- #1

- 54

- 1

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