I Violation of spin conservation in pion annihilation

  • I
  • Thread starter Thread starter Trollfaz
  • Start date Start date
Trollfaz
Messages
143
Reaction score
14
##\pi^\pm## are mutual particle-antiparticle pairs, while ##\pi^0## is it's own antiparticle. All has a spin of 0.
In any annihilation reaction of a particle ##x##, the equation is
$$x+\bar{x}\to \gamma+\gamma$$
Photons have no charge but a spin 1. I can see charges are conserved but spin is not since the total spin before the reaction is 0 while after the reaction is 2. Is there a solution to this or am I just wrong?
Sorry the LaTeX here behaves differently from the one on my computer.
 
Physics news on Phys.org
Trollfaz said:
Is there a solution to this or am I just wrong?
What's conserved in these reactions is the total angular momentum. For spin-0 pions in the center-of-mass frame, the reaction ##\pi+\bar{\pi}\rightarrow\gamma+\gamma## results in two spin-1 photons travelling in opposite directions with their spins antiparallel, so the total angular-momentum is zero both before and after the reaction.
 
Last edited:
  • Like
Likes ohwilleke, Vanadium 50 and PeroK
So p is your symbol for pion? The general symbol for pion is ##\pi## p represents proton
 
Trollfaz said:
So p is your symbol for pion? The general symbol for pion is ##\pi## p represents proton
Yes, and I've edited post #2 to reflect that notation. But same conservation reasoning also applies to protons and antiprotons: two spin-##\frac{1}{2}## particles annihilate to two spin-1 photons with antiparallel spins.
 
Trollfaz said:
In any annihilation reaction of a particle , the equation is
False.
Trollfaz said:
total spin before the reaction is 0 while after the reaction is 2.
False.
 
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

Replies
4
Views
2K
Replies
10
Views
3K
Replies
3
Views
5K
Replies
1
Views
2K
Replies
8
Views
3K
Replies
6
Views
2K
Back
Top