# Proton and Electron Collision

Hi,

i have the following exercise:
At the HERA acceleratro of DESY in Hamburg electrons and protons were accelerated to E_e=27.5 GeV and E_p = 920GeV and brought to collision.

A)How large is the reaction energy in the centre-of-mass system?
B) What is the velocity of the HERA centre-of-mass System in the lab system?

A) I think the CMS is defined by$$\vec{p}$$_e +$$\vec{p}$$_p=0, am i right? ($$\vec{p}$$ is the three-momentum.

Therefore, i can calculated the sum of the two four-momentum.
$$s=(p_p+p_e)^2= (E_p+E_e)^2$$
Reaction energy ist then sqrt(E).

B) How can i solve this?
Thank you.

the centre of mass equation is $$p^{2} = (E_{p}+E_{e})^{2} - (p_{p}+p_{e})^{2}$$ and you have to keep in mind direction for momenta

Well, this was my question.
Isn't the CMS defined by $$\vec{p}_p+\vec{p}_e=0$$?

no that's not how its defined its $$p_{cms} = 0$$ but that dosen't mean $$p_{e} + p_{p} = 0$$ it just means that we transport ourselves in the frame of the collision which is stationary and all the energy of the system is expressed as a mass.

Well, if you have just two particles..shouldn't it just be $$P_{CMS}=P_E+P_P$$? how do you define P_CMS?

$$p_{cms}$$ is the three momentum in the centre of mass frame and it is always 0. now the invariant quantity in any frame is the rest mass i.e. $$(m_{cms})^{2} = (m_{lab})^2$$. The rest mass in the lab frame is given by $$(m_{lab})^2 = (E_{p}+E_{e})^{2} - (\overline{p}_{p}+\overline{p}_{e})^{2}$$ and that is the origin of that formula. In one word $$\overline{p}_{cms} \neq \overline{p}_{p}+\overline{p}_{e}$$

Backing up a little...

Well, this was my question.
Isn't the CMS defined by $$\vec{p}_p+\vec{p}_e=0$$?

it might clarify things to know that

$$\vec{p}_{p\ lab} +\vec{p}_{e\ lab} \neq 0 \ ,$$

but in the center of mass frame,

$$\vec{p}_{p\ cm} +\vec{p}_{e\ cm} =0 \ .$$