For all alpha, the lowest energy orientation of the dumbbell would be that which maximizes negativity of the quadrupole contribution, since the quadrupole term will always be smaller in magnitude than the monopole term. As (2(cos(theta))^2-(sin(theta))^2) is equal to (2-3(sin(theta))^2), the extrema for this term are 2 and -1.

So, if alpha < 0, then theta such that this term is equal to 2 would make the quadrupole contribution most negative. Theta equal to 0 or pi, so aligned with the z axis.

If alpha=0, the orientation does not matter.

If alpha > 0, then theta such that this term is equal to -1 would make the quadrupole contribution most negative. Theta equal to pi/2 or 3pi/2, so aligned with the xy-plane.

• This topic was modified 9 months, 3 weeks ago by ajsheffler.
• This topic was modified 9 months, 3 weeks ago by ajsheffler.