[55e82f0] | 1 | import numpy as np |
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| 2 | from numpy import exp, sin, degrees, radians, pi, sqrt |
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| 3 | |
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| 4 | from sasmodels.weights import Dispersion as BaseDispersion |
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| 5 | |
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| 6 | class Dispersion(BaseDispersion): |
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| 7 | r""" |
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| 8 | Maier-Saupe dispersion on orientation (equal weights). |
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| 9 | |
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| 10 | .. math: |
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| 11 | |
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[35d2300] | 12 | w(\theta) = e^{a{\cos^2 \theta}} |
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[55e82f0] | 13 | |
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| 14 | This provides a close match to the gaussian distribution for |
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[35d2300] | 15 | low angles, but the tails are limited to $\pm 90^\circ$. For $a \ll 1$ |
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[55e82f0] | 16 | the distribution is approximately uniform. The usual polar coordinate |
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| 17 | projection applies, with $\theta$ weights scaled by $\cos \theta$ |
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| 18 | and $\phi$ weights unscaled. |
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| 19 | |
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| 20 | This is equivalent to a cyclic gaussian distribution |
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| 21 | $w(\theta) = e^{-sin^2(\theta)/(2\sigma^2)}. |
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| 22 | |
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| 23 | The $\theta$ points are spaced such that each interval has an |
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| 24 | equal contribution to the distribution. |
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| 25 | |
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| 26 | This works surprisingly poorly. Try:: |
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| 27 | |
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| 28 | $ sascomp cylinder -2d theta=45 phi=20 phi_pd_type=maier_saupe_eq \ |
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| 29 | phi_pd_n=100,1000 radius=50 length=2*radius -midq phi_pd=5 |
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| 30 | |
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| 31 | Leaving it here for others to improve. |
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| 32 | """ |
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| 33 | type = "maier_saupe_eq" |
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| 34 | default = dict(npts=35, width=1, nsigmas=None) |
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| 35 | |
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| 36 | # Note: center is always zero for orientation distributions |
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| 37 | def _weights(self, center, sigma, lb, ub): |
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[35d2300] | 38 | # use the width parameter as the value for Maier-Saupe "a" |
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| 39 | a = sigma |
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| 40 | sigma = 1./sqrt(2.*a) |
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[55e82f0] | 41 | |
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| 42 | # Create a lookup table for finding n points equally spaced |
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| 43 | # in the cumulative density function. |
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| 44 | # Limit width to +/-90 degrees. |
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| 45 | width = min(self.nsigmas*sigma, pi/2) |
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| 46 | xp = np.linspace(-width, width, max(self.npts*10, 100)) |
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| 47 | |
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| 48 | # Compute CDF. Since we normalized the sum of the weights to 1, |
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| 49 | # we can scale by an arbitrary scale factor c = exp(m) to get: |
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| 50 | # w = exp(m*cos(x)**2)/c = exp(-m*sin(x)**2) |
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[35d2300] | 51 | yp = np.cumsum(exp(-a*sin(xp)**2)) |
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[55e82f0] | 52 | yp /= yp[-1] |
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| 53 | |
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| 54 | # Find the mid-points of the equal-weighted intervals in the CDF |
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| 55 | y = np.linspace(0, 1, self.npts+2)[1:-1] |
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| 56 | x = np.interp(y, yp, xp) |
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| 57 | wx = np.ones(self.npts) |
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| 58 | |
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| 59 | # Truncate the distribution in case the parameter value is limited |
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| 60 | index = (x >= lb) & (x <= ub) |
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| 61 | x, wx = x[index], wx[index] |
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| 62 | |
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| 63 | return degrees(x), wx |
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