Solve Feynman Calculus Homework: Collision 1+2 -> 3+4

• kcirick
In summary, the given problem involves deriving the formula for the differential cross section of the collision 1 + 2 -> 3 + 4 in the lab frame, with particles 3 and 4 being massless and particle 2 at rest. Utilizing Fermi's Golden Rule for scattering and the energy delta function, the final formula is obtained as 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

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:
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!

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

kcirick said:

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!

1. What is the Feynman calculus homework about?

The Feynman calculus homework is a problem that involves solving a collision between two particles, labeled 1 and 2, and determining the resulting particles, labeled 3 and 4. This problem is commonly used in physics and has applications in fields such as particle physics and astrophysics.

2. How do I approach solving the Feynman calculus homework?

The first step in solving the Feynman calculus homework is to draw a Feynman diagram, which is a graphical representation of the particles involved in the collision and their interactions. Then, you need to use mathematical equations and principles, such as conservation of energy and momentum, to solve for the unknown variables.

3. What are the key concepts involved in solving the Feynman calculus homework?

The key concepts involved in solving the Feynman calculus homework include momentum, energy, and conservation laws. Momentum is the product of mass and velocity, and is conserved in a closed system. Energy is also conserved in a closed system, and can be transferred between particles during a collision. Conservation laws state that the total momentum and energy of a system before and after a collision must be equal.

4. Are there any tips for solving the Feynman calculus homework?

One tip for solving the Feynman calculus homework is to carefully label all the particles involved and their corresponding variables in your equations. It can also be helpful to break down the problem into smaller steps and solve each step individually. Additionally, double-checking your work and units can help ensure accuracy.

5. What are some real-life applications of the Feynman calculus homework?

The Feynman calculus homework has many real-life applications, particularly in the field of particle physics. It is used to understand and predict the behavior of subatomic particles during collisions, such as those in particle accelerators. This concept also has applications in astrophysics, where collisions between particles play a role in the formation of stars and galaxies.