# 4-momenta problem

1. Apr 6, 2013

### kudoushinichi88

1. The problem statement, all variables and given/known data
For a scattering problem A + A - > B + B in the Toy Theory,

Need to show

$$(p_4-p_2)^2-m_C^2c^2=p_4^2+p_2^2-2p_4\cdot p_2 = -\vec{\textbf{p}}^2(1-\cos\theta)$$

2. Relevant equations

this is assuming $m_A=m_B=m$ and $m_C=0$

3. The attempt at a solution
$$p_4^2+p_2^2=2m^2c^2$$

but i'm not sure how to do

$$p_4\cdot p_2$$

do i work with

$$\sqrt{\left( \frac{E_4^2}{c^2} - \vec{\textbf{p}}_4^2 \right) \left( \frac{E_2^2}{c^2} - \vec{\textbf{p}}_2^2 \right)}$$

?

2. Apr 6, 2013

### andrien

please mention what is mc.Also the expression so far is correct only in CM frame.also it seems that a factor of 2 is missing in p2(1-cosθ).

3. Apr 6, 2013

### Shinobii

It's been awhile since I have done this problem! Have you tried,

$$p_2 p_4 = \frac{E_2 E_4}{c^2} - \vec{p_2} \cdot \vec{p_4}$$

Last edited: Apr 6, 2013
4. Apr 6, 2013

### kudoushinichi88

Terribly sorry, I should've made my question clearer. All of the $p$s in my post refers 4-momenta while $\vec{\textbf{p}}$ refers to 3 momenta.

I'm working in the CM frame.

In this scattering reaction the subscripts refers to the particle in the reaction as follows: 1 + 2 -> 3 + 4

@andrien, you are right, I missed the factor 2. $m_c$ is mass of the mediating particle in the lowest order Feynman diagram of this particular scattering reaction. In this problem, I guess it could just be ignored.

@Shinobii, does your $\vec{p_2}$ and $\vec{p_4}$ denote 3-momenta or 4-momenta? Both particles 2 and 4 are not massless...

5. Apr 6, 2013

### BruceW

well, they are both 4-vectors in minkowski spacetime. So how would you normally take the inner product? (Maybe think of a general 4-vector first, then apply it to the specific vectors you have here)