Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Q^2 values

  1. May 16, 2008 #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?
  2. jcsd
  3. May 17, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    FOUR-momentum change Q^2 is.
  4. May 17, 2008 #3


    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook