I looked at events in which a BMU stub was associated with one
isolated BSU scintillator hit. Events with more than one hit in the
BSU or hits in the
cell adjacent were rejected. I then looked at the deviation of the
Z as found by time division from the Z of the center of the BSU paddle,
and asked how many were left after requiring that the deviation be less
than some distance. The results are shown
here for front and rear. I lose 10% with a 65cm cut in the front.
Since there is a substantial pileup for the rear scintillator Z distribution,
which isn't cut away until after about a 90cm cut, I use 85 cm cut as
the reference, and one again lose 10% after a 65cm cut. The paddles
themselves are about 164cm long, so an 92cm cut seems reasonable, and
gives about 96% efficiency in the front (presumably something similar
in the rear). TopDrawer source
Note that there are no timing cuts on these events: I did not demand
that the scintillator time agree with the BMU stub t0. This is thus
a first iteration.
In order to look at timing, first consider
the BSU time vs t0 plots for 4-hit stubs. The left column is
for events with front scintillator and the right column for those with
the rear scintillator. I interpret the wing in the upper right
plot as coming from late particles coming from quad splash or splash
back from the walls. Along the same line as the wing one sees a
cluster at about (60,120), which I interpret as early particles from
beam halo. The horizontal band seems to comprise BMU stubs which
are often only accidently associated with BSU hits, with the
exception of the region where the wing and band intersect.
Looking more closely at the time differences between the BMU stub
and the BSU time, fit the distributions
with a couple of gaussians. The distribution is for 4-hit stubs
with Z deviation from the center of the appropriate BSU of less than
65cm, BSU time less than 200 nsec and greater than 140 nsec, and the
time difference between -125 and 50 nsec. You can see that the front
peak is at -59.5nsec with a sigma of 5.68, and the rear at -63.6nsec
with a sigma of 5.61. 90% cuts would be at (-59.5 +- 9.32) and
(-63.6 +- 9.2) respectively.
Now these cuts take in some contamination from the other peak as
well. I used the Pari package to try to draw the contamination rate
as a function of number of sigmas from the peak at which we cut.
Unfortunately the PostScript version leaves off the tic marks that
the X-windows display has, so it is a bit hard to read. With that
warning, this is the front
contamination rate and this is
the rear contamination rate. If I use the 1.64 sigma cut mentioned
above to get 90% of the original signal, the front contamination is
7.1% and the rear contamination is 8.7%.
The chi-squared distribution is not sensible, so I look at the
chi vs t0 distribution. This certainly
has interesting aspects. No matter which side or layer we use, there
are two curves which must represent two different possible origins.
The calculation of the chi-squared does not adjust for the t0 as of
now--I must revisit that. The plot is made using 4-hit stubs only.
I'm not sure which of the two curves represents the prompt muons. Bob's
work suggests that halo comes in at 46nsec and real muons at 126, but
the minima of these curves represent t0's of 108.5 and 131. This means
that the fits described above are somewhat obsolete already.
