Calculating Reaction Energy and Velocity in a Proton-Electron Collision

In summary, the conversation revolves around the HERA accelerator in Hamburg where electrons and protons were accelerated to E_e=27.5 GeV and E_p = 920GeV and brought to collision. The first question asks about the reaction energy in the centre-of-mass system, which is calculated using the sum of the two four-momentum. The second question is about the velocity of the HERA centre-of-mass system in the lab system. The experts clarify that the CMS is not defined by \vec{p}_p+\vec{p}_e=0 but rather p_{cms} = 0, and explain the concept of rest mass in different frames. The summary concludes with the clarification that in the center of mass frame
  • #1
hesus
5
0
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[tex]\vec{p}[/tex]_e +[tex]\vec{p}[/tex]_p=0, am i right? ([tex]\vec{p}[/tex] is the three-momentum.

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

B) How can i solve this?
Thank you.
 
Physics news on Phys.org
  • #2
the centre of mass equation is [tex]p^{2} = (E_{p}+E_{e})^{2} - (p_{p}+p_{e})^{2}[/tex] and you have to keep in mind direction for momenta
 
  • #3
Well, this was my question.
Isn't the CMS defined by [tex] \vec{p}_p+\vec{p}_e=0[/tex]?
 
  • #4
no that's not how its defined its [tex]p_{cms} = 0[/tex] but that dosen't mean [tex]p_{e} + p_{p} = 0[/tex] 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.
 
  • #5
Well, if you have just two particles..shouldn't it just be [tex]P_{CMS}=P_E+P_P[/tex]? how do you define P_CMS?
 
  • #6
[tex]p_{cms}[/tex] 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. [tex](m_{cms})^{2} = (m_{lab})^2[/tex]. The rest mass in the lab frame is given by [tex] (m_{lab})^2 = (E_{p}+E_{e})^{2} - (\overline{p}_{p}+\overline{p}_{e})^{2}[/tex] and that is the origin of that formula. In one word [tex]\overline{p}_{cms} \neq \overline{p}_{p}+\overline{p}_{e}[/tex]
 
  • #7
Backing up a little...

hesus said:
Well, this was my question.
Isn't the CMS defined by [tex] \vec{p}_p+\vec{p}_e=0[/tex]?

it might clarify things to know that

[tex] \vec{p}_{p\ lab} +\vec{p}_{e\ lab} \neq 0 \ ,[/tex]

but in the center of mass frame,

[tex] \vec{p}_{p\ cm} +\vec{p}_{e\ cm} =0 \ .[/tex]
 

1. What is a proton-electron collision?

A proton-electron collision is a scientific phenomenon in which a positively charged proton and a negatively charged electron come into contact with each other, resulting in a transfer of energy and potentially creating new particles.

2. How do protons and electrons collide?

Protons and electrons can collide in various settings, such as in particle accelerators or in natural phenomena like lightning. In these collisions, the particles are accelerated to high speeds and then directed towards each other, resulting in a collision.

3. What happens during a proton-electron collision?

During a proton-electron collision, the particles interact through electromagnetic forces and can transfer energy to each other. This energy can then be used to create new particles or cause other changes in the particles involved.

4. What can we learn from studying proton-electron collisions?

Studying proton-electron collisions allows scientists to better understand the fundamental properties of matter and the forces that govern them. It also helps us understand the structure of atoms and the behavior of subatomic particles.

5. Are proton-electron collisions dangerous?

In controlled settings, such as in particle accelerators, proton-electron collisions are not dangerous. However, in natural phenomena like lightning, the energy released from these collisions can be harmful. It is important to take precautions and stay away from lightning strikes to avoid any potential harm.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
21
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
479
  • Advanced Physics Homework Help
Replies
5
Views
2K
Replies
21
Views
11K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
3
Views
1K
Replies
17
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
2K
  • Introductory Physics Homework Help
2
Replies
54
Views
8K
Back
Top