Momentum conservation in collisions.

[AlphaPhoton]

Hello.

I have a problem that is making me crazy. Consider the following collision

A + B \rightarrow C

which results in both particles (A and B) being destroyed and C being created.
I know the rest mass of all particles. Also, in the lab system, B is stationary and A is moving toward B. What is asked for is the momentum of particle A needed to allow for particle C to be created.

So I used the following squared four vectors:

(E_A + m_B, p_A)^2 = (m_C, 0)^2

with

E_A = \sqrt{(p_A^2 + m_A^2)}

and solved for p_A. What I got was different from what I got when I used energy conservation with

E_A + E_B = E_C \Leftrightarrow \sqrt{(p_A^2 + m_A^2)} + m_B = m_C

Then I realized that I completely neglected the momentum on the right side in my first approach, so instead I did

(E_A + m_B, p_A)^2 = (m_C, p_A)^2

(p_A on the right side due to momentum conversion)

and got to the same result as with the second approach. Which is correct?

To make things even more confusing, I remembered an example our professor gave us:

\gamma + p \rightarrow p + \pi_- + \pi_+

and he used the following squared four vectors:

(E_\gamma + m_p, p_\gamma)^2 = (m_p + 2*m_\pi, 0)^2

Note that he completely neglected the momentum on the right side. If neglecting the momentum was wrong in my first example, why is it correct here? I don't get it, please, someone enlighten me :)

Thanks in advance

 

[Meir Achuz]

The threshold energy to produce a final state is the square of the sum of the produced masses, as in the RHS of your profs equation.
(E_\gamma+m_p)^2-p_\gamma^2, the invariant he has on the left hand side equals the right hand side. Since the LHS is an invariant, its rest frame value equals its cm value where P=0. You can apply that to any case.

 

[AlphaPhoton]

The threshold energy to produce a final state is the square of the sum of the produced masses, as in the RHS of your profs equation.
(E_\gamma+m_p)^2-p_\gamma^2, the invariant he has on the left hand side equals the right hand side. Since the LHS is an invariant, its rest frame value equals its cm value where P=0. You can apply that to any case.

Okay, but what about my first example then? Can I use
<br /> (E_A + m_B, p_A)^2 = (m_C, 0)^2<br />
then with the same reasoning to calculate the needed momentum of particle A, and if so, why don't I get the same result if I use pure energy conservation laws?

 

[Meir Achuz]

So I used the following squared four vectors:

(E_A + m_B, p_A)^2 = (m_C, 0)^2

That equation gives the correct threshold lab energy E_A.

What I got was different from what I got when I used energy conservation with

E_A + E_B = E_C \Leftrightarrow \sqrt{(p_A^2 + m_A^2)} + m_B = m_C

This equation is wrong for the reason you mentioned. In the lab system, conservation of energy requires \sqrt{p_A^2+m_c^2} for E_C.

Note that he completely neglected the momentum on the right side.
[/QUOTE
He did not. The momentum in that case is the total cm momentum which is zero.