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Relativistic electromagnetism problem

  1. Jun 11, 2007 #1
    1. The problem statement, all variables and given/known data


    Hi to everyone! I would like you to help me for a problem of classical electrodynamics. I have to study the relativistic motion of a charged particle in a coulombian field with center in the origin of the cartesian axes. I have to study the case in which the initial velocity is purely radial.

    Can anyone give me some advices, or links where I can find this problem performed?

    Thank you very much!

    C.


    2. Relevant equations



    3. The attempt at a solution
    I'm sorry,but I have only the .tex version...if you can compile it...

    {}

    \documentclass[a4paper,8pt]{report}
    \usepackage[latin1]{inputenc}
    \usepackage[italian]{babel}
    \usepackage[dvips]{graphicx}
    \usepackage{vmargin}
    \setpapersize{A4}
    \setmarginsrb{22mm}{25mm}{22mm}{25mm}%
    {0mm}{10mm}{0mm}{10mm}


    \begin{document}











    \section{Secondo esercizio}

    \vspace{1cm}
    \Large{\bfseries{TESTO}}
    Considerare il moto (relativistico) di una particella carica in un campo coulombiano con centro nell'origine del sistema di riferimento. Studiare il caso in cui la velocità iniziale della particella è puramente radiale. Calcolare la legge oraria del moto. Illustrare il moto con grafici per diverse condizioni iniziali significative.


    \vspace{1cm}
    \Large{\bfseries{SVOLGIMENTO}}

    \vspace{0.5cm}



    Utilizziamo per lo svolgimento dell'esercizio il formalismo di Hamilton-Jacobi. L'Hamiltoniana relativistica del sistema in esame è:

    \begin{equation}
    \epsilon = \sqrt{c^{2}p^{2} + m^{2}c^{4}} + \frac{\alpha}{r} \ \ \ \ \ \ \ \ \ {\bf p}=m\gamma{\bf v}
    \end{equation}

    \begin{itemize}
    \item $\alpha > 0 \ \ \ \rightarrow \ \ \ $campo repulsivo
    \item $\alpha < 0 \ \ \ \rightarrow \ \ \ $campo attrattivo
    \end{itemize}

    Essendo la forza centrale, il moto avviene su un piano, e il momento angolare $p_{\varphi}$ viene conservato; infatti, se $\frac{d{\bf p}}{dt} = \frac{\alpha}{r^{2}}{\bf r}$ si ha:

    \vspace{5mm}
    $
    \left\{\begin{array}{l}
    \frac{\alpha}{r^{2}}{\bf r} \times {\bf r} \ = \ \frac{d{\bf p}}{dt} \times {\bf r} = 0 \\[2mm]
    {\bf p}\times \frac{d{\bf r}}{dt} \ =\ {\bf p}\times {\bf v}\ =\ 0
    \end{array}\right.
    $
    \vspace{5mm}

    e quindi\ $\frac{d}{dt}({\bf r}\times {\bf p})=0$.


    Scegliamo un sistema di riferimento in coordinate polari, \\

    $
    \left\{\begin{array}{l}
    x=r\cos\varphi \\[2mm]
    y = r\sin\varphi
    \end{array}\right.
    $

    \begin{equation}
    v^{2}=\dot{x}^{2}+\dot{y}^{2}= \dot{r}^{2} + r^{2}\dot{\varphi}^{2}
    \end{equation}

    La lagrangiana relativistica è:

    \begin{equation}
    \mathcal{L} = -mc^{2}\sqrt{1-\frac{v^{2}}{c^{2}}}-\frac{\alpha}{r} \ = \ -mc^{2}\sqrt{1-\frac{\dot{r}^{2}}{c^{2}}-\frac{r^{2}\dot{\varphi}^{2}}{c^{2}}} -\frac{\alpha}{r}
    \end{equation}


    I momenti possono essere calcolati facilmente dalla definizione:

    \begin{eqnarray}
    p_{r} \equiv \frac{\partial\mathcal{L}}{\partial\dot{r}}=m\gamma\dot{r}\\
    p_{\varphi} \equiv \frac{\partial\mathcal{L}}{\partial\dot{\varphi}}=m\gamma r^{2}\dot{\varphi}
    \end{eqnarray}

    Così otteniamo l'energia:

    \begin{equation}
    \epsilon =c\sqrt{m^{2}\gamma^{2}(\dot{r}^{2} + r^{2}\dot{\varphi}^{2} + m^{2}c^{2})} + \frac{\alpha}{r} = \mathcal{H}
    \end{equation}

    \begin{equation}
    \epsilon =c\sqrt{p^{r}_{2}+\frac{M^{2}}{r^{2}}+m^{2}c^{2}}+ \frac{\alpha}{r}
    \end{equation}

    Nel nostro caso la velocità iniziale è radiale, perciò si ha $M=0$. E' possibile adesso scrivere l'equazione di Hamilton-Jacobi



    \begin{equation}
    \frac{\partial S}{\partial t}+\mathcal{H}\left(r,\varphi,\frac{\partial S}{\partial r},\frac{\partial S}{\partial\varphi}\right)=0
    \end{equation}
    \begin{equation}
    \frac{\partial S}{\partial t}+c\sqrt{\left(\frac{\partial S}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial S}{\partial\varphi}\right)^{2}+m^{2}c^{2}}+\frac{\alpha}{r} =0
    \end{equation}

    $$\left(\frac{\partial S}{\partial t}+\frac{\alpha}{r}\right)^{2}=c^{2}\left[\left(\frac{\partial S}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial S}{\partial\varphi}\right)^{2}+m^{2}c^{2}\right]$$

    Poichè in ogni caso $\frac{\partial S}{\partial\varphi}=p_{\varphi}$ è costante, e poichè l'hamiltoniana non dipende direttamente dal tempo, la dipendenza di $S$ da $t$e da $\varphi$ è molto semplice:

    \begin{equation}
    S=-\epsilon t + M\varphi + f(r)
    \end{equation}

    $$\left(-\epsilon+\frac{\alpha}{r}\right)^{2} = c^{2}\left[\left(\frac{\partial f}{\partial r}\right)^{2}+\frac{M^{2}}{r^{2}}+m^{2}c^{2}\right]$$

    $$\frac{\partial f}{\partial r} = \pm\sqrt{\frac{1}{c^{2}}\left(-\epsilon+\frac{\alpha}{r}\right)^{2} - \frac{M^{2}}{r^{2}}- m^{2}c^{2}}
    $$

    Quindi, nel nostro caso:

    \begin{eqnarray}
    f(r)=\pm \int dr \sqrt{\frac{1}{c^{2}}\left(\epsilon - \frac{\alpha}{r}\right)^{2} - m^{2}c^{2}}\\
    \Rightarrow S=-\epsilon t \pm \int dr \sqrt{\frac{1}{c^{2}}\left(\epsilon - \frac{\alpha}{r}\right)^{2} - m^{2}c^{2}}
    \end{eqnarray}

    Adesso si può procedere alla soluzione delle equazioni $\beta_{i}=\frac{\partial S}{\partial\alpha_{i}}= cost$, che nel nostro caso si riducono all'unica equazione:


    \begin{equation}
    \frac{\partial S}{\partial\epsilon} \ =\ -t \pm\int dr\frac{\epsilon - \frac{\alpha}{r}}{\sqrt{\frac{1}{c^{2}}\left(\epsilon - \frac{\alpha}{r}\right)^{2} - m^{2}c^{2}}} \ =\ cost
    \end{equation}


    Questa equazione può essere risolta utilizzando l'identità:

    \begin{equation}
    \int dx \frac{1}{\sqrt{ax^{2}+bx +c}} = \frac{1}{\sqrt{a}}\log\left|2ax + b + 2 \sqrt{a(ax^{2}+bx +c)}\right| + cost
    \end{equation}











    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

    \end{document}
     
  2. jcsd
  3. Jun 12, 2007 #2

    StatusX

    User Avatar
    Homework Helper

    You apparently 1) speak english and 2) know english is what most people on this board speak (exclusively), so if you really want help, why not translate that?
     
  4. Jun 13, 2007 #3
    I thought that in that solutions the spoken parts are not necessary to understand it,because mathematical formulas are quite clear...sorry!
     
  5. Jun 13, 2007 #4

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [itex]\TeX[/itex]ified... (click to see how I enclosed the tex-source with [ tex ] [ /tex ] (no spaces))

    [tex]
    \section{Secondo esercizio}

    \vspace{1cm}
    \Large{\bfseries{TESTO}}
    Considerare il moto (relativistico) di una particella carica in un campo coulombiano con centro nell'origine del sistema di riferimento. Studiare il caso in cui la velocità iniziale della particella è puramente radiale. Calcolare la legge oraria del moto. Illustrare il moto con grafici per diverse condizioni iniziali significative.


    \vspace{1cm}
    \Large{\bfseries{SVOLGIMENTO}}

    \vspace{0.5cm}



    Utilizziamo per lo svolgimento dell'esercizio il formalismo di Hamilton-Jacobi. L'Hamiltoniana relativistica del sistema in esame è:

    \begin{equation}
    \epsilon = \sqrt{c^{2}p^{2} + m^{2}c^{4}} + \frac{\alpha}{r} \ \ \ \ \ \ \ \ \ {\bf p}=m\gamma{\bf v}
    \end{equation}

    \begin{itemize}
    \item $\alpha > 0 \ \ \ \rightarrow \ \ \ $campo repulsivo
    \item $\alpha < 0 \ \ \ \rightarrow \ \ \ $campo attrattivo
    \end{itemize}

    Essendo la forza centrale, il moto avviene su un piano, e il momento angolare $p_{\varphi}$ viene conservato; infatti, se $\frac{d{\bf p}}{dt} = \frac{\alpha}{r^{2}}{\bf r}$ si ha:

    \vspace{5mm}
    $
    \left\{\begin{array}{l}
    \frac{\alpha}{r^{2}}{\bf r} \times {\bf r} \ = \ \frac{d{\bf p}}{dt} \times {\bf r} = 0 \\[2mm]
    {\bf p}\times \frac{d{\bf r}}{dt} \ =\ {\bf p}\times {\bf v}\ =\ 0
    \end{array}\right.
    $
    \vspace{5mm}

    e quindi\ $\frac{d}{dt}({\bf r}\times {\bf p})=0$.


    Scegliamo un sistema di riferimento in coordinate polari, \\

    $
    \left\{\begin{array}{l}
    x=r\cos\varphi \\[2mm]
    y = r\sin\varphi
    \end{array}\right.
    $

    \begin{equation}
    v^{2}=\dot{x}^{2}+\dot{y}^{2}= \dot{r}^{2} + r^{2}\dot{\varphi}^{2}
    \end{equation}

    La lagrangiana relativistica è:

    \begin{equation}
    \mathcal{L} = -mc^{2}\sqrt{1-\frac{v^{2}}{c^{2}}}-\frac{\alpha}{r} \ = \ -mc^{2}\sqrt{1-\frac{\dot{r}^{2}}{c^{2}}-\frac{r^{2}\dot{\varphi}^{2}}{c^{2}}} -\frac{\alpha}{r}
    \end{equation}


    I momenti possono essere calcolati facilmente dalla definizione:

    \begin{eqnarray}
    p_{r} \equiv \frac{\partial\mathcal{L}}{\partial\dot{r}}=m\gamm a\dot{r}\\
    p_{\varphi} \equiv \frac{\partial\mathcal{L}}{\partial\dot{\varphi}}= m\gamma r^{2}\dot{\varphi}
    \end{eqnarray}

    Così otteniamo l'energia:

    \begin{equation}
    \epsilon =c\sqrt{m^{2}\gamma^{2}(\dot{r}^{2} + r^{2}\dot{\varphi}^{2} + m^{2}c^{2})} + \frac{\alpha}{r} = \mathcal{H}
    \end{equation}

    \begin{equation}
    \epsilon =c\sqrt{p^{r}_{2}+\frac{M^{2}}{r^{2}}+m^{2}c^{2}}+ \frac{\alpha}{r}
    \end{equation}

    Nel nostro caso la velocità iniziale è radiale, perciò si ha $M=0$. E' possibile adesso scrivere l'equazione di Hamilton-Jacobi



    \begin{equation}
    \frac{\partial S}{\partial t}+\mathcal{H}\left(r,\varphi,\frac{\partial S}{\partial r},\frac{\partial S}{\partial\varphi}\right)=0
    \end{equation}
    \begin{equation}
    \frac{\partial S}{\partial t}+c\sqrt{\left(\frac{\partial S}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial S}{\partial\varphi}\right)^{2}+m^{2}c^{2}}+\frac{\ alpha}{r} =0
    \end{equation}

    $$\left(\frac{\partial S}{\partial t}+\frac{\alpha}{r}\right)^{2}=c^{2}\left[\left(\frac{\partial S}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial S}{\partial\varphi}\right)^{2}+m^{2}c^{2}\right]$$

    Poichè in ogni caso $\frac{\partial S}{\partial\varphi}=p_{\varphi}$ è costante, e poichè l'hamiltoniana non dipende direttamente dal tempo, la dipendenza di $S$ da $t$e da $\varphi$ è molto semplice:

    \begin{equation}
    S=-\epsilon t + M\varphi + f(r)
    \end{equation}

    $$\left(-\epsilon+\frac{\alpha}{r}\right)^{2} = c^{2}\left[\left(\frac{\partial f}{\partial r}\right)^{2}+\frac{M^{2}}{r^{2}}+m^{2}c^{2}\right]$$

    $$\frac{\partial f}{\partial r} = \pm\sqrt{\frac{1}{c^{2}}\left(-\epsilon+\frac{\alpha}{r}\right)^{2} - \frac{M^{2}}{r^{2}}- m^{2}c^{2}}
    $$

    Quindi, nel nostro caso:

    \begin{eqnarray}
    f(r)=\pm \int dr \sqrt{\frac{1}{c^{2}}\left(\epsilon - \frac{\alpha}{r}\right)^{2} - m^{2}c^{2}}\\
    \Rightarrow S=-\epsilon t \pm \int dr \sqrt{\frac{1}{c^{2}}\left(\epsilon - \frac{\alpha}{r}\right)^{2} - m^{2}c^{2}}
    \end{eqnarray}

    Adesso si può procedere alla soluzione delle equazioni $\beta_{i}=\frac{\partial S}{\partial\alpha_{i}}= cost$, che nel nostro caso si riducono all'unica equazione:


    \begin{equation}
    \frac{\partial S}{\partial\epsilon} \ =\ -t \pm\int dr\frac{\epsilon - \frac{\alpha}{r}}{\sqrt{\frac{1}{c^{2}}\left(\epsi lon - \frac{\alpha}{r}\right)^{2} - m^{2}c^{2}}} \ =\ cost
    \end{equation}


    Questa equazione può essere risolta utilizzando l'identità:

    \begin{equation}
    \int dx \frac{1}{\sqrt{ax^{2}+bx +c}} = \frac{1}{\sqrt{a}}\log\left|2ax + b + 2 \sqrt{a(ax^{2}+bx +c)}\right| + cost
    \end{equation}











    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%

    [/tex]
     
    Last edited: Jun 13, 2007
  6. Jun 15, 2007 #5
    so nobody wants to help me?
     
  7. Jun 15, 2007 #6

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Maybe noboy can?

    and why write in spanish (?) if you want help?..
     
  8. Jun 15, 2007 #7
    i think it would be much better if you can translate your .tex file into English...:cool:
     
  9. Jun 15, 2007 #8

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think it's Italian.

    Here's a pass with Google:

    [tex]
    \section {According to exercise}

    \vspace {1cm}
    \Large {\bfseries {TEXT}}
    To consider the motion (relativistico) of a particle loaded in a coulomb field with center in the origin with the reference system. To study the case in which the speed it begins them of the particle is pure radial. To calculate the hour law of the motion. To illustrate the motion with diagrams for various conditions begins them meaningful.


    \vspace {1cm}
    \Large {\bfseries {DEVELOPMENT}}

    \vspace {0.5cm}



    We use for the development of the exercise the formalism of Hamilton-Jacobi. The relativistica Hamiltoniana of the system under investigation is:

    \begin{equation}
    \epsilon = \sqrt {c^ {2} p^ {2} + m^ {2} c^ {4}} + \frac {\alpha} {r} \\\\\\\\\{\bf p} =m \range {\bf v}
    \end{equation}

    \begin{itemize}
    \item $ \alpha > 0 \\\\rightarrow \\\$campo repulsivo
    \item $ \alpha < 0 \\\\rightarrow \\\$campo attractive
    \end{itemize}

    Being the force they centers, the motion happens on a plan, and angular moment $p_ {\varphi} $ comes conserved; in fact, if $ \frac {d {\bf p}} {dt} = \frac {\alpha} {r^ {2}} {\bf r} $ are had:

    \vspace {5mm}
    $
    \left \{\begin{Array} {l}
    \frac {\alpha} {r^ {2}} {\bf r} \times {\bf r} \= \\frac {d {\bf p}} {dt} \times {\bf r} = 0 \\[2mm]
    {\bf p} \times \frac {d {\bf r}} {dt} \= \{\bf p} \times {\bf v} \= \0
    \end{Array} \right.
    $
    \vspace {5mm}

    and therefore \$ \frac {d} {dt} ({\bf r} \times {\bf p}) =0$.


    We choose a system of reference in polar coordinates, \\

    $
    \left \{\begin{Array} {l}
    x=r \cos \varphi \\[2mm]
    y = r \sin \varphi
    \end{Array} \right.
    $

    \begin{equation}
    v^ {2} = \dot {x} ^ {2} + \dot {y} ^ {2} = \dot {r} ^ {2} + r^ {2} \dot {\varphi} ^ {2}
    \end{equation}

    The relativistica lagrangiana is:

    \begin{equation}
    \mathcal {L} = - mc^ {2} \sqrt {1- \frac {v^ {2}} {c^ {2}}} - \frac {\alpha} {r} \= \- mc^ {2} \sqrt {1- \frac {\dot {r} ^ {2}} {c^ {2}} - \frac {r^ {2} \dot {\varphi} ^ {2}} {c^ {2}}} - \frac {\alpha} {r}
    \end{equation}


    The moments can be calculate to you from the definition easy:

    \begin{eqnarray}
    p_ {r} \equiv \frac {\partial \mathcal {L}} {\partial \dot {r}} =m \gamm to \dot {r} \\
    p_ {\varphi} \equiv \frac {\partial \mathcal {L}} {\partial \dot {\varphi}} = m \range r^ {2} \dot {\varphi}
    \end{eqnarray}

    Therefore we obtain the energy:

    \begin{equation}
    \epsilon =c \sqrt {m^ {2} \gamma^ {2} (\dot {r} ^ {2} + r^ {2} \dot {\varphi} ^ {2} + m^ {2} c^ {2})} + \frac {\alpha} {r} = \mathcal {H}
    \end{equation}

    \begin{equation}
    \epsilon =c \sqrt {p^ {r} _ {2} + \frac {M^ {2}} {r^ {2}} +m^ {2} c^ {2}} + \frac {\alpha} {r}
    \end{equation}

    In our case the speed begins them is radial, therefore it is had $M=0$. E' possible now to write the equation of Hamilton-Jacobi



    \begin{equation}
    \frac {\partial S} {\partial t} + \mathcal {H} \left (r, \varphi, \frac {\partial S} {\partial r}, \frac {\partial S} {\partial \varphi} \right) =0
    \end{equation}
    \begin{equation}
    \frac {\partial S} {\partial t} +c \sqrt {\left (\frac {\partial S} {\partial r} \right) ^ {2} + \frac {1} {r^ {2}} \left (\frac {\partial S} {\partial \varphi} \right) ^ {2} +m^ {2} c^ {2}} + \frac {\alpha} {r} =0
    \end{equation}

    $$ \left (\frac {\partial S} {\partial t} + \frac {\alpha} {r} \right) ^ {2} =c^ {2} \left [\left (\frac {\partial S} {\partial r} \right) ^ {2} + \frac {1} {r^ {2}} \left (\frac {\partial S} {\partial \varphi} \right) ^ {2} +m^ {2} c^ {2} \right] $$

    Poichè in any case $ \frac {\partial S} {\partial \varphi} =p_ {\varphi} $ is constant, and poichè the hamiltoniana does not depend directly on the time, the dependency of $S$ from $t$e from $ \varphi$ is much simple one:

    \begin{equation}
    S=- \epsilon t + M \varphi + f (r)
    \end{equation}

    $$ \left (- \epsilon+ \frac {\alpha} {r} \right) ^ {2} = c^ {2} \left [\left (\frac {\partial f} {\partial r} \right) ^ {2} + \frac {M^ {2}} {r^ {2}} +m^ {2} c^ {2} \right] $$

    $$ \frac {\partial f} {\partial r} = \pm \sqrt {\frac {1} {c^ {2}} \left (- \epsilon+ \frac {\alpha} {r} \right) ^ {2} - \frac {M^ {2}} {r^ {2}} - m^ {2} c^ {2}}
    $$

    Therefore, in our case:

    \begin{eqnarray}
    f (r) = \pm \int dr \sqrt {\frac {1} {c^ {2}} \left (\epsilon - \frac {\alpha} {r} \right) ^ {2} - m^ {2} c^ {2}} \\
    \rightarrow S=- \epsilon t \pm \int dr \sqrt {\frac {1} {c^ {2}} \left (\epsilon - \frac {\alpha} {r} \right) ^ {2} - m^ {2} c^ {2}}
    \end{eqnarray}

    Now S can be proceeded to the solution of equations $ \beta_ {} = \frac {\partial} {\partial \alpha_ {the}} = cost$, than in our case they are reduced to the only equation:


    \begin{equation}
    \frac {\partial S} {\partial \epsilon} \= \- t \pm \int dr \frac {\epsilon - \frac {\alpha} {r}} {\sqrt {\frac {1} {c^ {2}} \left (\epsi lon - \frac {\alpha} {r} \right) ^ {2} - m^ {2} c^ {2}}} \= \cost
    \end{equation}


    This equation can be resolved using the identity:

    \begin{equation}
    \int dx \frac {1} {\sqrt {ax^ {2} +bx +c}} = \frac {1} {\sqrt {to}} \log \left|2ax + b + 2 \sqrt {to (ax^ {2} +bx +c)}\right| + cost
    \end{equation}
    [/tex]

    (Google messed up some of the LateX. No time [or real interest] to hunt for errors.)
     
    Last edited: Jun 15, 2007
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