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

1. May 23, 2010

### maverick280857

Hi,

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

The nonrelativistic Lagrangian of the system is

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

and the equations of motion are

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

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

2. May 23, 2010

### clem

"stationary position" means the position for which all time derivatives are zero.
This means $$x_0=q E_0/m\omega_0^2$$.

3. May 24, 2010

Thanks clem!