Calculating Reaction Energy and Velocity in a Proton-Electron Collision

[hesus]

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.

 

[sgd37]

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

 

[hesus]

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

 

[sgd37]

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.

 

[hesus]

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

 

[sgd37]

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}

 

[Phrak]

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 \ .