Form Factor:Pyramid

Equations

For pyramid of base edge-length 2R, and height H. The angle of the pyramid walls is $\alpha$. If $H<R/\tan \alpha$ then the pyramid is truncated (flat top).

• Volume $V_{pyr}={\frac {4}{3}}\tan(\alpha )\left[R^{3}-\left(R-{\frac {H}{\tan(\alpha )}}\right)^{3}\right]$
• Projected (xy) surface area $Sp_{pyr}=4R^{2}$

Form Factor Amplitude

$F_{pyr}(\mathbf {q} )={\frac {H}{q_{x}q_{y}}}\left({\begin{array}{l}\cos \left[(q_{x}-q_{y})R\right]K_{1}\\\,\,\,\,+\sin \left[(q_{x}-q_{y})R\right]K_{2}\\\,\,\,\,-\cos \left[(q_{x}+q_{y})R\right]K_{3}\\\,\,\,\,-\sin \left[(q_{x}+q_{y})R\right]K_{4}\end{array}}\right)$

where

${\begin{alignedat}{2}K_{1}&=\,\,+{\text{sinc}}(q_{1}H)e^{iq_{1}H}+\,\,{\text{sinc}}(q_{2}H)e^{-iq_{2}H}\\K_{2}&=-i{\text{sinc}}(q_{1}H)e^{iq_{1}H}+i{\text{sinc}}(q_{2}H)e^{-iq_{2}H}\\K_{3}&=\,\,+{\text{sinc}}(q_{3}H)e^{iq_{3}H}+\,\,{\text{sinc}}(q_{4}H)e^{-iq_{4}H}\\K_{4}&=-i{\text{sinc}}(q_{3}H)e^{iq_{3}H}+i{\text{sinc}}(q_{4}H)e^{-iq_{4}H}\end{alignedat}}$

${\begin{alignedat}{2}q_{1}={\frac {1}{2}}\left[{\frac {q_{x}-q_{y}}{\tan \alpha }}+q_{z}\right]&\,\,,\,\,\,\,&q_{2}={\frac {1}{2}}\left[{\frac {q_{x}-q_{y}}{\tan \alpha }}-q_{z}\right]\\q_{3}={\frac {1}{2}}\left[{\frac {q_{x}+q_{y}}{\tan \alpha }}+q_{z}\right]&\,\,,\,\,\,\,&q_{4}={\frac {1}{2}}\left[{\frac {q_{x}+q_{y}}{\tan \alpha }}-q_{z}\right]\\\end{alignedat}}$

Isotropic Form Factor Intensity

$P_{pyr}(q)=\left\{{\begin{array}{c l}{\frac {\Delta \rho ^{2}V_{pyr}^{2}}{(qR)^{6}}}???&\mathrm {when} \,\,q\neq 0\\4\pi \Delta \rho ^{2}V_{pyr}^{2}&\mathrm {when} \,\,q=0\\\end{array}}\right.$

Derivations

Form Factor

For a pyramid of base-edge-length 2R, side-angle $\alpha$, truncated at H (along z axis), we note that the in-plane size of the pyramid at height z is:

$R_{z}=R-{\frac {z}{\tan \alpha }}$

Integrating with Cartesian coordinates:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&=\int \limits _{V}e^{i\mathbf {q} \cdot \mathbf {r} }\mathrm {d} \mathbf {r} \\&=\int \limits _{z=0}^{H}\int \limits _{y=-R_{z}}^{+R_{z}}\int \limits _{x=-R_{z}}^{+R_{z}}e^{i(q_{x}x+q_{y}y+q_{z}z)}\mathrm {d} x\mathrm {d} y\mathrm {d} z\\&=\int \limits _{0}^{H}\left(\int \limits _{-R_{z}}^{+R_{z}}e^{iq_{x}x}\mathrm {d} x\right)\left(\int \limits _{-R_{z}}^{+R_{z}}e^{iq_{y}y}\mathrm {d} y\right)e^{iq_{z}z}\mathrm {d} z\end{alignedat}}$

A recurring integral is (c.f. cube form factor):

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

Which gives:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&=\int \limits _{0}^{H}\left(-2R_{z}\mathrm {sinc} (q_{x}R_{z})\right)\left(-2R_{z}\mathrm {sinc} (q_{y}R_{z})\right)e^{iq_{z}z}\mathrm {d} z\\&=4\int \limits _{0}^{H}R_{z}^{2}\mathrm {sinc} (q_{x}R_{z})\mathrm {sinc} (q_{y}R_{z})e^{iq_{z}z}\mathrm {d} z\end{alignedat}}$

This can be simplified automated solving. For a regular pyramid, we obtain:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&={\frac {4{\sqrt {2}}}{q_{x}q_{y}}}{\frac {\left({\begin{array}{l}-q_{y}\left(-q_{x}^{2}+q_{y}^{2}-2q_{z}^{2}\right)\cos(q_{y}R)\sin(q_{x}R)\\\,\,\,\,-q_{x}\cos(q_{x}R)\left(2i{\sqrt {2}}q_{y}q_{z}\cos(q_{y}R)+\left(q_{x}^{2}-q_{y}^{2}-2q_{z}^{2}\right)\sin(q_{y}R)\right)\\\,\,\,\,+i{\sqrt {2}}q_{z}\left(2e^{i{\sqrt {2}}q_{z}R}q_{x}q_{y}-\left(q_{x}^{2}+q_{y}^{2}-2q_{z}^{2}\right)\sin(q_{x}R)\sin(q_{y}R)\right)\end{array}}\right)}{q_{x}^{4}+(q_{y}^{2}-2q_{z}^{2})^{2}-2q_{x}^{2}(q_{y}^{2}+2q_{z}^{2})}}\end{alignedat}}$

Form Factor near q=0

qy

When $q_{y}=0$:

${\begin{alignedat}{2}q_{1}&=q_{3}\\q_{2}&=q_{4}\\K_{1}&=K_{3}\\K_{2}&=K_{4}\\\end{alignedat}}$

So:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&={\frac {H}{q_{x}q_{y}}}\left({\begin{array}{l}\cos \left[(q_{x}-q_{y})R\right]K_{1}\\\,\,\,\,+\sin \left[(q_{x}-q_{y})R\right]K_{2}\\\,\,\,\,-\cos \left[(q_{x}+q_{y})R\right]K_{3}\\\,\,\,\,-\sin \left[(q_{x}+q_{y})R\right]K_{4}\end{array}}\right)\\&={\frac {H}{q_{x}0}}\left({\begin{array}{l}\cos \left[q_{x}R\right]K_{1}\\\,\,\,\,+\sin \left[q_{x}R\right]K_{2}\\\,\,\,\,-\cos \left[q_{x}R\right]K_{1}\\\,\,\,\,-\sin \left[q_{x}R\right]K_{2}\end{array}}\right)\\&={\frac {H}{q_{x}}}{\frac {0}{0}}\\\end{alignedat}}$

qx

When $q_{x}=0$:

${\begin{alignedat}{2}q_{1}&=-q_{4}\\q_{2}&=-q_{3}\\K_{1}&=\,\,+{\text{sinc}}(+q_{1}H)e^{+iq_{1}H}+\,\,{\text{sinc}}(+q_{2}H)e^{-iq_{2}H}\\K_{2}&=-i{\text{sinc}}(+q_{1}H)e^{+iq_{1}H}+i{\text{sinc}}(+q_{2}H)e^{-iq_{2}H}\\K_{3}&=\,\,+{\text{sinc}}(-q_{2}H)e^{-iq_{2}H}+\,\,{\text{sinc}}(-q_{1}H)e^{+iq_{1}H}\\K_{4}&=-i{\text{sinc}}(+q_{2}H)e^{-iq_{2}H}+i{\text{sinc}}(-q_{1}H)e^{+iq_{1}H}\end{alignedat}}$

Since sinc is an even function:

${\begin{alignedat}{2}K_{1}&=\,\,+{\text{sinc}}(q_{1}H)e^{+iq_{1}H}+\,\,{\text{sinc}}(q_{2}H)e^{-iq_{2}H}=K_{3}\\K_{2}&=-i{\text{sinc}}(q_{1}H)e^{+iq_{1}H}+i{\text{sinc}}(q_{2}H)e^{-iq_{2}H}=K_{4}\\K_{3}&=\,\,+{\text{sinc}}(q_{2}H)e^{-iq_{2}H}+\,\,{\text{sinc}}(q_{1}H)e^{+iq_{1}H}=K_{1}\\K_{4}&=-i{\text{sinc}}(q_{2}H)e^{-iq_{2}H}+i{\text{sinc}}(q_{1}H)e^{+iq_{1}H}=K_{2}\end{alignedat}}$

And:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&={\frac {H}{0q_{y}}}\left({\begin{array}{l}\cos \left[-q_{y}R\right]K_{1}\\\,\,\,\,+\sin \left[-q_{y}R\right]K_{2}\\\,\,\,\,-\cos \left[+q_{y}R\right]K_{1}\\\,\,\,\,-\sin \left[+q_{y}R\right]K_{2}\end{array}}\right)\\&={\frac {H}{0q_{y}}}\left({\begin{array}{l}\cos \left[+q_{y}R\right]K_{1}\\\,\,\,\,-\sin \left[+q_{y}R\right]K_{2}\\\,\,\,\,-\cos \left[+q_{y}R\right]K_{1}\\\,\,\,\,-\sin \left[+q_{y}R\right]K_{2}\end{array}}\right)\\&={\frac {-2H}{0q_{y}}}\sin \left(q_{y}R\right)\left[-i{\text{sinc}}(q_{1}H)e^{+iq_{1}H}+i{\text{sinc}}(q_{2}H)e^{-iq_{2}H}\right]\\&={\frac {2iH\sin(q_{y}R)}{0q_{y}}}\left[{\text{sinc}}(q_{1}H)\left(\cos(+iq_{1}H)-i\sin(+iq_{1}H)\right)-{\text{sinc}}(q_{2}H)\left(\cos(-iq_{2}H)-i\sin(-iq_{2}H)\right)\right]\\\end{alignedat}}$

qz

When $q_{z}=0$:

${\begin{alignedat}{2}q_{1}&=q_{2}\\q_{3}&=q_{4}\\\end{alignedat}}$

So:

${\begin{alignedat}{2}K_{1}&=\,\,+{\text{sinc}}(q_{1}H)e^{iq_{1}H}+\,\,{\text{sinc}}(q_{1}H)e^{-iq_{1}H}\\K_{2}&=-i{\text{sinc}}(q_{1}H)e^{iq_{1}H}+i{\text{sinc}}(q_{1}H)e^{-iq_{1}H}\\K_{3}&=\,\,+{\text{sinc}}(q_{3}H)e^{iq_{3}H}+\,\,{\text{sinc}}(q_{3}H)e^{-iq_{3}H}\\K_{4}&=-i{\text{sinc}}(q_{3}H)e^{iq_{3}H}+i{\text{sinc}}(q_{3}H)e^{-iq_{3}H}\end{alignedat}}$

q

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 q=0} :

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} q_1 & = q_2 = q_3 = q_4 = 0\\ \end{alignat} }

So:

${\begin{alignedat}{3}K_{1}&=+1+1&=2\\K_{2}&=-i+i&=0\\K_{3}&=+1+1&=2\\K_{4}&=-i+i&=0\end{alignedat}}$

And:

$F_{pyr}(\mathbf {q} )={\frac {H}{0\times 0}}\left({\begin{array}{l}\cos \left[(0)R\right]2\\\,\,\,\,+\sin \left[(0)R\right]0\\\,\,\,\,-\cos \left[(0)R\right]2\\\,\,\,\,-\sin \left[(0)R\right]0\end{array}}\right)$

qx and qy

When $q_{x}=q_{y}=0$:

${\begin{alignedat}{3}q_{1}&=q_{3}&=+{\frac {q_{z}}{2}}\\q_{2}&=q_{4}&=-{\frac {q_{z}}{2}}\\\end{alignedat}}$

${\begin{alignedat}{2}K_{1}&=\,\,+{\text{sinc}}(+q_{z}H/2)e^{+iq_{z}H/2}+\,\,{\text{sinc}}(-q_{z}H/2)e^{+iq_{z}H/2}\\K_{2}&=-i{\text{sinc}}(+q_{z}H/2)e^{+iq_{z}H/2}+i{\text{sinc}}(-q_{z}H/2)e^{+iq_{z}H/2}\\K_{3}&=\,\,+{\text{sinc}}(+q_{z}H/2)e^{+iq_{z}H/2}+\,\,{\text{sinc}}(-q_{z}H/2)e^{+iq_{z}H/2}=K_{1}\\K_{4}&=-i{\text{sinc}}(+q_{z}H/2)e^{+iq_{z}H/2}+i{\text{sinc}}(-q_{z}H/2)e^{+iq_{z}H/2}=K_{2}\end{alignedat}}$

So:

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} 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} }

To analyze the behavior in the limit of 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 q_x} and 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 q_y} , we consider the limit of 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 q\to0} where 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 q_x=q_y=q} . We replace the trigonometric functions by their expansions near zero (keeping only the first two terms):

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_{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} }

Note that since 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 \mathrm{sinc}} is symmetric 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 K_2=K_4=0} . 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 q_x} and 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 q_y} are small (but not zero and not necessarily equal), many of the above arguments still apply. It remains that 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 K_2 \approx K_4 \approx 0} , and:

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_{(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} }

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_{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} }