Conservation of Four Momentum Problem

[ChrisJ]

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]

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 :)

 

[mfb]

What is the relative velocity of C and D at the minimal energy for this reaction?

Something went wrong with the formatting (missing line breaks?), but what you got at the end is correct, it just needs a last simplification step.

 

[vela]

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!

Also remember that the product of the two four-vectors is invariant, so you can calculate it in the center-of-mass frame instead of the lab frame.

 

[ChrisJ]

Also remember that the product of the two four-vectors is invariant, so you can calculate it in the center-of-mass frame instead of the lab frame.

Ok, thank you. Yeah I knew it was invariant but maybe did not fully understand the principle or implications of it. I assumed that the question set's it up so that it is from particle ##B## 's rest frame, and that is how I got ##E_a m_B## from the inner product of ##\textbf{p}_A## and ##\textbf{p}_B## since ##B## has no 3-velocity . And I assumed if I computed that bit in B's rest frame then the other bits of the equation had too to? Or is it that because 4-momentum is invariant I can calculate that one bits from B's rest frame, and then compute the other bits from the CoM frame of ##C## and ##D##?

What is the relative velocity of C and D at the minimal energy for this reaction?

Something went wrong with the formatting (missing line breaks?), but what you got at the end is correct, it just needs a last simplification step.

Thanks for replying, the formatting looks fine for me? All the line breaks are coded and are working properly, well, according to my browser at least :) . The relative velocity would be zero wouldn’t it? I.e. moving at same velocity.

 

[mfb]

After reloading, the line breaks are now shown to me as well. Probably a weird one-time error.

The relative velocity is zero, right, this should give you the product of the two momentum vectors, especially if you use vela's hint. If something is invariant under Lorentz transformations, it does not matter in which reference frame you evaluate it - that is exactly what the invariance tells you.

 

[ChrisJ]

it does not matter in which reference frame you evaluate it - that is exactly what the invariance tells you.

Ok yeah I get the final answer now, thanks both. :)