Form Factor:Cube

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Cube.png

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_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

        & \mathrm{when} \,\, q\neq0\\
        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_xR) \mathrm{sinc}(q_yR) \mathrm{sinc}(q_zR)

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\neq0
\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(q a \sin\alpha\cos\beta) }{ q a \sin\alpha\cos\beta }
    \frac { \sin(q b \sin\alpha\cos\beta) }{ q b \sin\alpha\sin\beta }
    \frac { \sin(q c \cos\alpha) }{ q c \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(q a \sin\alpha\cos\beta) }{ \sin\alpha\cos\beta }
    \frac { \sin(q a \sin\alpha\cos\beta) }{ \sin\alpha\sin\beta }
    \frac { \sin(q a \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{alignat}{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^{i q_x x} \mathrm{d}x \int_{-R}^{R} e^{i q_y y} \mathrm{d}y  \int_{-R}^{R} e^{i q_z z} \mathrm{d}z 
\end{alignat}

Each integral is of the same form:



\begin{alignat}{2}

f_{cube,x}(q_x) & = \int_{-R}^{R} e^{i q_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{alignat}

Which gives:


\begin{alignat}{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) \\
 & = 2^3R^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{alignat}

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{alignat}{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{4 V_{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{alignat}

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


\begin{alignat}{2}
\left\langle F_{cube}(\mathbf{q}) \right\rangle_{\mathrm{iso}}

 & = - \frac{32 V_{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{alignat}

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{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:


\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:


\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 \phi=0

The integrand of the \phi-integral becomes:


\begin{alignat}{2}

           \left(
           \frac{\sin(q_xR)\sin(q_yR)}
                { \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{alignat}

For small \phi, the various \sin(\phi) can be replaced by \phi, and the various \cos(\phi) can be replaced by 1:


\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 \phi). The part of the \phi-integral near \phi=0 has the contribution:


\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:


\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}