[Halld-cal] Fwd: Re: MC Michel Spectra selection cuts

Elton Smith elton at jlab.org
Wed Nov 4 17:46:15 EST 2015


-------- Forwarded Message --------
Subject: 	Re: [Halld-cal] MC Michel Spectra selection cuts
Date: 	Wed, 4 Nov 2015 10:53:47 -0500
From: 	Elton Smith <elton at jlab.org>
To: 	wmcginle at andrew.cmu.edu, semenov at jlab.org
CC: 	elton at jlab.org, Andrei Semenov <andrei.semenov at uregina.ca>, Zisis 
Papandreou <Zisis at uregina.ca>, Mark Dalton <dalton at jlab.org>



Hi Andrei and Will,

In order to calibrate using the Michel electrons, it seems the most
useful quantity is the full energy spectrum (Ed and Emc below). But
since the energy is distributed over several cells, this complicates the
gain assignment. The energies in each cell are denoted by Ed_i and
Emc_i, for data and MC, respectively. Here is an outline (for
comments/feedback) for a strategy to determine the gain factors
iteratively.

We can write the total energy for the data as follows:

Ed = Sum_k (g0_k * Ed_k) = g0_i*Ed_i + Sum_j.ne.i (g0_j * Ed_j); [Note:
Ed has units of MeV, Ed_i has units of fADC channels]

where i is the channel that we wish to calibrate. The 'g0' constants
denote the current value to be optimized. It also makes sense to choose
the channel i that has the largest amount of energy deposition. Of
particular interest is the faction of energy deposited in channel i:

Rd_i = g0_i*Ed_i/Ed

We can also compute the same quantities in MC, where the gain factors
are set to unity

Emc = Sum_k (Emc_k) = Emc_i + Sum_j.ne.i (Emc_j), and correspondingly
[Note: Emc and Emc_i have units of MeV]

Rmc_i = Emc_i/Emc

Suppose we plot Ed, Ed_i for bins in Rd_i (e.g. 0.4 < Rd_i < 0.6, 0.6 <
Rd_i < 0.8, 0.8 < Rd_i < 1.0), and correspondingly for the MC data.

(1)   If we assume that on average the channels for j.ne.i have the
correct calibration, i.e. <Sum_j.ne.i (g0_j * Ed_j)> = < Sum_j.ne.i
(Emc_j)>, then

<Emc> - <Ed> = <Emc_i> - c*g0_i <Ed_i> = 0, where g_i = c*g0_i is the
correct gain factor for channel i. Then

c = <Emc_i> / (g0_i * <Ed_i>)

Of course, assumption (1) above may not be correct in general, but the
conclusions still follow for the case that Rd_i = Rmc_i = 1 (i.e. sums
j.ne.i are zero).

Now, since we have plotted Emc_i and Ed_i in bins of Rd_i and Rmc_i, one
play plot c vs R_i. The correct gain factor will be given by g_i =
c(R_i=1) * g0_i.

The process may be repeated with updated gain factors, and if they are
converging, then the value of c should become independent of R_i.

If we agree that this procedure makes sense, Andrei should produce the
plots of Emc_i for different bins in Rmc_i and Will would generate the
corresponding histograms for the data and complete this procedure. Of
course all other selection cuts, which have been discussed, should be
common for both data and MC.

Suggestions/comments?

Elton.




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