# Form Factor:Pyramid

## Equations

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

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

### Form Factor Amplitude

${\displaystyle F_{pyr}(\mathbf {q} )={\frac {H}{q_{x}q_{y}}}\left({\begin{array}{l}\cos \left[(q_{x}-q_{y})R\right]K_{1}\\\,\,\,\,+\sin \left[(q_{x}-q_{y})R\right]K_{2}\\\,\,\,\,-\cos \left[(q_{x}+q_{y})R\right]K_{3}\\\,\,\,\,-\sin \left[(q_{x}+q_{y})R\right]K_{4}\end{array}}\right)}$
where
{\displaystyle {\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}}}
{\displaystyle {\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 ${\displaystyle \alpha }$, truncated at H (along z axis), we note that the in-plane size of the pyramid at height z is:

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

Integrating with Cartesian coordinates:

{\displaystyle {\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):

{\displaystyle {\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:

{\displaystyle {\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:

{\displaystyle {\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 ${\displaystyle q_{y}=0}$:

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

So:

{\displaystyle {\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 ${\displaystyle q_{x}=0}$:

{\displaystyle {\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:

{\displaystyle {\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:

{\displaystyle {\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 ${\displaystyle q_{z}=0}$:

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

So:

{\displaystyle {\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 ${\displaystyle q=0}$:

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

So:

{\displaystyle {\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:

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

#### qx and qy

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

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

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

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

{\displaystyle {\begin{alignedat}{2}\lim _{(q_{x},q_{y})\to 0}F_{pyr}(\mathbf {q} )&={\frac {HK_{1}}{q_{x}q_{y}}}\left(\cos \left[(q_{x}-q_{y})R\right]-\cos \left[(q_{x}+q_{y})R\right]\right)\\&={\frac {HK_{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 {HK_{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 {HR^{2}K_{1}}{2q_{x}q_{y}}}\left((q_{x}+q_{y})^{2}-(q_{x}-q_{y})^{2}\right)\\\end{alignedat}}}

### Isotropic Form Factor Intensity

To average over all possible orientations, we note:

${\displaystyle \mathbf {q} =(q_{x},q_{y},q_{z})=(-q\sin \theta \cos \phi ,q\sin \theta \sin \phi ,q\cos \theta )}$

and use:

{\displaystyle {\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:

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

This makes the height:

{\displaystyle {\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, ${\displaystyle \alpha }$ is:

{\displaystyle {\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:

{\displaystyle {\begin{alignedat}{2}\cos(45^{\circ })&={\frac {R}{H}}\\H&={\frac {R}{1/{\sqrt {2}}}}\\&={\sqrt {2}}R\end{alignedat}}}

### Surface Area

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

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

The base is simply:

{\displaystyle {\begin{alignedat}{2}S_{base}&=2R\times 2R\\&=4R^{2}\end{alignedat}}}

Total:

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

### Volume

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

{\displaystyle {\begin{alignedat}{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{alignedat}}}