Madame Wu Experiment: Doubts & Reasoning

[Federica]

Hi all,

I have some doubts regarding the experiment of Madame Wu. I know a strong magnetic field is used to polarise the $^{60}Co$ nuclei, then we have an anthracene scintillator on the top of the sample which will detect the electron produced in the decay: $^{60}Co \rightarrow ^{60}Ni^{**} + e^{-} + \bar{\nu}_{e}$. Two others scintillators will detect the two gammas produced by the de-excitation of $^{60}Ni^{**}$, one is set at 90° to the field (the 'equatorial counter') and the other one at 0° ('the polar counter').

Now, since I suppose the spins are parallel to the field, I would expect the polar counter to detect more photons than the equatorial one (as long as we have polarisation), but the opposite happens. And since the anti-neutrino is right-handed, the electron should be left-handed, which means the anthracene detector should detect more electrons when the field is on the downside. But once again, the opposite happens.

Can someone help me to understand what's wrong with my reasoning?

 

[mfb]

Which direction receives more photons depends on the direction of the magnetic field. Not that it would matter: Any preference is parity violation.

 

[Aleolomorfo]

Hello everybody!

I join the conversation since I also have some doubts about Wu experiment.

Any preference is parity violation.

Yes, you are right, but if the emission is in the opposite direction of the spin the weak interactions obey V-A structure, otherwise V+A... So, the direction of the emission is important.

From the attached graphs it seems there is a contradiction. Let's consider the graph about the beta asymmetry and the situation "B up" (a).
If the field is upward, the spin is upward. Since V-A structure implies the emission of the electrons opposite to the spin, I would expect to count less frequently than the situation in which the filed in downward (remember that in Wu experiment the scintillator is over the source). But the graph shows the opposite.

Where does my reasoning fail?

 

[mfb]

I'm not sure where you got these graphs from because the publication has it in the opposite direction, and it is quite clear:

The sign of the asymmetry coefficient, $\alpha$, is negative, that is, the emission of beta particles is more favored in the direction opposite to that of the nuclear spin.

 

[Federica]

I'm not sure where you got these graphs from because the publication has it in the opposite direction, and it is quite clear:

Those graphs are taken from "Introduction to Elementary Particle Physics", A. Bettini. Thank you for your prompt reply, it seems there's an error on the book then.