I'm reading an article describing a particle detector. I did not understand the following things in it.
"The drift chamber (of this detector) was described as having a dE/dX resolution that is better than 6%. & a momentum resolution which is better than 0.5% for charged tracks with a momentum of 1 GeV/C."

Could anyone please explain to me in simple terms what they mean by these two sentences?

  • $\begingroup$ I would appreciate it. if you can help out with this question as well :) @dukwon $\endgroup$ – the phoenix Nov 11 '16 at 17:15

$dE/dx$ is the rate of energy loss (over distance) from a charged particle in some material. If you measure $dE/dx$ as a function of momentum ($p$), this can be used as a form of particle identification, because different types of particles follow different curves in this distribution.

dE/dx vs p

Resolution of some variable $q$ is usually quoted as $\frac{\Delta q}{q}$ where $\Delta q$ is e.g. the width of a Gaussian distribution of the deviation between the true and measured quantity (taken from simulation or otherwise from some calibration or other).

So your sentences evaluate to:

$$\frac{\Delta (dE/dx)}{dE/dx}<0.06$$


$$\frac{\Delta p}{p}<0.005 \; \left(\text{for } p>1\text{ GeV}\right)$$

  • $\begingroup$ thank you very much for your answer. what does this tell us about the detector? does it identify charged particles precisely if delta(dE/dx)/(dE/dx)<0.06 ? @dukwon $\endgroup$ – the phoenix Dec 29 '20 at 8:13
  • $\begingroup$ Those sound like pretty good numbers, yes. Better than the ALICE TPC, for example. $\endgroup$ – dukwon Dec 30 '20 at 12:33

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