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:

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

and use:

${\begin{alignedat}{2}P_{cube}(q)&=\int \limits _{S}|F_{cube}(\mathbf {q} )|^{2}\mathrm {d} \mathbf {s} \\&=\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }|F_{cube}(-q\sin \theta \cos \phi ,q\sin \theta \sin \phi ,q\cos \theta )|^{2}\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\&=V_{cube}^{2}\int _{0}^{2\pi }\int _{0}^{\pi }|\mathrm {sinc} (q_{x}R)\mathrm {sinc} (q_{y}R)\mathrm {sinc} (q_{z}R)|^{2}\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\&=V_{cube}^{2}\int _{0}^{\pi }\sin \theta \left({\frac {\sin(q\cos(\theta )R)}{q\cos(\theta )R}}\right)^{2}\int _{0}^{2\pi }\left({\frac {\sin(-q\sin(\theta )\cos(\phi )R)}{-q\sin(\theta )\cos(\phi )R}}\right)^{2}\left({\frac {\sin(q\sin(\theta )\sin(\phi )R)}{q\sin(\theta )\sin(\phi )R}}\right)^{2}\mathrm {d} \phi \mathrm {d} \theta \\&={\frac {V_{cube}^{2}}{(qR)^{6}}}\int _{0}^{\pi }{\frac {\sin \theta \sin ^{2}(q\cos(\theta )R)}{\cos ^{2}(\theta )\sin ^{2}(\theta )\sin ^{2}(\theta )}}\int _{0}^{2\pi }{\frac {\sin ^{2}(-q\sin(\theta )\cos(\phi )R)\sin ^{2}(q\sin(\theta )\sin(\phi )R)}{\cos ^{2}(\phi )\sin ^{2}(\phi )}}\mathrm {d} \phi \mathrm {d} \theta \\&={\frac {V_{cube}^{2}}{(qR)^{6}}}\int _{0}^{\pi }{\frac {\sin ^{2}(q\cos(\theta )R)}{\sin ^{3}(\theta )\cos ^{2}(\theta )}}\int _{0}^{2\pi }{\frac {\sin ^{2}(-q\sin(\theta )\cos(\phi )R)\sin ^{2}(q\sin(\theta )\sin(\phi )R)}{({\frac {1}{2}}\sin(2\phi ))^{2}}}\mathrm {d} \phi \mathrm {d} \theta \\\end{alignedat}}$

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

${\begin{alignedat}{2}P_{cube}(q)&={\frac {V_{cube}^{2}}{(qR)^{6}}}\int _{0}^{\pi }{\frac {\sin ^{2}(q_{z}R)}{\sin ^{3}(\theta )\cos ^{2}(\theta )}}\int _{0}^{2\pi }{\frac {\sin ^{2}(q_{x}R)\sin ^{2}(q_{y}R)}{({\frac {1}{2}}\sin(2\phi ))^{2}}}\mathrm {d} \phi \mathrm {d} \theta \\&={\frac {2^{2}V_{cube}^{2}}{(qR)^{6}}}\int _{0}^{\pi }{\frac {\sin ^{2}(q_{z}R)}{\sin(\theta )({\frac {1}{2}}\sin(2\theta ))^{2}}}\int _{0}^{2\pi }{\frac {\sin ^{2}(q_{x}R)\sin ^{2}(q_{y}R)}{(\sin(2\phi ))^{2}}}\mathrm {d} \phi \mathrm {d} \theta \\&={\frac {16V_{cube}^{2}}{(qR)^{6}}}\int _{0}^{\pi }{\frac {1}{\sin \theta }}\left({\frac {\sin(q_{z}R)}{\sin(2\theta )}}\right)^{2}\int _{0}^{2\pi }\left({\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin(2\phi )}}\right)^{2}\mathrm {d} \phi \mathrm {d} \theta \\\end{alignedat}}$

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

${\begin{alignedat}{2}P_{cube}(q)&={\frac {128V_{cube}^{2}}{(qR)^{6}}}\int _{0}^{\pi /2}{\frac {1}{\sin \theta }}\left({\frac {\sin(q_{z}R)}{\sin(2\theta )}}\right)^{2}\int _{0}^{\pi /2}\left({\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin(2\phi )}}\right)^{2}\mathrm {d} \phi \mathrm {d} \theta \\\end{alignedat}}$

Isotropic Form Factor Intensity contribution when $\phi$=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 $\phi$, the various $\sin(\phi )$ can be replaced by $\phi$, and the various $\cos(\phi )$ can be replaced by $1$:

${\begin{alignedat}{2}\lim _{\phi \to 0}\left({\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin(2\phi )}}\right)^{2}&=\left({\frac {\sin(-q\sin(\theta )R)\sin(q\sin(\theta )\phi R)}{2\phi }}\right)^{2}\\&=\left({\frac {\sin(-q\sin(\theta )R)q\sin(\theta )\phi R}{2\phi }}\right)^{2}\\&=\left({\frac {\sin(-q\sin(\theta )R)q\sin(\theta )R}{2}}\right)^{2}\\\end{alignedat}}$

Which is a constant (with respect to $\phi$). The part of the $\phi$-integral near $\phi =0$ has the contribution:

${\begin{alignedat}{2}\int _{\phi =0}^{\phi =0+\delta }\left({\frac {\sin(q_{x}R)\sin(q_{y}R)}{\sin(2\phi )}}\right)^{2}\mathrm {d} \phi &=\left({\frac {\sin(-q\sin(\theta )R)q\sin(\theta )R}{2}}\right)^{2}\int _{\phi =0}^{\phi =0+\delta }\mathrm {d} \phi \\&=\left({\frac {\sin(-q\sin(\theta )R)q\sin(\theta )R}{2}}\right)^{2}\delta \\\end{alignedat}}$

Isotropic Form Factor Intensity at q=0

At very small q:

${\begin{alignedat}{2}P_{cube}(0)&=V_{cube}^{2}\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\&=4\pi V_{cube}^{2}\\\end{alignedat}}$