## Abstract

It is shown that, contrary to classical theory, the circular halos need not be caused by randomly oriented crystals. Furthermore, if Brownian motion is the disorienting mechanism then the circular halos cannot be caused by the randomly oriented crystals, which are too small to produce a reasonably sharp diffraction pattern. However, the circular halos can be caused by crystals that are in the region where there is a transition between randomness and high orientation. These crystals have diameters between about 12 and 40 *μ*m. Larger crystals produce the parhelia and tangent arcs. It is shown that the 46° halo is rare because it can be produced only by solid columns, and then for only a restricted range of sun heights.

© 1979 Optical Society of America

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### Equations (17)

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(1)
$$r=c\left\{1+\u220a\left[{r}^{3}\frac{{\partial}^{2}}{\partial {z}^{2}}\left(1/r\right)\right]\right\},$$
(2)
$$r=c\left[1+\u220a\left(\frac{3{z}^{2}}{{r}^{2}}-1\right)\right],$$
(4)
$$E(\theta )=-q{\mathit{\int}}_{\pi /2}^{\pi /2+\theta}sin2\psi d\psi =Q{sin}^{2}\theta ,$$
(5)
$$E(\theta )=-q{\mathit{\int}}_{0}^{\theta}sin2\psi d\psi =Q{sin}^{2}\theta ,$$
(6)
$$Q=(87/40)\pi \hspace{0.17em}\rho \hspace{0.17em}{c}^{3}{\upsilon}^{2}|\u220a|.$$
(7)
$$P(c,\theta )=A(c)exp[-E(\theta )/kT],$$
(8)
$$1=A(c){\mathit{\int}}_{0}^{\pi /2}exp[-E(\theta )/kT]d\theta .$$
(9)
$$0.5={\mathit{\int}}_{0}^{{\theta}_{.5}}P(c,\theta )d\theta .$$
(10)
$$I={[0.37DL(sinx/x)]}^{2},$$
(11)
$$x=\pi (0.37D/\mathrm{\lambda})sin\beta ,$$
(12)
$$[I/n(D)]\propto {D}^{4}\hspace{0.17em}P(D,\theta ).$$
(13)
$$cos\theta =sin\alpha cosh$$
(14)
$$sin\theta =sin\alpha cosh.$$
(15)
$$sin{\theta}^{\prime}=sin\beta sinh-cos\beta coshcos\alpha ,$$
(16)
$$sin\theta =cos\beta sinh+sin\beta coshcos\alpha .$$
(17)
$$Q/kT\propto {c}^{7}|\u220a|\propto {L}^{7}{(1+2D/L)}^{6}|1-D/L|$$