So I am trying to get an estimate of the electromagnetic calorimeter resolution at LHCb, and I have found this online:

enter image description here

But I have no idea of what it means.

Can anyone explain what the last part represents?

My signal is a $B_d$ mass with rest mass $m = 5,279.53$ $ MeV/c^2 $

  • $\begingroup$ BTW -- the tag [accelerator-physics] is for the physics of accelerators, not physics done by taking the beams. $\endgroup$ Nov 22 '14 at 21:44

The uncertainty in any particular measurement is $\sigma_E$. Resolution for these devices is almost always stated in relative terms as here, but take it like this because it depends on the energy measured.

So just multiply by the energy. That is, express your signal in $\mathrm{GeV}$ and then find $$ \begin{align} \sigma_E = \left(\frac{0.1}{\sqrt{E}} \oplus 0.01\right) E \end{align} $$

I've not seen this notation before (I'm not a colider guy) but I suspect that the $\oplus$ mean "add in quadrature", so $$ \begin{align} \sigma_E = \sqrt{ \left(\frac{0.1}{\sqrt{E}}\right)^2 + (0.01)^2} E \end{align} $$

Now, you say that your signal is a particular heavy meson mass, but that isn't what the calorimeter sees. The instrument sees several decay products. You have to figure the energy resolution for each peak seen by the calorimeter, use that and PID to construct a four-vector for each decay product (with uncertainties) and then reconstruct the four vector of the meson (still with uncertainties) and from that the measured mass.

You might suspect that the uncertainties are going to pile up and you'd be right.

  • $\begingroup$ So would the $E$ here be my signal? The rest energy of the heavy meson? $\endgroup$
    – SuperCiocia
    Nov 23 '14 at 15:19
  • $\begingroup$ When you look at a (segmented) calorimeter, you do apply some process to select a bunch of hits which you are treating as belonging together. Then you add up the energy those hits represent and that sum is your $E$. The selection process is important and for a calorimeter in a colider experiment it is unlikely to be "just take it all" the way it might be in a low rate experiment. $\endgroup$ Nov 23 '14 at 16:31

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