# 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 $F_{sphere}(q) = \left\{ \begin{array}{c l} 3 \Delta\rho V_{sphere} \frac{ \sin(qR)-qR \cos(qR) }{ (qR)^3 } & \mathrm{when} \,\, q\neq0\\ \Delta\rho V_{sphere} & \mathrm{when} \,\, q=0 \\ \end{array} \right.$

### Isotropic Form Factor Intensity $P_{sphere}(q) = \left\{ \begin{array}{c l} 36 \pi \Delta\rho^2 V_{sphere}^2 \frac{ (\sin(qR)-qR \cos(qR))^2 }{ (qR)^6 } & \mathrm{when} \,\, q\neq0\\ 4\pi \Delta\rho^2 V_{sphere}^2 & \mathrm{when} \,\, q=0\\ \end{array} \right.$

## Sources

#### NCNR $P(q)=\frac{ \rm{scale} }{ V }\left[ \frac{ 3V(\Delta\rho)( \sin(qr)-qr \cos(qr)) }{ (qr)^3 } \right]^2 + \rm{background}$
• 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 $F(\mathbf{q}, r)=4\pi r^3 \frac{ \sin(qr)-qr \cos(qr) }{ (qr)^3 } \exp{(i q_z r)}$ $V = \frac{4}{3} \pi r^3 , S = \pi r^2$
• 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}