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Elementary argument against longitudinal photons?

  1. Dec 30, 2009 #1

    bcrowell

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    Can anyone provide an elementary argument for why the zero mass of the photon means that it can't be longitudinally polarized?

    I know of at least one non-elementary argument, which is that if you write down the Proca equation, it only has longitudinally polarized solutions if the mass is nonzero. This is fine, but seems to me to be very indirect.

    I know of a couple of elementary arguments that prove the converse: that you can't put constraints on the polarization or helicity of a particle unless it has zero mass. (E.g., +1 helicity could be -1 helicity in a frame where you were catching up with the particle. Or you could go into the particle's rest frame, in which case there's no preferred direction, so it has to have all three axes of polarization available.)

    What I'm looking for is an argument that proves the original statement, not its converse, and that's more elementary than solving the Proca equation.

    It would be interesting to know whether there's a similar reason that the graviton can't have zero helicity.

    Thanks in advance!

    -Ben
     
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  3. Dec 31, 2009 #2

    clem

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    I am not sure that a zero mass particle can't have zero helicity.
    Deriving that result from a specific equation does not prove it in general.
     
  4. Dec 31, 2009 #3

    diazona

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    The photon is specifically zero-mass, spin-1, and therefore is governed by the Proca equation. So if the equation says it can't have zero helicity and zero mass, then it can't have zero helicity and zero mass.

    But actually, I'm interested in an answer to this question as well. Several times in various classes I've taken, I've seen mathematical derivations of the fact that massless particles have only two spin states, but none of them have ever really made sense to me. So if there's an intuitive way to understand it, I'd really like to know.
     
  5. Dec 31, 2009 #4

    Vanadium 50

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    Well, you might find this hand-wavy too.

    Let's think about a classical EM wave. If I allow a longitudinal polarization, that means the E-field is moving along the direction of motion. This won't satisfy Maxwell's equations. (One way to see this is that the Poynting vector needs to be simultaneously parallel with and perpendicular to E)

    When I quantize things, this feature persists.
     
  6. Dec 31, 2009 #5

    blechman

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    Massless fields that have vector indices in 4 dimensions cannot have a spin-0 component. A photon is an example of this, as is the graviton (if you try to do the naive quantum gravity).

    The argument goes like this: If a particle is massless, the 4 momentum MUST have two nonzero components, for example:

    [tex]k^\mu= (k;k,0,0)[/tex]

    for a photon moving in the +x direction. The equations of motion (Maxwell's equations in this case) require

    [tex]k^\mu\epsilon_\mu=0[/tex]

    where [itex]\epsilon^\mu[/itex] is the spin vector. A longitudinal photon (spin-0) has its spin along the direction of motion, and so would have

    [tex]\epsilon^\mu=(0;1,0,0)[/tex]

    This violates the previous equation.

    Notice that if the particle were massive, we can go to the rest frame where

    [tex]k^\mu= (k;0,0,0)[/tex]

    Now there is no problem with a longitudinal mode. The problem lies in the fact that massless particles have no rest frames, and always move at the speed of light.

    Those who are more technically oriented can also argue in terms of the "Little Group" and how the particle momentum transforms under rotations, etc. That is the rigorous way to do it, but hopefully this argument is enough.

    Hope that helps!
     
  7. Dec 31, 2009 #6

    blechman

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    You mean MAXWELL equation, not Proca. Proca assumes NONzero mass where there COULD be a longitudinal mode.
     
  8. Dec 31, 2009 #7

    Vanadium 50

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    You have this backwards - you are an unfortunate victim of the most confusing terminology since Ben Franklin decided what was positive and what was negative charge. A transverse photon has its spin along its direction of motion, and a longitudinal photon would have its spin transverse to its direction of motion.

    "Longitudinal" and "transverse" by convention refer to the direction the electric field points, not the helicity direction.

    However, I believe your derivation is still correct - if epsilon is not the spin vector, and instead it's the polarization vector.
     
  9. Dec 31, 2009 #8

    diazona

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    My understanding is that the Proca equation governs both massive and massless particles of spin 1. It reduces to Maxwell's equations when you set [itex]m = 0[/itex]. (And I double-checked this in Griffiths' particle physics book)
     
  10. Dec 31, 2009 #9

    bcrowell

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    Thanks for the interesting reply, blechman. But when you say:

    This is essentially what I'm asking about. I understand that the equations of motion (Maxwell's equations, or the Proca equation in the special case of m=0) require this, but I'm wondering if there's a way to see that more directly, without getting into grungy details.
     
  11. Dec 31, 2009 #10

    blechman

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    It's semantics, I suppose. But massless spin-1 (abelian) gauge theory is "Maxwell", while massive is "Proca" - I have never heard of anyone talking about "massless Proca fields" in practice.

    The better (and more appropriate) way to think about it is to consider the "Little Group" - but this is what I would call "grungy details" so that's why I didn't want to bring it up. It has to do with the behavior of the field under rotations.

    Quickly: the Little Group for a massive spin-1 particle is SO(3), which is the usual quantum angular momentum group, and so spin-1 should have three polarization states. But for MASSLESS fields, the little group is SO(2), which has a much different structure, and in particular does not allow for the longitudinal polarization.

    I would consider this approach, although more formally correct (and generalizable to higher dimensions and spins), to be much more about "grungy details" than my description. Although, as I said, it's the "correct" way to think about it.

    I'll leave you to look up "Little Group" in your favorite textbook or internet site and we can take it from there.
     
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