CMS published for educational purposes (caveat) 100.000 dimuon events:


As one easily sees, column 4 shows the first muon's energy, column 12 the second's. Although the first energy is not higher in general (the data are not ordered),the distribution of energies is asymmetric. The first muons energy is higher on average, as one can easily see with the following Mathematica code: (download the first file dimuon.csv to your working directory)

qwe = Import["dimuon.csv", "CSV"]; 

wer = Drop[qwe, 1];    (* select the events labelled "GG"*)

rty = Select[wer, ((#[[1]] == "GG")) &]; ert = Transpose[rty];

energy1 = ert[[4]]; energy2 = ert[[12]];
relationOfEnergies = energy1/(energy1 + energy2);

bcEnergies = BinCounts[relationOfEnergies, {0, 1, .01}];

BarChart[bcEnergies, Ticks -> {Table[i, {i, 0, 100, 10}], Automatic}];

Here is the result ("GG" events oly), the first energy as a fraction of both muon's energies:

relation of 1st muon's energy to both muon's energy

Could have a trivial reason, just curiosity...


I guess that the peak comes from events in which the muons have the same origin (a decay of a highly boosted Z-boson, for example) so they have almost the same energies; in other events, the two muons come pretty much from different places of the collision.

The asymmetry indicates that an ordering of the two muons was imposed in the whole dataset. It wasn't an ordering according to the energy (because the histogram would vanish in 1/2 of the range) but it was an ordering according to a quantity that is correlated with the energy so that it's more likely for the first muon in this ordering to carry a higher energy than the second one. Ordering according to the transverse energy or something like that comes in mind. An experienced experimenter/phenomenologist would probably be able to say exactly what was done.

  • $\begingroup$ Makes sense along these lines. If I select the high energy events with rtz = Select[wer, ((#[[1]] == "GG") && (#[[4]] >= 46) && (#[[12]] >= 46)) &]; then the angles phi point in almost opposite directions. The transversal momentum can be checked by replacing the respective line above with energy1 = ert[[8]]; energy2 = ert[[16]]; It shows a more symmetric distribution, though not ordered either. $\endgroup$ – ClassicalPhysicist Aug 17 '13 at 22:08

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