 Small-angle width of nematic orientation

For a narrow nematic liquid crystal orientation distribution about its unoriented nematic director, expand $S=⟨(3{\cos }^2\theta -1)/2⟩=1-3⟨{\theta }^2⟩/2+O(⟨{\theta }^4⟩)$. This proves the displayed root-mean-square width. It does not determine the full distribution or a full width at half maximum. If the two small transverse tilt components are independent equal-variance Gaussians with variance ${\sigma }^2$, then $⟨{\theta }^2⟩\simeq 2{\sigma }^2$, so $\sigma \simeq \sqrt{(1-S)/3}$.