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Urgent help

  1. Nov 16, 2004 #1
    Hi, got an exam tomorrow for particle physics and need a question answered concerning helicity.

    How would it be possible to determine if a photon had a non-zero rest mass, based on measured helicity values?

  2. jcsd
  3. Nov 16, 2004 #2
    Well, restmass (which is always referred to as just mass, so don't bother to make the distinction :wink: ) mixes up the two different possible chiralities. This is a result of the famous "tau-theta-puzzle" in QFT. To make a long story short : in QFT all elementary particles are massless because the two chiralities each correspond to a different kind of fundamental "particle-property". I mean, in QFT particles are classified based upon their chirality because a left handed chiral particle does not behave in the same way as a right hand chiral particle. So mass has to be zero (before the spontaneous breakdown of symmetry) because otherwise one particle possesses the TWO chiralities at once (because of the mass-term in the Lagrangian) and thus it is NOT fundamental. So basically photons are always massless because they do not interact with the Higgs-field. They contain a distinct chirality-value and thus not different values for ONE photon. Just look at the V-A-current in the elektroweak-theory : only left handed fermions (or right handed anti-fermions) are involved in weak charged currents.

  4. Nov 16, 2004 #3
    OK, i understand the your point, but im only in second year undergraduate physics at university so we havent covered a lot of QFT. Basically due to parity violation for the Weak Interaction, participating fermions are emitted and absorbed predominantly in negative helicity states (and positive for anti-fermions), the probability a fermion's in a negative helicity state can be found by

    P=0.5(1+v/c) where v is the particle's speed
    and for a fermion in a positive helicity

    So from here i see that a massless particle has zero probability of being in a positive helicity. I am just wondering if this same argument can be used to deduce a photon's helicity.
  5. Nov 16, 2004 #4
    You are correct...it is gonna be something like that...

  6. Nov 16, 2004 #5
    Thanks for that, feeling a bit relieved for tomorrow's exam now!

  7. Nov 16, 2004 #6
    good luck man...

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