Why are heilicty and chirality equivalent for massless particles?by Superposed_Cat Tags: chirality, equivalent, heilicty, massless, particles 

#1
Dec1713, 03:23 PM

P: 257

Salutations, question:




#2
Dec1713, 05:26 PM

Mentor
P: 10,767

Or, in other words, the spin is always in the direction of motion or against, and this does not change. 



#4
Dec1713, 06:19 PM

P: 367

Why are heilicty and chirality equivalent for massless particles?
does it change for massive particles? The chirality does not coincide with helicity, because chirality symmetry is explicitly broken




#6
Dec1713, 07:03 PM

Mentor
P: 15,570

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
Dec1813, 04:52 AM

Mentor
P: 15,570

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.




#9
Dec1813, 05:01 AM

P: 367





#10
Dec1813, 07:00 AM

P: 257

V50 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 



#11
Dec1813, 07:05 AM

Sci Advisor
Thanks
P: 3,846

Chirality (aka handedness) = ±1 is the eigenvalue of γ^{5}. It's a Lorentz invariant, but for a particle with nonzero m it's not a constant of the motion, i.e. [H, γ^{5}] ≠ 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 PeskinSchroeder and use the Weyl representation, in which γ^{5} is diagonal, [tex]\gamma^5 = \left(\begin{array}{cc}1&0\\0&1\end{array}\right)[/tex] The eigenfunctions of γ^{5} corresponding to ±1 are denoted ψ_{R} and ψ_{L}, respectively. In terms of ψ_{L} and ψ_{R} 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 ψ_{R} and ψ_{L} are also eigenstates of helicity, σ·p = ±E. 



#12
Dec1813, 10:17 AM

P: 1,196





#14
Dec1813, 05:04 PM

P: 308

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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