diff --git a/Lectures_my/KT2_2017/ex1/sheet01.pdf b/Lectures_my/KT2_2017/ex1/sheet01.pdf index e600ae7..c1a4109 100644 --- a/Lectures_my/KT2_2017/ex1/sheet01.pdf +++ b/Lectures_my/KT2_2017/ex1/sheet01.pdf Binary files differ diff --git a/Lectures_my/KT2_2017/ex1/sheet01.tex b/Lectures_my/KT2_2017/ex1/sheet01.tex index 83c608d..7e54b82 100644 --- a/Lectures_my/KT2_2017/ex1/sheet01.tex +++ b/Lectures_my/KT2_2017/ex1/sheet01.tex @@ -7,8 +7,8 @@ \newcommand{\sheetnr}{1} -\newcommand{\issued}{21.09.2016} -\newcommand{\dueUni}{28.09.2016 16:00} +\newcommand{\issued}{09.03.2017} +\newcommand{\dueUni}{22.03.2017} %\newcommand{\version}{1} % if you need to release a corrected version, uncomment and increase this counter \newcommand{\dd}[1][]{\text{d}^{#1}} @@ -19,147 +19,92 @@ \startsheet -\setcounter{exercise}{0} -\exercise[3]{Conservation laws} +\setcounter{exercise}{1} +\exercise[4.5]{Conservation laws} -One other quantum number that we did not discuss is Izospin. +One other quantum number that we did not discuss is Izospin. The izospin is a symmetry that you can exchange the quarks $u$ and $d$ without changing the strong interactions. Historically it was introduced by Heisenberg to ''unify'' proton and neutron. Both of them have the izospin $I=\frac{1}{2}$. The 3rd component of the izospin are however different: $I_3 = \pm \frac{1}{2}$. The strong interactions conserve the $I$ and the $I_3$\footnote{This is valid up to quark mass effects.}. The EM interactions conserve the $I_3$ component but not the $I$. The weak iterations do not conserve both the $I$ and the $I_3$. -To install a virtual Linux machine on Linux, MacOS and Windows, follow these steps: -% -\begin{subtasks} -% -\task Follow this link \url{http://www.physik.uzh.ch/data/mchrzasz/Teaching/MC2016/}, download the zip file “VM\_MC2016.zip" and unzip the folder. -% -\task Follow this link \url{https://www.virtualbox.org/wiki/Downloads} and download {\it virtualbox} (select “VirtualBox 5.1.6 for OS X hosts"). -% -\task Install {\it virtualbox}, following the instructions of the installer. - \begin{center} - \begin{minipage}{0.45\textwidth} - \includegraphics[width=0.97 \textwidth]{virtual_box_1.png} - \end{minipage} - \begin{minipage}{0.45\textwidth} - \includegraphics[width=0.97 \textwidth]{virtual_box_3.png} - \end{minipage} - \end{center} -% -\task Open {\it virtualbox} and press the “New" button to start the installation of a new virtual machine. - \begin{center} - \begin{minipage}{0.45\textwidth} - \includegraphics[width=0.97 \textwidth]{instal_1.png} - \end{minipage} - \end{center} -% -\task Start the installation following the suggestions of the installer, then when required select “use an existing virtual hard disk file" and press "choose a virtual hard disk". - \begin{center} - \begin{minipage}{0.45\textwidth} - \includegraphics[width=0.97 \textwidth]{instal_2.png} - \end{minipage} - \begin{minipage}{0.45\textwidth} - \includegraphics[width=0.97 \textwidth]{instal_3.png} - \end{minipage} - \end{center} -% -\task Browse to the “VM\_MC2016" folder and select “Ubuntu 64-bit.vmdk". Then press “Create". -Note: please keep nonetheless all the other images present in the folder. - \begin{center} - \begin{minipage}{0.45\textwidth} - \includegraphics[width=0.97 \textwidth]{instal_4.png} - \end{minipage} - \begin{minipage}{0.45\textwidth} - \includegraphics[width=0.97 \textwidth]{instal_5.png} - \end{minipage} - \end{center} -% -\task If the installation go smooth, you should end up with a correctly installed virtual machine present in {\it virtualbox} main menu. - \begin{center} - \begin{minipage}{0.45\textwidth} - \includegraphics[width=0.97 \textwidth]{instal_6.png} - \end{minipage} - \end{center} -% -\task Once accessed to the virtual machine, these are the needed username and password: +In summary Tab.~\ref{tab:1} presents the quantum numbers and which iteration conserves them. + +\begin{table}[h!] \begin{center} -\begin{tabular}{l l} -username: & excellent student -\\ -password: & student + +\begin{tabular}{|l||c|c|c|} +\hline +Quantum Number & Strong int. & EM int. & Weak int. \\ \hline +Charge Q & Y & Y & Y\\ \hline +Baryon Number B & Y & Y & Y \\ \hline +Lepton number L & Y & Y & Y \\ \hline +Lepton family/flavour number $L_i$ & Y & Y & Y \\ \hline +Izospin I & Y & N & N \\ \hline +3rd component Izospin I$_3$ & Y & Y & N \\ \hline +S,C,B,T & Y & Y & N \\ \hline +Parity P & Y & Y & N \\ \hline +Charge conjugate C & Y & Y & N \\ \hline +CP, T & Y & Y & N $\mathcal{O}(10^{-3})$ \\ \hline \end{tabular} +\caption{Conservation of quantum numbers for different forces. Y=YES, N=NO.\label{tab:1}} \end{center} -% -\end{subtasks} +\end{table} - - - - - - - - -\turnpage - - -\exercise[3]{Floating point representation} - -Let us label the numbers in decimal representation with a subscript ‘$10$', in binary representation with a subscript ‘$2$', in hexadecimal representation with a subscript ‘$16$'. - -\begin{subtasks} - -\task Convert the following integer decimal numbers in hexadecimal and binary representations: - $12_{10}$, $53_{10}$, $123_{10}$, $431_{10}$. (0.75 Pt.) -%$12$, $53$, $123$, $431$. - -\task Convert the following binary and hexadecimal integer numbers in decimal representation: $10011_{2}$, $1101_{2}$, $A2_{16}$, $1AD_{16}$. (0.75 Pt.) - -\task In {\it binary16}, a number is represented using 16 digits in the following way: - \begin{center} - \includegraphics[width=0.25 \textwidth]{binary_half.png} - \end{center} -In some detail: +Based on the table above please indicate which forces is responsible for which interaction: \begin{itemize} -\item[-] The first bit represent the sign: $0 \leftrightarrow +$ and $1 \leftrightarrow -$. -\item[-] The exponent is represented with 5 bits. Thus it can be an integer number from 0 to 31. The values 0 (i.e.~00000) and 31 (i.e.~11111) are reserved for special numbers (NaN, infinity, subnormal numbers). -We are thus left with values from 1 to 30. \emph{By convention}, the exponent of the floating number is the number represented by these 5 digits \emph{minus} a bias, in this case 15 (example: if we want to store $2^{-3}$, the exponent is $-3$ and thus we should store $-3+15=12_{10}=01100_2$ in that slot). In this way we can store exponents from $-14$ (stored as $00001_2=1_{10}$) to $+15$ (stored as $11110_2=30_{10}$). -\item[-] The remaining 10 bits are left for the mantissa. \emph{By convention} (with the exception of subnormal numbers), the first significant digit (the one before the dot) is always 1 and is not stored. Thus, if the mantissa stored is $1000000000$, one should read it as $1.100_{2}=1.5_{10}$. +\item $\pi^- p \to \pi^- \pi^+ n$ +\item $\gamma p \to \pi^{+} n$ +\item $\nu_{\mu} n \to \mu^- p$ +\item $\pi^0 \to e^- e^- e^+ e^+$ +\item $p \bar{p} \to \pi^- \pi^+ \pi^0$ +\item $\tau^- \to \pi^- \nu_{\tau}$ +\item $D \to K^+ \pi^- \pi^+$ +\item $\pi^- \to \pi^0 e^- \nu_e$ +\item $\Lambda_0 p \to K^- p p$ \end{itemize} -% - -Try to convert into {\it binary16} the following decimal floating numbers: $0.3125_{10}$, $-431_{10}$. (0.5 Pt.) - -\task With respect to the results of the problem ({\it c}), convert to hexadecimal representation the two bytes representing your floating number and discuss the difference in the storage of your float between big endian and little endian. (0.5 Pt.) - -\task Discuss in some detail why, using a machine working with {\it half precision} (i.e.~{\it binary16}), one would get the following results: -\begin{center} - \begin{tabular}{l @{$\qquad\qquad\qquad$} l @{$\qquad\qquad\qquad$} l} - INPUT A: & INPUT B: & INPUT C: - \\ - float a ,b, c; & float a; & float a, b, c; - \\ - a = 2050; & a = 2050; & a = 2050; - \\ - b = 1; & for (int i=0; i<10; i++) & b = 0; - \\ - c = a+b; &~~~~~~ a = a+1; & for (int i=0; i<10; i++) - \\ - & &~~~~~~b = b+1; - \\ - OUTPUT A: & OUTPUT B: & c = a+b; - \\ - c = 2050; & a = 2050; & - \\ - & & OUTPUT C: - \\ - & & c = 2060; - \end{tabular} -\end{center} -% -Write down the binary representation of relevant intermediary results. (0.5 Pts.) -\end{subtasks} + +\setcounter{exercise}{2} +\exercise[2]{Conservation laws 2} + +Show that mezon that decays to a pair $\pi^+ \pi^-$ via strong interactions has $C=P= (-1)^J$, where J is the total angular momentum. + + +\setcounter{exercise}{3} +\exercise[4]{Conservation laws 3} + +We know that mezons $f_2(1275)$ $(J=2)$ and $\rho(769)$ $(J=1)$ are decaying via strong interactions to two charged pions. Which of the processes is inpossible in the EM interactions: $\rho \to \pi^0 \gamma$ or $f_2^0 \to \pi^0 \gamma$. Which of the decays is forbidden in all interactions: $\rho \to \pi^0 \pi^0$, $f_2 \to \pi^0 \pi^0$. + + +\setcounter{exercise}{4} +\exercise[2]{Conservation laws 4} + +We have proton and antiproton in the S state. Why reaction of $p \bar{p} \to \pi^0 \pi^0$ cannot proceed via strong interactions? + +\setcounter{exercise}{5} +\exercise[16]{Pion decay} + +Please calculate the matrix element of the dacay $\pi^- \to \mu \nu_{\mu}$. The form factor for the pion has the form of $F^{\mu} = p_{\mu} f_{\pi}$, where $f_{\pi}$ is so called pion decay constant and is calculated on lattice to be $f_{\pi}=130~\rm MeV$. Using the matrix element calculate the $\Gamma$. Calculate also the $\Gamma$ for the $\pi^- \to e \nu_{e}$. Why is the electron mode different then the muon one? + +\setcounter{exercise}{6} +\exercise[16]{Muon decay} + +Calculate the matrix element for the dacay of the muon: $\mu^- \to e^- \nu_{\mu} \bar{\nu_e}$. Using ''Golden Rule'' calculate the calculate the $\Gamma$ and the lifetime of the muon. + +\setcounter{exercise}{7} +\exercise[12]{Muon decay simulation} + +Please simulate the muon decay from exercise 6 using ROOT. Please assume for the moment flat phase space (aka matrix element =1). The example can be found: + +\url{https://root.cern.ch/root/html/tutorials/physics/PhaseSpace.C.html} + +having simulate this decay please calculate the electron energy in the muon central of mass and draw it for your simulated events. Simulate at least 100.000 events. + + + +\clearpage +\iffalse \begin{solution} \begin{subtasks} @@ -441,7 +386,7 @@ - +\fi \end{document}