# High pT and high Q^2 in deep inelastic hadronic collisions.

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When reading about high energy collisions (for example proton-proton collisions at LHC), I always find that deep inelastic scattering (DIS) means high transverse momentum (pT). So that usually at the LHC experiments one look at high pT jets of particles. What is the relation between DIS and pT?

I also find the relation $Q \sim p_T$, which, for me, is hard to demonstrate. Can someone please shed light on this? Moreover, I found statements about the interaction time $\tau$, which is related to Q and pT in this way: $\tau \sim \frac{1}{Q} \sim \frac{1}{p_T}$, where again $Q\sim p_T$ is assumed. To understand, I would really like someone to define Q for a proton-proton (parton-parton) interaction. Is this Q the same if I take an s channel or a t channel process? Thanks!

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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

answered Jun 15, 2016 by (1,885 points)
edited Jun 16, 2016 by anna v
Hi @anna, thanks for your answer. What I don't understand is why all people say: "At LHC we look at high $p_T$ events, as these events are due to deep inelastic scattering processes". This statement works for a process with a photon exchanged in the $t$ channel. But take a Drell-Yan process: $q\bar{q}\rightarrow \gamma^*\rightarrow q\bar{q}$, which happens in the $s$ channel. In this case the two out-coming quarks (which I'll see as two jets) won't have a preferred direction (i.e. they won't be necessarily at high $p_T$). Am I wrong?

This post imported from StackExchange Physics at 2016-06-16 13:40 (UTC), posted by SE-user Marco
One chases high ptransverse events because they are the true picture of the center of mass interaction, the transverse component does not change with the Lorenz transformation. It is sufficient for an interesting event to have a high pt because it shows high break up at the cms. Identifying the pt with Q2 is a mathematical definition which in terms of Feynman diagrams may be useful or not.

This post imported from StackExchange Physics at 2016-06-16 13:40 (UTC), posted by SE-user anna v

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