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I've been reading into the concepts of chirality & helicity and often I find a statement that chirality is Lorentz invariant in contrast to helicity (which of course depends on the frame). BUT I don't see in which way chirality IS Lorentz invariant.

For massless particles things are easy, of course. In Weyl notation for a left chiral particle, you have some Weyl bispinor [itex]\left( \begin{array}{c} \Phi \\ 0 \end{array} \right)[/itex] with a Weyl spinor [itex]\Phi[/itex] depending on the energy and spin orientation of the particle. Since boosting to another frame means multiplying the bispinor with a diagonal matrix, it stays in this form: a left chiral particle is left chiral in every frame. Since the Dirac representation of the Lorentz group is reducible into left and right chiral states, it is not surprising that they don't mix.

But for a massive particle things seems strange to me. Suspect we have an electron. Then the positive frequence solution of the dirac eq. for momentum [itex]\vec{p}[/itex]=0 and spin up in z-direction is [itex]u\left(p=0\right) = \sqrt{m}\left( \begin{array}{c} \xi \\ \xi \end{array} \right)[/itex] with the spinor [itex]\xi = \left( \begin{array}{c} 1 \\ 0 \end{array} \right)[/itex]. Obviously [itex]\frac{\left(1-\gamma^{5}\right)}{2}u[/itex] and [itex]\frac{\left(1+\gamma^{5}\right)}{2}u[/itex] are of the same size; the mixing angle between the left and right chiral amount of the electron is 45 degrees. But boosting this bispinor in z-direction gives [itex]u\left(p\right) = \left( \begin{array}{c} \sqrt{E-p^{3}}\xi \\ \sqrt{E+p^{3}}\xi \end{array} \right)[/itex]. In the limit [itex]p^{3}[/itex]→∞ this becomes [itex]u\left(p\right) = \left( \begin{array}{c} 0 \\ \sqrt{2E}\xi \end{array} \right)[/itex]. So an electron which has equal amount of left and right chirality in one frame is fully right chiral in another!

How can we then state that chirality is a Lorentz invariant concept?

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# Lorentz invariance of chirality

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