I try to calculate the scattering amplitude of a process $A + A \rightarrow B + B $ in lowest order contribution.

enter image description here

For one part of the amplitude $M$ i got: $$M=\frac{g^2}{(p_4-p_2)^2-m_C^2}$$
We are working in the center of mass frame and we set all masses equal and give the virtual particel zero mass: $$m_A=m_B=m\\ m_C=0$$ Then:$$M=\frac{g^2}{(p_4-p_2)^2}$$ $p_2$ and $p_4$ are four momentum vectors. So therefore $$(p_4-p_2)^2=p_4^2+p_2^2-2p_2\cdot p_4=2m^2-2\cdot E_2\cdot E_4+2\cdot \vec p_2 \cdot \vec p_4$$ My Question now:
How do we get from this result to $-2(\vec p_1)^2(1-\cos\theta)$
(where $\vec p_1$ ist the incident momentum of particle 1)? enter image description here

I know, that $\vec p_1 = -\vec p_2$ and similarly for $p_3$ and $p_4$, but $E_2$ and $E_4$ are not the same, since $p_2\neq p_4$ an the cosine comes from the dot product of the space momentum.


1 Answer 1


The answer goes like this:

Due to energy momentum conservation $$(p_4-p_2)^2=(p_1-p_3)^2=2m^2-2E_1E_3+2\vec{p}_1\cdot\vec{p}_3$$ Then, one can write $\vec{p}_1=p\hat{x}$ and $$\vec{p}_3=p\cos\theta\hat{x}+p\sin\theta\hat{y}$$

Also, in order to get the result you need, I think it is nescessary to assume that $m=0$ and that the collision is elastic, yielding $$(p_4-p_2)^2=(p_1-p_3)^2=0-2E_1E_3+2\vec{p}_1\cdot\vec{p}_3= -2p^2+2p^2\cos\theta=-2p^2(1-\cos\theta)$$ where I have used that $\vec{p}_1\cdot\vec{p}_3=(p\hat{x})\cdot(p\cos\theta\hat{x}+p\sin\theta\hat{y})=p^2\cos\theta$ and that $E_1=E_3=p$.

If there are any questions, please let me know.


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