# Denominator scattering amplitude

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

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 $$E_1=E_3=p$$.