kcirick
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Homework Statement
I am pretty sure it's been done many times before, but I can't seem to figure it out:
Consider the collision 1 + 2 -> 3 + 4 in the lab frame (2 at rest), with particles 3 and 4 massless. Derive the forumla for the differential cross section
Homework Equations
We have Fermi's Golden Rule for scattering:
d\sigma = \left|M\right|^{2}\frac{\hbar^{2} S}{4\sqrt{\left(p_1.p_2\right)^{2}-\left(m_{1}m_{2} c^{2}\right)^{2}}} \left(\frac{cd^{3}p_{3}}{\left(2\pi\right)^{3}2E_{3}}\right) \left(\frac{cd^{3}p_{4}}{\left(2\pi\right)^{3}2E_{4}}\right) X \left(2\pi\right)^{4}\delta^{4}\left(p_1+p_2-p_3-p_4\right)
(My god it took a while to type that out!)
The Attempt at a Solution
I start by figuring out the dot product p_{1}.p_{2}. We get m_2 \left|p_{1}\right| c
So what we have is:
d\sigma = \left(\frac{\hbar}{8\pi}\right)^{2} \frac{S\left|M\right|^{2}}{m_2 \left|p_{1}\right| c} \frac{d^{3}p_{3}d^{3}p{4}}{\left|p_3\right|\left|p_4\right|} \delta\left(\frac{E_{1}}{c}+m_{2}c-\left|p_3\right|-\left|p_4\right|\right) \delta^{3}\left(p_{1}-p_{3}-p_{4}\right)
From here on, I don't quite understand. In the textbook we use (Griffiths), it says to integrate p_{4} which replaces it with p_{1}-p_{3}. So the formula will look like:
d\sigma = \left(\frac{\hbar}{8\pi}\right)^{2} \frac{S\left|M\right|^{2}}{m_2 \left|p_{1}\right| c} \frac{\delta\left(\frac{E_{1}}{c}+m_{2}c-\left|p_3\right|-\left|p_{1}-p_{3}\right|\right) }{\left|p_3\right|\left|p_{1}-p_{3}\right|} d^{3}p_{3}
Now we let:
d^{3}p_{3}=\left|p_{3}\right|^{2}d\left|p_{3}\right|d\Omega
where d\Omega=sin\theta d\theta d\phi
...And somehow we should get the right answer:
\frac{d\sigma}{d\Omega} = \left(\frac{\hbar}{8\pi}\right)^{2} \frac{S\left|M\right|^{2}\left|p_{3}\right|}{m_2 \left|p_{1}\right| \left(E_{1}+m_{2}c^{2}-\left|p_{1}\right|ccos\theta\right)}
Can someone help me out? Thanks!
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