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Massive particle has a specific chirality 
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#1
Feb1612, 02:13 AM

P: 283

What does the author mean here when he says
Is the author thus alluding to Majorana spinors here? Or, which massive fields do have specific chirality? And what do people mean when they say chirality is a Lorentz invariant concept, though it mixes in the Dirac spinors? thanks EDIT: And yes, both Dirac and Majorana spinors break chiral symmetry! Again, how can you say that massive spinors have specific chirality? 


#2
Feb1612, 04:00 AM

Sci Advisor
P: 3,571

Hm, a Dirac spinor can have definite chirality at a given time, but it won't be a solution of the time independent Dirac equation. Not a problem in principle.



#3
Feb1612, 12:17 PM

P: 283

So when I Lorentz transform a massive leftchiral state, it stays a leftchiral state.
Whereas time evolving it with respect to a equation of motion (e.g. Dirac equation), might turn it into a rightchiral state. Correct? 


#4
Feb1612, 02:44 PM

P: 2

Massive particle has a specific chirality
I'll take you through a small derivation of the dirac equation, a famous one. [tex]\partial_{t} \psi + \alpha^i \partial^i \psi = \beta m \psi[/tex] Move everything to the left hand side [tex]\partial_{t} \psi + \alpha^i \partial^i \psi  \beta m \psi = 0[/tex] Now all you do is multiply the entire equation by [tex]\psi^{*}[/tex] to obtain the action [tex]\psi^{*}(\partial_{t} \psi + \alpha^i \partial^i \psi  \beta m \psi) = \mathcal{L}[/tex] And produces the Langrangian. It is still zero, but it is a langrangian. This equation describes how to move one particle from one point to another. You might even think of it describing the Langrangian of a possible fragment of a world line. Now we will revert to using gammanotation which will express the covariant language. When you take [tex]\psi^{*}[/tex] and multiply it by [tex]\beta[/tex] you get [tex]\bar{\psi}[/tex]. So another way to write this is by saying [tex]\bar{\psi} \beta \partial_t \psi + \bar{\psi} \beta \alpha_i \partial_i\psi + m \bar{\psi}\psi[/tex] We can change the configuration of this expression in terms of new symbols. [tex]\gamma^{0}[/tex] is the gamma notation in respect to time, we can see the coefficient of beta is the derivative taken with respect to time and [tex]\beta \alpha_i[/tex] as [tex]\gamma_i[/tex]. We end up with [tex]\bar{\psi} (\gamma^{\mu}\partial_{mu} + m)\psi =\{ \bar{\psi} \gamma^{0} \partial_t \psi + \bar{\psi} \gamma^i \partial_i \psi + m \bar{\psi}\psi \}[/tex] There is what is called the fifth dirac matrix from this point. I'll assume you'll know that [tex]\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^5[/tex]. It is gamma 5 which is concerned with righthandedness and lefthandedness which in the technical term means, Chirality which has Eigenvalues of either +1 or 1. 


#5
Feb1712, 02:02 AM

Sci Advisor
P: 3,571




#6
Feb1712, 06:02 AM

P: 283

Thanks!!



#7
Feb1712, 04:36 PM

PF Gold
P: 354

the 2 components Weyl spinors are those with a defined chirality.
I had this answer in another forum: "A (right or left) chiral fermion is an irreducible representation of the Lorentz group. There is thus no Lorentz transformation that can convert it into another fermion of opposite chirality." Could somebody develop this? 


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