1
$\begingroup$

In 1956 Lee and Yang proposed parity violation of the weak interactions to explain the $\theta-\tau$ puzzle.

The following year, 1957, Madam Wu and collaborators found that in the $\beta$ decay of Cobalt 60, there was an excess of electrons in the direction opposite to that of the nuclear spin, which was explained by requiring parity violation.

Now, my question is the following: why is it said that the violation is maximal, since in the experiments only the 60% of the electrons and not 100% were emitted in the direction opposite to that of the nuclear spin?

I just can't find anywhere any explanation for this. Isn't it more natural to say that the parity is partially violated and in the standard model lagrangian add a coefficient which encapsulate this asymmetry?

|cite|improve this question
$\endgroup$
1
$\begingroup$

From experiment as mentioned in Wu's paper the electrons are preferentially emitted in the direction opposite to nuclear spin. My speculative answer is that it is difficult to carry out this experiment so from systematic uncertainties it is not 100%.

However, the answer to your question might boil down to theoretical grounds rather than experimental. In this paper "A Priori Definition of Maximal CP Violation" there is a definition of Maximal parity violation as follows. "Maximal" parity violation means that the vector and axial vector currents occur with equal normalization and equal coupling constants in the fundamental Lagrangian of weak interactions. So, basically a term of the form V-A inside the Lagrangian. Such a formalism has been used to explain the data from weak interactions.

PS - This is a great question and I read the "Experimental Test of Parity Conservation in Beta Decay" paper by Wu without any hint about why the parity violation is Maximal. It would be great if some expert can answer this question.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ for example in this paper it says that the vector-axial vector coupling ratio of the weak currents to the nucleons is not 1: hitoshi.berkeley.edu/129a/cahn-goldhaber/chapter6.pdf This makes me think that this should be true even between quark.. $\endgroup$ – Kevin De Notariis Feb 20 '19 at 18:23
  • $\begingroup$ Yes, I remember reading studies with K-meson and B-meson decays which showed that. They approximate it as V-(1+$\epsilon$)A $\endgroup$ – Invariance Feb 20 '19 at 19:34
  • $\begingroup$ Indeed, "maximal" is shorthand for V-A in a fundamental vertex so, in practical terms , $1-\gamma_5$; you can't get more parity violation than that. However, even though W interactions violate parity maximally, the quark currents are "dressed" (modified) by the strong interactions, and the effective interactions in the hadron amplitudes are modified/soured. Neutrinos always violate parity maximally. Note the Z boson does not violate parity maximally by dint of Weinberg mixing. $\endgroup$ – Cosmas Zachos Dec 17 '19 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.