# A Lagrangian of a monopole (Einstein notation is used)

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1. May 4, 2016

### IanBerkman

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

The EL equation is:
$$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x_i}} = \frac{\partial\mathcal{L}}{\partial x_i}$$

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

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

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.

Ian

2. May 4, 2016

### stevendaryl

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

$\nabla \cdot \vec{B} \propto \rho_m$ the monopole density
$\nabla \cdot (\nabla \times \vec{A}) = 0$

3. May 4, 2016

### stevendaryl

Staff Emeritus
But also,

$\frac{\partial \mathcal{L}}{\partial \dot{r}} = m \dot{r} + e \vec{A}$

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