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A Lagrangian of a monopole (Einstein notation is used)

  1. May 4, 2016 #1
    Hi everyone,

    I am trying to calculate the equation of motion of a charged particle in the field of a monopole.

    The magnetic field of a monopole of strength g is given by:
    [tex] \textbf{B} = g \frac{\textbf{r}}{r^3} [/tex]
    And the Lagrangian by:
    [tex] \mathcal{L} = \frac{m\dot{\textbf{r}}^2}{2} + e\textbf{A}\cdot \dot{\textbf{r}}[/tex]
    Where e is the electric charge and A is the vector potential of the magnetic field [tex] \textbf{B} = \nabla \times \textbf{A} [/tex]
    The equations of motion (1) should become:
    [tex] m \ddot{\textbf{r}} = e(\dot{\textbf{r}} \times \textbf{B}) [/tex]

    The EL equation is:
    [tex] \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x_i}} = \frac{\partial\mathcal{L}}{\partial x_i}[/tex]

    In Einstein notation, I get the corresponding terms:
    [tex] \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x_i}} = m\ddot{x_i}\\
    \frac{\partial\mathcal{L}}{\partial x_i} = e \dot{x_j}\partial_i A_j[/tex]

    However, this does not correspond to the equations of motion given by equation (1) translated into an Einstein notation:
    [tex] m\ddot{x_i} = e \dot{x_j}\partial_i A_j - e\dot{x_j}\partial_j A_i [/tex]
    Since this equation includes the extra term:
    [tex]- e\dot{x_j}\partial_j A_i [/tex]
    Which corresponds to:
    [tex] -e(\dot{\textbf{r}} \cdot \nabla)\textbf{A} [/tex]

    I have the feeling everything is alright, but this extra term becomes zero since the vector potential only has components perpendicular to r, however, I cannot find a way to prove it mathematically.

    It is quite some work to type all the steps in between, and I tried to stay as clear as possible.
    If there are any questions or if something is not clear, I would gladly like to answer them.

    Thanks in advance,
    Ian
     
  2. jcsd
  3. May 4, 2016 #2

    stevendaryl

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    I'm not exactly sure what you're doing, but you can't describe a monopole using [itex]\vec{B} = \nabla \times \vec{A}[/itex]. That's because

    [itex]\nabla \cdot \vec{B} \propto \rho_m[/itex] the monopole density
    [itex]\nabla \cdot (\nabla \times \vec{A}) = 0[/itex]
     
  4. May 4, 2016 #3

    stevendaryl

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    But also,

    [itex]\frac{\partial \mathcal{L}}{\partial \dot{r}} = m \dot{r} + e \vec{A}[/itex]

    So [itex]\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{r}} = m \ddot{r} + e \frac{d}{dt} \vec{A}[/itex], not just [itex]m \ddot{r}[/itex]
     
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