# Feynman Calculus

## 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!

Last edited:

Related Advanced Physics Homework Help News on Phys.org
Hm... Did I not follow the PF guideline correctly, or is this too long (and/or boring) of a question?

It's frustrating because it seems like I'm only a couple of steps from getting the correct answer, so please help me out if you can. Thanks!

Meir Achuz
Homework Helper
Gold Member
Write |p1-p3| as sqrt{p1^1+p3^2-2p1p3cos\theta}.
Then use \delta[f(x)]=\delta(x)/[df/dx]

nrqed
Homework Helper
Gold Member

## 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:

$$d^{3}p_{3}=\left|p_{3}\right|^{2}d\left|p_{3}\right|d\Omega$$
Ok, now you still have to integrate over $p_3$ using the energy delta function. Did you do that?

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!

I got to the right answer by substituting:

$$E = \left|p_3\right| +\left(\left|p_1\right|^2 + \left|p_3\right|^2 - 2\left|p_1\right|\left|p_3\right| cos\theta\right)^{1/2}$$

Then:

$$dE = \frac{E-\left|p_1\right| cos\theta}{E-\left|p_3\right|}d\left|p_3\right|$$

Then using definition of the delta function as par Meir Achuz, I got safely to the right answer. Thank you to both of you for your help!