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

  1. Nov 9, 2014 #1
    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. jcsd
  3. Nov 9, 2014 #2

    TSny

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    Homework Helper
    Gold Member

    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
  4. Nov 10, 2014 #3
    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##
     
  5. Nov 10, 2014 #4

    TSny

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    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
  6. Nov 10, 2014 #5
    Aha! Perfectly clear now. Thank you very much!
     
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