What is transverse energy? Why we use transverse total energy instead of energy and transverse momentum in place of Total momentum in the particle detectors?


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Transverse momentum, $\vec{p}_T$, is the momentum of an object transverse to the beam. Transverse energy is defined as $E_T = \sqrt{m^2+p_T^2}$ for an object with mass $m$ and transverse momentum $p_T$.

The initial longitudinal momentum in a parton collision is unknown, because the partons that make up a proton share the momentum. We do know, however, that the initial transverse momentum was zero. So we look for missing transverse momentum, defined $E_T^\textrm{miss} = -\sum_i \left|\vec{p}_T(i)\right| \equiv -\sum_{i}p_{T}\left(i\right)$ for visible particles $i$. Finding missing transverse momentum would indicate that new, unaccounted for particle(s) had escaped the detector.

Confusingly, $E_T^\textrm{miss} = -\sum_i p_T(i)$ is commonly called missing transverse energy or MET. Missing transverse energy is equivalent to missing transverse momentum only if the missing particle(s) were massless.

Also, events in which the products have large transverse momentum are more likely to be genuine, interesting events.

  • $\begingroup$ I am afraid sth doesn't make sense: Clearly, $E_{\text{T}}^{\text{miss}}$ is a scalar, but the RH side of your equation is a vector.. Maybe it should read $E_{\text{T}}^{\text{miss}} = -\sum_{i}\left| p_{T}\left( i\right)\right| \equiv -\sum_{i}p_{T}\left( i\right)$? $\endgroup$
    – user248824
    Jan 25, 2021 at 19:53
  • $\begingroup$ indeed, should probably take norm of RHS of those expressions, will fix later $\endgroup$
    – innisfree
    Jan 27, 2021 at 5:43
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    $\begingroup$ i believe it’s still incorrect: a perfectly fine collision with whole final state detected should have no missing energy, but if you have any transverse momentum and you sum the absolute value the sum will be non-zero. i would say that you want to take absolute value after taking the sum $\endgroup$
    – Annibale
    Jul 26, 2021 at 10:14

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