Form Factor:Sphere

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Sphere.png

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\neq0\\
        \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\neq0\\
        4\pi \Delta\rho^2 V_{sphere}^2
        & \mathrm{when} \,\, q=0\\
    \end{array}
    
\right.
Sphere form factor.png

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{(i q_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{alignat}{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{alignat}

Where the form factor is:


\begin{alignat}{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{alignat}

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{alignat}{2}
\mathbf{q}\cdot\mathbf{r} & =q_x x + q_y y + q_z z \\
 & = q_z z \\
 & = q r \cos\theta
\end{alignat}

And so:


\begin{alignat}{2}
F_{sphere}(q) & = \int_{0}^{2\pi} \mathrm{d}\phi \int_{0}^{\pi}\int_{0}^{R} e^{i q r \cos\theta } r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta  \\
 & = [2\pi]  \int_{0}^{\pi}\int_{0}^{R} ( \cos(q r \cos\theta) + i \sin(q r \cos\theta)  ) r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta
\end{alignat}

A simple variable substitution:

u = q r \cos\theta
\mathrm{d}u = - q r \sin\theta \mathrm{d}\theta

Yields:


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

Using the fact that:

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

We integrate:


\begin{alignat}{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^3R^3} - \frac{qR\cos(qR)}{q^3R^3} \right]  \\
 & = \frac{3\times4\pi R^3}{3}  \left[ \frac{\sin(qR)-qRcos(qR)}{q^3R^3} \right]  \\
 & = 3 V_{sphere} \frac{  \sin(qR)-qR \cos(qR) }{ (qR)^3 }


\end{alignat}

Form Factor at q=0

At very small q:


\begin{alignat}{2}

\lim_{q\to0}F_{sphere}(q)
 & = \frac{3 V_{sphere}}{R^3} \lim_{q\to0} \frac{  \sin(qR)- qR \cos(qR) }{ q^3 } \\
 & = 4\pi \lim_{q\to0} \frac{  \sin(qR)- qR \cos(qR) }{ q^3 } \\

 & = 4\pi \lim_{q\to0} \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\to0}
                     \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\to0}
                     R^3\left( \frac{1}{2}-\frac{1}{6}\right)
                     + O(q^2)
                     \\

 & = 4\pi \lim_{q\to0}
                     \frac{R^3}{3}
                     + O(q^2)
                     \\

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


\end{alignat}

Isotropic Form Factor Intensity

To average over all possible orientations, we use:


\begin{alignat}{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{alignat}

For a sphere:


\begin{alignat}{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| 3 V_{sphere} \frac{  \sin(qR)-qR \cos(qR) }{ (qR)^3 } \right|^2 \sin\theta\mathrm{d}\theta\mathrm{d}\phi
\end{alignat}

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


\begin{alignat}{2}
P_{sphere}(q) & = \left( 3 V_{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] \\
 & =  9 V_{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{alignat}

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{alignat}{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{alignat}

For large q, the -q R term dominates the numerator:


\begin{alignat}{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{alignat}

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


\begin{alignat}{2}
\lim_{q \rightarrow \infty} P_{sphere}(q) 
 & \approx   \frac{  64 \pi^3 R^2  }{ q^4 } \\
\end{alignat}