Form Factor:Cube

Equations

For cubes of edge-length 2R (volume $V_{cube}=(2R)^{3}$):

Form Factor Amplitude

$F_{cube}(\mathbf {q} )=\left\{{\begin{array}{c l}\Delta \rho V_{cube}\mathrm {sinc} (q_{x}R)\mathrm {sinc} (q_{y}R)\mathrm {sinc} (q_{z}R)&\mathrm {when} \,\,\mathbf {q} \neq (0,0,0)\\\Delta \rho V_{cube}&\mathrm {when} \,\,\mathbf {q} =(0,0,0)\\\end{array}}\right.$

Isotropic Form Factor Intensity

$P_{cube}(q)=\left\{{\begin{array}{c l}{\frac {16\Delta \rho ^{2}V_{cube}^{2}}{(qR)^{6}}}\int _{0}^{\pi }{\frac {1}{\sin \theta }}\left({\frac {\sin(q_{z}R)}{\sin(2\theta )}}\right)^{2}\int _{0}^{2\pi }\left({\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin(2\phi )}}\right)^{2}\mathrm {d} \phi \mathrm {d} \theta &\mathrm {when} \,\,q\neq 0\\4\pi \Delta \rho ^{2}V_{cube}^{2}&\mathrm {when} \,\,q=0\\\end{array}}\right.$

Sources

Byeongdu Lee (APS)

From Supplementary Information of: Matthew R. Jones, Robert J. Macfarlane, Byeongdu Lee, Jian Zhang, Kaylie L. Young, Andrew J. Senesi, and Chad A. Mirkin "DNA-nanoparticle superlattices formed from anisotropic building blocks" Nature Materials 9, 913-917, 2010. doi: 10.1038/nmat2870

$F_{cube}(\mathbf {q} )=V_{cube}\mathrm {sinc} (q_{x}R)\mathrm {sinc} (q_{y}R)\mathrm {sinc} (q_{z}R)$

Where 2R is the edge length of the cube, such that the volume is:

$V_{cube}=\left(2R\right)^{3}$

and sinc is the unnormalized sinc function:

$\mathrm {sinc} (x)=\left\{{\begin{array}{c l}1&\mathrm {when} \,\,x=0\\{\frac {\sin x}{x}}&\mathrm {when} \,\,x\neq 0\end{array}}\right.$

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 For a rectangular parallelepipedon with edges a, b, and c:

$P(q,a,b,c)={\frac {2}{\pi }}\int _{0}^{\pi /2}\int _{0}^{\pi /2}{\frac {\sin(qa\sin \alpha \cos \beta )}{qa\sin \alpha \cos \beta }}{\frac {\sin(qb\sin \alpha \cos \beta )}{qb\sin \alpha \sin \beta }}{\frac {\sin(qc\cos \alpha )}{qc\cos \alpha }}\sin \alpha \mathrm {d} \alpha \mathrm {d} \beta$

For a cube of edge length a this would be:

$P_{cube}(q,a)={\frac {2}{\pi q^{3}a^{3}}}\int _{0}^{\pi /2}\int _{0}^{\pi /2}{\frac {\sin(qa\sin \alpha \cos \beta )}{\sin \alpha \cos \beta }}{\frac {\sin(qa\sin \alpha \cos \beta )}{\sin \alpha \sin \beta }}{\frac {\sin(qa\cos \alpha )}{\cos \alpha }}\sin \alpha \mathrm {d} \alpha \mathrm {d} \beta$

Derivations

Form Factor

For a cube of edge-length 2R, the volume is:

$V_{cube}=\left(2R\right)^{3}$

We integrate over the interior of the cube, using Cartesian coordinates:

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

$\mathbf {r} =(x,y,z)$

$\mathbf {q} \cdot \mathbf {r} =q_{x}x+q_{y}y+q_{z}z$

Such that:

${\begin{alignedat}{2}F_{cube}(\mathbf {q} )&=\int \limits _{V}e^{i\mathbf {q} \cdot \mathbf {r} }\mathrm {d} \mathbf {r} \\&=\int _{z=-R}^{R}\int _{y=-R}^{R}\int _{x=-R}^{R}e^{i(q_{x}x+q_{y}y+q_{z}z)}\mathrm {d} x\mathrm {d} y\mathrm {d} z\\&=\int _{-R}^{R}e^{iq_{x}x}\mathrm {d} x\int _{-R}^{R}e^{iq_{y}y}\mathrm {d} y\int _{-R}^{R}e^{iq_{z}z}\mathrm {d} z\end{alignedat}}$

Each integral is of the same form:

${\begin{alignedat}{2}f_{cube,x}(q_{x})&=\int _{-R}^{R}e^{iq_{x}x}\mathrm {d} x\\&=\int _{-R}^{R}\left[\cos(q_{x}x)+i\sin(q_{x}x)\right]\mathrm {d} x\\&=\left[{\frac {-1}{q_{x}}}\sin(q_{x}x)+{\frac {i}{q_{x}}}\cos(q_{x}x)\right]_{x=-R}^{R}\\&=\left[{\frac {-1}{q_{x}}}\sin(q_{x}R)+{\frac {i}{q_{x}}}\cos(q_{x}R)-{\frac {-1}{q_{x}}}\sin(-q_{x}R)-{\frac {i}{q_{x}}}\cos(-q_{x}R)\right]\\&=\left[-{\frac {1}{q_{x}}}\sin(q_{x}R)-{\frac {1}{q_{x}}}\sin(q_{x}R)\right]\\&=-{\frac {2}{q_{x}}}\sin(q_{x}R)\\\end{alignedat}}$

Which gives:

${\begin{alignedat}{2}F_{cube}(\mathbf {q} )&={\frac {2}{q_{x}}}\sin(q_{x}R){\frac {2}{q_{y}}}\sin(q_{y}R){\frac {2}{q_{z}}}\sin(q_{z}R)\\&=2^{3}R^{3}{\frac {\sin(q_{x}R)}{q_{x}R}}{\frac {\sin(q_{y}R)}{q_{y}R}}{\frac {\sin(q_{z}R)}{q_{z}R}}\\&=V_{cube}\mathrm {sinc} (q_{x}R)\mathrm {sinc} (q_{y}R)\mathrm {sinc} (q_{z}R)\end{alignedat}}$

Form Factor at q=0

At small q:

$F_{cube}\left(0\right)=V_{cube}$

Isotropic Form Factor

To average over all possible orientations, we note:

$\mathbf {q} =(q_{x},q_{y},q_{z})=(-q\sin \theta \cos \phi ,q\sin \theta \sin \phi ,q\cos \theta )$

and use:

${\begin{alignedat}{2}\left\langle F_{cube}(\mathbf {q} )\right\rangle _{\mathrm {iso} }&=\int \limits _{S}F_{cube}(\mathbf {q} )\mathrm {d} \mathbf {s} \\&=\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }F_{cube}(-q\sin \theta \cos \phi ,q\sin \theta \sin \phi ,q\cos \theta )\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\&=V_{cube}\int _{0}^{2\pi }\int _{0}^{\pi }\mathrm {sinc} (q_{x}R)\mathrm {sinc} (q_{y}R)\mathrm {sinc} (q_{z}R)\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\&=V_{cube}\int _{0}^{\pi }\sin \theta \left({\frac {\sin(q_{z}R)}{q_{z}R}}\right)\int _{0}^{2\pi }\left({\frac {\sin(q_{x}R)}{q_{x}R}}\right)\left({\frac {\sin(q_{y}R)}{q_{y}R}}\right)\mathrm {d} \phi \mathrm {d} \theta \\&={\frac {V_{cube}}{(qR)^{3}}}\int _{0}^{\pi }{\frac {\sin \theta \sin(q_{z}R)}{\cos \theta }}\int _{0}^{2\pi }{\frac {\sin(q_{x}R)\sin(q_{y}R)}{-\sin \theta \cos \phi \sin \theta \sin \phi }}\mathrm {d} \phi \mathrm {d} \theta \\&=-{\frac {V_{cube}}{(qR)^{3}}}\int _{0}^{\pi }{\frac {\sin(q_{z}R)}{\sin \theta \cos \theta }}\int _{0}^{2\pi }{\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin \phi \cos \phi }}\mathrm {d} \phi \mathrm {d} \theta \\&=-{\frac {4V_{cube}}{(qR)^{3}}}\int _{0}^{\pi }{\frac {\sin(q_{z}R)}{\sin(2\theta )}}\int _{0}^{2\pi }{\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin(2\phi )}}\mathrm {d} \phi \mathrm {d} \theta \\\end{alignedat}}$

From symmetry, it is sufficient to integrate over only one of the eight octants:

${\begin{alignedat}{2}\left\langle F_{cube}(\mathbf {q} )\right\rangle _{\mathrm {iso} }&=-{\frac {32V_{cube}}{(qR)^{3}}}\int _{0}^{\pi /2}{\frac {\sin(q_{z}R)}{\sin(2\theta )}}\int _{0}^{\pi /2}{\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin(2\phi )}}\mathrm {d} \phi \mathrm {d} \theta \\\end{alignedat}}$

Isotropic Form Factor Intensity

To average over all possible orientations, we note:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{q}=(q_x,q_y,q_z)=(-q\sin\theta\cos\phi,q\sin\theta\sin\phi,q\cos\theta)}

and use:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} P_{cube}(q) & = \int\limits_{S} | F_{cube}(\mathbf{q}) |^2 \mathrm{d}\mathbf{s} \\ & = \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\pi} | F_{cube}(-q\sin\theta\cos\phi,q\sin\theta\sin\phi,q\cos\theta)|^2 \sin\theta\mathrm{d}\theta\mathrm{d}\phi \\ & = V_{cube}^2 \int_{0}^{2\pi}\int_{0}^{\pi} | \mathrm{sinc}(q_x R) \mathrm{sinc}(q_y R) \mathrm{sinc}(q_z R) |^2 \sin\theta\mathrm{d}\theta\mathrm{d}\phi \\ & = V_{cube}^2 \int_{0}^{\pi} \sin\theta \left( \frac{\sin(q\cos(\theta)R)}{q \cos(\theta)R} \right)^2 \int_{0}^{2\pi} \left( \frac{\sin(-q\sin(\theta)\cos(\phi)R)}{-q \sin(\theta)\cos(\phi)R} \right)^2 \left( \frac{\sin(q\sin(\theta)\sin(\phi)R)}{q \sin(\theta)\sin(\phi)R} \right)^2 \mathrm{d}\phi \mathrm{d}\theta \\ & = \frac{V_{cube}^2}{ (qR)^6 } \int_{0}^{\pi} \frac{\sin\theta \sin^2(q\cos(\theta)R)}{\cos^2(\theta)\sin^2(\theta)\sin^2(\theta)} \int_{0}^{2\pi} \frac{\sin^2(-q\sin(\theta)\cos(\phi)R)\sin^2(q\sin(\theta)\sin(\phi)R)} {\cos^2(\phi)\sin^2(\phi)} \mathrm{d}\phi \mathrm{d}\theta \\ & = \frac{V_{cube}^2}{ (qR)^6 } \int_{0}^{\pi} \frac{ \sin^2(q\cos(\theta)R) }{ \sin^3(\theta)\cos^2(\theta) } \int_{0}^{2\pi} \frac{\sin^2(-q\sin(\theta)\cos(\phi)R)\sin^2(q\sin(\theta)\sin(\phi)R)} { ( \frac{1}{2} \sin(2\phi) )^2 } \mathrm{d}\phi \mathrm{d}\theta \\ \end{alignat} }

Solving integrals that involve nested trigonometric functions is not generally possible. However we can simplify in preparation for performing the integrals numerically:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} P_{cube}(q) & = \frac{V_{cube}^2}{ (qR)^6 } \int_{0}^{\pi} \frac{ \sin^2(q_zR) }{ \sin^3(\theta)\cos^2(\theta) } \int_{0}^{2\pi} \frac{\sin^2(q_xR)\sin^2(q_yR)} { ( \frac{1}{2} \sin(2\phi) )^2 } \mathrm{d}\phi \mathrm{d}\theta \\ & = \frac{2^2 V_{cube}^2}{ (qR)^6 } \int_{0}^{\pi} \frac{ \sin^2(q_zR) }{ \sin(\theta)(\frac{1}{2}\sin(2\theta) )^2 } \int_{0}^{2\pi} \frac{\sin^2(q_xR)\sin^2(q_yR)} { ( \sin(2\phi) )^2 } \mathrm{d}\phi \mathrm{d}\theta \\ & = \frac{16 V_{cube}^2}{ (qR)^6 } \int_{0}^{\pi} \frac{1}{\sin\theta}\left( \frac{ \sin(q_zR) }{ \sin(2\theta) } \right)^2 \int_{0}^{2\pi} \left( \frac{\sin(q_xR)\sin(q_yR)} { \sin(2\phi) } \right)^2 \mathrm{d}\phi \mathrm{d}\theta \\ \end{alignat} }

From symmetry, it is sufficient to integrate over only one of the eight octants:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} P_{cube}(q) & = \frac{128 V_{cube}^2}{ (qR)^6 } \int_{0}^{\pi/2} \frac{1}{\sin\theta}\left( \frac{ \sin(q_zR) }{ \sin(2\theta) } \right)^2 \int_{0}^{\pi/2} \left( \frac{\sin(q_xR)\sin(q_yR)} { \sin(2\phi) } \right)^2 \mathrm{d}\phi \mathrm{d}\theta \\ \end{alignat} }

Isotropic Form Factor Intensity contribution when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} =0

The integrand of the $\phi$-integral becomes:

${\begin{alignedat}{2}\left({\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin(2\phi )}}\right)^{2}&=\left({\frac {\sin(-q\sin(\theta )\cos(\phi )R)\sin(q\sin(\theta )\sin(\phi )R)}{\sin(2\phi )}}\right)^{2}\\&=\left({\frac {\sin(-q\sin(\theta )\cos(\phi )R)\sin(q\sin(\theta )\sin(\phi )R)}{2\sin(\phi )\cos(\phi )}}\right)^{2}\\\end{alignedat}}$

For small Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} , the various Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(\phi)} can be replaced by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} , and the various Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos(\phi)} can be replaced by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \lim_{\phi\to0} \left( \frac{\sin(q_xR)\sin(q_yR)} { \sin(2\phi) } \right)^2 & = \left( \frac{\sin(-q \sin(\theta)R)\sin(q \sin(\theta) \phi R)} { 2 \phi } \right)^2 \\ & = \left( \frac{\sin(-q \sin(\theta)R) q \sin(\theta) \phi R} { 2 \phi } \right)^2 \\ & = \left( \frac{\sin(-q \sin(\theta)R) q \sin(\theta) R} { 2 } \right)^2 \\ \end{alignat} }

Which is a constant (with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} ). The part of the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} -integral near Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi=0} has the contribution:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \int_{\phi=0}^{\phi=0+\delta} \left( \frac{\sin(q_xR)\sin(q_yR)} { \sin(2\phi) } \right)^2 \mathrm{d}\phi & = \left( \frac{\sin(-q \sin(\theta)R) q \sin(\theta) R} { 2 } \right)^2 \int_{\phi=0}^{\phi=0+\delta} \mathrm{d}\phi \\ & = \left( \frac{\sin(-q \sin(\theta)R) q \sin(\theta) R} { 2 } \right)^2 \delta \\ \end{alignat} }

Isotropic Form Factor Intensity at q=0

At very small q:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} P_{cube}(0) & = V_{cube}^2 \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\pi}\sin\theta\mathrm{d}\theta\mathrm{d}\phi \\ & = 4\pi V_{cube}^2 \\ \end{alignat} }