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Weak Interactions violate Lorentz?

  1. Dec 26, 2014 #1

    ChrisVer

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    I was wondering.
    For fermions someone can find that the Lorentz group is isomorphic to [itex]SU_L(2) \times SU_R(2) [/itex].
    However in the Standard Model there is only left-handed neutrinos interacting with the rest of matter. If Lorentz was not partially violated, wouldn't someone expect for the left and right-fermions interacting in the same way?
    Or am I missing something?
     
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  3. Dec 26, 2014 #2

    Vanadium 50

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    Why do you think a gamma5 violates Lorentz invariance? And if your answer is "it doesn't", my next question will be "Why do you think a 1+gamma5 violates Lorentz invariance?"
     
  4. Dec 26, 2014 #3

    ChrisVer

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    Neither does.. But why then the SM doesn't have right-handed fermions as it has left-handed ones [and in particular neutrinos]?
    The distinction appears when somebody said that weak interactions only couple to the left-handed neutrinos... So I'm saying that for some reason somebody breaks the symmetry between left and right ( which is the Lorentz symmetry)...
    If Lorentz symmetry was still held, then someone could write the same interactions for the right handed as he did for the left handed, with the same couplings. Weak interactions however, concerning neutrinos, change this symmetry, and tells you "ah, the left is the "nice one" whereas the right vanishes"
     
  5. Dec 26, 2014 #4

    Vanadium 50

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    No, it's not. That's parity. The SM breaks parity, but it does not break Lorentz invariance.
     
  6. Dec 27, 2014 #5

    vanhees71

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    That's it: The weak interaction breaks P, C, CP, and due to the CPT theorem then also T invariance. These are all discrete symmetries of space-time, and there's no need for them to hold in nature. The only thing that's really important on a very fundamental ground is that the causality structure of space-time holds, and for this it's sufficient that the variation of the action is invariant under the part of the Poincare group that is simply connected with the neutral element of that group, and that's the proper orthochroneous Lorentz transform.

    As soon as parity (spatial-reflection symmetry) is violated, a spin-1/2 particle needs not be represented necessarily by a Dirac spinor but also a Weyl spinor is sufficient, and the two kinds of Weyl spinors corresponding to the irreducible representations of the covering group of the Lorentz group, SL(2,C), labelled as the (1/2,0) and (0,1/2) representations, correspond to the left- and right-handed parts (particles of definite chirality) of the Dirac spinor ##(1\mp \gamma^5)/2 \psi##.

    In the standard model the weak interaction has the famous structure (vector current minus axial vector current), and thus only the left-handed part of the neutrinos and the right-handed part of the antineutrinos couples to the "massive leptons". In the approximation of massless neutrinos, there are only left-handed neutrinos and right-handed anti-neutrinos in the game.

    For massive neutrinos, there is also mixing as in the quark sector, i.e., the flavor-eigenstates are not the mass eigenstates, which gives rise to neutrino oscillations.
     
  7. Dec 27, 2014 #6


    Right; just so I am clear - as the OP somewhat alludes, the Standard Model is *not* in fact invariant under *all* Lorentz transformations, only the proper orthochronous ones, that is, SO+(1,3), not SO(1,3).
     
  8. Dec 29, 2014 #7

    dextercioby

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    That is correct. That's why people call the proper inhomogenous Lorentz group (or the proper Poincare group) the fundamental symmetry group of flat 1+3 spacetime. The C,P,T discrete symmetries are thus purely of phenomenologic nature. However, as soon as one turns to serious (Wightman) QFT, then TCP (all three at the same time) is another symmetry of the SM as well, along with the proper Poincare one. These 2 symmetries limit the bare Lagrangian densities, along with known experimental facts.
     
    Last edited: Dec 29, 2014
  9. Dec 29, 2014 #8

    Orodruin

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    Just to chime in a bit for clarification: Even for massive neutrinos, there is no a priori need to introduce right-handed neutrinos into the game even if it is one (rather attractive) option. Lepton and quark mixing is related to a missmatch of the up and down type fermion mass matrices, which can also happen for pure Majorana neutrinos (e.g., in a type-II seesaw).
     
  10. Dec 29, 2014 #9

    arivero

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    So, perhaps the question is "Why does QCD preserve parity"? We can always argue that U(1) electromagnetism preserves parity by design, because it is the unbroken piece of a non-preserving interaction. But QCD?
     
  11. Dec 29, 2014 #10

    Vanadium 50

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    The U(1) is not really the "unbroken piece" - it's a mix of the w3 and the bino, and as such is every bit as much a part of the breaking as the Z.

    If you start with Weyl fields and put the left multiplets of the quarks in an SU(3) color triplet, if you want your theory to be anomaly free and to leave the leptons colorless, you need to put the right multiplets into the same triplet (within a rotation, but that rotation can be phased away by redefining the triplet basis states).
     
  12. Dec 29, 2014 #11

    arivero

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    So, for instance, QCD with only left quarks is anomalous? Is this true for every SU(N)?
     
  13. Dec 29, 2014 #12

    Vanadium 50

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    No, but if you have both handednesses, they need to either carry the same charges (like real QCD) or be neutral (like a wacky chiral QCD).
     
  14. Dec 29, 2014 #13

    arivero

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    I am a bit lost. Are you speaking of electric charge or colour charge?. My question is if it is possible to have a QCD where the strong coupling for right-handed quarks is not the same that for left-handed quarks. Is such coupling anomalous, or were you just speaking of electric charge?


    EDIT: From table III here, it would seem that the complex N-dim representation of SU(N) is always anomalous, but I am not sure if I am reading it correctly. http://inspirehep.net/record/11544?ln=es
     
    Last edited: Dec 29, 2014
  15. Dec 29, 2014 #14

    Vanadium 50

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    You have only two choices. Either the left and right-handed versions contain the same color charges, or only one does and the other is color-neutral. The reason is that the sum of the charges for every interaction needs to sum to zero over the all the fields in the theory.
     
  16. Dec 30, 2014 #15

    vanhees71

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    Sure. I hope that there will be clear evidence whether neutrinos are Majorana or Dirac fermions soon. I think KATRIN should get into this business 2015 too in addition to the mass measurement with the tritium decay spectrum endpoint, right?
     
  17. Dec 30, 2014 #16

    arivero

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    I am still a bit lost. Which is the "sum of charges" for colour charge, and how is it that it is zero in the two cases you mention?
     
  18. Dec 30, 2014 #17

    Vanadium 50

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    Arivero, this is not really a side point anymore but is growing to a full-scale thread derailment. It probably deserves its own thread. Further, it's clear that your are finding what I am saying confusing, so it's unlikely that another message from me will help.
     
  19. Dec 30, 2014 #18

    ChrisVer

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    the sum is taken over the generators (or charges) in these cases [as it's taken for the hypercharges in U(1)]...
     
  20. Dec 30, 2014 #19

    arivero

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    No, no, it is the same thread. Once we have fixed, in the few first comments, that what the OP calls "to violate Lorentz" is really parity violation, then the question becomes "why can the electroweak theory violate parity?". And the answer seems to be that it can because there are complex representations which are, by a fortunate combination of charges, anomaly free. To understand it better, we ask if there are some theories that, having complex representations, can not violate parity. It is not a derailment; it is further study of the OP question.
     
    Last edited: Dec 30, 2014
  21. Dec 31, 2014 #20

    ChrisVer

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    So, to clear this up, do we have the breaking of [itex]SU(2)_L \times SU(2)_R [/itex] by the violation of parity?
     
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