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Boundary Conditions on a Penning Trap

  1. Dec 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a charged particle, of mass m and charge q, confined in a device called a Penning Trap. In this device, there is a quadrupole electric field described in cartesian coordinates by the potential

    Phi[x,y,z] = U0 (2z^2 - x^2 - y^2) / (r0^2 + 2z0^2)

    Where U0 is the constant electrode potential, r0 and z0 are the radial and vertical extension of the device. There is also an uniform magnetic field through OZ of intensity B0. Determine the equations of motion for the particle and represent the respective trajectory, using Mathematica.

    2. Relevant equations

    Lorentz Force


    3. The attempt at a solution

    So, basically what I did was convert the potential to cylindrical coordinates and write it as such:

    Phi[r, θ, z] = U0 (2z^2 - r^2)/4d^2, where d 4d^2 is the denominator in the cartesian potential.

    Then I split the movement into radial (r) and axial (z). For axial movement, there'd be no influence from the magnetic field (it's parallel to it), so using the Lorentz force we get a simple harmonic motion:


    For the radial component, there IS influence from the magnetic field. Lorentz force (the cross product simplifies since the radial component is perpendicular to the field) and simplification into two frequencies (q U0 / (md^2) for wz and q B0 / (mc) for wc:


    But now I'm not sure what boundary conditions to apply to simplify the problem. The next step is graphing a trajectory of a particle in this Penning trap but I'm a bit lost on what to do next.

    Any direction is appreciated. Thank you!
  2. jcsd
  3. Dec 16, 2013 #2


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    There's something that doesn't look right to me here.

    In the first line that gives the expression for F[##\rho##[t]], doesn't the last term ##\rho## '[t] B0/c represent a component of force in the azimuthal direction rather than the radial direction?
  4. Dec 16, 2013 #3
    That's one thing I'm confused about too. In the place I read about, (it's the second link in my "Relevant equations"), they put the equation of movement in like that. I tried figuring it out but I'm not really sure about it.

    So the force would be perpendicular to the particle's tangential velocity and the field, which means it'd be in the direction of radial velocity, no? That's how I percieved it. Since it's Cross[B, v] by definition.
  5. Dec 16, 2013 #4


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    Look at equation 2.11 in your link and note that ρ is a vector in the x-y plane. The second term of the equation is a cross product of vectors.
  6. Dec 16, 2013 #5


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