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

This can be computed numerically.

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 $q=0$:

${\begin{alignedat}{2}q_{1}&=q_{2}=q_{3}=q_{4}=0\\\end{alignedat}}$

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:

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

To analyze the behavior in the limit of small $q_{x}$ and $q_{y}$, we consider the limit of $q\to 0$ where $q_{x}=q_{y}=q$. We replace the trigonometric functions by their expansions near zero (keeping only the first two terms):

${\begin{alignedat}{2}\lim _{q\to 0}F_{pyr}(\mathbf {q} )&={\frac {H}{qq}}\left({\begin{array}{l}\cos \left[(q-q)R\right]K_{1}\\\,\,\,\,+\sin \left[(q-q)R\right]K_{2}\\\,\,\,\,-\cos \left[(q+q)R\right]K_{1}\\\,\,\,\,-\sin \left[(q+q)R\right]K_{2}\end{array}}\right)\\&={\frac {H}{q^{2}}}\left({\begin{array}{l}\left[1-{\frac {((q-q)R)^{2}}{2!}}+\cdots \right]K_{1}\\\,\,\,\,+\left[(q-q)R-{\frac {((q-q)R)^{3}}{3!}}+\cdots \right]K_{2}\\\,\,\,\,-\left[1-{\frac {((q+q)R)^{2}}{2!}}+\cdots \right]K_{1}\\\,\,\,\,-\left[(q+q)R-{\frac {((q-q)R)^{3}}{3!}}+\cdots \right]K_{2}\end{array}}\right)\\&={\frac {H}{q^{2}}}\left({\begin{array}{l}\left[1-{\frac {((q-q)R)^{2}}{2!}}-1+{\frac {((q+q)R)^{2}}{2!}}\right]K_{1}\\\,\,\,\,+\left[(q-q)R-{\frac {((q-q)R)^{3}}{3!}}-(q+q)R+{\frac {((q-q)R)^{3}}{3!}}\right]K_{2}\\\end{array}}\right)\\&={\frac {H}{q^{2}}}\left({\begin{array}{l}\left[{\frac {((2q)R)^{2}}{2!}}-{\frac {((q-q)R)^{2}}{2!}}\right]K_{1}\\\,\,\,\,+\left[(q-q)R-(2q)R\right]K_{2}\\\end{array}}\right)\\&={\frac {(2qR)^{2}}{2!}}{\frac {HK_{1}}{q^{2}}}-{\frac {((q-q)R)^{2}}{2!}}{\frac {HK_{1}}{q^{2}}}+(q-q)R{\frac {HK_{2}}{q^{2}}}-2qR{\frac {HK_{2}}{q^{2}}}\\&={\frac {4R^{2}HK_{1}}{2}}-{\frac {R^{2}HK_{1}}{2}}{\frac {(q-q)^{2}}{q^{2}}}+RHK_{2}{\frac {(q-q)}{q^{2}}}-{\frac {2RHK_{2}}{q}}\\&=2R^{2}HK_{1}\end{alignedat}}$

Note that since $\mathrm {sinc}$ is symmetric $K_{2}=K_{4}=0$. When $q_{x}$ and $q_{y}$ are small (but not zero and not necessarily equal), many of the above arguments still apply. It remains that $K_{2}\approx K_{4}\approx 0$, 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:

${\begin{alignedat}{2}P_{pyr}(q)&=\int \limits _{S}|F_{pyr}(\mathbf {q} )|^{2}\mathrm {d} \mathbf {s} \\&=\int _{\phi =0}^{2\pi }\int _{\theta =0}^{\pi }\left|{\frac {H}{q_{x}q_{y}}}\left({\begin{array}{l}\cos \left[(q_{x}-q_{y})R\right]K_{1}\\\,\,\,\,+\sin \left[(q_{x}-q_{y})R\right]K_{2}\\\,\,\,\,-\cos \left[(q_{x}+q_{y})R\right]K_{3}\\\,\,\,\,-\sin \left[(q_{x}+q_{y})R\right]K_{4}\end{array}}\right)\right|^{2}\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\&={\frac {H^{2}}{q^{2}}}\int _{0}^{2\pi }\int _{0}^{\pi }{\frac {1}{\sin ^{4}\theta \sin ^{2}\phi \cos ^{2}\phi }}\left|\left({\begin{array}{l}\cos \left[(q_{x}-q_{y})R\right]K_{1}\\\,\,\,\,+\sin \left[(q_{x}-q_{y})R\right]K_{2}\\\,\,\,\,-\cos \left[(q_{x}+q_{y})R\right]K_{3}\\\,\,\,\,-\sin \left[(q_{x}+q_{y})R\right]K_{4}\end{array}}\right)\right|^{2}\sin \theta \mathrm {d} \theta \mathrm {d} \phi \\\end{alignedat}}$

Regular Pyramid

A regular pyramid (half of an octahedron) has faces that are equilateral triangles (each vertex is 60°). The 'corner-to-edge' distance along each triangular face is then:

$d_{face,c-e}=R\tan(60^{\circ })={\sqrt {3}}R$

This makes the height:

${\begin{alignedat}{2}(d_{face,c-e})^{2}&=(H)^{2}+(R)^{2}\\H^{2}&=(d_{face,c-e})^{2}-(R)^{2}\\H&={\sqrt {({\sqrt {3}}R)^{2}-(R)^{2}}}\\&={\sqrt {3R^{2}-R^{2}}}\\&={\sqrt {2}}R\\\end{alignedat}}$

So that the pyramid face angle, $\alpha$ is:

${\begin{alignedat}{2}\tan(\alpha )&={\frac {H}{R}}\\\alpha &=\arctan \left({\frac {{\sqrt {2}}R}{R}}\right)\\&=\arctan({\sqrt {2}})\\&\approx 0.9553\\&\approx 54.75^{\circ }\end{alignedat}}$

The square base of the pyramid has edges of length 2R. The distance from the center of the square to any corner is H, such 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 \begin{alignat}{2} \cos(45^{\circ}) & = \frac{R}{H} \\ H & = \frac{R}{ 1/\sqrt{2} } \\ & = \sqrt{2} R \end{alignat} }

Surface Area

For a non-truncated, regular pyramid, each face is an equilateral triangle (each vertex is 60°). So each face:

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} S_{face} & = 2 \times \left( \frac{ R R \tan(60^{\circ}) }{2} \right) \\ & = R^2 \sqrt{3} \end{alignat} }

The base is simply:

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} S_{base} & = 2 R \times 2 R \\ & = 4 R^2 \end{alignat} }

Total:

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} S_{pyr} & = 4 \times R^2 \sqrt{3} + 4 R^2 \\ & = 4(1 + \sqrt{3}) R^2 \end{alignat} }

Volume

For a regular pyramid, the height 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 H=\sqrt{2}R} 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 \tan(\alpha)=H/R = \sqrt{2}} :

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} V_{pyr} & = \frac{4}{3} \tan (\alpha) \left[ R^3 - \left( R - \frac{H}{ \tan (\alpha)} \right)^3 \right] \\ & = \frac{4}{3} \sqrt{2} \left[ R^3 - \left( R - \frac{ \sqrt{2} R }{ \sqrt{2}} \right)^3 \right] \\ & = \frac{4\sqrt{2}}{3} R^3 \\ \end{alignat} }