Understanding the Wu Experiment: ParityV and Gamma-Ray Anisotropy Explained

  • Thread starter Thread starter ChrisVer
  • Start date Start date
  • Tags Tags
    Experiment
ChrisVer
Science Advisor
Messages
3,372
Reaction score
465
Hi, a fast question while I was reading the results of the Wu experiment:
http://iktp.tu-dresden.de/uploads/media/Experimental_test_of_parity_in_beta_decay_-_Wu.pdf
It says that "the observed gamma-ray anisotropy was used as a measure of polarization and effectively temperature"
However, I don't understand why there should be any gamma-ray anisotropy for the EM transition of Nickel-60* to its ground state. Any explanation for that?
 
Physics news on Phys.org
If you go back one paragraph, you will see the first time this is mentioned, there is a a little number 2. This indicates a reference, and if you are unclear about this, you should look at that reference. (Ambler et al., who is one of the authors of this paper)

In that paper, it is explained that there are two gamma rays in Co-60, and a) they have a characteristic angular distribution between them, and b) this distribution also depends on the spin axis of the Co-60 nucleus. If you want to calculate this, it's what you get from a 5+ nucleus transitioning to a 0+ nucleus by emission of two photons.
 
I don't have access at the moment in the references. :frown:
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top