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This might seem basic, but it is a bit confusing. You hear about missing transverse energy a lot in SUSY searches due to the LSP which cannot be detected.

Let's say I have the 4-vector for the LSP. How does one go about calculating missing transverse energy the 4-vector?

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You don't actually get a 4-vector for anything in particular. You get a 4-vector for the-sum-of-unobserved-stuff where the unobserved stuff may included neutrinos as well as any neutral beyond-the-standard-model things you could be searching for.

In an ideal world you would get this 4-vector by totaling up the detected particles and subtracting them from the total for the incident particles.

Alas, this is not a perfect world, and a lot of stuff is generally lost down the beam-pipes. That's why we restrict ourselves to the transverse observables. However, that means that the energy expected in the transverse channels is not simply given by the incident particles.

I'm not sure where they usually go from there (not doing collider physics), but you do know that the expected, total transverse momentum is zero, and that means you can find the missing transverse momentum. If you then assume a mass for the unobserved particle (perhaps having chosen events for which there is no obvious source of neutrinos), you can then figure the energy.

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  • $\begingroup$ I am assuming I perform the calculation theoretically. Then, I can get the exact 4-vector of the outgoing LSP. How would I convert that to the MET variable? How is MET defined? Intuitively, I would just use px py components, but it is missing transverse ENERGY, not missing transverse momentum. $\endgroup$
    – user788171
    Oct 23, 2012 at 21:18
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    $\begingroup$ @user788171 I assume you would simple project the energy along the momentum vector, but you say "in SUSY searches" which implies that someone must be thinking about an experimental component, and that introduces some additional difficulties as I've outlines. $\endgroup$
    – dmckee --- ex-moderator kitten
    Oct 23, 2012 at 21:25
  • $\begingroup$ To leading order I think you just assume the mass of the particle is negligible so that $E = pc$. There is certainly precedent for particle physicists saying "energy" when they mean "momentum." $\endgroup$
    – David Z
    Oct 24, 2012 at 3:20
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    $\begingroup$ In a lepton collider, the total energy of the colliding particles is known, so the missing energy is a good experimental signature. In a hadron collider, you don't know the longitudinal momentum of the partons involved in the collision, hence you know only the sum of the transverse momentum. $\endgroup$
    – pfnuesel
    Feb 1, 2014 at 21:45
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You measure the "Sum of transverse calorimeter depositions" Then you expect them to be zero in a full azimuth calculation. If not, the remainder is the missing transverse energy

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    $\begingroup$ Not quite, muons don't deposit much energy in the calorimeters, but you have to take them into account as well. $\endgroup$
    – pfnuesel
    Feb 1, 2014 at 21:41

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