Parity as a kinematic property?

metroplex021
Messages
148
Reaction score
0
Weird question, but does anyone have any feelings on whether parity can be classified as a kinematic property? It doesn't scale with energy and so in that sense doesn't seem to be classifiable as a dynamic property, nor do objects interact through it; but parity is of course violated by the weak interaction, meaning that -- unlike spin and mass -- it is not conserved unlike the other kinematic quantities. I appreciate that a rose by any other name would smell as sweet, but I'd like to hear your thoughts!
 
Physics news on Phys.org
First of all, you should probably notice that mass is not conserved in high energy interactions because of the famous E=mc^2. What is conserved it the total energy (including the rest mass). Same thing is valid for the spin, what is conserved is the total angular momentum J=L+S.
What what concerns your question I would say that no, parity is not a kinematical property. Parity is an intrinsic property of each particle and it's not related to its status of motion. However, it is not conserved by all interaction because it doesn't come from a universal symmetry of nature. Total energy and total angular momentum are always conserved because they arise from the invariance of our universe under time translations and spatial rotation, which are always symmetries of our world. The is no such universal symmetry for parity.
 
violation in the general QM sense is a result of some interacting hamiltonian H' (so that H_{tot}=H_0+H') which does not commute with the operator of parity:
[P,H]= [P,H'] \ne 0
I don't understand how then you are trying to categorize it as a kinematic property (or not).
 
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}...
Back
Top