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

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maverick280857
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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.
 
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"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].
 
clem said:
"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].

Thanks clem!