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Why is the photon massless?

  1. May 27, 2008 #1
    The Higgs mechanism can give gauge bosons mass. However, in the electroweak theory, only W and Z get masses, not the photon. Is there any fundamental reason why the photon should remain massless?

    Considering the experimental limit of the photon mass (2. 10^-16 eV), is there any theory that will predict, or accomodate, finite photon masses?
     
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  3. May 27, 2008 #2

    arivero

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    I find myself also puzzling about this. The breaking is SU(2)xU(1) --> U(1), but given that in principle you can give separate masses to Z and W, should it be better to consider
    SU(2)xU(1) ---> U(1)xU(1) ----> U(1)
    Or is it
    SU(2)xU(1) ----> 1 x U(1) ----> U(1)?
    Or is it
    SU(2)xU(1)-----> SU(2) x 1 ----> U(1)?

    GSW adds a feature: the mass of Z is to be higher than the mass of W. But Z is not W3, so I do not see if it is a real restriction in the breaking pattern; in any case it seems that we could restore W and still keep Z, and then really SU(2)xU(1) ---> SU(2) x 1. This hints about how the GSW model was discovered: a first idea is simply to mix electromagnetism and weak charged decay into a single SU(2) group, but it does not work.
     
  4. May 27, 2008 #3

    malawi_glenn

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    the photon does not interact via the weak force, and hence it should not interact with the Higgs field?
     
  5. May 27, 2008 #4

    Vanadium 50

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    The photon is the linear combination of w3 and B that doesn't couple to the Higgs. The beauty of the Higgs mechanism is that you are guaranteed one combination that doesn't couple, so you are guaranteed three massive bosons and one massless boson. (This is the same guarantee)

    Experimentally could the photon be found to be massive? Of course - although we know that the mass is really, really small. But there already is a theory of electrodynamics with massive photons, called the Proca theory. As you might imagine, in the limit where the photon is almost massless, this approaches the E&M theory we all know and love.
     
  6. May 27, 2008 #5

    arivero

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    But, can we scheme a Higgs mechanism where neither of the two combinations of w3 and B get mass? In GSW minimal, I'd say that it is not possible, but not sure about beyond minimal...

    Another interesting practice is to scheme so that Z is massive but neither W nor the photon are.

    Third, what about the case where the massless boson is not the photon, ie it violates parity?
     
  7. May 27, 2008 #6

    Vanadium 50

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    Sure, so long as you leave the w1 and w2 massless as well. :wink:

    More generally, I don't think this works. It's not just that neither combination of the w3 and B gets mass, it's that no combination gets a mass. I'm now free to define the "photon" and the "Z" any way I want: I've added a new continuous symmetry that wasn't there to begin with. One way to maybe look at it is that I have two U(1) fields in this situation, starting with SU(2) x U(1). Where did this other U(1) come from? Not from the SU(2). So I don't think the proper symmetries are there for this to work.

    I am even more sure this is impossible. To do this, I need to put the Higgs in a hypercharge singlet, rather than a weak isospin doublet - essentially, I make the B massive, and leave the w1, w2 and w3 massive. The problem with that is that while the w's carry weak isospin, the B doesn't carry hypercharge, it only couples to it. I need a non-Abelian group, but the B is a U(1).


    Isn't this automatic?
     
  8. May 27, 2008 #7

    arivero

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    Not sure. A part of the purpose of my questions were to try to understand why did we predict the Z neutral current. In principle you could try SU(2) naive and look for a way to put mass only into the W. Problems are, 1) that you need to avoid a mass for W3 and 2) that even if you get it to work, the original SU(2) is chiral or whatever you call the weak interaction.

    So you add an extra U(1), also chiral, hoping to combine B with W3 and to extract a nice right EM photon out of two wrongs.
     
  9. May 27, 2008 #8
    You can answer the question of which gauge bosons remain massless in spontaneous symmetry breaking unambiguously and exactly. To do this you need know: (1) the gauge symmetry group G (which determines all the gauge bosons), (2) the Higgs multiplet H (which may be reducible) (3) The self interacting scalar potential V (which may in fact be invariant under a group G' larger than G.) In the Standard theory of electro-weak interactions G is SU(2)XU(1) (hence 4 gauge bosons), H is just a doublet=(2, 0) of Higgs in the minimal Higgs scheme, and V is a quartic potential with no cubic terms which is invariant under SU(2)xU(1)XZ2, where Z2 is the discrete symmetry (fi--> -fi). The rest is just group-theory gymnastics. No more physics is needed. Given those three above, how the Higgs can break the symmetry is uniquely defined. (Here we assume that if Higgs can break the symmetry, it will.) After the symmetry breaking, let U be the unbroken symmetry group. If U is another Lie Group, all gauge bosons corresponding to its generators will remain massless. If U is a discrete group then all gauge bosons will become massive. How many (and which) gauge bosons remain massless is solely a function of what the group U is. These massless gauge bosons can carry any charge (not broken by the Higgs vacuum expectation value.) However, U has to be a group!!! The generators of U cannot be an arbitrary subset of of the generators of G, they have to form a subgroup.

    So, back to the Minimal Standard Model: H breaks SU(2)XU(1) down to U(1)', which has a single generator. Therefore, there will be precisely one massless gauge boson, and three massive ones. We simply rename the gauge boson corresponding to U1)' as "photon". If you wonder if we can ever have two massless gauge bosons, the answer is no. For that to happen , the final unbroken gauge group has to be U(1)XU(1). No Higgs multiplet of G will do that (even reducible ones.) Likewise, you can't have three massless gauge bosons either. (For that U must be SU(2)' or U(1)xU(1)xU(1), neither of which is possible for any Higgs multiplet.) However, it is possible to break the group completely so that all four gauge bosons (including the "photon") become massive. For that, H needs to be reducible (i.e. multiple doublets.)
     
  10. May 28, 2008 #9

    arivero

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    Hi, I have enjoyed this post, it put me to think! It is the kind of posting we need here in the net.

    U(1)xU(1)xU(1)? It is probably a mistake due to the fast answer, or perhaps a purposeful example... Of course when the group is not a subgroup of SU(2)xU(1), I do not expect the Higgs to produce it. And I do not see how U(1)^3 can be a subgroup, please someone correct me here if I am wrong. On other hand, I am open to mechanisms breaking into objects which are not a group (eg a q-group) but can not remember any.

    I am a bit surprised by the other example, that SU(2)x1 can not be produced via a Higgs mechanism. Are we telling that it is a very particular circumstance or that the Higgs can not break abelian groups? This can not be very general, because SU(2)xU(1) can break to U(1) and also totally, so it is is a sort of counterexample where U(1) can be totally broken.

    On other hand, I think we can get two massless gauge bosons, can we? The breaking of SU(2) into U(1) was the theme of Glashow-Giorgi 1972 model (not the GUT one), so the same trick should drive SU(2)xU(1) down to U(1)xU(1).

    Hmm we are answering the OP of the thread: the non minimal standard model accomodates massive photons?!
     
    Last edited: May 28, 2008
  11. May 28, 2008 #10

    Haelfix

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    If you want a massive photon you need to use the Proca or Stueckelberg action, otherwise the photon is trivially massless by demanding manifest local gauge invariance.
     
  12. May 28, 2008 #11

    arivero

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    Stueckelberg action, according wikipedia,
    uses a affine represntation of the Higgs field.

    (But also the wikipedia article seems to disagree between sections, about if this mechanism is a Higgs mechanism or a different one)
     
    Last edited: May 28, 2008
  13. May 28, 2008 #12
    Thank you for an illuminating reply.

    So, the fact that the photon is massless is associated mainly to the fact that the Higgs multiplet is a doublet.

    But then, what does this mean? Is there some higher theory (supersymmetry, superstrings ...), which restricts the representations of the Gauge group to which the Higgs field can belong?
     
  14. May 28, 2008 #13
    You are absolutely right on U(1)XU(1)XU(1). While I was emphasizing that no Higgs multiplet can generate it, I forgot that it is irrelevant because it is not a subgroup of SU(2)XU(1). All the more reason that 3 massless gauge bosons are not possible.

    On the issue of two massless gauge bosons, it is not possible in the usual electro-weak sense. But again, I stand corrected because there is a bizarre way. If you want to get U(1)xU(1) by breaking SU(2) down to U(1) (with a doublet), this is possible as you observed. But then, you have to leave the original U(1) (of SU(2)XU(1)) also unbroken. The only way to achieve this is to prevent the Higgs doublet (or any other multiplet) from coupling to U(1). In other words you will get U(1)xU(1) unbroken, if the Higgs couples to SU(2) only, not to the entire gauge group SU(2)xU(1). In this this case, the U(1) never mixes with SU(2), and stands as a separate unbroken entity. The photon and the weak couplings are no longer related by the usual relations after the symmetry breaking. It is no longer an electro-weak theory.
     
  15. May 28, 2008 #14

    arivero

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    Still, these bizarre cases SU(2)x1 and U(1)xU(1) are very interesting, to me. They correspond to actions in the two-dimensional sphere and torus, respectively. So when suplemented with SU(3) --which acts in the 4-sphere-- they correspond to Kaluza Klein theories in 6 extra dimensions.

    I am not sure if we can claim that U(1)xU(1) is not an electroweak theory.
     
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