[Hps-ecal] Opening angle from Ecal

[Garcon Michel]

Hi Holly,

                Nice study. It is a bit counter-intuitive to call Phi a (projected) horizontal angle and Theta a vertical angle, but this is a minor detail.

                One does expect indeed dPhi/dx = dTheta/dx ~ 1/D ~ 1/1.4 m ~ 0.007 radian/cm . In the vertical projection, the trajectory is a  straight line whatever the energy (for an ideal field), which explains that you find no energy dependence for Theta. In the horizontal projection, your line in the first plot Phi vs x at 1 GeV is the same as the light blue in Fig. 31 of HPS-Note-2014-001 (one can see also in this figure that the slope does not depend on energy). So everything checks nicely.

                I would expect B = 0 when E goes to infinity (straight line -> x = 0 when Phi =0). But you are right to do whatever empirical fit works in our range.

                                                                              Best regards,
                                                                                              Michel.

PS: by cluster energy, do you mean deposited energy in ECal or real energy? I would imagine the latter since you generate particles at fixed energies; then better not call it cluster energy, could be confusing (see Eqs 1 and 2 of the Note).

_______________
Michel Garçon
Irfu/SPhN
CEA-Saclay, bât. 703
91191 Gif-sur-Yvette cedex
France
Tél.: +33 1 69 08 86 23
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