Is it possible the decay of rho neutral meson to two pions?

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SUMMARY

The decay of the neutral rho meson to two pions is indeed possible through strong interaction, as confirmed by the reference to Yung Kuo's "Problems and Solutions in Atomic, Nuclear and Particle Physics." Conservation laws for energy, momentum, angular momentum, electric charge, C-parity, parity, and isospin must be satisfied. The discussion highlights a critical misunderstanding regarding angular momentum conservation, emphasizing the necessity to include orbital angular momentum in the analysis to resolve discrepancies in angular momentum and parity calculations.

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¿Is it possible the decay of rho neutral meson to two pions?

Homework Statement



We initially have the neutral rho meson -----> pion+ + pion-

¿Is this reaction possible? ¿What are the conserved quantities?¿Which is the interaction involved in the process?

Homework Equations



Conservation laws of energy, momentum, angular momentum, electric charge, C-parity, parity,Isospin, cp symmetry.

Here we have the properties of the involved particles:

http://en.wikipedia.org/wiki/Rho_meson

http://en.wikipedia.org/wiki/Pion

The Attempt at a Solution



Well i am using the book of yung kuo "Problems and solutions in atomic, nuclear and particle physics", it is the 3018 problem.

The book says that the decay is possible, by strong interaction, and that means that every symmetry is conserved, there isn't any violation.

The problem is the next:

-for the angular momentum i have found that for the rho meson is 1, and for the pions + and - is 0.

So we have 1---->0 + 0, the composition of angular momentum says that in the second part of the equality is not possible to find J=1, so I think that conservation of angular momentun is violated ¿what am I doing wrong?.

-for the parity we have (-1 of the rho) -----> (-1 of the pion+ and -1 of the pion-)

So multipliying we have -1 = 1 so it is violated, I think that perhaps the mistake is that the parity of the pion- is oppossite to the parity of his antiparticle(pion+), but it is not listed on the table of the wikipedia ¿is this the mistake?.

¿The parity of an antiparticle is the same as the one of the original particle? .

-for the C-parity we have that the rho meson has a -1, but for the pions is not defined ¿is it violated?.
 
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Don't forget the pions can have relative orbital angular momentum.

Also, this page will probably help you figure out what's going on wrt charge parity.

http://en.wikipedia.org/wiki/C_parity
 


vela said:
Don't forget the pions can have relative orbital angular momentum.

Also, this page will probably help you figure out what's going on wrt charge parity.

http://en.wikipedia.org/wiki/C_parity

Yes, i know. The total parity, is multiplicative, intrinsic parity must be multiplied by (-1) times the angular momentum, but the problem statement does not say anything about the orbital angular momentum, so I made the calculations considering only internal properties.

¿Do you consider that it is a mistake?.

Thanks for the answer.:smile:
 


Yes, it's a mistake. You already noted that if you only consider the spins, you can't conserve angular momentum. You have to include orbital angular momentum in your analysis.
 


The problem to me is the next, the formula of the parity of a meson is:

Pm= Pa*Pb* (-1)L=(-1)L+1

¿What is the value of L?

The problem statement does not say anything about the orbital angular momentum of the final particles ¿how can i find out its value?. I am completely lost at this point ¿should I invent the L values :biggrin: ?
 


Use conservation of angular momentum.
 


vela said:
Use conservation of angular momentum.

Ok.

For the rho meson J=L+S, with S=1, for the pion+ J=L (because S=0, and L is the orbital angular momentum of one pion over the other).

But I don't know if the rho meson is in a excited state, so I don't know his L.

And I don't know, another time, what is the final L value so I think that it is undetermined :frown:

I guess that something fails with this reasoning, but I don't know what is it.
 

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