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A related post was What is polarisation, spin, helicity, chirality and parity?

In $m_\mu<<E$ region, $m_\mu$ could be treated as massless particles and the conservation of helicity indicated the polarized cross section dependent on $\theta$.

In $m_\mu\approx E$ region, $m_\mu$ could not be ignored but that the conservation of spin made the polarized cross section independent on $\theta$.

However, notice that in both case, $m_e<<E$ the electron was in the relativistic region, and it was a bit hard to see why in one case the computation used helicity and in another case the computation used spin, where the treatment of the electron's amplitude was not explained.

Why not use the spin conservation for $m_\mu<<E$ and helicity conservation for $m_\mu\approx E$? What happened to the helicity and spin conservation in the $e^+e^-\rightarrow \mu^+ \mu^-$?

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    $\begingroup$ Linked. $\endgroup$
    – Cosmas Zachos
    Commented Oct 23, 2021 at 15:47
  • $\begingroup$ @CosmasZachos Peskin chapter 5 $\endgroup$
    – ShoutOutAndCalculate
    Commented Oct 23, 2021 at 22:10
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    $\begingroup$ But... m𝜇 <𝐸 is below threshold, as P&S remind you! the ruction can't go for lack of energy. In fact, at threshold, the cross section vanishes. They also explain that associating helicity with chirality breaks down in the non relativistic case, near threshold. Keep reading... $\endgroup$
    – Cosmas Zachos
    Commented Oct 23, 2021 at 22:50
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    $\begingroup$ You learned that in the NR limit there is no association between helicity and chirality for the muons. He reminds you (5.12) → (5.33) that your chimerical m𝜇 <<𝐸 so, m𝜇 <𝐸 cannot support muon production, and, at threshold, the cross section vanishes. At that very point, the muons are stationary, so θ is not defined. At slightly higher energies, spin can point any which way, as in the figure, so the direction of the muon momenta is decoupled from it, as per the properties of the Dirac equation solutions. $\endgroup$
    – Cosmas Zachos
    Commented Oct 25, 2021 at 14:08
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    $\begingroup$ These are issues you brought up, and I'm telling you what his (5.12) → (5.33) mean. He need not define threshold and its implications for you: aren't they evident? $\endgroup$
    – Cosmas Zachos
    Commented Oct 25, 2021 at 17:54

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