How exactly do you calculate the direction of the missing transverse energy? This paper (arXiv:1412.2641), for example, makes use of it to get some cuts. Adding to this, how can you correlate this to the 4-momentum of the neutrinos? Is the MET (or missing 4 momentum?) just the sum of all of the final neutrinos?
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$\begingroup$ Keep in mind that the energy of particles in an interaction depend on the inertial frame. In colliders, the laboratory frame differs from the center of mass frame by a transformation that changes the energies of all particles , but the momenta change only in the beam direction, Due to the conservation of momentum the transverse momenta are the same in the cms and the lab. So the transverse values give a possibility to measure the true center of mass behavior of particles. This is not useful for individual interactions, but makes it statitistically possible to get average behaviors at cms $\endgroup$– anna vSep 7, 2021 at 4:17
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MET is not just the sum of the four-momentum of the neutrinos, no. It is opposite of the sum of the three-momenta of the neutrinos (and potentially any other invisible particles), projected onto the plane transverse to the beam direction.
It is impossible to determine the sum of the energy of the neutrinos or their momentum in a direction parallel to the beam axis. Protons are composite particles, and in a given interaction only part of their momenta actually contributes. The effect of this is that the interaction energy and total momentum are both unknown. Because we don't know what we started with, we don't know how much four-momentum is missing.
The only thing that is known is that there should be little to no net momentum perpendicular to the beam axis. So in a hadron collider like the LHC, we can only talk about missing transverse momentum.
