source: sasmodels/sasmodels/models/ellipsoid.py @ 92708d8

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Last change on this file since 92708d8 was 92708d8, checked in by Paul Kienzle <pkienzle@…>, 7 years ago

re-enable ellipsoid ER

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1# ellipsoid model
2# Note: model title and parameter table are inserted automatically
3r"""
4The form factor is normalized by the particle volume
5
6Definition
7----------
8
9The output of the 2D scattering intensity function for oriented ellipsoids
10is given by (Feigin, 1987)
11
12.. math::
13
14    P(q,\alpha) = \frac{\text{scale}}{V} F^2(q,\alpha) + \text{background}
15
16where
17
18.. math::
19
20    F(q,\alpha) = \Delta \rho V \frac{3(\sin qr  - qr \cos qr)}{(qr)^3}
21
22for
23
24.. math::
25
26    r = \left[ R_e^2 \sin^2 \alpha + R_p^2 \cos^2 \alpha \right]^{1/2}
27
28
29$\alpha$ is the angle between the axis of the ellipsoid and $\vec q$,
30$V = (4/3)\pi R_pR_e^2$ is the volume of the ellipsoid, $R_p$ is the polar
31radius along the rotational axis of the ellipsoid, $R_e$ is the equatorial
32radius perpendicular to the rotational axis of the ellipsoid and
33$\Delta \rho$ (contrast) is the scattering length density difference between
34the scatterer and the solvent.
35
36For randomly oriented particles use the orientational average,
37
38.. math::
39
40   \langle F^2(q) \rangle = \int_{0}^{\pi/2}{F^2(q,\alpha)\sin(\alpha)\,d\alpha}
41
42
43computed via substitution of $u=\sin(\alpha)$, $du=\cos(\alpha)\,d\alpha$ as
44
45.. math::
46
47    \langle F^2(q) \rangle = \int_0^1{F^2(q, u)\,du}
48
49with
50
51.. math::
52
53    r = R_e \left[ 1 + u^2\left(R_p^2/R_e^2 - 1\right)\right]^{1/2}
54
55To provide easy access to the orientation of the ellipsoid, we define
56the rotation axis of the ellipsoid using two angles $\theta$ and $\phi$.
57These angles are defined in the
58:ref:`cylinder orientation figure <cylinder-angle-definition>`.
59For the ellipsoid, $\theta$ is the angle between the rotational axis
60and the $z$ -axis in the $xz$ plane followed by a rotation by $\phi$
61in the $xy$ plane.
62
63NB: The 2nd virial coefficient of the solid ellipsoid is calculated based
64on the $R_p$ and $R_e$ values, and used as the effective radius for
65$S(q)$ when $P(q) \cdot S(q)$ is applied.
66
67
68The $\theta$ and $\phi$ parameters are not used for the 1D output.
69
70
71
72Validation
73----------
74
75Validation of the code was done by comparing the output of the 1D model
76to the output of the software provided by the NIST (Kline, 2006).
77
78The implementation of the intensity for fully oriented ellipsoids was
79validated by averaging the 2D output using a uniform distribution
80$p(\theta,\phi) = 1.0$ and comparing with the output of the 1D calculation.
81
82
83.. _ellipsoid-comparison-2d:
84
85.. figure:: img/ellipsoid_comparison_2d.jpg
86
87    Comparison of the intensity for uniformly distributed ellipsoids
88    calculated from our 2D model and the intensity from the NIST SANS
89    analysis software. The parameters used were: *scale* = 1.0,
90    *radius_polar* = 20 |Ang|, *radius_equatorial* = 400 |Ang|,
91    *contrast* = 3e-6 |Ang^-2|, and *background* = 0.0 |cm^-1|.
92
93The discrepancy above $q$ = 0.3 |cm^-1| is due to the way the form factors
94are calculated in the c-library provided by NIST. A numerical integration
95has to be performed to obtain $P(q)$ for randomly oriented particles.
96The NIST software performs that integration with a 76-point Gaussian
97quadrature rule, which will become imprecise at high $q$ where the amplitude
98varies quickly as a function of $q$. The SasView result shown has been
99obtained by summing over 501 equidistant points. Our result was found
100to be stable over the range of $q$ shown for a number of points higher
101than 500.
102
103Model was also tested against the triaxial ellipsoid model with equal major
104and minor equatorial radii.  It is also consistent with the cyclinder model
105with polar radius equal to length and equatorial radius equal to radius.
106
107References
108----------
109
110L A Feigin and D I Svergun.
111*Structure Analysis by Small-Angle X-Ray and Neutron Scattering*,
112Plenum Press, New York, 1987.
113
114A. Isihara. J. Chem. Phys. 18(1950) 1446-1449
115
116Authorship and Verification
117----------------------------
118
119* **Author:** NIST IGOR/DANSE **Date:** pre 2010
120* **Converted to sasmodels by:** Helen Park **Date:** July 9, 2014
121* **Last Modified by:** Paul Kienzle **Date:** March 22, 2017
122"""
123from __future__ import division
124
125from numpy import inf, sin, cos, pi
126
127name = "ellipsoid"
128title = "Ellipsoid of revolution with uniform scattering length density."
129
130description = """\
131P(q.alpha)= scale*f(q)^2 + background, where f(q)= 3*(sld
132        - sld_solvent)*V*[sin(q*r(Rp,Re,alpha))
133        -q*r*cos(qr(Rp,Re,alpha))]
134        /[qr(Rp,Re,alpha)]^3"
135
136     r(Rp,Re,alpha)= [Re^(2)*(sin(alpha))^2
137        + Rp^(2)*(cos(alpha))^2]^(1/2)
138
139        sld: SLD of the ellipsoid
140        sld_solvent: SLD of the solvent
141        V: volume of the ellipsoid
142        Rp: polar radius of the ellipsoid
143        Re: equatorial radius of the ellipsoid
144"""
145category = "shape:ellipsoid"
146
147#             ["name", "units", default, [lower, upper], "type","description"],
148parameters = [["sld", "1e-6/Ang^2", 4, [-inf, inf], "sld",
149               "Ellipsoid scattering length density"],
150              ["sld_solvent", "1e-6/Ang^2", 1, [-inf, inf], "sld",
151               "Solvent scattering length density"],
152              ["radius_polar", "Ang", 20, [0, inf], "volume",
153               "Polar radius"],
154              ["radius_equatorial", "Ang", 400, [0, inf], "volume",
155               "Equatorial radius"],
156              ["theta", "degrees", 60, [-360, 360], "orientation",
157               "ellipsoid axis to beam angle"],
158              ["phi", "degrees", 60, [-360, 360], "orientation",
159               "rotation about beam"],
160             ]
161
162source = ["lib/sas_3j1x_x.c", "lib/gauss76.c", "ellipsoid.c"]
163
164def ER(radius_polar, radius_equatorial):
165    import numpy as np
166    # see equation (26) in A.Isihara, J.Chem.Phys. 18(1950)1446-1449
167    ee = np.empty_like(radius_polar)
168    idx = radius_polar > radius_equatorial
169    ee[idx] = (radius_polar[idx] ** 2 - radius_equatorial[idx] ** 2) / radius_polar[idx] ** 2
170    idx = radius_polar < radius_equatorial
171    ee[idx] = (radius_equatorial[idx] ** 2 - radius_polar[idx] ** 2) / radius_equatorial[idx] ** 2
172    idx = radius_polar == radius_equatorial
173    ee[idx] = 2 * radius_polar[idx]
174    valid = (radius_polar * radius_equatorial != 0)
175    bd = 1.0 - ee[valid]
176    e1 = np.sqrt(ee[valid])
177    b1 = 1.0 + np.arcsin(e1) / (e1 * np.sqrt(bd))
178    bL = (1.0 + e1) / (1.0 - e1)
179    b2 = 1.0 + bd / 2 / e1 * np.log(bL)
180    delta = 0.75 * b1 * b2
181
182    ddd = np.zeros_like(radius_polar)
183    ddd[valid] = 2.0 * (delta + 1.0) * radius_polar * radius_equatorial ** 2
184    return 0.5 * ddd ** (1.0 / 3.0)
185
186def random():
187    import numpy as np
188    V = 10**np.random.uniform(5, 12)
189    radius_polar = 10**np.random.uniform(1.3, 4)
190    radius_equatorial = np.sqrt(V/radius_polar) # ignore 4/3 pi
191    pars = dict(
192        #background=0, sld=0, sld_solvent=1,
193        radius_polar=radius_polar,
194        radius_equatorial=radius_equatorial,
195    )
196    return pars
197
198demo = dict(scale=1, background=0,
199            sld=6, sld_solvent=1,
200            radius_polar=50, radius_equatorial=30,
201            theta=30, phi=15,
202            radius_polar_pd=.2, radius_polar_pd_n=15,
203            radius_equatorial_pd=.2, radius_equatorial_pd_n=15,
204            theta_pd=15, theta_pd_n=45,
205            phi_pd=15, phi_pd_n=1)
206q = 0.1
207# april 6 2017, rkh add unit tests, NOT compared with any other calc method, assume correct!
208qx = q*cos(pi/6.0)
209qy = q*sin(pi/6.0)
210tests = [[{}, 0.05, 54.8525847025],
211        [{'theta':80., 'phi':10.}, (qx, qy), 1.74134670026 ],
212        ]
213del qx, qy  # not necessary to delete, but cleaner
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