Q^2 values

  • Thread starter jdstokes
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
malawi_glenn
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FOUR-momentum change Q^2 is.
 
  • #3
pam
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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.
 

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