Calculating the Field from the Potential — Question 5

Question 5

A smooth field zero can be stable in one direction and unstable in another

A vacuum-region potential is $V=A(x^2+y^2-2z^2)$ with A>0. A positive particle q of mass m is near the origin.

1. Find E and the linear equations of motion.

2. Find transverse oscillation frequencies and the axial exponential growth rate; explain why this is not a stable three-dimensional trap.

Question 5 - Solution

A smooth field zero can be stable in one direction and unstable in another

(a) Method and calculation

$$\mathbf E=-2Ax\hat{\mathbf x}-2Ay\hat{\mathbf y}+4Az\hat{\mathbf z}.$$$$m\ddot x=-2qAx,\quad m\ddot y=-2qAy,\quad m\ddot z=4qAz.$$

(b) Interpretation and further calculation

$$\omega_x=\omega_y=\sqrt{2qA/m},\quad z=C_+e^{\gamma t}+C_-e^{-\gamma t},\quad\gamma=\sqrt{4qA/m}.$$$$\nabla^2V=2A+2A-4A=0.$$The axial instability accompanies transverse confinement.

Physical check

A field zero alone gives equilibrium, while curvature signs determine stability in each direction.