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Stationary Position of a 3D Harmonic Oscillator in a constant EM field

  1. May 23, 2010 #1
    Hi,

    I have to find the 'stationary position' of a particle of mass [itex]m[/itex] and charge [itex]q[/itex] which moves in an isotropic 3D harmonic oscillator with natural frequency [itex]\omega_{0}[/itex], in a region containing a uniform electric field [itex]\boldsymbol{E} = E_{0}\hat{x}[/itex] and a uniform magnetic field [itex]\boldsymbol{B} = B_{0}\hat{z}[/itex].

    The nonrelativistic Lagrangian of the system is

    [tex]L = \frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2) - \frac{1}{2}m\omega^2(x^2+y^2+z^2) + qE_{0}x[/tex]

    and the equations of motion are

    [tex]\ddot{x}+\omega_{0}^2 x - \frac{qB_{0}}{mc}\dot{y} = \frac{qE_{0}}{m}[/tex]
    [tex]\ddot{y}+\omega_{0}^2 y + \frac{qB_{0}}{mc}\dot{x} = 0[/tex]
    [tex]\ddot{z}+\omega_{0}^2 z = 0[/tex]

    What does "stationary position" mean here? Is it the point where [itex]\ddot{x} = \ddot{y} = \ddot{z} = 0[/itex]? The next part of the question asks to find the equations of motions for oscillations about this position, and the normal modes.

    Thanks in advance.
     
  2. jcsd
  3. May 23, 2010 #2

    clem

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    Science Advisor

    "stationary position" means the position for which all time derivatives are zero.
    This means [tex]x_0=q E_0/m\omega_0^2[/tex].
     
  4. May 24, 2010 #3
    Thanks clem!
     
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