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2014-10-15 11:39:19 +09:00
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2 changed files with 58 additions and 4 deletions

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@@ -382,7 +382,8 @@ the cut on protons is estimated to be small compared to the statistical ones.
\subsection{Corrections for the number of protons}
\label{sub:corrections_for_the_number_of_protons}
The protons spectra observed by the silicon detectors have been modified by
the energy loss inside the target so correction (or unfolding) is necessary.
the energy loss inside the target so correction (also called unfolding, or
reconstruction) is necessary.
The unfolding, essentially, is finding a response function that relates proton's
true energy and measured value. This can be done in MC simulation by generating
protons with a spatial distribution as close as possible to the real
@@ -413,4 +414,57 @@ method is implemented.
\label{fig:al100_resp_matrices}
\end{figure}
After training the unfolding code is applied on the measured spectra from the
left and right arms. The unfolded proton spectra
left and right arms. The unfolded proton spectra in \cref{fig:al100_unfold}
reasonably reflect the distribution of initial protons which is off-centred to
the right arm. The path length to the left arm is longer so less protons at
energy lower than \SI{5}{\MeV} could reach the detectors. The sharp low-energy
cut off on the right arm is consistent with the Coulomb barrier for protons,
which is \SI{4.1}{\MeV} for protons emitted from $^{27}$Mg.
Comparing the reconstructed spectra from \SIrange{5}{8}{\MeV}, the protons
yields are consistent with each other:
\begin{align}
N_{\textrm{p reco. left}} &= (110.9 \pm 2.0)\times 10^3\\
N_{\textrm{p reco. right}} &= (110.2 \pm 2.3)\times 10^3
\end{align}
Therefore, the number of emitted protons is taken as average value:
\begin{equation}
N_{\textrm{p reco.}} = (110.6 \pm 2.2) \times 10^3
\end{equation}
\begin{figure}[htb]
\centering
\includegraphics[width=0.85\textwidth]{figs/al100_unfold}
\caption{Unfolded proton spectra from the 100-\si{\um} aluminium target.}
\label{fig:al100_unfold}
\end{figure}
\subsection{Number of nuclear captures}
\label{sub:number_of_nuclear_captures}
\begin{figure}[htb]
\centering
\includegraphics[width=0.85\textwidth]{figs/al100_ge_spec}
\caption{X-ray spectrum from the aluminium target, the characteristic
$(2p-1s)$ line shows up at 346.67~keV}
\label{fig:al100_ge_spec}
\end{figure}
The X-ray spectrum on the germanium detector is shown on
\cref{fig:al100_ge_spec}. Fitting the double peaks on top of a first-order
polynomial gives the X-ray peak area of $5903.5 \pm 109.2$. With the same
procedure as in the case of the active target, the number stopped muons and
the number of nuclear captures are:
\begin{align}
N_{\mu \textrm{ stopped}} &= (1.57 \pm 0.05)\times 10^7\\
N_{\mu \textrm{ nucl. cap.}} &= (9.57\pm 0.31)\times 10^6
\end{align}
The proton emission rate in the range from \SIrange{5}{8}{\MeV} is therefore:
\begin{equation}
R_{\textrm{p}} = \frac{110.6\times 10^3}{9.57\times 10^6} = 1.16\times
10^{-2}
\label{eq:proton_rate_al}
\end{equation}
\subsection{Uncertainties of the emission rate}
\label{sub:uncertainties_of_the_emission_rate}

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@@ -32,8 +32,8 @@ for the COMET experiment}
%\input{chapters/chap1_intro}
%\input{chapters/chap2_mu_e_conv}
%\input{chapters/chap3_comet}
%\input{chapters/chap4_alcap_phys}
%\input{chapters/chap5_alcap_setup}
\input{chapters/chap4_alcap_phys}
\input{chapters/chap5_alcap_setup}
\input{chapters/chap6_analysis}
%\input{chapters/chap7_results}