Comparing Q^2 Values for Colliding Beam Processes

  • Context: Graduate 
  • Thread starter Thread starter jdstokes
  • Start date Start date
  • Tags Tags
    Beam
Click For Summary
SUMMARY

The discussion focuses on the computation of Q^2 values for colliding beam processes, specifically in the context of high-energy proton and positron interactions. The formula provided for Q^2 is Q^2 = (\vec{p}_p + \vec{p}_e - p_e')^2 = (E_p - E_e)^2 + E'^2 - 2(E_p-E_e)E'\cos\theta. It is clarified that Q^2 represents the negative square of the electron's four-momentum transfer, emphasizing that the proton's momentum does not factor into this definition. The invariance of Q^2 across different collision types is also confirmed.

PREREQUISITES
  • Understanding of four-momentum in particle physics
  • Familiarity with high-energy collision processes
  • Knowledge of conservation laws in particle interactions
  • Basic grasp of scattering angles and energy calculations
NEXT STEPS
  • Study the concept of four-momentum in detail
  • Learn about fixed-target vs. colliding beam experiments
  • Investigate the implications of Q^2 invariance in particle physics
  • Explore advanced scattering theories and calculations
USEFUL FOR

Physicists, particularly those specializing in particle physics, researchers involved in high-energy collision experiments, and students seeking to deepen their understanding of Q^2 values in scattering processes.

jdstokes
Messages
520
Reaction score
1
I'm a little bit confused about the concept of Q^2. For fixed-target processes, Q is the momentum change of the incident particle.

How does one compute the Q^2 values for colliding beam processes?

If I have a high energy proton E_p incident on a positron E_e (head on) which scatters the positron at an angle to the proton direction with known scattering energy E', then is the Q^2 value given by

[itex]Q^2 = (\vec{p}_p + \vec{p}_e - p_e')^2 = (E_p - E_e)^2 + E'^2 - 2(E_p-E_e)E'\cos\theta[/itex]?

Consveration of momentum implies that

[itex]\vec{p}_p + \vec{p}_e = \vec{p}_p' + \vec{p}_e'[/itex] so I have defined Q as the momentum of the scattered proton. Is this the conventional definition of Q?
 
Physics news on Phys.org
FOUR-momentum change Q^2 is.
 
Q^2 is the negative of the square of the electron's 4-momentum (not 3-momentum) transfer. The proton doesn't enter. Since Q^2 is an invariant, it is the same in colliding beam as in fixed target.
 

Similar threads

Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
7K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K