# Lagarangian of a charged particle in a magnetic field (magnetic monopole)

1. Nov 9, 2014

### Jansen

1. The problem statement, all variables and given/known data
Show that the Lagrangian

$$L = \frac{m}{2}(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\dot{\phi}^2sin^2\theta) - qg\dot{\phi}cos\theta$$

describes the motion of a charged particle (mass $m$ charge $q$) in the magnetic field $\vec{B} = g\vec{r}/r^3$ ($g$ is a constant). Find the first integrals of the equations of motion.
2. Relevant equations
$L=T-V$, $T=\frac{m}{2} \dot{\vec{x}}^2$ and for non conservative $\vec{F}$ we may still use an effective 'potential energy' provided it satisfies $$F_\alpha=\frac{d}{dt}\left(\frac{\partial V}{\partial \dot{q}_\alpha} \right) - \frac{\partial V}{\partial q_\alpha}$$

3. The attempt at a solution
First thing that grabs me is that this is a diverging magnetic field. g represents some sort of magnetic charge, I suppose, maybe there is a prefactor in g too. I do not know. So, I guess that means there is no vector potential. I mean, $\vec{\nabla} \times \vec{B} = \vec{0} \Rightarrow \exists \Phi_m \ni \vec{B} = -\vec{\nabla}\Phi_m$
Here, $\Phi_m=-\frac{g}{r}$. But the Lorentz force that results from such a field is not conservative.
I figure there could be 2 approaches:

1) The lazy way. We see from the given L that the first part of the sum, $\frac{m}{2}(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\dot{\phi}^2sin^2\theta)=T$. Then we already have, $V=qg\dot{\phi}cos\theta$ We need simply show that it satisfies $F_\alpha=\frac{d}{dt}\left(\frac{\partial V}{\partial \dot{q}_\alpha} \right) - \frac{\partial V}{\partial q_\alpha}$. But I get: $$\vec{F}=q\dot{\vec{x}} \times \vec{B} = q(\dot{r} \hat r + r\dot{\theta} \hat \theta + r\dot{\phi}sin\theta \hat \phi) \times \frac{g}{r^2} \hat r = q(\frac{g \dot{\phi}}{r} sin \theta \hat \theta - \frac{g \dot{\theta}}{r} \hat \phi)$$
But $$\frac{d}{dt}\frac{\partial}{\partial \dot \theta}(gq\dot \phi cos\theta)= 0$$ and $$\frac{\partial}{\partial \theta}(gq\dot \phi cos\theta) = -gqsin\theta \dot \phi \neq F_\theta$$

Am I doing the calculus wrong? I don't understand if I have a 1/r factor in the force, this should appear somehow in the derivatives of the given potential. But how? Am I assuming an incorrect expression for the force in the first place? Do forces on charged particles due to magnetic fields from magnetic monopoles behave in some strange way I am not thinking about?

2) Derive the potential function and end up with the suggested expression. This is how I would prefer to do it, but I don't know quite where to start. Taking the line integral over some arbitrary curve will not help, I think, because the Lorentz force is not conservative. Yes, so if any one has a hint on this method I would really appreciate it. The examples of how to set up the equation of motion of a particle due to an electromagnetic force in my classical mechanics book just give a potential function and don't describe the way to obtain the potential.

I feel obliged to say this is my first post and I hope that my formatting isn't too painful and that I didn't break any major rules. Thanks!

2. Nov 9, 2014

### TSny

Hello, Jansen. Welcome to PF!

The forces on the left of this equation are so-called "generalized forces" and are often written $Q_\alpha$. Thus, it might be better to write $$Q_\alpha=\frac{d}{dt}\left(\frac{\partial V}{\partial \dot{q}_\alpha} \right) - \frac{\partial V}{\partial q_\alpha}$$
See http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node76.html
or http://en.wikipedia.org/wiki/Generalized_forces#Generalized_forces

Generalized forces need not have the same dimensions as "true" forces.

If the generalized coordinates are the spherical coordinates $q_1 = r$, $q_2 =\theta$, and $q_3 = \phi$, then see if you can show that, in general, $Q_1 = F_r$, $Q_2 = r \cdot F_{\theta}$, and $Q_3 = r \sin \theta \cdot F_{\phi}$

Apply these relations to your specific problem and see if you can get it to work out.

Last edited: Nov 9, 2014
3. Nov 10, 2014

### Jansen

Yes, I see this works. I am not too sure why this is so. According to the UT reference provided $Q_i = \sum_{j} F_j \frac{\partial x_j}{\partial q_i}$. Perhaps my interpretation is wrong. But I think that means (since I am going from spherical to spherical coordinates): $$Q_1 = F_r \frac{\partial r}{\partial r} + F_\theta \frac{\partial \theta}{\partial r} + F_\phi \frac{\partial \phi}{\partial r}$$ $$Q_2= F_r \frac{\partial r}{\partial \theta} + F_\theta \frac{\partial \theta}{\partial \theta} + F_\phi \frac{\partial \phi}{\partial \theta}$$ $$Q_3= F_r \frac{\partial r}{\partial \phi} + F_\theta \frac{\partial \theta}{\partial \phi} + F_\phi \frac{\partial \phi}{\partial \phi}$$

If I have done that right I guess my trouble is in spherical coordinates. I do not see how $\frac{\partial \theta}{\partial \theta} = r$ and similarly for phi $\partial_\phi \phi = rsin\theta$

4. Nov 10, 2014

### TSny

In the expression $Q_i = \sum_{j} F_j \frac{\partial x_j}{\partial q_i}$, the $x_j$ are Cartesian coordinates and the $F_j$ are Cartesian components of the force, So, $$Q_1 = F_x \frac{\partial x}{\partial r} + F_y \frac{\partial y}{\partial r} + F_z \frac{\partial z}{\partial r}$$ etc.

However, there is an easier way to work out the $Q_i$ using the fact that the $Q_i$ are defined so that under an infinitesimal displacement, the work done is $dW = \sum_{i} Q_i dq_i$.

As an expample, consider an infinitesimal displacement in the $\hat{\theta}$ direction. Then the work is $dW = Q_2 d\theta$.

But this can also be expressed in terms of the "actual" force $F_{\theta}$ acting in the $\hat{\theta}$ direction: $dW = F_{\theta} ds$, where $ds$ is the "actual" distance moved in space when $\theta$ increases by $d\theta$.

Think about how to express $ds$ in terms of $d\theta$. Then deduce the relation between $Q_2$and $F_{\theta}$ by requiring $Q_2 d\theta = F_{\theta} ds$.

Last edited: Nov 10, 2014
5. Nov 10, 2014

### Jansen

Aha! Perfectly clear now. Thank you very much!