In experimental papers such as http://arxiv.org/abs/1603.02248 (page 5, section 3.2) there are requirements that the transverse momenta of certain particles are greater than a certain minimum amount in order to be selected for further analysis.

Why is there are requirement like this? Pseudorapidity bounds ensure the particles arrive in the detectors so what does this do?


2 Answers 2


The only way to get high-$p_T$ particles is if the parton collision has a large invariant mass. If your signal is a heavy particle, then $p_T$ cuts are a good way to reduce background.


The clearest framework to study and interaction is the center of mass system. To go to the center of mass one has a Lorentz transformation. All vector components in the direction of the velocity vector entering the transformation are changed by the transformation. The components perpendicular to the transformation vector are not affected.

Thus p transverse is the same either at the center of mass or in the laboratory system. So it is a snapshot at the time of running the experiment, of what is going on at the center of mass without having to go to the details of the event. The same is true as a first filter in the subsequent analysis for picking up subsets of data that have events of interest.

In experiments one searches for heavier and heavier mass resonances of particles in the interaction, searching for new physics. These will be decaying into high momentum fragments and thus a high p transverse cut picks at the raw data level events that have a probability of coming from the decay of a supersymmetric particle, for example.

  • $\begingroup$ What I don't understand is that I assumed that the pp collision experiments were done where both the protons have the same energy, thus I would not expect any boost in the CM frame, which should leave any momentum unchanged in the conversion from the CM to LAB frame. $\endgroup$ Mar 28, 2023 at 12:25
  • $\begingroup$ @KaanGüven In colliders yes, but there exist many experiments where a beam of protons interacts with protons of a target at rest in in the laboratory system, before colliders became the dominant tool. $\endgroup$
    – anna v
    Mar 28, 2023 at 16:29
  • $\begingroup$ Okay. I just saw it in a paper on CMS today which was used quite in the same way as the owner of the post used in the question. Thank you for the clarification. $\endgroup$ Mar 28, 2023 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.