269 lines
7.9 KiB
TeX
269 lines
7.9 KiB
TeX
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\documentclass[12pt]{beamer}
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\usetheme{Copenhagen}
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\usepackage[utf8]{inputenc}
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\usepackage[spanish]{babel}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{graphicx}
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\usepackage{tikz}
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\usepackage{appendixnumberbeamer}
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%\setbeamerfont{page number in head}{size=\large}
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%\setbeamertemplate{footline}{Diapositiva}
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\setbeamertemplate{footline}[frame number]
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\newcommand{\backupbegin}{
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\newcounter{finalframe}
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\setcounter{finalframe}{\value{framenumber}}
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}
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\newcommand{\backupend}{
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\setcounter{framenumber}{\value{finalframe}}
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}
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\author{Física Nuclear y subnuclear }
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\title{Introducción}
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%\setbeamercovered{transparent}
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%\setbeamertemplate{navigation symbols}{}
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%\logo{}
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%\institute{}
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%\date{}
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%\subject{}
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\begin{document}
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\begin{frame}
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\titlepage
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\end{frame}
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%\begin{frame}{Contenido}
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% \tableofcontents
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%\end{frame}
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\begin{frame}{Comparando unidades}
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\begin{itemize}
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\item Longitud de Plank: $1.6162\times 10^{-35} m$ \footnote{\url{https://physics.nist.gov/cgi-bin/cuu/Value?plkl}}
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\item Radio de un cuark: $\leq 1\times 10^{-18}m$
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\item Radio nuclear: $\approx 1\times 1\times 10^{-15}m $
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\item Radio del átomo: $\approx 1\times 10^{-10}m$
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\item Grosor de un cabello: $\approx 8\times 10^{-5}m$
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\end{itemize}
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\end{frame}
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\begin{frame}{Prefijos para magnitudes}
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\begin{table}[ht!]
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\begin{tabular}{|lll|lll|}
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\hline
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Potencia & Nombre & Símbolo & Potencia & Nombre & Símbolo \\
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\hline
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$10^1$ & deca & da & $10^{-1}$ & deci & d \\
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$10^2$ & hecto & h & $10^{-2}$ & centi & c \\
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$10^3$ & kilo & k & $10^{-3}$ & mili & m \\
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$10^6$ & mega & M & $10^{-6}$ & micro & $\mu$ \\
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$10^9$ & giga & G & $10^{-9}$ & nano & n \\
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$10^{12}$ & tera & T & $10^{-12}$ & pico & p \\
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$10^{15}$ & pate & P & $10^{-15}$ & femto & f \\
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$10^{18}$ & exa & E & $10^{-18}$ & atto & a \\
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\hline
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\end{tabular}
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\end{table}
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\end{frame}
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\begin{frame}{Unidades}
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\begin{table}[ht!]
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\begin{tabular}{lll}
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Cantidad & Unidad & Abreviatura \\
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Longitud & metro & $m$ \\
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Tiempo & segundos & $s$ \\
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Energía & electron volts & $eV$ \\
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Masa & & $eV/c^2$ \\
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Momento & & $eV/c$
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\end{tabular}
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\end{table}
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\end{frame}
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\begin{frame}{¿$eV/c$ y $eV/c^2$?}
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\begin{itemize}
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\item $1 eV = 1.6\times 10^{-19}J$
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\item $E^2 = p^2c^2 + m^2c^4$
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\end{itemize}
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\end{frame}
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\begin{frame}{Propiedades relativistas}
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Partículas dentro del formalismo cuántico y relativista
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\begin{align*}
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p =& \gamma mv \\
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E^2 =& p^2c^2 + m^2c^4
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\end{align*}
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\end{frame}
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\begin{frame}{Propiedades relativistas II}
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\begin{align*}
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E^2=& {\gamma}^2m^2v^2c^2 + m^2c^4 \\
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=& {\gamma}^2m^2(\frac{v^2}{c^2})c^4 + m^2c^4 \\
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=& {\gamma}^2m^2{\beta}^2c^4 + m^2c^4 \\
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=& ({\gamma}^2{\beta}^2 + 1)m^2c^4 \\
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=& {\gamma}^2 m^2 c^4
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\end{align*}
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\end{frame}
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\begin{frame}{Dispersión de Rutherford}
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\begin{figure}[ht!]
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\begin{center}
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\includegraphics[width=0.7\linewidth]{rutherford.jpg}
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\caption{Arreglo experimental para la dispersión de Rutherford. Imagen adaptada a partir de \href{https://commons.wikimedia.org/w/index.php?curid=36736367}{``File:Peliculafinadeouro.jpg''} por \href{https://commons.wikimedia.org/w/index.php?title=User:Costa_Isa_14&action=edit&redlink=1}{Costa Isa 14} con una licencia \href{https://creativecommons.org/licenses/by-sa/4.0?ref=ccsearch&atype=rich}{CC BY-SA 4.0}}
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\label{fig:rute}
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\end{center}
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\end{figure}
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\end{frame}
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\begin{frame}{Cinemática clásica}
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\begin{align*}
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\frac{1}{2}m_{\alpha}v_0^2 =& \frac{1}{2}m_{\alpha}v_{\alpha}^2 + \frac{1}{2} m_t v_t^2 \\
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v_0^2 =& v_{\alpha}^2 + \frac{m_t}{m_{\alpha}}v_t^2
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\end{align*}
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Usando $m_{\alpha}v_0 = m_{\alpha}v_{\alpha} + m_t v_t$
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\begin{equation*}
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v_t^2 \left( 1-\frac{m_t}{m_{\alpha}} \right) = 2\overrightarrow{v_{\alpha}} \cdot \overrightarrow{v_t}.
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\end{equation*}
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\end{frame}
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\begin{frame}{¿Qué nos está haciendo falta?}
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Interación:
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\begin{equation*}
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V(r)=\frac{ZZ'e^2}{r}
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\end{equation*}
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\begin{figure}[ht!]
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\begin{center}
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\includegraphics[width=0.7\linewidth]{dispersion.eps}
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\label{fig:disp}
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\end{center}
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\end{figure}
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\end{frame}
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\begin{frame}{Analizando}
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Imaginemos muy lejos:
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\begin{align*}
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E =& \frac{1}{2}mv_0^2 \notag \\
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v_0 =& \sqrt{\frac{2E}{m}}
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\end{align*}
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Conservación de momento angular
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\begin{align*}
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\ell =& m v_0 b \\
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\frac{d\omega}{dt} =& \frac{\ell}{mr^2}
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\end{align*}
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\end{frame}
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\begin{frame}{Energía total}
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\begin{align*}
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E =& \frac{1}{2}m {\left( \frac{dr}{dt} \right)}^2 + \frac{1}{2}m r^2 {\left( \frac{d\omega}{dt} \right)}^2 + V(r) \notag \\
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=& \frac{1}{2}m {\left( \frac{dr}{dt} \right)}^2 + \frac{1}{2}m r^2 {\left( \frac{\ell}{mr^2} \right)}^2 + V(r) \notag \\
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\frac{dr}{dt} =& -\left[ \frac{2}{m}\left( E-V(r)-\frac{\ell^2}{2mr^2}\right) \right]^{\frac{1}{2}}
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\end{align*}
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\end{frame}
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\begin{frame}{Velocidad radial}
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Introducimos la $\ell$ en términos del parámetro de impacto
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\begin{equation*}
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\frac{dr}{dt} = -\frac{\ell}{mrb}\left[ r^2\left( 1-\frac{V(r)}{E} \right) -b^2\right]^{\frac{1}{2}}
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\end{equation*}
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Manipulando la velocidad angular
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\begin{align}
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d\omega =& \frac{\ell}{mr^2}dt = \frac{\ell}{mr^2}\frac{dt}{dr}dr \notag \\
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=& -\frac{\ell}{mr^2}\frac{dr}{\frac{\ell}{mrb}\left[ r^2\left( 1-\frac{V(r)}{E}\right) -b^2\right]^{\frac{1}{2}}} \notag \\
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=& -\frac{bdr}{r\left[ r^2\left( 1-\frac{V(r)}{E}\right)-b^2\right]^{\frac{1}{2}}}
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\label{ec:chanch2}
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\end{align}
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\end{frame}
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\begin{frame}{Ya casi}
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Metemos la física al integrar
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\begin{align}
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\int_0^{\omega_0} d\omega =& -\int_{\infty}^{r_0} \frac{bdr}{r\left[ r^2\left( 1-\frac{V(r)}{E}\right)-b^2\right]^{\frac{1}{2}}} \notag \\
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\omega_0 =& b \int_{r_0}^{\infty} \frac{dr}{r\left[ r^2\left( 1-\frac{V(r)}{E}\right)-b^2\right]^{\frac{1}{2}}}
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\label{ec:intom}
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\end{align}
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El púnto de mínima distancia, donde la $\frac{dr}{dt}$ se hace cero:
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\begin{align}
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E-V(r)-\frac{\ell^2}{2mr^2} =& 0 \notag \\
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r^2\left( 1-\frac{V(r)}{E} \right) -b^2 =& 0
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\end{align}
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\end{frame}
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\begin{frame}{El final}
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Haciendo un cambio de variable e integral
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\begin{align}
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\omega_0 =& b \int_{r_0}^{\infty} \frac{dr}{r\left[ r^2\left( 1-\frac{ZZ'e^2}{Er}\right)-b^2\right]^{\frac{1}{2}}} \notag \\
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\theta = \pi - 2\omega =& \pi - 2b \int_{r_0}^{\infty} \frac{dr}{r\left[ r^2\left( 1-\frac{ZZ'e^2}{Er}\right)-b^2\right]^{\frac{1}{2}}},
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\end{align}
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Llegaremos a un término
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\begin{equation*}
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b = \frac{ZZ'e^2}{2E}cot\frac{\theta}{2}
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\end{equation*}
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\end{frame}
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\begin{frame}{Sección eficaz I}
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\begin{itemize}
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\item No es una sola partícula, son un bonche
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\item Densidad de partículas $N_0$ ($\frac{part.}{tiempo \times \text{área}}$)
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\item Parámetro de impacto de $b$ a $b+db$
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\item Dispersadas de $\theta$ a $\theta-d\theta$
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\item Ángulo sólido $2\pi N_0bdb$ (part. dispersadas/ tiempo)
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\item $\Delta \sigma = 2\pi bdb$
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\end{itemize}
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\end{frame}
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\begin{frame}{Sección eficaz}
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\begin{align*}
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\Delta \sigma(\theta,\phi) =& b\ db\ d\phi \notag \\
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\Delta \sigma(\theta,\phi) =& -\frac{d\sigma}{d\Omega} (\theta,\phi) d\Omega = -\frac{d\sigma}{d\Omega}(\theta,\phi) sen\theta d\theta d\phi.
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\end{align*}
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Se llega
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\begin{equation*}
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\frac{d\sigma}{d\Omega} (\theta) = \left( \frac{ZZ'e^2}{4E} \right)^2 \frac{1}{sen^4 \theta}
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\end{equation*}
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\end{frame}
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\begin{frame}{Camino libre medio}
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\newtheorem{defi}{Definición}
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\begin{defi}
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El camino libre medio $\lambda$ es la distancia promedio que viaja una partícula entre colisiones dentro de un medio material.
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\end{defi}
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\begin{equation*}
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\lambda = \frac{1}{n\sigma}
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\end{equation*}
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Coeficiente de atenuación
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\begin{equation*}
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\mu = n\sigma
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\end{equation*}
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\end{frame}
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\backupbegin
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\section*{Apéndices}
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\begin{frame}[noframenumbering]{}
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\end{frame}
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\backupend
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\end{document}
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