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enter image description here

I'm not 100% sure. but I think that the width of a particle could change depending on the decay channel. for example: The J/psi's mass resolution could be 40 MeV or in other cases 8 MeV.

So I would like to know :

  1. What actually contributes to the width of a particle?.
  2. Why some particles are really narrow and some other are wide?
  3. Do we expect to see the same particle's mass resolution in the Monte Carlo simulation as the one we will see in the real experiment Data?

I hope someone will be able to answer my questions. Thanks in advance.

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    $\begingroup$ You seem to be using 'mass resolution' interchangeably with 'width'. Mass resolution is how precisely a detector can measure a particular invariant mass. Width is a property of a resonance. Decay channels have partial widths. The branching fraction of a decay is the partial width divided by the (total) width. $\endgroup$ – dukwon Nov 5 '16 at 13:32
  • $\begingroup$ When you say "width", do you mean a decay width? Please try to make your question accessible to others not familiar with your exact terminology. $\endgroup$ – ACuriousMind Nov 5 '16 at 15:07
  • $\begingroup$ To emphasize what @dukwon said, widths are physics properties of a decay or a decay channel, while resolution are detector properties and combine the width with a contribution from the limitations of the apparatus. $\endgroup$ – dmckee Nov 5 '16 at 16:00
  • $\begingroup$ @dukwon Thank you for your reply. You're right, I'm confused between the two. If I look up J/psi in the PDG, I can find its width which is related to it's lifetime. However what I'm referring to now is the (Sigma) of a particle that we get after a Gaussian fit. ( I added a picture to my question). Now if this "Sigma" is related to the precision of a detector, does it mean it will stay unchanged for all decay channels measured in the same detector? If we measure it in another detector? $\endgroup$ – the phoenix Nov 6 '16 at 10:55
  • $\begingroup$ @dmckee thank you for your reply. Please see my last comment, you're more than welcome if you have anything more to add. $\endgroup$ – the phoenix Nov 6 '16 at 10:57
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This is e+e- interactions versus energy (sqrt(s)) from the particle data book, fig 49.5 :

e+e-cross

Many of the interesting experimental measurements that led to the quark model are in this plot.

Particles like protons, electrons muons have a fixed mass, no width, measured by the "length" of the four vector. Measurement errors will introduce a statistical indeterminacy which can be fitted with the statistically defined gaussian. In the plot though, we see practically a delta function for the J/psi of the plot in the question, because the scale is different. Omega and rho have a large width which is not gaussian, Y is very narrow and the Z is also non gaussian.

What defines the intrinsic width is the type of interaction entering in the possible decays of the resonances and whether strong interaction decays are suppressed or not. The J/ψ is a good example:

It has a rest mass of 3.0969 GeV/c2, just above that of the ηc (2.9836 GeV/c2), and a mean lifetime of 7.2×10^−21 s. This lifetime was about a thousand times longer than expected.

....

Hadronic decay modes of J/ψ are strongly suppressed because of the OZI Rule. This effect strongly increases the lifetime of the particle and thereby gives it its very narrow decay width of just 93.2±2.1 keV. Because of this strong suppression, electromagnetic decays begin to compete with hadronic decays. This is why the J/ψ has a significant branching fraction to leptons.

The width of the Z also is interesting,and is composed out of all the partial widths in the channels it can decay,page 2 in this link.

In conclusion, the intrinsic interaction width ,( calculable by the standard model), of resonances has to be folded with the experimental error widths ,as explained in the comments to the question. When the intrinsic width is very small , the experimental error dominates and is a gaussian.

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  • $\begingroup$ Anna thank you very much for your awesome answer :) $\endgroup$ – the phoenix Nov 11 '16 at 16:53

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