#LyX 2.5 created this file. For more info see https://www.lyx.org/ \lyxformat 643 \begin_document \begin_header \save_transient_properties true \origin unavailable \textclass article \use_default_options true \maintain_unincluded_children no \language spanish \language_package default \inputencoding utf8 \fontencoding auto \font_roman "default" "default" \font_sans "default" "default" \font_typewriter "default" "default" \font_math "auto" "auto" \font_default_family sfdefault \use_non_tex_fonts false \font_sc false \font_roman_osf false \font_sans_osf false \font_typewriter_osf false \font_sf_scale 100 100 \font_tt_scale 100 100 \use_microtype false \use_dash_ligatures true \graphics default \default_output_format default \output_sync 0 \bibtex_command default \index_command default \float_placement class \float_alignment class \paperfontsize default \spacing single \use_hyperref true \pdf_title "Actividad 2" \pdf_author "Jesús María Mora Mur" \pdf_subject "Mecánica Clásica" \pdf_bookmarks true \pdf_bookmarksnumbered false \pdf_bookmarksopen false \pdf_bookmarksopenlevel 1 \pdf_breaklinks false \pdf_pdfborder false \pdf_colorlinks false \pdf_backref false \pdf_pdfusetitle true \papersize default \use_geometry false \use_package amsmath 1 \use_package amssymb 1 \use_package cancel 1 \use_package esint 1 \use_package mathdots 1 \use_package mathtools 1 \use_package mhchem 1 \use_package stackrel 1 \use_package stmaryrd 1 \use_package undertilde 1 \cite_engine basic \cite_engine_type default \biblio_style plain \use_bibtopic false \use_indices false \paperorientation portrait \suppress_date false \justification default \crossref_package refstyle \use_formatted_ref 0 \use_minted 0 \use_lineno 0 \backgroundcolor none \fontcolor none \notefontcolor lightgray \boxbgcolor red \table_border_color default \table_odd_row_color default \table_even_row_color default \table_alt_row_colors_start 1 \index Index \shortcut idx \color #008000 \end_index \secnumdepth 3 \tocdepth 3 \paragraph_separation indent \paragraph_indentation default \is_math_indent 0 \math_numbering_side default \quotes_style english \dynamic_quotes 0 \papercolumns 1 \papersides 1 \paperpagestyle default \tablestyle default \tracking_changes false \output_changes false \change_bars false \postpone_fragile_content true \html_math_output 0 \html_css_as_file 0 \html_be_strict false \docbook_table_output 0 \docbook_mathml_prefix 1 \docbook_mathml_version 0 \end_header \begin_body \begin_layout Title Actividad 2: Boletín de problemas general. \end_layout \begin_layout Author Jesús María Mora Mur \end_layout \begin_layout Date \begin_inset ERT status open \begin_layout Plain Layout \backslash today \end_layout \end_inset \end_layout \begin_layout Section Primer ejercicio. \end_layout \begin_layout Standard Las ecuaciones de Euler-Lagrange son las siguientes para las coordenadas generalizadas: \end_layout \begin_layout Standard \begin_inset Formula \[ q_{i}=\left\{ x_{1},y_{1},x_{2},y_{2}\right\} \] \end_inset \end_layout \begin_layout Standard \begin_inset Formula \[ \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\partial\mathcal{L}}{\partial\dot{q_{i}}}\right)-\frac{\partial\mathcal{L}}{\partial q_{i}}=0 \] \end_inset \end_layout \begin_layout Standard obteniendo las 4 siguientes: \end_layout \begin_layout Standard \begin_inset Formula \[ m_{1}\ddot{x_{1}}+\frac{\partial V}{\partial r}\cdot\frac{x_{1}-x_{2}}{r}=0 \] \end_inset \begin_inset Formula \[ m_{1}\ddot{y_{1}}+\frac{\partial V}{\partial r}\cdot\frac{y_{1}-y_{2}}{r}=0 \] \end_inset \begin_inset Formula \[ m_{1}\ddot{x_{2}}-\frac{\partial V}{\partial r}\cdot\frac{x_{1}-x_{2}}{r}=0 \] \end_inset \begin_inset Formula \[ m_{1}\ddot{y_{2}}-\frac{\partial V}{\partial r}\cdot\frac{y_{1}-y_{2}}{r}=0 \] \end_inset \end_layout \begin_layout Standard Al ser el potencial \begin_inset Formula $\mathrm{V}(r)$ \end_inset la única parte de la función lagrangiana dependiente de las coordenadas no derivadas. \end_layout \begin_layout Section Segundo ejercicio. \end_layout \begin_layout Standard Necesitamos en primer lugar los momentos conjugados: \end_layout \begin_layout Standard \begin_inset Formula \[ \begin{cases} p_{1x}=\frac{\partial\mathcal{L}}{\partial\dot{x_{1}}}=m_{1}\dot{x_{1}}\\ p_{2x}=\frac{\partial\mathcal{L}}{\partial\dot{x_{2}}}=m_{2}\dot{x_{2}}\\ p_{1y}=\frac{\partial\mathcal{L}}{\partial\dot{y_{1}}}=m_{1}\dot{y_{1}}\\ p_{2y}=\frac{\partial\mathcal{L}}{\partial\dot{y_{2}}}=m_{2}\dot{y_{2}} \end{cases} \] \end_inset \end_layout \begin_layout Standard Formamos el hamiltoniano despajando las velocidades: \end_layout \begin_layout Standard \begin_inset Formula \[ \mathcal{H}=\sum p_{i}\cdot q_{i}-\mathcal{L}=\frac{p^{2}_{1x}+p^{2}_{1y}}{2m_{1}}+\frac{p^{2}_{2x}+p^{2}_{2y}}{2m_{2}}+V(r) \] \end_inset \end_layout \begin_layout Standard Las ecuaciones de Hamilton quedan de la forma siguiente: \end_layout \begin_layout Standard \begin_inset Formula \[ \begin{cases} \dot{x_{1}}=\frac{\partial\mathcal{H}}{\partial p_{1x}}=\frac{p_{1x}}{m_{1}}\\ \dot{x_{2}}=\frac{\partial\mathcal{H}}{\partial p_{2x}}=\frac{p_{2x}}{m_{2}}\\ \dot{y_{1}}=\frac{\partial\mathcal{H}}{\partial p_{1y}}=\frac{p_{1y}}{m_{1}}\\ \dot{y_{2}}=\frac{\partial\mathcal{H}}{\partial p_{2y}}=\frac{p_{2y}}{m_{2}} \end{cases} \] \end_inset \end_layout \begin_layout Standard para las velocidades. Por otro lado, las fuerzas son: \end_layout \begin_layout Standard \begin_inset Formula \[ \begin{cases} \dot{p_{1x}}=-\frac{\partial\mathcal{H}}{\partial x_{1}}=-\frac{dV}{dr}\frac{x_{1}-x_{2}}{r}\\ \dot{p_{2x}}=-\frac{\partial\mathcal{H}}{\partial x_{2}}=\frac{dV}{dr}\frac{x_{1}-x_{2}}{r}\\ \dot{p_{1y}}=-\frac{\partial\mathcal{H}}{\partial y_{1}}=-\frac{dV}{dr}\frac{y_{1}-y_{2}}{r}\\ \dot{p_{2y}}=-\frac{\partial\mathcal{H}}{\partial y_{2}}=\frac{dV}{dr}\frac{y_{1}-y_{2}}{r} \end{cases} \] \end_inset \end_layout \begin_layout Standard Utilizando el mismo procedimiento que en el ejercicio anterior. \end_layout \begin_layout Section Tercer ejercicio. \end_layout \begin_layout Standard Para calcular que son constantes de movimiento veremos como la derivada temporal para cada magnitud se anula: \end_layout \begin_layout Standard \begin_inset Formula \[ \frac{dP_{x}}{dt}=\dot{p_{1x}}+\dot{p_{2x}}=\frac{dV}{dr}\frac{x_{1}-x_{2}}{r}-\frac{dV}{dr}\frac{x_{1}-x_{2}}{r}=0 \] \end_inset \end_layout \begin_layout Standard \begin_inset Formula \[ \frac{dP_{y}}{dt}=\dot{p_{1y}}+\dot{p_{2y}}=\frac{dV}{dr}\frac{y_{1}-y_{2}}{r}-\frac{dV}{dr}\frac{y_{1}-y_{2}}{r}=0 \] \end_inset \end_layout \begin_layout Standard \begin_inset Formula \[ \frac{dK_{x}}{dt}=m_{1}\dot{x_{1}}+m_{2}\dot{x_{2}}-P_{x}-t\frac{dP_{x}}{dt}=P_{x}-P_{x}-0=0 \] \end_inset \begin_inset Formula \[ \frac{dK_{y}}{dt}=m_{1}\dot{y_{1}}+m_{2}\dot{y_{2}}-P_{y}-t\frac{dP_{y}}{dt}=P_{y}-P_{y}-0=0 \] \end_inset \end_layout \begin_layout Standard Conservándose pues las magnitudes. \end_layout \begin_layout Section Cuarto ejercicio \end_layout \begin_layout Standard El corchete de Poisson se calcula como sigue para dos funciones \begin_inset Formula $f$ \end_inset y \begin_inset Formula $g$ \end_inset dependientes de las coordenadas generalizadas \begin_inset Formula $q$ \end_inset y \begin_inset Formula $p$ \end_inset : \end_layout \begin_layout Standard \begin_inset Formula \[ \left\{ f,g\right\} =\frac{\partial f}{\partial q_{i}}\frac{\partial g}{\partial p_{i}}-\frac{\partial f}{\partial p_{i}}\frac{\partial g}{\partial q_{i}} \] \end_inset \end_layout \begin_layout Standard Para los casos que nos ocupan: \end_layout \begin_layout Standard \begin_inset Formula \[ \left\{ K_{x},P_{x}\right\} =\left\{ m_{1}x_{1}+m_{2}x_{2},p_{1x}+p_{2x}\right\} \] \end_inset \end_layout \begin_layout Standard eliminando el término temporal de \begin_inset Formula $K_{x}$ \end_inset al anularse el corchete de Poisson \begin_inset Formula $\left\{ P_{x},P_{x}\right\} $ \end_inset . Aplicando propiedades: \end_layout \begin_layout Standard \begin_inset Formula \[ \left\{ K_{x},P_{x}\right\} =m_{1}\left\{ x_{1},p_{1x}+p_{2x}\right\} +m_{2}\left\{ x_{2},p_{1x}+p_{2x}\right\} \] \end_inset \end_layout \begin_layout Standard Ambos corchetes dan 1, por lo que el resultado es \begin_inset Formula $m_{1}+m_{2}$ \end_inset . \end_layout \begin_layout Standard Para el siguiente caso, \begin_inset Formula $\left\{ K_{y},\mathcal{H}\right\} $ \end_inset , eliminando el término temporal de \begin_inset Formula $K_{y}$ \end_inset : \end_layout \begin_layout Standard \begin_inset Formula \[ \left\{ K_{y},\mathcal{H}\right\} =\left\{ m_{1}y_{1}+m_{2}y_{2},\mathcal{H}\right\} =m_{1}\left\{ y_{1},\mathcal{H}\right\} +m_{2}\left\{ y_{2},\mathcal{H}\right\} =m_{1}\dot{y_{1}}+m_{2}\dot{y_{2}}=P_{y} \] \end_inset \end_layout \begin_layout Standard Para el último caso, \begin_inset Formula $\left\{ J,P_{y}\right\} $ \end_inset , consideramos que \begin_inset Formula $J=x_{1}p_{1y}+x_{2}p_{2y}-y_{1}p_{1x}-y_{2}p_{2x}$ \end_inset \end_layout \begin_layout Standard Los términos en \begin_inset Formula $y$ \end_inset de los momentos no contribuyen, así que se simplifica el corchete a: \end_layout \begin_layout Standard \begin_inset Formula \[ \left\{ y_{i},P_{y}\right\} p_{ix} \] \end_inset \end_layout \begin_layout Standard Como el corchete en cuestión da \begin_inset Formula $1$ \end_inset , la suma dará el momento total \begin_inset Formula $P_{x}$ \end_inset . \end_layout \end_body \end_document