KevinYager at 21:59, 29 June 2014

2014-06-29T21:59:29Z

KevinYager: /* Isotropic Form Factor Intensity */

2014-06-13T21:02:26Z

‎Isotropic Form Factor Intensity

KevinYager at 21:00, 13 June 2014

2014-06-13T21:00:42Z

KevinYager: Created page with "==Equations== For pyramid of base edge-length 2''R'', and height ''H''. The angle of the pyramid walls is $\alpha$ . If $H < R/ \tan\alpha$ then the pyram..."

2014-06-13T20:59:59Z

Created page with "==Equations== For pyramid of base edge-length 2''R'', and height ''H''. The angle of the pyramid walls is <math>\alpha</math>. If <math>H < R/ \tan\alpha</math> then the pyram..."

New page

==Equations==
For pyramid of base edge-length 2''R'', and height ''H''. The angle of the pyramid walls is <math>\alpha</math>. If <math>H < R/ \tan\alpha</math> then the pyramid is truncated (flat top).
* Volume <math>V_{pyr} = \frac{4}{3} \tan (\alpha) \left[ R^3 - \left( R - \frac{H}{ \tan (\alpha)} \right)^3 \right]</math>
* Projected (''xy'') surface area <math>Sp_{pyr} = 4R^2</math>
===Form Factor Amplitude===
::<math>
F_{pyr}(\mathbf{q}) = \frac{H}{q_x q_y} \left(

\begin{array}{l}

\cos\left[ (q_x-q_y)R \right] K_1 \\
\,\,\,\, + \sin\left[ (q_x-q_y)R \right] K_2 \\
\,\,\,\, - \cos\left[ (q_x+q_y)R \right] K_3 \\
\,\,\,\, - \sin\left[ (q_x+q_y)R \right] K_4

\end{array}

\right)
</math>
:where
:::<math>
\begin{alignat}{2}
K_1 & = \,\, +\text{sinc}(q_1 H) e^{i q_1 H} + \,\, \text{sinc}(q_2 H)e^{-iq_2 H} \\
K_2 & = -i\text{sinc}(q_1 H) e^{i q_1 H} + i\text{sinc}(q_2 H)e^{-iq_2 H} \\
K_3 & = \,\, +\text{sinc}(q_3 H) e^{i q_3 H} + \,\, \text{sinc}(q_4 H)e^{-iq_4 H} \\
K_4 & = -i\text{sinc}(q_3 H) e^{i q_3 H} + i\text{sinc}(q_4 H)e^{-iq_4 H}
\end{alignat}
</math>
:::<math>
\begin{alignat}{2}
q_1 = \frac{1}{2}\left[ \frac{q_x - q_y}{\tan\alpha} + q_z \right] & \,\, , \,\,\,\, & q_2 = \frac{1}{2}\left[ \frac{q_x - q_y}{\tan\alpha} - q_z \right] \\
q_3 = \frac{1}{2}\left[ \frac{q_x + q_y}{\tan\alpha} + q_z \right] & \,\, , \,\,\,\, & q_4 = \frac{1}{2}\left[ \frac{q_x + q_y}{\tan\alpha} - q_z \right] \\
\end{alignat}
</math>

===Isotropic Form Factor Intensity===

::<math>
P_{pyr}(q) = \left\{

\begin{array}{c l}

\frac{\Delta\rho^2 V_{pyr}^2}{ (qR)^6 } ???

& \mathrm{when} \,\, q\neq0\\
4\pi \Delta\rho^2 V_{pyr}^2
& \mathrm{when} \,\, q=0 \\
\end{array}

\right.
</math>

==Derivations==
===Form Factor===
For a pyramid of base-edge-length 2''R'', side-angle <math>\alpha</math>, truncated at ''H'' (along ''z'' axis), we note that the in-plane size of the pyramid at height ''z'' is:
:<math> R_z = R - \frac{ z }{ \tan \alpha }</math>
Integrating with Cartesian coordinates:
:<math>
\begin{alignat}{2}

F_{pyr}(\mathbf{q}) & = \int\limits_V e^{i \mathbf{q} \cdot \mathbf{r} } \mathrm{d}\mathbf{r} \\

& = \int\limits_{z=0}^{H}\int\limits_{y=-R_z}^{+R_z}\int\limits_{x=-R_z}^{+R_z} e^{i (q_x x + q_y y + q_z z) } \mathrm{d}x \mathrm{d}y \mathrm{d}z \\
& = \int\limits_{0}^{H} \left( \int\limits_{-R_z}^{+R_z} e^{i q_x x} \mathrm{d}x \right) \left( \int\limits_{-R_z}^{+R_z} e^{i q_y y} \mathrm{d}y \right) e^{i q_z z} \mathrm{d}z
\end{alignat}
</math>
A recurring integral is (c.f. [[Form_Factor:Cube#Derivations|cube form factor]]):
::<math>

\begin{alignat}{2}

f_{x}(q_x) & = \int_{-R_z}^{R_z} e^{i q_x x} \mathrm{d}x \\
& = \int_{-R_z}^{R_z} \left[\cos(q_x x) + i \sin(q_x x)\right] \mathrm{d}x \\
& = -\frac{2}{q_x}\sin(q_x R_z) \\
& = -2 R_z\mathrm{sinc}(q_x R_z) \\
\end{alignat}
</math>
Which gives:
:<math>
\begin{alignat}{2}

F_{pyr}(\mathbf{q})
& = \int\limits_{0}^{H} \left( -2 R_z\mathrm{sinc}(q_x R_z) \right) \left( -2 R_z\mathrm{sinc}(q_y R_z) \right) e^{i q_z z} \mathrm{d}z \\
& = 4 \int\limits_{0}^{H} R_z^2 \mathrm{sinc}(q_x R_z) \mathrm{sinc}(q_y R_z) e^{i q_z z} \mathrm{d}z
\end{alignat}
</math>
This can be simplified [http://en.wikipedia.org/wiki/Computer_algebra_system automated solving]. For a regular pyramid, we obtain:
:<math>
\begin{alignat}{2}
F_{pyr}(\mathbf{q})

& = \frac{ 4 \sqrt{2} }{q_x q_y} \frac{

\left(
\begin{array}{l}

-q_y \left(-q_x^2+q_y^2-2 q_z^2\right) \cos(q_y R) \sin(q_x R) \\
\,\,\,\, -q_x \cos(q_x R) \left(2 i \sqrt{2} q_y q_z \cos(q_y R) +\left(q_x^2-q_y^2-2 q_z^2\right) \sin(q_y R)\right) \\
\,\,\,\, +i \sqrt{2} q_z \left(2 e^{i \sqrt{2} q_z R} q_x q_y-\left(q_x^2+q_y^2-2 q_z^2\right) \sin(q_x R) \sin(q_y R)\right)

\end{array}
\right)

}
{
q_x^4 + (q_y^2 - 2 q_z^2)^2 - 2 q_x^2 (q_y^2 + 2 q_z^2)
}

\end{alignat}
</math>

===Form Factor near ''q''=0===
====qy====
When <math>q_y=0</math>:
::<math>
\begin{alignat}{2}
q_1 & = q_3 \\
q_2 & = q_4 \\
K_1 & = K_3 \\
K_2 & = K_4 \\
\end{alignat}
</math>
So:
:<math>
\begin{alignat}{2}

F_{pyr}(\mathbf{q}) & = \frac{H}{q_x q_y} \left(

\begin{array}{l}

\cos\left[ (q_x-q_y)R \right] K_1 \\
\,\,\,\, + \sin\left[ (q_x-q_y)R \right] K_2 \\
\,\,\,\, - \cos\left[ (q_x+q_y)R \right] K_3 \\
\,\,\,\, - \sin\left[ (q_x+q_y)R \right] K_4

\end{array}\right) \\

& = \frac{H}{q_x 0} \left(

\begin{array}{l}

\cos\left[ q_x R \right] K_1 \\
\,\,\,\, + \sin\left[ q_x R \right] K_2 \\
\,\,\,\, - \cos\left[ q_x R \right] K_1 \\
\,\,\,\, - \sin\left[ q_x R \right] K_2

\end{array}\right) \\

& = \frac{H}{q_x } \frac{0}{0} \\

\end{alignat}
</math>
====qx====
When <math>q_x=0</math>:
::<math>
\begin{alignat}{2}
q_1 & = - q_4 \\
q_2 & = - q_3 \\
K_1 & = \,\, +\text{sinc}(+q_1 H) e^{+i q_1 H} + \,\, \text{sinc}(+q_2 H)e^{-iq_2 H} \\
K_2 & = -i\text{sinc}(+q_1 H) e^{+i q_1 H} + i\text{sinc}(+q_2 H)e^{-iq_2 H} \\
K_3 & = \,\, +\text{sinc}(-q_2 H) e^{-i q_2 H} + \,\, \text{sinc}(-q_1 H)e^{+iq_1 H} \\
K_4 & = -i\text{sinc}(+q_2 H) e^{-i q_2 H} + i\text{sinc}(-q_1 H)e^{+iq_1 H}

\end{alignat}
</math>
Since [http://en.wikipedia.org/wiki/Sinc_function sinc] is an [http://en.wikipedia.org/wiki/Even_function#Even_functions even function]:
::<math>
\begin{alignat}{2}
K_1 & = \,\, +\text{sinc}(q_1 H) e^{+i q_1 H} + \,\, \text{sinc}(q_2 H)e^{-iq_2 H} = K_3 \\
K_2 & = -i\text{sinc}(q_1 H) e^{+i q_1 H} + i\text{sinc}(q_2 H)e^{-iq_2 H} = K_4 \\
K_3 & = \,\, +\text{sinc}(q_2 H) e^{-i q_2 H} + \,\, \text{sinc}(q_1 H)e^{+iq_1 H} = K_1 \\
K_4 & = -i\text{sinc}(q_2 H) e^{-i q_2 H} + i\text{sinc}(q_1 H)e^{+iq_1 H} = K_2
\end{alignat}
</math>
And:
:<math>
\begin{alignat}{2}

F_{pyr}(\mathbf{q}) & = \frac{H}{0 q_y} \left(

\begin{array}{l}

\cos\left[ -q_yR \right] K_1 \\
\,\,\,\, + \sin\left[ -q_yR \right] K_2 \\
\,\,\,\, - \cos\left[ +q_yR \right] K_1 \\
\,\,\,\, - \sin\left[ +q_yR \right] K_2

\end{array}\right) \\

& = \frac{H}{0 q_y} \left(

\begin{array}{l}

\cos\left[ +q_yR \right] K_1 \\
\,\,\,\, - \sin\left[ +q_yR \right] K_2 \\
\,\,\,\, - \cos\left[ +q_yR \right] K_1 \\
\,\,\,\, - \sin\left[ +q_yR \right] K_2

\end{array}\right) \\

& = \frac{-2 H}{0 q_y} \sin\left( q_yR \right)
\left[
-i \text{sinc}(q_1 H) e^{+i q_1 H} + i\text{sinc}(q_2 H)e^{-iq_2 H}
\right]

\\

& = \frac{2 i H \sin( q_y R )}{0 q_y}
\left[
\text{sinc}(q_1 H) \left( \cos(+i q_1 H) - i \sin(+i q_1 H) \right)
- \text{sinc}(q_2 H) \left( \cos(-i q_2 H) - i \sin(-i q_2 H) \right)
\right] \\

\end{alignat}
</math>
====qz====
When <math>q_z=0</math>:
::<math>
\begin{alignat}{2}
q_1 & = q_2 \\
q_3 & = q_4 \\
\end{alignat}
</math>
So:
::<math>
\begin{alignat}{2}
K_1 & = \,\, +\text{sinc}(q_1 H) e^{i q_1 H} + \,\, \text{sinc}(q_1 H)e^{-iq_1 H} \\
K_2 & = -i\text{sinc}(q_1 H) e^{i q_1 H} + i\text{sinc}(q_1 H)e^{-iq_1 H} \\
K_3 & = \,\, +\text{sinc}(q_3 H) e^{i q_3 H} + \,\, \text{sinc}(q_3 H)e^{-iq_3 H} \\
K_4 & = -i\text{sinc}(q_3 H) e^{i q_3 H} + i\text{sinc}(q_3 H)e^{-iq_3 H}
\end{alignat}
</math>
====q====
When <math>q=0</math>:

::<math>
\begin{alignat}{2}
q_1 & = q_2 = q_3 = q_4 = 0\\
\end{alignat}
</math>
So:
::<math>
\begin{alignat}{3}
K_1 & = +1+1& = 2 \\
K_2 & = -i + i & = 0 \\
K_3 & = +1 + 1 & = 2 \\
K_4 & = -i + i & = 0
\end{alignat}
</math>
And:
::<math>
F_{pyr}(\mathbf{q}) = \frac{H}{0 \times 0} \left(

\begin{array}{l}

\cos\left[ (0)R \right] 2 \\
\,\,\,\, + \sin\left[ (0)R \right] 0 \\
\,\,\,\, - \cos\left[ (0)R \right] 2 \\
\,\,\,\, - \sin\left[ (0)R \right] 0

\end{array}

\right)
</math>
====qx and qy====
When <math>q_x=q_y=0</math>:
:::<math>
\begin{alignat}{3}
q_1 & = q_3 & = +\frac{q_z}{2} \\
q_2 & = q_4 & = -\frac{q_z}{2} \\
\end{alignat}
</math>
:::<math>
\begin{alignat}{2}
K_1 & = \,\, +\text{sinc}(+q_z H/2) e^{+i q_z H/2} + \,\, \text{sinc}(-q_z H/2)e^{+iq_z H/2} \\
K_2 & = -i\text{sinc}(+q_z H/2) e^{+i q_z H/2} + i\text{sinc}(-q_z H/2)e^{+iq_z H/2} \\
K_3 & = \,\, +\text{sinc}(+q_z H/2) e^{+i q_z H/2} + \,\, \text{sinc}(-q_z H/2)e^{+iq_z H/2} = K_1 \\
K_4 & = -i\text{sinc}(+q_z H/2) e^{+i q_z H/2} + i\text{sinc}(-q_z H/2)e^{+iq_z H/2} = K_2
\end{alignat}
</math>
So:
::<math>
\begin{alignat}{2}
F_{pyr}(\mathbf{q})
& = \frac{H}{q_x q_y} \left(

\begin{array}{l}

\cos\left[ (q_x-q_y)R \right] K_1 \\
\,\,\,\, + \sin\left[ (q_x-q_y)R \right] K_2 \\
\,\,\,\, - \cos\left[ (q_x+q_y)R \right] K_1 \\
\,\,\,\, - \sin\left[ (q_x+q_y)R \right] K_2

\end{array}
\right) \\
\end{alignat}
</math>

To analyze the behavior in the limit of small <math>q_x</math> and <math>q_y</math>, we consider the limit of <math>q\to0</math> where <math>q_x=q_y=q</math>. We replace the trigonometric functions by their expansions near zero (keeping only the first two terms):
::<math>
\begin{alignat}{2}
\lim_{q\to0} F_{pyr}(\mathbf{q})
& = \frac{H}{q q} \left(

\begin{array}{l}

\cos\left[ (q-q)R \right] K_1 \\
\,\,\,\, + \sin\left[ (q-q)R \right] K_2 \\
\,\,\,\, - \cos\left[ (q+q)R \right] K_1 \\
\,\,\,\, - \sin\left[ (q+q)R \right] K_2

\end{array}
\right) \\

& = \frac{H}{q^2} \left(

\begin{array}{l}

\left[ 1 - \frac{ ((q-q)R)^2 }{2!} + \cdots \right] K_1 \\
\,\,\,\, + \left[ (q-q)R - \frac{((q-q)R)^3}{3!} + \cdots \right] K_2 \\
\,\,\,\, - \left[ 1 - \frac{ ((q+q)R)^2}{2!} + \cdots \right] K_1 \\
\,\,\,\, - \left[ (q+q)R - \frac{((q-q)R)^3}{3!} + \cdots \right] K_2

\end{array}
\right) \\

& = \frac{H}{q^2} \left(

\begin{array}{l}
\left[ 1 - \frac{ ((q-q)R)^2 }{2!} - 1 + \frac{ ((q+q)R)^2}{2!} \right] K_1 \\
\,\,\,\, + \left[ (q-q)R - \frac{((q-q)R)^3}{3!} - (q+q)R + \frac{((q-q)R)^3}{3!}\right] K_2 \\
\end{array}
\right) \\

& = \frac{H}{q^2} \left(

\begin{array}{l}
\left[ \frac{ ((2q)R)^2}{2!} - \frac{ ((q-q)R)^2 }{2!} \right] K_1 \\
\,\,\,\, + \left[ (q-q)R - (2q)R \right] K_2 \\
\end{array}
\right) \\

& =
\frac{ (2qR)^2}{2!}\frac{H K_1}{q^2} - \frac{ ((q-q)R)^2 }{2!}\frac{H K_1}{q^2}
+ (q-q)R \frac{H K_2}{q^2} - 2qR \frac{H K_2}{q^2} \\

& =
\frac{ 4R^2 H K_1}{2} - \frac{ R^2 H K_1}{2}\frac{(q-q)^2}{q^2}
+ R H K_2\frac{(q-q)}{q^2} - \frac{2 R H K_2}{q} \\

& =
2R^2 H K_1

\end{alignat}
</math>
Note that since <math>\mathrm{sinc}</math> is symmetric <math>K_2=K_4=0</math>. When <math>q_x</math> and <math>q_y</math> are small (but not zero and not necessarily equal), many of the above arguments still apply. It remains that <math>K_2 \approx K_4 \approx 0</math>, and:
::<math>
\begin{alignat}{2}
\lim_{(q_x,q_y)\to0} F_{pyr}(\mathbf{q})

& = \frac{H K_1}{q_x q_y} \left(
\cos\left[ (q_x-q_y)R \right]
- \cos\left[ (q_x+q_y)R \right]
\right) \\

& = \frac{H K_1}{q_x q_y} \left(
\left[ 1 - \frac{ ((q_x-q_y)R)^2}{2!} + \cdots \right]
- \left[ 1 - \frac{((q_x+q_y)R)^2}{2!} + \cdots \right]
\right) \\

& = \frac{H K_1}{q_x q_y} \left(
\frac{(q_x+q_y)^2 R^2}{2!} - \frac{(q_x-q_y)^2 R^2}{2!}
\right) \\

& = \frac{H R^2 K_1}{2 q_x q_y} \left(
(q_x+q_y)^2 - (q_x-q_y)^2
\right) \\

\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_{pyr}(q) & = \int\limits_{S} | F_{pyr}(\mathbf{q}) |^2 \mathrm{d}\mathbf{s} \\
& = \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\pi} \left|

\frac{H}{q_x q_y} \left(

\begin{array}{l}

\cos\left[ (q_x-q_y)R \right] K_1 \\
\,\,\,\, + \sin\left[ (q_x-q_y)R \right] K_2 \\
\,\,\,\, - \cos\left[ (q_x+q_y)R \right] K_3 \\
\,\,\,\, - \sin\left[ (q_x+q_y)R \right] K_4

\end{array}

\right)

\right|^2 \sin\theta\mathrm{d}\theta\mathrm{d}\phi \\

& = \frac{H^2}{q^2} \int_{0}^{2\pi}\int_{0}^{\pi}

\frac{1}{\sin^4\theta \sin^2\phi\cos^2\phi}
\left|

\left(

\begin{array}{l}

\cos\left[ (q_x-q_y)R \right] K_1 \\
\,\,\,\, + \sin\left[ (q_x-q_y)R \right] K_2 \\
\,\,\,\, - \cos\left[ (q_x+q_y)R \right] K_3 \\
\,\,\,\, - \sin\left[ (q_x+q_y)R \right] K_4

\end{array}

\right)

\right|^2 \sin\theta\mathrm{d}\theta\mathrm{d}\phi \\

\end{alignat}
</math>

← Older revision		Revision as of 21:59, 29 June 2014
Line 490:		Line 490:
	\end{alignat}		\end{alignat}
	</math>		</math>
		+
		+	==See Also==
		+	[[Form Factor:Octahedron]]

@@ Line 35: / Line 35: @@
 ===Isotropic Form Factor Intensity===
+This can be computed numerically.
 ==Derivations==

@@ Line 430: / Line 430: @@
 \right|^2 \sin\theta\mathrm{d}\theta\mathrm{d}\phi \\
 \end{alignat}
 </math>

Form Factor:Pyramid - Revision history

KevinYager at 21:59, 29 June 2014

KevinYager: /* Isotropic Form Factor Intensity */

KevinYager at 21:00, 13 June 2014

KevinYager: Created page with "==Equations== For pyramid of base edge-length 2''R'', and height ''H''. The angle of the pyramid walls is \alpha. If H < R/ \tan\alpha then the pyram..."

KevinYager: Created page with "==Equations== For pyramid of base edge-length 2''R'', and height ''H''. The angle of the pyramid walls is $\alpha$ . If $H < R/ \tan\alpha$ then the pyram..."