KevinYager at 21:07, 13 June 2014

2014-06-13T21:07:49Z

68.194.136.6: Created page with "==Equations== For cubes of edge-length 2''R'' (volume $V_{cube}=(2R)^3$ ): ===Form Factor Amplitude=== :: $F_{cube}(\mathbf{q}) = \left\{ \begin{array..."$

2014-06-05T02:00:11Z

Created page with "==Equations== For cubes of edge-length 2''R'' (volume <math>V_{cube}=(2R)^3</math>): ===Form Factor Amplitude=== ::<math> F_{cube}(\mathbf{q}) = \left\{ \begin{array..."

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==Equations==
For cubes of edge-length 2''R'' (volume <math>V_{cube}=(2R)^3</math>):
===Form Factor Amplitude===
::<math>
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.
</math>
===Isotropic Form Factor Intensity===

::<math>
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.
</math>

==Sources==
====Byeongdu Lee (APS)====
From [http://www.nature.com/nmat/journal/v9/n11/extref/nmat2870-s1.pdf Supplementary Information] of: Matthew R. Jones, Robert J. Macfarlane, Byeongdu Lee, Jian Zhang, Kaylie L. Young, Andrew J. Senesi, and Chad A. Mirkin "[http://www.nature.com/nmat/journal/v9/n11/full/nmat2870.html DNA-nanoparticle superlattices formed from anisotropic building blocks]" Nature Materials '''9''', 913-917, '''2010'''. [http://dx.doi.org/10.1038/nmat2870 doi: 10.1038/nmat2870]
:<math>
F_{cube}(\mathbf{q}) = V_{cube} \mathrm{sinc}(q_xR) \mathrm{sinc}(q_yR) \mathrm{sinc}(q_zR)
</math>
Where ''2R'' is the edge length of the cube, such that the volume is:
::<math> V_{cube} = \left(2R \right)^3 </math>
and sinc is the [http://en.wikipedia.org/wiki/Sinc_function unnormalized sinc function]:
::<math>
\mathrm{sinc}(x) = \left\{
\begin{array}{c l}
1 & \mathrm{when} \,\, x=0 \\
\frac{\sin x}{x} & \mathrm{when} \,\, x\neq0
\end{array}
\right.
</math>
====Pedersen====
From Pedersen review, [http://linkinghub.elsevier.com/retrieve/pii/S0001868697003126 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. [http://dx.doi.org/10.1016/S0001-8686(97)00312-6 doi: 10.1016/S0001-8686(97)00312-6]
For a rectangular [http://en.wikipedia.org/wiki/Parallelepiped parallelepipedon] with edges ''a'', ''b'', and ''c'':
:<math>
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
</math>
For a cube of edge length ''a'' this would be:
:<math>
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
</math>

==Derivations==
===Form Factor===
For a cube of edge-length 2''R'', the volume is:
::<math> V_{cube} = \left(2R \right)^3 </math>
We integrate over the interior of the cube, using [http://en.wikipedia.org/wiki/Cartesian_coordinate_system Cartesian coordinates]:
::<math>\mathbf{q} = (q_x, q_y, q_z)</math>
::<math>\mathbf{r} = (x, y, z)</math>
::<math>\mathbf{q}\cdot\mathbf{r} = q_x x + q_y y + q_z z</math>
Such that:
:<math>
\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}
</math>
Each integral is of the same form:
::<math>

\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}
</math>
Which gives:
:<math>
\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}
</math>
===Form Factor at ''q''=0===
At small ''q'':
:<math>
F_{cube}\left(0\right) = V_{cube}
</math>

===Isotropic Form Factor===
To average over all possible orientations, we note:
::<math>\mathbf{q}=(q_x,q_y,q_z)=(-q\sin\theta\cos\phi,q\sin\theta\sin\phi,q\cos\theta)</math>
and use:
:<math>
\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}
</math>
From symmetry, it is sufficient to integrate over only one of the eight octants:
:<math>
\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}
</math>

===Isotropic Form Factor Intensity===
To average over all possible orientations, we note:
::<math>\mathbf{q}=(q_x,q_y,q_z)=(-q\sin\theta\cos\phi,q\sin\theta\sin\phi,q\cos\theta)</math>
and use:
:<math>
\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}
</math>
Solving integrals that involve nested trigonometric functions is not generally possible. However we can simplify in preparation for performing the integrals numerically:
:<math>
\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}
</math>

From symmetry, it is sufficient to integrate over only one of the eight octants:
:<math>
\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}
</math>

===Isotropic Form Factor Intensity contribution when <math>\phi</math>=0===
The integrand of the <math>\phi</math>-integral becomes:
:<math>
\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}
</math>
For small <math>\phi</math>, the various <math>\sin(\phi)</math> can be replaced by <math>\phi</math>, and the various <math>\cos(\phi)</math> can be replaced by <math>1</math>:

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

:<math>
\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}
</math>

===Isotropic Form Factor Intensity at ''q''=0===
At very small ''q'':
:<math>
\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}
</math>

← Older revision		Revision as of 21:07, 13 June 2014
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	==Equations==		==Equations==
	For cubes of edge-length 2''R'' (volume <math>V_{cube}=(2R)^3</math>):		For cubes of edge-length 2''R'' (volume <math>V_{cube}=(2R)^3</math>):

Form Factor:Cube - Revision history

KevinYager at 21:07, 13 June 2014

68.194.136.6: Created page with "==Equations== For cubes of edge-length 2''R'' (volume V_{cube}=(2R)^3): ===Form Factor Amplitude=== :: F_{cube}(\mathbf{q}) = \left\{ \begin{array..."

68.194.136.6: Created page with "==Equations== For cubes of edge-length 2''R'' (volume $V_{cube}=(2R)^3$ ): ===Form Factor Amplitude=== :: $F_{cube}(\mathbf{q}) = \left\{ \begin{array..."$