- This topic has 0 replies, 1 voice, and was last updated 3 years, 1 month ago by
.
Viewing 1 post (of 1 total)
-
Posts
-
For alpha larger than zero, the sign is negative. This means we need to look for the maxima of 2 cos^2(x) – sin^2(x), which are x=0° and x=180° -> parallel
For alpha smaller than zero, the sign is positive, so to obtain the lowest contribution we must find the minima of 2 cos^2(x) – sin^2(x) which are at x=90° and x=270° -> orthogonal
For alpha equal to zero there is no quadrupole contributions so all orientations have (up to the quadrupole term) the same energy
Viewing 1 post (of 1 total)
- You must be logged in to reply to this topic.