- #1
ChrisJ
- 70
- 3
This is not coursework, it is from a past paper (which we don’t have solutions for) and am doing preparation for this years exam.
1. Homework Statement
Particle ##A## with energy ##E_A## hits particle ##B## (at rest), producing particles ##C## and ##D## in the reaction
[tex]
A+B \longrightarrow C + D
[/tex]
Show that the minimum values of ##E_A## for this reaction , in terms of the particles masses ##m_A##,##m_B##,##m_C## and ##m_D## is given as
[tex]
E_A = \frac{(m_C+m_D)^2-m_A^2-m_B^2}{2m_B}
[/tex]
##\textbf{p} \cdot \textbf{p} = m^2 ##
##\textbf{p}=(E,|\vec{p}|) = (\gamma m , \gamma m \vec{v})##
As the relevant equation I posted suggests, the problem is in natural units and using the Minkoswki metric with the (+,-,-,-) sign convention.
I have done most of it, but there is one last bit which I can't see how it comes about. So far I have done..
[tex]
\textbf{p}_A + \textbf{p}_B = \textbf{p}_C + \textbf{p}_D \\
\left( \textbf{p}_A + \textbf{p}_B \right) \cdot \left( \textbf{p}_A + \textbf{p}_B \right) = \left( \textbf{p}_C + \textbf{p}_D \right) \cdot \left( \textbf{p}_C + \textbf{p}_D \right) \\
m_A^2 + m_B^2 +2(\textbf{p}_A \cdot \textbf{p}_B) = m_C^2 + m_D^2 +2(\textbf{p}_C \cdot \textbf{p}_D)
[/tex]
And since particle ##B## is at rest ##\textbf{p}_A \cdot \textbf{p}_B = E_A m_B ##, therefore..
[tex]
m_A^2 + m_B^2 +2 E_a m_B = m_C^2 + m_D^2 +2(\textbf{p}_C \cdot \textbf{p}_D) \\
\therefore E_A = \frac{m_C^2 + m_D^2 + 2(\textbf{p}_C \cdot \textbf{p}_D) - m_A^2 - m_B^2}{2 m_B}
[/tex]
And from looking at result in the question and expanding the squared term on top this suggests that ##(\textbf{p}_C \cdot \textbf{p}_D) = m_C m_D ## and I cannot see how!
I tried doing the dot product, but the only way I can see how that works is if both ##C## and ##D## have zero three-velocity, and I can't see how that could be from the question if so, and if not, I can't see how it comes about.
Any help/advice is much appreciated!
Thanks :)
1. Homework Statement
Particle ##A## with energy ##E_A## hits particle ##B## (at rest), producing particles ##C## and ##D## in the reaction
[tex]
A+B \longrightarrow C + D
[/tex]
Show that the minimum values of ##E_A## for this reaction , in terms of the particles masses ##m_A##,##m_B##,##m_C## and ##m_D## is given as
[tex]
E_A = \frac{(m_C+m_D)^2-m_A^2-m_B^2}{2m_B}
[/tex]
Homework Equations
##\textbf{p} \cdot \textbf{p} = m^2 ##
##\textbf{p}=(E,|\vec{p}|) = (\gamma m , \gamma m \vec{v})##
The Attempt at a Solution
As the relevant equation I posted suggests, the problem is in natural units and using the Minkoswki metric with the (+,-,-,-) sign convention.
I have done most of it, but there is one last bit which I can't see how it comes about. So far I have done..
[tex]
\textbf{p}_A + \textbf{p}_B = \textbf{p}_C + \textbf{p}_D \\
\left( \textbf{p}_A + \textbf{p}_B \right) \cdot \left( \textbf{p}_A + \textbf{p}_B \right) = \left( \textbf{p}_C + \textbf{p}_D \right) \cdot \left( \textbf{p}_C + \textbf{p}_D \right) \\
m_A^2 + m_B^2 +2(\textbf{p}_A \cdot \textbf{p}_B) = m_C^2 + m_D^2 +2(\textbf{p}_C \cdot \textbf{p}_D)
[/tex]
And since particle ##B## is at rest ##\textbf{p}_A \cdot \textbf{p}_B = E_A m_B ##, therefore..
[tex]
m_A^2 + m_B^2 +2 E_a m_B = m_C^2 + m_D^2 +2(\textbf{p}_C \cdot \textbf{p}_D) \\
\therefore E_A = \frac{m_C^2 + m_D^2 + 2(\textbf{p}_C \cdot \textbf{p}_D) - m_A^2 - m_B^2}{2 m_B}
[/tex]
And from looking at result in the question and expanding the squared term on top this suggests that ##(\textbf{p}_C \cdot \textbf{p}_D) = m_C m_D ## and I cannot see how!
I tried doing the dot product, but the only way I can see how that works is if both ##C## and ##D## have zero three-velocity, and I can't see how that could be from the question if so, and if not, I can't see how it comes about.
Any help/advice is much appreciated!
Thanks :)