Form Factor:Sphere

This page provides the equations for calculating the form factor of a sphere (including derivations).

Equations

For spheres of radius R (volume $V_{sphere}=4\pi R^{3}/3$):

Form Factor Amplitude

$F_{sphere}(q)=\left\{{\begin{array}{c l}3\Delta \rho V_{sphere}{\frac {\sin(qR)-qR\cos(qR)}{(qR)^{3}}}&\mathrm {when} \,\,q\neq 0\\\Delta \rho V_{sphere}&\mathrm {when} \,\,q=0\\\end{array}}\right.$

Isotropic Form Factor Intensity

$P_{sphere}(q)=\left\{{\begin{array}{c l}36\pi \Delta \rho ^{2}V_{sphere}^{2}{\frac {(\sin(qR)-qR\cos(qR))^{2}}{(qR)^{6}}}&\mathrm {when} \,\,q\neq 0\\4\pi \Delta \rho ^{2}V_{sphere}^{2}&\mathrm {when} \,\,q=0\\\end{array}}\right.$

Sources

NCNR

From NCNR SANS Models documentation:

$P(q)={\frac {\rm {scale}}{V}}\left[{\frac {3V(\Delta \rho )(\sin(qr)-qr\cos(qr))}{(qr)^{3}}}\right]^{2}+{\rm {background}}$

• Parameters:
  1. ${\rm {scale}}$ : Intensity scaling
  2. $r$ : sphere radius (Å)
  3. $\Delta \rho$ : scattering contrast (Å⁻²), $\Delta \rho =SLD_{core}-SLD_{solvent}$
  4. ${\rm {background}}$ : incoherent background (cm⁻¹)

Pedersen

From Pedersen review, Analysis of small-angle scattering data from colloids and polymer solutions: modeling and least-squares fitting Jan Skov Pedersen, Advances in Colloid and Interface Science 1997, 70, 171. doi: 10.1016/S0001-8686(97)00312-6

$F(q,r)={\frac {3\left[\sin(qr)-qr\cos(qr)\right]}{(qr)^{3}}}$

• Parameters:
  1. $r$ : sphere radius (Å)

IsGISAXS

From IsGISAXS, Born form factors:

$F(\mathbf {q} ,r)=4\pi r^{3}{\frac {\sin(qr)-qr\cos(qr)}{(qr)^{3}}}\exp {(iq_{z}r)}$

$V={\frac {4}{3}}\pi r^{3},S=\pi r^{2}$

• Parameters:
  1. $r$ : sphere radius (Å)

Code

    def sphere(self, q, r, scale=1.0, contrast=0.1, background=0.0):
        
        V = (4/3)*numpy.pi*(r**3)

        return (scale/V)*(( 3*V*contrast*(sin(q*r)-q*r*cos(q*r) )/( (q*r)**3 ) )**2) + background
        

Derivations

Form Factor

For a sphere of radius R, the volume is:

$V_{sphere}={\frac {4}{3}}\pi R^{3}$

We can use a spherical coordinates, where $\theta$ denotes the angle with respect to the $+q_{z}$ axis, and $\phi$ is the in-plane angle (i.e. with respect to the $+x$ axis):

${\begin{alignedat}{2}&\mathbf {r} =(x,y,z)=(r\sin \theta \cos \phi ,r\sin \theta \sin \phi ,r\cos \theta )\\&\mathbf {q} =(q_{x},q_{y},q_{z})\\&q=|\mathbf {q} |^{2}={\sqrt {q_{x}^{2}+q_{y}^{2}+q_{z}^{2}}}\end{alignedat}}$

Where the form factor is:

${\begin{alignedat}{2}F_{sphere}(\mathbf {q} )&=\int \limits _{V}e^{i\mathbf {q} \cdot \mathbf {r} }\mathrm {d} \mathbf {r} \\&=\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }\int _{r=0}^{R}e^{i\mathbf {q} \cdot \mathbf {r} }r^{2}\mathrm {d} r\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\\end{alignedat}}$

We take advantage of spherical symmetry. E.g. we can rotate any q onto a particular axis, such as $q_{z}$. So that:

$\mathbf {q} =(0,0,q_{z})$

${\begin{alignedat}{2}\mathbf {q} \cdot \mathbf {r} &=q_{x}x+q_{y}y+q_{z}z\\&=q_{z}z\\&=qr\cos \theta \end{alignedat}}$

And so:

${\begin{alignedat}{2}F_{sphere}(q)&=\int _{0}^{2\pi }\mathrm {d} \phi \int _{0}^{\pi }\int _{0}^{R}e^{iqr\cos \theta }r^{2}\mathrm {d} r\sin \theta \mathrm {d} \theta \\&=[2\pi ]\int _{0}^{\pi }\int _{0}^{R}(\cos(qr\cos \theta )+i\sin(qr\cos \theta ))r^{2}\mathrm {d} r\sin \theta \mathrm {d} \theta \end{alignedat}}$

A simple variable substitution:

$u=qr\cos \theta$

$\mathrm {d} u=-qr\sin \theta \mathrm {d} \theta$

Yields:

${\begin{alignedat}{2}F_{sphere}(q)&=2\pi \int _{0}^{R}r^{2}\left[\int _{0}^{\pi }(\cos(u)+i\sin(u)){\frac {-\mathrm {d} u}{qr}}\right]\mathrm {d} r\\&=2\pi \int _{0}^{R}r^{2}{\frac {-1}{qr}}\left[\sin(u)-i\cos(u)\right]_{\theta =0}^{\pi }\mathrm {d} r\\&=2\pi \int _{0}^{R}{\frac {-r}{q}}\left[\sin(qr\cos \theta )-i\cos(qr\cos \theta )\right]_{\theta =0}^{\pi }\mathrm {d} r\\&=2\pi \int _{0}^{R}{\frac {-r}{q}}\left[\sin(-qr)-i\cos(-qr)-\sin(qr)+i\cos(qr)\right]\mathrm {d} r\\&=2\pi \int _{0}^{R}{\frac {r}{q}}\left[2\sin(qr)\right]\mathrm {d} r\\&={\frac {4\pi }{q}}\int _{0}^{R}r\sin(qr)\mathrm {d} r\\\end{alignedat}}$

Using the fact that:

$\int x\sin ax\;dx={\frac {\sin ax}{a^{2}}}-{\frac {x\cos ax}{a}}+C\,\!$

We integrate:

${\begin{alignedat}{2}F_{sphere}(q)&={\frac {4\pi }{q}}\left[{\frac {\sin(qr)}{q^{2}}}-{\frac {r\cos(qr)}{q}}\right]_{r=0}^{R}\\&={\frac {4\pi }{q}}\left[{\frac {\sin(qR)}{q^{2}}}-{\frac {R\cos(qR)}{q}}-{\frac {\sin(0)}{q^{2}}}+{\frac {0\cos(q0)}{q}}\right]\\&={\frac {4\pi }{1}}\left[{\frac {\sin(qR)}{q^{3}}}-{\frac {R\cos(qR)}{q^{2}}}-0+0\right]\\&=4\pi \left[{\frac {\sin(qR)}{q^{3}}}-{\frac {R\cos(qR)}{q^{2}}}\right]\\&=4\pi R^{3}\left[{\frac {\sin(qR)}{q^{3}R^{3}}}-{\frac {qR\cos(qR)}{q^{3}R^{3}}}\right]\\&={\frac {3\times 4\pi R^{3}}{3}}\left[{\frac {\sin(qR)-qRcos(qR)}{q^{3}R^{3}}}\right]\\&=3V_{sphere}{\frac {\sin(qR)-qR\cos(qR)}{(qR)^{3}}}\end{alignedat}}$

Form Factor at q=0

At very small q:

${\begin{alignedat}{2}\lim _{q\to 0}F_{sphere}(q)&={\frac {3V_{sphere}}{R^{3}}}\lim _{q\to 0}{\frac {\sin(qR)-qR\cos(qR)}{q^{3}}}\\&=4\pi \lim _{q\to 0}{\frac {\sin(qR)-qR\cos(qR)}{q^{3}}}\\&=4\pi \lim _{q\to 0}{\frac {1}{q^{3}}}\left[qR-{\frac {(qR)^{3}}{3!}}+...\right]-{\frac {R}{q^{2}}}\left[1-{\frac {(qR)^{2}}{2!}}+...\right]\\&=4\pi \lim _{q\to 0}{\frac {R}{q^{2}}}-{\frac {R^{3}}{3!}}+{\frac {O((qR)^{5})}{q^{3}}}-{\frac {R}{q^{2}}}+{\frac {R^{3}}{2!}}+{\frac {R\times O((qR)^{4})}{q^{2}}}\\&=4\pi \lim _{q\to 0}R^{3}\left({\frac {1}{2}}-{\frac {1}{6}}\right)+O(q^{2})\\&=4\pi \lim _{q\to 0}{\frac {R^{3}}{3}}+O(q^{2})\\&={\frac {4\pi R^{3}}{3}}\\&=V_{sphere}\\\end{alignedat}}$

Isotropic Form Factor Intensity

To average over all possible orientations, we use:

${\begin{alignedat}{2}P(q)&=\int \limits _{S}|F(\mathbf {q} )|^{2}\mathrm {d} \mathbf {s} \\&=\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }|F(-q\sin \theta \cos \phi ,q\sin \theta \sin \phi ,q\cos \theta )|^{2}\sin \theta \mathrm {d} \theta \mathrm {d} \phi \end{alignedat}}$

For a sphere:

${\begin{alignedat}{2}P_{sphere}(q)&=\int _{0}^{2\pi }\int _{0}^{\pi }|F_{sphere}(q)|^{2}\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\&=\int _{0}^{2\pi }\int _{0}^{\pi }\left|3V_{sphere}{\frac {\sin(qR)-qR\cos(qR)}{(qR)^{3}}}\right|^{2}\sin \theta \mathrm {d} \theta \mathrm {d} \phi \end{alignedat}}$

Note that the spherical symmetry guarantees that the integrand does not depend on $\phi$ or $\theta$:

${\begin{alignedat}{2}P_{sphere}(q)&=\left(3V_{sphere}{\frac {\sin(qR)-qR\cos(qR)}{(qR)^{3}}}\right)^{2}\int _{0}^{2\pi }\int _{0}^{\pi }\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\&=3^{2}V_{sphere}^{2}\left({\frac {\sin(qR)-qR\cos(qR)}{(qR)^{3}}}\right)^{2}\left[\int _{0}^{2\pi }\mathrm {d} \phi \right]\left[\int _{0}^{\pi }\sin \theta \mathrm {d} \theta \right]\\&=9V_{sphere}^{2}{\frac {(\sin(qR)-qR\cos(qR))^{2}}{(qR)^{6}}}\left[2\pi \right]\left[2\right]\\&=36\pi V_{sphere}^{2}{\frac {(\sin(qR)-qR\cos(qR))^{2}}{(qR)^{6}}}\\\end{alignedat}}$

Isotropic Form Factor Intensity at q=0

At q=0, we expect:

$P_{sphere}\left(0\right)=4\pi V_{sphere}^{2}$

Isotropic Form Factor Intensity at large q

Note that:

${\begin{alignedat}{2}P_{sphere}(q)&=36\pi V_{sphere}^{2}{\frac {(\sin(qR)-qR\cos(qR))^{2}}{(qR)^{6}}}\\&=36\pi \left({\frac {4\pi R^{3}}{3}}\right)^{2}{\frac {(\sin(qR)-qR\cos(qR))^{2}}{q^{6}R^{6}}}\\&=64\pi ^{3}{\frac {(\sin(qR)-qR\cos(qR))^{2}}{q^{6}}}\\\end{alignedat}}$

For large q, the $-qR$ term dominates the numerator:

${\begin{alignedat}{2}\lim _{q\rightarrow \infty }P_{sphere}(q)&=\lim _{q\rightarrow \infty }64\pi ^{3}{\frac {(\sin(qR)-qR\cos(qR))^{2}}{q^{6}}}\\&=\lim _{q\rightarrow \infty }64\pi ^{3}{\frac {q^{2}R^{2}\cos ^{2}(qR)}{q^{6}}}\\&=64\pi ^{3}R^{2}\lim _{q\rightarrow \infty }{\frac {\cos ^{2}(qR)}{q^{4}}}\\\end{alignedat}}$

The oscillation of the numerator is overwhelmed by the decay of the denominator:

${\begin{alignedat}{2}\lim _{q\rightarrow \infty }P_{sphere}(q)&\approx {\frac {64\pi ^{3}R^{2}}{q^{4}}}\\\end{alignedat}}$