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Why are heilicty and chirality equivalent for massless particles?

  1. Dec 17, 2013 #1
    Salutations, question:
    ~wikipedia. I'm assuming that's because (correct me if i'm wrong) those particles would travel at c and you could not overtake them so their chirality and helicity are equal.But if a photon was coming towards you then zoomed past you surely the chirality would change because you now see it going away from you?
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  3. Dec 17, 2013 #2


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    It is now going away, but the spin also changed its radial direction: if it was oriented towards you before, it will be oriented away from you afterwards, and vice versa.
    Or, in other words, the spin is always in the direction of motion or against, and this does not change.
  4. Dec 17, 2013 #3
    But then why does it change for massive particles?
  5. Dec 17, 2013 #4


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    does it change for massive particles? The chirality does not coincide with helicity, because chirality symmetry is explicitly broken
  6. Dec 17, 2013 #5
    helicity sorry. why does for massive particles?
  7. Dec 17, 2013 #6

    Vanadium 50

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    Please don't take this the wrong way, but do you know what either helicity or chirality are? From your questions, it doesn't look like you do.
  8. Dec 18, 2013 #7
    No worries I take no offence. I just no what i read on the wiki.
  9. Dec 18, 2013 #8

    Vanadium 50

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    Usually people understand one and are confused about the other. (And answering the question hinges on finding which one they understand) But if you don't know what either of them are, it will be difficult to explain the difference.
  10. Dec 18, 2013 #9


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    In fact, can't your just write your state as a superposition of helicity eigenstates, and then look at how it transforms under a lorentz transformation to see that?
  11. Dec 18, 2013 #10
    V-50 I was wondering why they said you can't change your reference frame because you can't overtake it. Surely if it flyies past you it is the same as overtaking it?
    P.S. thanks chrisver
  12. Dec 18, 2013 #11


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    Chirality (aka handedness) = ±1 is the eigenvalue of γ5. It's a Lorentz invariant, but for a particle with nonzero m it's not a constant of the motion, i.e. [H, γ5] ≠ 0.

    Helicity = ±1 is the eigenvalue of σ·p/|p|. For a free particle it's a constant of the motion, but clearly not a Lorentz invariant.

    The easiest way to see the relation between the two is to follow Peskin-Schroeder and use the Weyl representation, in which γ5 is diagonal,

    [tex]\gamma^5 = \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)[/tex]
    The eigenfunctions of γ5 corresponding to ±1 are denoted ψR and ψL, respectively. In terms of ψL and ψR the Dirac equation is
    [tex]\left(\begin{array}{cc}-m&E + \sigma \cdot p\\E - \sigma \cdot p&m\end{array}\right) \left(\begin{array}{c}\psi_L\\\psi_R\end{array}\right) = 0[/tex]
    and so for m = 0 we have that ψR and ψL are also eigenstates of helicity, σ·p = ±E.
  13. Dec 18, 2013 #12
    No, when you overtake a particle, you're moving faster than the particle and in your reference frame the particle is seen moving the opposite direction reversing its helicity. When a particle overtakes you you still see the particle moving in the same direction with no helicity reversal.
  14. Dec 18, 2013 #13
    Thanks all esp. dauto
  15. Dec 18, 2013 #14
    And perhaps to make it even more obvious, there is no preferred "centre" of a reference frame, i.e. nothing explicitly stating where "you" are relative to the particle. x=0 is not special. So nothing you measure about the particle is going to change as it zooms past some arbitrary location, except it's position of course.

    Going to a different frame via a Lorentz boost is a different story though (note again that you don't have to "actually" "overtake" the particle, since "you" are not anywhere in the math)
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