Dalitz plots & mass distributions

[JonReeDO]

Homework Statement

We are given a reaction $$X_{1} + X_{2} \rightarrow Y_{1} + Y_{2} + Y_{3}$$. The quantities $$m_{Y_{1}Y_{2}}^2$$ and $$m_{Y_{2}Y_{3}}^2$$ are plotted in a Dalitz plot. $$Y_{1}$$ + $$Y_{2}$$ resonate at a fixed mass $$m_{Y_{1}Y_{2}}$$.

Show how this resonance leads to a mass distribution for the "wrong pairing" of particles $$m_{Y_{2}Y_{3}}^2$$ where $$m_{Y_{2}Y_{3}}^2 = A + Bcos\theta$$. $$A$$ and $$B$$ are constants and $$\theta$$ is the angle between the outgoing $$Y_{3}$$ and either $$Y_{1}$$ or $$Y_{2}$$. What are $$A$$ and $$B$$?

Homework Equations

I'm not sure ... I think these are relevant:

In a 3-body system, we have, by definition, $$m_{13}^2 = (P_{1} + P_{3})^2 = (E_{1} + E_{3})^2 - (\vec{p_{1}} + \vec{p_{3}})^2 = m_{1}^2 + m_{3}^2 + 2(E_{1}E_{3} - p_{1}p_{3}cos\theta)$$. So here's the $$\theta$$ between two particles.

The Dalitz plot plots $$m_{Y_{1}Y_{2}}^2$$ against $$m_{Y_{2}Y_{3}}^2$$.

Comparing the particles going in and coming out, we can find the energy release $$Q$$, which is equal to the kinetic energies of the final state particles $$T_{i}$$.

The Attempt at a Solution

I'm trying to show that $$m_{Y_{2}Y_{3}}^2 = A + Bcos\theta$$ ... but I'm not understanding exactly where these values come from. I assume I need to somehow find $$m_{Y_{2}Y_{3}}^2$$ in terms of the various variables above -- how do I do that? I'll also have $$\theta$$ in terms of the fixed masses -- and momenta and energies?

I don't think that'll give me enough to find $$A$$ and $$B$$, except in terms of so many variables that I'm sure this approach is wrong. I'm probably quite a bit off the mark, but what am I missing here?

Thanks in advance for any help.

 

[Jhero]

Thank you for your post. It seems like you are on the right track with your attempt at a solution. Let me try to explain it in a bit more detail.

First, let's define the variables we are working with. We have three particles in the final state, Y_{1}, Y_{2}, and Y_{3}. We know that the mass of Y_{1} and Y_{2} together resonates at a fixed mass, which we will call m_{Y_{1}Y_{2}}. This means that the energy of this pair of particles is constant, and we can write it as E_{Y_{1}Y_{2}} = \sqrt{m_{Y_{1}Y_{2}}^2 + \vec{p_{Y_{1}}}^2} + \sqrt{m_{Y_{1}Y_{2}}^2 + \vec{p_{Y_{2}}}^2}. This energy is also equal to the energy of the initial state particles, X_{1} and X_{2}, which we can write as E_{X_{1}X_{2}} = \sqrt{m_{X_{1}X_{2}}^2 + \vec{p_{X_{1}}}^2} + \sqrt{m_{X_{1}X_{2}}^2 + \vec{p_{X_{2}}}^2}.

Now, let's consider the third particle, Y_{3}. This particle has a mass m_{Y_{3}} and a momentum \vec{p_{Y_{3}}}. We can use the conservation of momentum and energy to write the energy and momentum of Y_{3} in terms of the other particles: E_{Y_{3}} = E_{X_{1}X_{2}} - E_{Y_{1}Y_{2}} and \vec{p_{Y_{3}}} = \vec{p_{X_{1}}} + \vec{p_{X_{2}}} - \vec{p_{Y_{1}}} - \vec{p_{Y_{2}}}.

Now, we can use these expressions to write m_{Y_{2}Y_{3}}^2 = (E_{Y_{3}} + E_{Y_{2}})^2 - (\vec{p_{Y_{3}}} + \vec{p_{Y_{2}}})^2