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- \title{Probing background with \\ Method of Moments for $\PBzero \to \PKstar \mu \mu$}
- \author{\underline{Marcin Chrzaszcz}$^{1,2}$}
- \date{\today}
-
- \begin{document}
-
- {
- \institute{$^1$ University of Zurich, $^2$ Institute of Nuclear Physics}
- \setbeamertemplate{footline}{}
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- \logo{
- \vspace{2 mm}
- \includegraphics[height=1cm,keepaspectratio]{images/uzh.jpg}~
- \includegraphics[height=1cm,keepaspectratio]{images/ifj.png}}
-
- \titlepage
- \end{frame}
- }
- \institute{UZH,IFJ}
-
- \section[Outline]{}
- \begin{frame}
- \tableofcontents
- \end{frame}
-
- \section{Reminder}
- \begin{frame}\frametitle{Plan}
- Method of moments:
- \begin{enumerate}
- \item Last meeting showed how orthogonality of the does magic for method of moments.
- \item Using toy MC (experimental math) checked the errors estimates.
- \item Checked that it does not suffer from boundary conditions.
- \item Many thanks to Tom for checking all my calculations.
- \end{enumerate}
- For today:
- \begin{enumerate}
- \item How this method behaves in terms of background?
- \end{enumerate}
-
- \end{frame}
- \section{Theory introduction}
- \begin{frame}\frametitle{What do we start with}
- {~}
- Let's assume for simplicity we have our pdf:
- \begin{multline}
- \dfrac{d^4\Gamma}{ \Gamma dq^2 dcos\theta_k dcos\theta_l d\phi}=\dfrac{9}{32\pi}( \dfrac{3}{4} (1-F_l) \sin^2 \theta_k + F_l\cos^2 \theta_k + ( \dfrac{1}{4}(1-Fl)\sin^2 \theta_k \\ - F_l\cos^2) cos 2\theta_l + S_3 \sin^2 \theta_k \sin^2 \theta_l \cos2\phi + S_4 \sin2 \theta_k \sin \theta_l \cos\phi +\\ S_5 \sin2 \theta_k \sin \theta_l \cos \phi + (S_{6s} \sin^2 \theta_k) \cos \theta_l + \\ S_7 \sin 2\theta_k \sin \theta_l \sin \phi + S_8 \sin 2 \theta_k \sin 2 \theta_l \sin phi + S_9 sin^2 \theta_k \sin^2 \theta_l \sin 2 \phi)
- \end{multline}
- What did we assume:
- \begin{itemize}
- \item $S_{1x}$,~$J_{2x}$ can be parametrized by $F_l$.
- \item $S_{6c}=0$.
- \item In short what was in the paper.
- \end{itemize}
- \end{frame}
-
-
- \begin{frame}\frametitle{Obtained moments 1}
- {~}
- Lets see how this works in practice:
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi } \sin^2 \theta_k = \dfrac{2}{5}(2-F_l)
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi } \cos^2 \theta_k = \dfrac{1}{5}(1+F_l)
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi } \sin^2 \theta_k \cos 2\theta_l = -\dfrac{2}{25}(2+F_l)
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi} \cos^2 \theta_k \cos 2\theta_l = -\dfrac{1}{25}(1+8F_l)
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi} \sin^2 \theta_k \cos \theta_l = \dfrac{2 S_{6s} }{5}
- \end{equation}
-
- \end{frame}
-
-
- \begin{frame}\frametitle{Obtained moments 2}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi} \cos^2 \theta_k \cos \theta_l = \dfrac{S_{6s} }{10}
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi }sin^2 \theta_k sin^2 \theta_l cos 2 \phi= \dfrac{8 S_3 }{25}
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi } sin 2 \theta_k sin 2 \theta_k cos\phi= \dfrac{8 S_4 }{25}
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi }sin2\theta_k sin\theta_l cos\phi = \dfrac{2 S_5 }{5}
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi} sin 2 \theta_k sin \theta_l sin \phi = \dfrac{2 S_7 }{5}
- \end{equation}
- \begin{equation}
- \dfrac{d^3\Gamma}{\Gamma dcos\theta_k dcos\theta_l d\phi} sin 2 \theta_k sin2 \theta_l sin \phi = \dfrac{8 S_8 }{25}
- \end{equation}
-
- \end{frame}
-
- \section{Background regions}
-
- \begin{frame}\frametitle{Studied background region}
- {~}
-
- \begin{columns}
-
- \column{2.5in}
- \begin{itemize}
- \item Defined $\PBzero$ mass bins: $1:(5,5,15)\cup 2:(5.15,5.22) \cup 3:(5.35,5.5) \cup 4:(5.5,6)~GeV$
- \begin{enumerate}
- \item Region 5:$(5.35, 6)$
- \item Region 6:$(5, 5.22)$
- \end{enumerate}
- \item use the old $q^2$ bins:
- \begin{itemize}
- \item 0:$0, 2$
- \item 1:$2,4.3$
- \item 2:$4.3, 8,6$
- \item 3:$10.1, 12.9$
- \item 4:$14.2, 16$
- \item 5:$16,19$
- \end{itemize}
- \end{itemize}
-
- Please remember the numbers, we will need then later on.
- \column{2.5in}
- \includegraphics[scale=0.22]{plots2/PLOT.png}
- \end{columns}
-
-
- \end{frame}
-
- \section{Results}
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- {~}
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- \section{Summary}
- \begin{frame}\frametitle{Summary}
- {~}
- \begin{itemize}
- \item Background moments are effectivelly 0
- \item Apart from $F_l$ and $S_6$
- \item $S_6$ is sizeable at the left hand sideband for certain bins, evidence of partially reconstructed semileptonic decays?
- \item Because the moments are small, they should have small effect on the final result :)
- \end{itemize}
-
- Wish list:
- \begin{itemize}
- \item Repeat the same with smaller $q^2$ bins.
- \item Optimise the binning in $q^2$ taking into account background systematics and error on signal
- \item Do unfolding.
- \end{itemize}
- \end{frame}
-
-
- \end{document}