velocity is an empirical question. There is nothing in our current understanding of elementary particles, i.e., the Standard Model, which forbids photons to have a non-zero (however tiny) mass.

The reason is that nothing forbids a mass for an Abelian gauge theory, and the photon is the gauge boson of the Abelian U(1) gauge group in the standard model. While for a non-Abelian gauge theory the only way to give the gauge bosons a mass, is the celebrated Higgs-Brout-Englert-Hagen-Kibble-Guralnik-Anderson-et-al mechanism (Nobel prize to Higgs and Englert this year), without destroying the consisency of the whole model, you can give a mass to the gauge boson of an Abelian U(1) gauge field in the naive way, by adding a mass term for the gauge field. This, is however somewhat unelegant, because it leads to the Proca approach to massive spin-1 fields, leading to a not manifestly renormalizable Lagrangian. It's more elegant to use the Stueckelberg approach, by introducing an additional scalar "ghost field". Then you can formulate an explicitly gauge invariant Lagrangian with a massive gauge boson. The only constraint is that the coupling of the gauge field to the matter fields is gauge invariant, e.g., by minimal coupling to a corresponding conserved current. Similarly to the Faddeev-Popov ghost field also the Stueckelberg ghost field decouples in linear gauges and thus is not observable.

Further it should be stressed that contrary to some postings in this thread, a first-quantization formulation of relativistic quantum mechanics is plagued not only with severe inconsistencies in the formalism but also disproven by observations. The reason is that in relativistic quantum mechanics there is no model for interacting particles that has a stable ground state and at the same time a conserved particle number, i.e., one-particle wave functions do not build a complete Hilbert space. That's why we use quantum field theory to describe relativistic particles: It's the natural formalism to describe many-body problems, and in relativistic quantum theory there are no single-body solutions (except for non-interacting particles which are not very interesting because they are unobservable; they are only important in the sense of "asymptotic free states", which are necessary in the QFT formalism to define observable quantities in the sense of the S matrix, i.e., cross sections, decay rates, etc.).

12. Dec 23, 2013

### ChrisVer

My problem on that is when you get the masses out of the Higg's mechanism you get a massless gauge boson describing the photon without making any assumptions... it was just taken out the maths...
To be more describtive you get the masses for the $W^{+-}$'s, and then a mass matrix for $W^{3}$ and $B$, which after diagonalizing you get the massless "photon" $A$ and the massive $Z$

Last edited: Dec 23, 2013
13. Dec 23, 2013

### bcrowell

Staff Emeritus
Not being able to find a good model for something isn't the same as disproving it observationally. Phys Rev Letters published an experimental paper in 2003 with the title "New Experimental Limit on the Photon Rest Mass with a Rotating Torsion Balance" (see #4). So according to the peer reviewers and editors at PRL, a nonzero photon mass didn't seem to have been "disproven by observations" as of 2003. Fundamentally, if you want to claim that a number has been proved experimentally to be *exactly* zero, you have a very tall burden of proof. Measurements don't measure *any* number with infinite precision. More generally, the long list of references I gave in #4 shows that this is still an area of active research, albeit not one that employs physicists by the boatload or one in which anyone seriously expects non-null results to come soon. A more accurate statement might be that as far as I know, nobody at the present time has proposed any good motivation for expecting a non-null result. This puts nonzero photon mass in a different category from speculations about physics beyond the standard model for which theorists have given strong motivation, e.g., supersymmetry, large hidden dimensions, or dispersion of the vacuum at the Planck scale.

Last edited: Dec 23, 2013
14. Dec 23, 2013

### ChrisVer

Also photons in Superconducting matterials seem to attain an "effective" mass... but I didnt want to get into that

15. Dec 26, 2013

### K^2

Apparently, some theories predict that in addition to a massless photon, there should be a massive one as well. I really don't know any details on that, but I know there are some experiments over at JLab looking for a massive photon.

16. Dec 26, 2013

### vanhees71

I have not said that a non-zero photon mass is disproven. To the contrary, it's no problem to give an abelian gauge field a mass.

What's disproven is the possibility to formulate relativistic quantum theory in terms of a first-quantization formulation, because the creation and annihilation of particles is very common in nowadays particle accelerators :-).

17. Dec 26, 2013

### vanhees71

Of course, the Standard Model of the electroweak interaction is formulated under the assumption that photons are massless. Together with a plethora of other empirical inputs ("V minus A" structure of the weak current, non flavor-changing neutral currents (at tree level), quarks with three colors as extra-degrees of freedom with the 1/3 and 2/3 charges, leading to the vanishing of anomalies of the gauge group, etc.) this determines the structure of the model pretty completely. Thus the model is built from the phenomenology, including the symmetry-breaking pattern $\mathrm{SU}(2)_{\text{weak}} \times \mathrm{U}(1)_{\text{weak hypercharge}} \rightarrow \mathrm{U}(1)_{\text{electromagnetic}}$. This pattern implies the masslessness of the photon, which is put in as a phenomenological fact.

This, of course, does not prove the masslessness of the photon. If a tiny photon mass is found, we have to reformulate the model such that it includes this finite photon mass in such way to keep the theory consistent.

An example is the (indirect) discovery of finite neutrino masses. In the usual standard model the neutrinos are considered massless, but that's only an approximation, as we know from the observation of neutrino-flavor oscillations, which can only happen when at least two of the three neutrinos are massive. Today, we don't know the precise values of the neutrino masses but only upper bounds.

18. Dec 26, 2013

### ChrisVer

Well the neutrinos were dealt with "normally" through the seesaw mechanism, in which you play with them as you did the quarks for the CKM matrix.
I guess I am not getting in which step in electroweak symmetry breaking somebody inputs the massless photon. From my post I'm pointing out that I'm missing it. For the neutrinos for example, you got 0 masses because (empirically) you dropped the right handed neutrinos... :/

19. Dec 27, 2013

### PhilDSP

Could you be more specific about the options of accounting for mass in a non-Abelian gauge theory? Do you mean that within the Standard Model the only way to give the gauge bosons a mass is via the celebrated HBEHKGA mechanism?

By the way, the idea that a photon has a very tiny mass was one important impetus behind the development of de Broglie's theory. It is not required, I believe, for his theory to be valid. But it would be good to study experimentally the details of what happens to the wavelength of a photon as its energy approaches zero.