0
$\begingroup$

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?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

$dE/dx$ 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:

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

and

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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
    Commented Dec 29, 2020 at 8:13
  • $\begingroup$ Those sound like pretty good numbers, yes. Better than the ALICE TPC, for example. $\endgroup$
    – dukwon
    Commented Dec 30, 2020 at 12:33

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.