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Velocity Dependent Potential

  1. Jun 21, 2009 #1

    CNX

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    1. The problem statement, all variables and given/known data

    Consider a particle of mass m and charge q that moves in an E-field [itex]\vec{E}=\frac{E_0}{r}\hat{r}[/itex] and a uniform magnetic field [itex]\vec{B}=B_0\hat{k}[/itex]. Find the scalar potential and show the vector potential is given by [itex]\vec{A}=\frac{1}{2}B_0 r \hat{\theta}[/itex]. Then obtain the Lagrange equations of motion and identify the conserved quantities

    2. Relevant equations

    Lagrange equations

    3. The attempt at a solution

    Using cylindrical coords,

    [tex]L=T-V=\frac{1}{2}m \left ( \dot{r}^2+\dot{\theta}^2+\dot{z}^2 \right) - e\phi + e\vec{v} \cdot \vec{A}[/tex]

    [tex]L = \frac{1}{2}m \left ( \dot{r}^2+\dot{\theta}^2+\dot{z}^2 \right) - eE_0\ln(r) + \frac{1}{2}e\dot{\theta}B_0r[/tex]

    Using the Lagrange equation,

    [tex]0 = m\ddot{r} + eE_0\frac{1}{r} - \frac{1}{2}e\dot{\theta}B_0[/tex]

    [tex]0=m\ddot{\theta}[/tex]

    [tex]0=m\ddot{z}[/itex]

    Correct?
     
  2. jcsd
  3. Jun 22, 2009 #2

    gabbagabbahey

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    It took me a minute to realize that you are using [itex](r,\theta,z)[/itex] as your cylindrical coordinates....ewww :yuck:.... that makes things confusing since one usually uses [itex]\vec{r}[/itex] to represent the position of the particle, and hence it is more natural for [itex]r[/itex] to be used as the spherical polar coordinate corresponding to the distance from the origin so that [itex]\vec{r}=r\hat{r}[/itex]. In this case however, [itex]\vec{r}=r\hat{r}+z\hat{z}[/itex] which is not very aesthetic....but, I digress....

    It is always a good idea to check your units...does the angular term in [itex]\vec{v}=\dot{r}\hat{r}+\dot{\theta}\hat{\theta}+\dot{z}\hat{z}[/itex] have the correct units?

    Looks like you have a sign error here...remember, [itex]\vec{E}=-\vec{\nabla}}\phi[/itex]

    Be careful with this equation, remember that it is a full time derivative (not a partial derivative) in the Euler-Lagrange equation:
    [tex]\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}\neq m\ddot{\theta}[/tex]
     
    Last edited: Jun 22, 2009
  4. Jun 23, 2009 #3

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    My attempt at corrections (Note the change in coordinate labels):

    Using cylindrical coords [itex](\rho, \theta, z)[/itex],

    [tex]L=T-V=\frac{1}{2}m \left ( \dot{\rho}^2+\rho\dot{\theta}^2+\dot{z}^2 \right) - e\phi + e\vec{v} \cdot \vec{A}[/tex]

    [tex]L = \frac{1}{2}m \left ( \dot{\rho}^2+\rho\dot{\theta}^2+\dot{z}^2 \right) + eE_0\ln(\rho) + \frac{1}{2}e\dot{\theta}B_0 \rho^2[/tex]

    Using the Lagrange equation,

    [tex]0 = m\ddot{\rho} - m\rho\ddot{\theta}^2 - eE_0\frac{1}{\rho} - e\dot{\theta}B_0\rho[/tex]

    [tex]0=m\rho^2 \ddot{\theta} + 2m\rho\dot{\rho}\dot{\theta} + e B_0\rho\dot{\rho}[/tex]

    [tex]0=m\ddot{z}[/itex]
     
    Last edited: Jun 23, 2009
  5. Jun 23, 2009 #4

    gabbagabbahey

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    Surely you meant to put some brackets in there:

    [tex]L=T-V=\frac{1}{2}m \left ( \dot{\rho}^2+(\rho\dot{\theta})^2+\dot{z}^2 \right) - e\phi + e\vec{v} \cdot \vec{A}[/tex]

    [tex]L = \frac{1}{2}m \left ( \dot{\rho}^2+(\rho\dot{\theta})^2+\dot{z}^2 \right) + eE_0\ln(\rho) + \frac{1}{2}e\dot{\theta}B_0 \rho^2[/tex]

    The first one should be:

    [tex]0 = m\ddot{\rho} - m\rho\dot{\theta}^2 - eE_0\frac{1}{\rho} - e\dot{\theta}B_0\rho[/tex]

    Other than that everything looks good :smile:.....You can double check your answers by comparing them to what you get from the Lorentz Force law; you should get the same equations of motion.

    How about the conserved quantities....what are you getting for those?
     
  6. Jun 23, 2009 #5

    CNX

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    Thanks. Wouldn't [itex]\dot{z}[/itex] be the conserved quantity?
     
  7. Jun 23, 2009 #6

    gabbagabbahey

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    Sure, the axial component of the velocity ( [itex]\dot{z}[/itex] ) or the axial momentum ( [itex]p_z=m\dot{z}[/itex] ) is conserved, but is that the only conserved quantity?....How does one typically go about finding the conserved quantities in Lagrangian or Hamiltonian dynamics?
     
  8. Jun 24, 2009 #7

    CNX

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    Thanks I got them
     
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