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Nonzero photon mass

  1. Dec 28, 2009 #1

    bcrowell

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    We used to think neutrinos were massless, and therefore traveled at c. Now we know that at least some types of neutrinos have mass, and travel at less than c. What about the photon?

    Section I.2 of Jackson has a description of the empirical limits on the mass of the photon. (This is the introductory chapter, letter "I", not chapter 1.) I have the 2nd ed., so this would be as of 1975. As of 1975, the 1/r2 form of the Coulomb force law had been tested to very high precision at scales of 10^-2 m to 10^7 m. (Section 12.9 of Jackson also has a lengthy description of why one particular method for testing for a nonzero photon mass doesn't work.) More recently, searches for dispersion of the vacuum have had negative results: http://arxiv.org/abs/0908.1832 , but this is at short wavelengths, whereas I assume the effects of massive photons would be felt at extremely long wavelengths.

    Now suppose that someone does an experiment tomorrow that finds that photons have a mass that's nonzero, although very small. Then photons would have the same status as neutrinos: particles that almost always travel very close to c, because they are hardly ever produced with energies small enough to make v significantly less than c. You would also be able to have real, longitudinally polarized photons, although I guess they'd be hard to manipulate in practice, since they'd have to have such low energies.

    Would this have any especially important implications?

    Do massive photons produce any technical problems in QED? Mandl and Shaw have the following remark, which I'm not sophisticated enough to understand, on p. 183: "...this [convergence] factor does not provide a suitable regularization procedure for QED, as it does not ensure zero rest mass for the real physical photon, nor the related gauge invariance of the theory." Does this mean that gauge invariance is somehow logically related to zero mass? From my limited understanding of gauge invariance in particle physics, one expects it to occur for any bosonic field, but I don't see how a nonzero mass affects anything.

    I'm inclined to believe that this would not have any particularly deep implications for relativity, but I'm open to arguments to the contrary.
     
  2. jcsd
  3. Dec 29, 2009 #2
    Hmmm....If we find that there is non-zero mass for the photon, I guess it does have deep implications for fundamental physics......as of now, there is nothing that says that you cannot have a particle with zero rest mass...but if you find that photon, the archetypical massless particle, is massive...then one has to start wondering whether there is something that requires a particle to have a non-zero rest mass...


    As far as relativity is concerned....we know that the ultimate velocity c cannot be attained by any massive particle....so if the photon is proved to be massive and we keep the basic structure of relativity, we have to stop identifying c with the velocity of light....

    It also means that one can also think of going into a frame in which the photon is at rest...and also that the velocity of light is frame dependent...which also means that the Maxwell's equations in their current form cannot survive...as the velocity of light predicted is not frame-dependent....

    Thus, massiveness of the photon attacks the very root of Einstein's intent in building up relativity..which was to bring electrodynamics into harmony with mechanics..and also calls into question many of his derivations... as he routinely used the fact that the speed of light is a constant...
     
  4. Dec 29, 2009 #3

    Vanadium 50

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    A massive photon breaks gauge invariance and permits charge non-conservation.
     
  5. Dec 29, 2009 #4
    I wonder if someone has an estimate of the upper bound for the photon mass one gets from the constraints that the observable universe has zero total charge. This is a very important constraint, since the tiniest charge will change cosmological evolution drastically.
     
  6. Dec 29, 2009 #5
    You can write down a theory for the massive photon in which you have a conserved current, see e.g. here for details:

    http://arxiv.org/abs/hep-ph/0306245
     
  7. Dec 29, 2009 #6

    bcrowell

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    I don't think it would affect those derivations at all. It would just mean that the c in those derivations would not be interpretable as the speed of light.
     
  8. Dec 29, 2009 #7

    bcrowell

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    Thanks, all, for the replies. There are two things I don't understand here. (1) There seems to be a general rule that only zero-mass bosons allow gauge symmetry. Why? (2) The Higgs gives a mechanism for wriggling out of #1, so that the W and Z can have mass. Why can't the same mechanism be applied in order to allow the photon to have mass?
     
  9. Dec 29, 2009 #8

    Vanadium 50

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    There are several ways to look at the answer to #1. The one I like the best goes as follows:

    • If I change the wave function by a phase [tex]\phi[/tex], all physics remains unchanged.
    • By the same token, I can change it by a varying phase, [tex]\phi(\bf(x},t)[/tex]. This is the principle of Local Gauge Invariance.
    • Doing so adds terms depending on [tex]\phi(\bf(x},t)[/tex] in the equations of motion, because the Schroedinger and Dirac equations involve derivatives.
    • To fix this, I add a new field [tex]A_\nu(\bf(x},t)[/tex] and change the derivative to [tex]\partial_\nu - ieA_\nu[/tex].
    • This balances the [tex]\phi(\bf(x},t)[/tex] term if and only if excitations in the [tex]A_\nu(\bf(x},t)[/tex] field (i.e. the gauge bosons) have exactly two helicity states. Any more, and I will have overcorrected, and be in the same trouble I was in before.
    • A massless spin-1 particle has two spin states. Therefore, you need massless bosons to keep Gauge Invariance.

    As for why the Higgs mechanism doesn't affect the photon, you are guaranteed to have one linear combination of the original SU(2) and U(1) fields that doesn't couple to the Higgs and as such remains massless. Essentially, the symmetries of the system leave you one conserved quantum number, and that needs to couple to a long-range force.
     
  10. Dec 29, 2009 #9

    bcrowell

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    Thanks, Vanadium 50. That's very helpful.
     
  11. Dec 29, 2009 #10
    This is not true, it is straightforward to write down a gauge invariant theory for a massive photon.
     
  12. Dec 29, 2009 #11
    We know from GRB observations that gamma ray bursts are accompanied by nearly simultaneous optical counterparts.

    http://en.wikipedia.org/wiki/File:Grb080319B_flux_curve.jpg

    GRB 080319B shows that optical and gamma ray photons from the source 7.5 billion light-years away arrive to Earth within 100 seconds from each other. That lets us set the upper limit on photon mass ~ E(optical) * sqrt(dt / t) or somewhere around [itex]10^{-7}[/itex] eV. That seems to be a strong limit already.

    Some extremely stringent limits (< [itex]10^{-18}[/itex] eV) can be found in PDG.
     
  13. Dec 29, 2009 #12

    Vanadium 50

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    True, but...

    If I do the simplest thing and write down a Lagrangian like

    [tex]-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}+\frac{1}{2}m^2 A_\mu A^\mu [/tex]

    this won't be Gauge invariant for exactly the reason I said: you'll get an overcorrection from the third piece.

    What I need to do instead is to use a derivative coupling:

    [tex]-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}+\frac{1}{2}m^2 A_\mu A^\mu +
    \frac{1}{2}m^2 (\partial_\mu \phi)^2 -mA_\mu \partial^\mu \phi[/tex]
     
  14. Dec 30, 2009 #13
    I agree....sorry for the loose statement..
     
  15. Dec 31, 2009 #14
    About the lack of charge conservation... in a superconductor, the gauge symmetry is (spontaneously?) broken and the photon gets a mass. Wouldn't your statement imply that (electric...?) charge isn't conserved in a superconductor?
     
  16. Dec 31, 2009 #15
    It would if you were to write down the wrong effective Langrangian. :biggrin:
     
  17. Dec 31, 2009 #16

    Vanadium 50

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    I am not an expert in superconductivity, but I believe that is the case. There's an old paper by Nambu on this. Note that "charge nonconservation" means that the continuity equation of the supercurrent is only valid statistically - it does not mean electrons vanish in superconductors, never to be heard from again.
     
  18. Dec 31, 2009 #17

    blechman

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    In fact, a "massive photon" does NOT mean (by itself) that charge is no longer conserved! I actually wrote an incorrect post on this many months ago, so now is my chance to fix it!

    The trick is that even though a mass term does break the gauge invariance, it does NOT break the GLOBAL charge invariance (where the "gauge parameter is no longer dependent on space-time). Therefore, Noether's theorem still implies that charge is conserved, despite the photon mass. This is the so called "Stukelberg trick" of massive abelian gauge theories. I might have spelled that wrong...

    This is unique to ABELIAN gauge theories. NON-ABELIAN theories (such as that which describes the W and Z) this is NOT true! The reason has to do with the additional self-couplings of the gauge bosons (and the fact that the gauge bosons themselves have charge, unlike the photon which is neutral). In that case, the only way to keep things under control is to "spontaneously break" the symmetry, a.k.a. the Higgs mechanism.

    As to superconductivity: in that case it IS true that charge is no longer conserved in the low-energy regime. This is not DIRECTLY because the photon has a mass (although there is an indirect relation), but because the order parameter (Cooper pair wavefunction) carries charge. In the effective theory where the Cooper pairs have collapsed to the ground state and can be removed from the dynamics of the theory, the "vacuum state" has charge. Therefore, charge can be "created" (a Cooper pair dissociates "creating" two electrons) or "destroyed" (two electrons reform a Cooper pair and collapse into the vacuum).

    Of course, in the grander sense, charge is obviously conserved. But the same argument can be used in Newtonian physics: any external force violates momentum conservation (dp/dt is not zero!). But in reality, when you consider the dynamics of the external force, then TOTAL momentum is conserved. But in practice there is no reason to do that! Same argument here.
     
  19. Jan 1, 2010 #18
    I don't know how much you can make of this. They were referring to a specific factor, equation (9.9). The paragraph
    continues:
     
  20. Jan 2, 2010 #19

    blechman

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    At the risk of hijacking this thread, I feel the need to clear the air and fix several wrong posts I have made in the past (including the above quote). I promise I'll limit myself to one post unless people have questions. :wink:

    The issue of massive gauge fields is a little tricky. The problem is that there are actually TWO kinds of symmetries that are relevant: the gauge (aka "local") symmetry and a global symmetry. Obviously, any gauge symmetry CONTAINS a global symmetry, since if the gauge parameter can depend arbitrarily on spacetime, then surely it can be independent of spacetime!

    Now it is often said that "charge conservation follows from gauge invariance and Noether's theorem" but this is not in fact literally true. It is, in fact, the GLOBAL part of the gauge symmetry that implies charge conservation! Therefore breaking the "gauge" symmetry while leaving the global part in tact does not do anything to charge conservation. This is true for nonabelian as well as abelian theories! I leave the proof to the reader that a mass term does not break the global symmetry at all. Therefore, massive vector bosons do not violate charge conservation at all, at least, not on their own.

    But there is another problem with gauge theories: it has been shown that the LONGITUDINAL component of the vector boson becomes strongly coupled in the UV. This means that the theory of a massive gauge boson, which has a longitudinal component, is not renormalizable!

    You might have heard the statement that "gauge symmetry is not a symmetry, but a redundancy". This is where that distinction pays dividends: the LOCAL part of the symmetry means (after some very nontrivial math) that the longitudinal components of a massless vector boson decouple, so that they never cause any problems with physical S matrix elements.

    So the question becomes: what about massive vector bosons, which have longitudinal components that do not decouple? Well, it was shown in the 1970s (and the 1999 Nobel Prize was awarded for it) that if the gauge symmetry is spontaneously broken, then the longitudinal component STILL decouples and the theory is still renormalizable.

    Now here's the rub: if the theory is spontaneously broken, it means that the vacuum has charge. This breaks the global symmetry, and NOW the charge conservation rule is violated!

    So in summary: a massive gauge boson alone does not mean that charge is not conserved. However, a massive gauge boson described by a renormalizable theory does!

    OK, I'm sorry to go off-topic, but I made many bad posts in the last year, and I hope this clears the air. Hope it is clear and it helps!
     
    Last edited: Jan 2, 2010
  21. Jan 2, 2010 #20

    bcrowell

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    Thanks, blechman, for post #19 -- very informative!

    I thought I'd post some references here, since they might be useful to others.

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

    Luo et al., “New Experimental Limit on the Photon Rest Mass with a Rotating Torsion Balance”, Phys. Rev. Lett, 90, no. 8, 081801 (2003)

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

    Luo seems to be the best available upper limit on the mass, but the interpretation is controversial. The Lakes paper, which is available online, is a similar, earlier experiment. Goldhaber and Nieto give a good overview of the theory, plus a review of measurements as of 1971.

    Correlating some of what blechman said with what I've learned from these papers: If the photon has mass, then the electromagnetic potential A becomes observable, in the sense that it has an energy density associated with it, and in Maxwell's equations it contributes to the source terms. However, if you make a global change to the potential, e.g., by adding a constant to A0, that would be unobservable, because a constant global energy density has no observable consequences, nor does a constant global charge density.
     
    Last edited: Jan 2, 2010
  22. Jan 2, 2010 #21

    blechman

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    Thanks for those references, bcrowell!

    As to the above quote, however, I don't follow it. I am not quite sure what you mean by "observable." The field [itex]A_\mu[/itex] carries energy/momentum even in the massless case! Further, it is not true that the mass term contributes to the source term: there would have to be a cubic or higher interaction for that to happen, and a photon mass is only quadratic.

    Also, shifting [itex]A^0[/itex] by a constant is not an example of a global symmetry: in fact, [itex]A_\mu[/itex] should not change at all under a global symmetry (only nonabelian gauge bosons like the gluon or W/Z bosons would shift under the global symmetry). Your example is actually an ordinary (local) gauge transformation

    [tex]A_\mu\longrightarrow A_\mu +\partial_\mu\alpha[/tex]

    where [itex]\alpha={\rm const}\times t[/itex].
     
  23. Jan 3, 2010 #22
    Hmm? Something does not add up here. [tex]\phi[/tex] appears to be a scalar field of dimension one. The term [tex]mA_\mu \partial^\mu \phi[/tex] confirms this (in other words the total dimension is four as it should be.) But the term [tex]m^2 (\partial_\mu \phi)^2 [/tex] has dimension six, and it makes no sense. Perhaps you meant to leave out the factor [tex]m^2[/tex] here? In any case, the statement that this new Lagrangian restores Gauge invariance is not valid, until you tell us how [tex]\phi[/tex] transforms under a gauge transformation. (I assume [tex]A_\mu[/tex] transforms in the usual manner.)

    I believe that the violation of the (usual) local gauge invariance introduced by the photon mass term cannot be compensated by adding a scalar field. (However, you can still have a globally conserved electric charge.) Please first correct the Lagrangian above so that each term has dimension four. After that, if you still think it is Gauge invariant, please show us explicitly how that is so. I could not find a simple gauge transformation property for [tex]\phi[/tex] that will do the trick. [Perhaps you are also changing the rule for [tex]A_\mu[/tex] as well?]
     
  24. Jan 3, 2010 #23

    blechman

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    The scalar field is dimensionless in this choice of normalization, and transforms inhomogeneously:

    [tex]\phi(x)\rightarrow \phi(x) + \Lambda(x)[/tex]
    [tex]A_\mu(x)\rightarrow A_\mu(x)+\partial_\mu\Lambda(x)[/tex]

    and then gauge invariance is restored (with m replaced with m^2 in the cross term). Equivalently, you can drop the m^2 in the kinetic term of the scalar, and make it dimension-1 as you say. But then the transformation law for the scalar becomes:

    [tex]\phi(x)\rightarrow \phi(x) + m\Lambda(x)[/tex]

    That works too.

    This is the so-called "Stukelberg trick" I alluded to above (also known as the "Green-Schwartz Mechanism" in string theory). The scalar field is effectively the longitudinal mode of the massive photon. In the Higgs formalism language, it is the goldstone boson.

    See above. And remember also that overall 4-divergences don't bother me, because they always drop out of the action after an integration by parts.
     
    Last edited: Jan 3, 2010
  25. Jan 3, 2010 #24

    bcrowell

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    If the whole universe is filled with a uniform background of charge, then you don't get any electrical forces as a result. E.g., this is why the Dirac sea doesn't create an electric field.


    Only its derivatives do. In the massless case, a constant A has an energy density of zero. When the photon has mass, the energy density becomes [itex][E^2+H^2+\mu^2(A^2+V^2)]/8\pi[/itex], so even if A's derivatives vanish (so that E and H vanish), you still have an energy density.


    Goldhaber and Nieto write Maxwell's equations like this:
    [tex]
    \nabla \cdot E = 4 \pi \rho - \mu^2 V
    [/tex]
    [tex]
    \nabla \times H = (4 \pi/c) J - \mu^2 A
    [/tex]

    Okay. I guess I didn't understand correctly what you meant by a global symmetry. Could you give an example?

    Thanks!

    -Ben
     
  26. Jan 3, 2010 #25

    blechman

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    OK, I'll concede that.


    Well, write the equations any way you want, but when I look for the "free field equations", that corresponds to the quadratic pieces in the action. In particular: I would have included the mass terms on the LEFT hand side, incorporated them into the free-field propagator (Green's Function), and then solved them with external sources [itex]\rho,J[/itex]. But I guess it depends on what limiting cases you want to take...

    By "global symmetry" I mean that the gauge parameter does not have spacetime dependence.

    In QED, this means that the gauge field does not transform at all, while the matter fields (electron, etc) would transform with a (constant) phase. In the case ordinary QED, this is just electron number symmetry.

    In nonabelian gauge theories, however, the vector potential would transform under a constant gauge transformation, since:

    [tex]A_\mu^a(x)\rightarrow A_\mu^a(x)+D_\mu\alpha^a[/tex]

    where

    [tex]D_\mu\alpha^a=\partial_\mu\alpha^a+if^{abc}A_\mu^b\alpha^c[/tex]
     
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