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\begin{document}
\title{On the Measurement of BCAL Photoelectron Statistics}
\author{R.T. Jones \\
University of Connecticut, Storrs, CT}
\date{September 28, 2007}
\maketitle
\vspace{2cm}
\begin{abstract}
The most important factor that determines the time resolution of the GlueX
barrel calorimeter is the number of detected photoelectrons per GeV deposited
by a charged particle.  Generally accepted values for scintillation yield
and collection efficiency in the BCAL lead to an estimate of 120 detected
photoelectrons per end for a 4~cm path length of a MIP at normal incidence.
Analysis of cosmic ray data has recently produced a much lower estimate for
this figure of about 25 pe/cm.  The purpose of this note is to examine the
assumptions that are entailed in the cosmic ray data analysis.
\end{abstract}

\newpage

\section{Photoelectron Yield}

The method that has been used until recently for estimating photoelectron
statistics in the BCAL is illustrated in Table~\ref{tab:est} for a readout
segment that contains 4 cm of a MIP track.  This number is for the 
geometric mean of the yield on the two ends of the calorimeter and so is
independent of the position of the hit along the BCAL module.  The estimate
includes the photon detection efficiency, but does not take into account
any loss at the coupling between the fiber and the end of the phototube.

\begin{table}
\caption{\label{tab:est}Estimated photoelectron yield per end of a BCAL
module sector that subtends 4~cm of a MIP track.}
\begin{center}
\begin{tabular}{lrl}
factor & value & \\
\hline
path length inside active scintillator & 1.86 & cm \\
specific energy deposition & 2.0 & MeV/cm \\
scintillation light yield & 8000 & /MeV \\
fiber capture fraction & 9.6 & \% \\
attenuation inside fiber & 0.43 & \\
fiber-detector coupling factor & 100 & \% \\
photoelectron detection efficiency & 10 & \% \\
\hline
total expected yield & 120 \\
\hline\hline
\end{tabular}
\end{center}
\end{table}

\section{Cosmic Ray Data}

A check of this estimate with empirical data has been carried out by the
Regina group~\cite{gluex-845}.  They examine cosmic rays that pass at close
to normal incidence through a sector of a BCAL module, while depositing
negligible energy in the sectors on either side.  The fluctuations in the
magnitude of the pulse heights from event to event are attributed to the
combination of sampling and Poisson statistics, while the differences between
the two ends for a single event are attributed to Poisson statistics alone.
The comparison between the two ends is done by observing the ratio of the
pulse heights from the north and south ends.  Attenuation effects on the
ratio for cosmic tracks at different positions along the module are 
correctly taken into account.  The result is $22\pm 2$ photoelectrons.

One way to explain the apparent discrepancy between measured and expected
photoelectron yields is to revise the figures in Table~\ref{tab:est}.  The most
obvious weakness is in the fiber - detector coupling factor of 100\%.  It
would be easy to imagine that this factor is more like 75\%.  It has
been suggested~\cite{lolos1} that the scintillation yield of the fiber is
only 2000/MeV instead of the nominal 8000/MeV.  Then if the fiber - detector
coupling efficiency is reduced to 75\% the discrepancy is fully resolved.
In view of the radical jump implied by this measurement for this broadly
accepted property of scintillator, it is worth while considering critically
the assumptions entailed in the cosmic ray data analysis.

\section{Alternative Explanation}

The goal in what follows is to examine carefully what is entailed in
replacing the 100\% fiber + light guide coupling efficiency with a more
realistic model.  Realistically one cannot assume that the average light
from every fiber is uniformly attenuated by 75\%.  Some fibers will couple
with 95\% efficiency and some that lie near the edges of the light guides
acceptance function may couple with 25\% efficiency or even less in the
corners.  Let us assign a coupling efficiency factor $c_n(i)$ and $c_s(i)$
for the north and south ends of fiber $i$, respectively.  There are
correlations between the $c$ factors for the two ends of a fiber but
because of geometric imperfections they will not be identical.  Accounting
for this leads to the fact that sampling fluctuations to feed into
differences in the north and south light yields that go beyond mere
Poisson statistics.  That is, the north/south ratio can no longer be
interpreted in terms of Poisson statistics alone.

The proposed model builds on the analysis presented in Ref.~\cite{gluex-845}.
The analysis begins with a sample of MIP tracks that leave an ionization
trail 4~cm long in the volume subtended by a particular readout segment.
Following the notation used by those authors, the symbol $R$ represents the
average value of the north/south photoelectron yields, taking into account
the exponential attenuation factor.  Let the symbol $p_n$ [$p_s$] represent
the expected number of photoelectrons seen at the north [south]
ends, which will later be smeared with Poisson statistics to generate
the pulse height spectrum predicted by the model.
\begin{eqnarray}
p_n &=& y\sqrt{R}\sum_i^n c_n(i)\, \varepsilon(i) \\
p_s &=& y\frac{1}{\sqrt{R}}\sum_i^n c_s(i)\, \varepsilon(i) \nonumber
\end{eqnarray}
The ideal yield factor $y$ corresponds to the product of factors in 
Table~\ref{tab:est} excluding the top two lines and assuming ideal coupling
at the fiber-readout interface.  The actual coupling efficiency is carried
by the $c_{n,s}$ factors.  The symbol $\varepsilon(i)$ represents the
amount of energy deposited in fiber $i$ in MeV.  The sum is over the fibers
that lie along the track in the cosmic ray event. There are roughly 30. 

The dominant source of fluctuations in $p_n$ and $p_s$ from event to event
under fixed trigger conditions is sampling fluctuations, ie.\ differences
in how much total ionization energy is deposited and how that energy is
allocated between lead, epoxy, and scintillator.  In fact this is what
drives the statistical term in the calorimeter energy resolution formula,
\begin{equation}
\label{eq:eres}
\sigma(\varepsilon) = a \sqrt{\varepsilon}
\end{equation}
where I let $a=0.06$~GeV$^{\frac{1}{2}}$.  Eq.~\ref{eq:eres} is justified 
by the following argument.  It is known to hold for an entire shower, if
$\varepsilon$ represents the total shower energy \footnote{Actually there is a small floor term that should also be included,
but it has negligible effect at low energy and can be neglected.}.
All of the scintillation light in the calorimeter comes from
ionization by electrons and positrons in the shower.  Electrons deposit
most of their ionization energy in the MIP region.  Therefore the calorimeter
response to a single passing high energy cosmic ray mimics its response to
an electromagnetic shower that deposits the same amount of energy
\footnote{Cosmic ray events also have so-called Landau fluctuations in the
total energy loss in the calorimeter that are not present in electromagnetic
showers.  This means that Eq.~\ref{eq:eres} underestimates the variances of
$p_n$ and $p_s$ in a MIP sample.  However, those fluctuations are completely
correlated between north and south ends and do not affect the argument being
made with this model.}.
But the energy deposition fluctuations in different segments of the track 
are independent.  Suppose that the total energy from a passing MIP seen in
a readout segment were made of a sum of the energies from many smaller
track segments as in $\varepsilon = \sum_i \varepsilon(i)$ and all of the
$\varepsilon(i)$ are independent.  Then it follows that Eq.~\ref{eq:eres}
also holds for the individual $\varepsilon(i)$ with the same value for $a$.
\footnote{The Poisson distribution also obeys this decomposition property,
which has led some to assume that Eq.~\ref{eq:eres} is a consequence of
some Poisson process underlying sampling fluctuations.  This is not a
sound argument.  The form of Eq.~\ref{eq:eres} is not a unique signature
of Poisson statistics, but holds for all statistical systems that are
self-similar within some range of scales.}
Therefore, Eq.~\ref{eq:eres} holds equally for the energy deposited in
all of the fibers in a readout sector and that seen by a single fiber.

Using Eq.~\ref{eq:eres}, the variance of $p_n$ due to sampling fluctuations is
\begin{equation}
V(p_n) = y^2 R a^2 \left<c_n^2\right> E
       + y^2 R n \left<\varepsilon^2\right> V(c_n)
\end{equation}
where variations in the coupling coefficients $c_n$ from one track to the
next are explicitly taken into account.  A similar expression applies to
the south end, with $R$ replaced with $1/R$ and $c_n$ replaced with $c_s$.
After several steps, this reduces to the following form.
\begin{equation}
V(p_n) = y^2 R\ a^2\,\overline{c}\, ^2\, E
       + 2 y^2 R \left(a^2 + \frac{E}{2n}\right) E\, V(c_n)
\label{eq:vpn}
\end{equation}
In the last term $a^2=3.6$~MeV whereas $E=4$~MeV and $n=30$ so the final
term proportional to $E/n$ can be dropped.  The first term in Eq.~\ref{eq:vpn}
takes into account the rescaling of the expected number of photoelectrons
due to losses at the fiber-readout coupling, while the last term estimates
how non-homogeneities in that coupling amplify the sampling fluctuations
with respect to the ideal coupling case.  This is very interesting because
this exact effect can also be mocked up by modeling these fluctuations as
due to Poisson photon statistics instead.  That is, if one assumed that
$V(c_n)$ were zero and introduced instead the extra Poisson fluctuations
by artificially reducing the photoelectron statistics that model should fit
the data equally well.

To see what magnitude of variation in the $c_n$ would be required to
explain the cosmic ray data in a way that is consistent with the numbers
in Table~\ref{tab:est}, consider the following simple model for the
frequency distribution of the $c_n$.
\begin{equation}
f(c_n) = 30\, c_n^4\,(1-c_n) 
\label{eq:fcn}
\end{equation}
This distribution is peaked at 80\% and has an average of 75\%.  Its variance
is 2.6\%.  The first term in Eq.~\ref{eq:vpn} is completely correlated
between north and south ends and so does not contribute to the spread in
the ratio.  To get an idea of the magnitude of the contribution from the
second term, I begin by assuming that the $c_n$ and $c_s$ are completely
uncorrelated, just like they were in the Poisson model.  This leads to an
uncorrelated error $\sigma(p_n)/p_n = \sqrt{V(c_n)}/\overline{c_n} = 0.22$
which corresponds to the Poisson statistics of 22 photoelectrons.  This
is just about exactly the number of photoelectrons that were reported in
Ref.~\cite{gluex-845} assuming perfectly uniform readout!  This precision
cannot be taken seriously, given the arbitrariness of the model.  In fact,
there are correlations between the $c_n$ and the corresponding $c_s$ and
Eq.~\ref{eq:fcn} is only a guess for their distribution.  In addition,
there actually are Poisson statistics to be added in the model presented
here, on top of the sampling fluctuations in Eq.~\ref{eq:vpn}.  The only
significant result emerging from this study is that non-uniformities in
the fiber-readout coupling do cause sampling fluctuations to feed into
the spread of the north/south pulse height ratios, and that reasonable
assumptions for the non-uniformities lead to corrections that are large
enough to potentially explain the cosmic ray data without a dramatic
reduction in the photon yield of the scintillator.

\section{Conclusions}

This author needs to be convince that the figure
of 8000 scintillation photoelectrons/MeV in plastic scintillator is
really wrong by a factor 4.  This number refers
to the part of the spectrum that is transmitted by the plastic, not the
original VUV emission.  Most of the photons from the original scintillation
process are in the VUV region and do not go further than a few cm of the
source, but the scintillator is
doped with a waveshifter that converts most of these into the blue region
where the scintillator is relatively transparent.  It is these blue photons
that are counted as 8000 per MeV deposited.  Their absorption takes place on
a length scale of order meters, and is described an exponential absorption
length.  If the yield is really 2000 instead of 8000 then this implies
that 3/4 of the photons were absorbed within a few mm of the source, so
close that they are not seen in the measured exponential attenuation curve
that goes down to a few cm.

A detailed model of the observed width of the distribution of the
ratio of the pulse heights at the two ends of a BCAL module exposed to
cosmic rays is proposed that takes into account potential non-uniformities
in the coupling between the fibers and the readout at the two ends.  Making
reasonable assumptions about the coupling in this model leads to fluctuations
in the ratio that are large enough to explain the cosmic ray data before
Poisson statistics are included.  Including Poisson statistics at the level
expected on the basis of 8000 photons/MeV does not appreciably affect this
result.  This does not vindicate any particular interpretation; either model
can explain the data.  However it does show that the cosmic ray data analysis
does not reliably extract the number of photoelectrons in the cosmic ray data
because it entails unwarranted assumptions about the uniformity of the
readout on the two ends.  The only thing that can be reliably said is
that 22 photoelectrons is a {\em lower bound} on the yield in the cosmic
ray data.

\begin{thebibliography}{d}
\bibitem{gluex-845} A.~Semenov, Z.~Papandreou, ``Analysis of Amplitude
Information from 2006 BCAL Cosmics Runs'', {\em gluex-doc-845} 2007.
\bibitem{070928a} A.~Semenov, Z.~Papandreou, ``Analysis of Amplitude
Information from 2006 BCAL Cosmics Runs, version 070928a'',
{\em unpublished} 2007.
\bibitem{lolos1} G. Lolos, {\em private communication}.
\end{thebibliography}

\end{document}
