- #1
Manojg
- 47
- 0
Hi,
I have a question about decay angle. For example,
Let us consider the decay
[tex]\Lambda \rightarrow p + \pi^{-}[/tex]
Here, [tex]\Lambda[/tex] and [tex] p [/tex] are spin 1/2 and [tex]\pi^{-}[/tex] is spin 0. So, spin angular momentum is conserved. So, should not the cosine of angular distribution of [tex] p [/tex] ( or [tex] \pi^{-} [/tex] ) in center of mass frame of [tex] \Lambda [/tex] be flat?
If the spin angular momentum is not conserved ( like in [tex] \rho \rightarrow \pi^{+} \pi^{-} [/tex] ) then total angular momentum will be conserved because pions system has orbital angular momentum. So, the cosine of angular distribution of one of the pions in center of mass frame of [tex] \rho [/tex] is not flat.
So, if the spin angular momentum is conserved and there is no orbital angular momentum then should not the cosine of angular distribution of one of the decay product ( two particle decay) in mother reference frame be flat?
Thanks.
I have a question about decay angle. For example,
Let us consider the decay
[tex]\Lambda \rightarrow p + \pi^{-}[/tex]
Here, [tex]\Lambda[/tex] and [tex] p [/tex] are spin 1/2 and [tex]\pi^{-}[/tex] is spin 0. So, spin angular momentum is conserved. So, should not the cosine of angular distribution of [tex] p [/tex] ( or [tex] \pi^{-} [/tex] ) in center of mass frame of [tex] \Lambda [/tex] be flat?
If the spin angular momentum is not conserved ( like in [tex] \rho \rightarrow \pi^{+} \pi^{-} [/tex] ) then total angular momentum will be conserved because pions system has orbital angular momentum. So, the cosine of angular distribution of one of the pions in center of mass frame of [tex] \rho [/tex] is not flat.
So, if the spin angular momentum is conserved and there is no orbital angular momentum then should not the cosine of angular distribution of one of the decay product ( two particle decay) in mother reference frame be flat?
Thanks.