# Massive particle has a specific chirality

by Lapidus
Tags: chirality, massive, particle, specific
P: 290
What does the author mean here when he says

 However, a massive particle has a specific chirality. A massive left-chiral particle may have either left- or right-helicity depending on your reference frame relative to the particle. In all reference frames the particle will still be left-chiral, no matter what helicity it is.
How does a massive particle have a specific chirality? I learned that the only massive single chiral fields are the ones with Majorana mass. Dirac fields are a mix of left-chiral and right chiral fields, they do not have a specific chirality.

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?
 Sci Advisor P: 3,633 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.
 P: 290 So when I Lorentz transform a massive left-chiral state, it stays a left-chiral state. Whereas time evolving it with respect to a equation of motion (e.g. Dirac equation), might turn it into a right-chiral state. Correct?
P: 2
Massive particle has a specific chirality

 Quote by Lapidus So when I Lorentz transform a massive left-chiral state, it stays a left-chiral state. Whereas time evolving it with respect to a equation of motion (e.g. Dirac equation), might turn it into a right-chiral state. Correct?
Do you know in principle how chirality enters the Dirac equation? How is your covariant notation?

I'll take you through a small derivation of the dirac equation, a famous one.

$$\partial_{t} \psi + \alpha^i \partial^i \psi = \beta m \psi$$

Move everything to the left hand side

$$\partial_{t} \psi + \alpha^i \partial^i \psi - \beta m \psi = 0$$

Now all you do is multiply the entire equation by $$\psi^{*}$$ to obtain the action

$$\psi^{*}(\partial_{t} \psi + \alpha^i \partial^i \psi - \beta m \psi) = \mathcal{L}$$

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 gamma-notation which will express the covariant language. When you take $$\psi^{*}$$ and multiply it by $$\beta$$ you get $$\bar{\psi}$$. So another way to write this is by saying

$$\bar{\psi} \beta \partial_t \psi + \bar{\psi} \beta \alpha_i \partial_i\psi + m \bar{\psi}\psi$$

We can change the configuration of this expression in terms of new symbols.

$$\gamma^{0}$$ is the gamma notation in respect to time, we can see the coefficient of beta is the derivative taken with respect to time and $$\beta \alpha_i$$ as $$\gamma_i$$. We end up with

$$\bar{\psi} (\gamma^{\mu}\partial_{mu} + m)\psi =\{ \bar{\psi} \gamma^{0} \partial_t \psi + \bar{\psi} \gamma^i \partial_i \psi + m \bar{\psi}\psi \}$$

There is what is called the fifth dirac matrix from this point. I'll assume you'll know that $$\gamma^0 \gamma^1 \gamma^2 \gamma^3 = \gamma^5$$. It is gamma 5 which is concerned with right-handedness and left-handedness which in the technical term means, Chirality which has Eigenvalues of either +1 or -1.