Q^2 is the Mandelstam variable t, i.e. the four momentum transfer squared where the s channel is the x axis in the feynman diagram.
where p1 p2 are incoming.
In the same link it is seen that at the relativistic limit :
The dot product for t (Q^2) is p transverse to the incoming beam direction of p1 , p2 are incoming.
etc.
Is this Q the same if I take an s channel or a t channel process
No, Q is only defined in the t channel, with incoming p1 p2.
p transverse to the beam is a very useful variable, as it does not change with the Lorenz transformation to the Center of Mass System. It is an easy trigger variable for experiments that indicates the size of the mass break up at the CMS . The higher the p transverse of the event the more deeply inelastic the specific event ( think Rutherford scattering). The approximation with Q allows comparison estimates with appropriate analytic expressions of crossections.
This post imported from StackExchange Physics at 2016-06-16 13:40 (UTC), posted by SE-user anna v