Form Factor:Pyramid

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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{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}

\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}

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{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}

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



\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}

Which gives:


\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}

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


\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}

Form Factor near q=0

qy

When q_y=0:


\begin{alignat}{2}
q_1 & = q_3 \\
q_2 & = q_4 \\
K_1 & = K_3 \\
K_2 & = K_4 \\
\end{alignat}

So:


\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}

qx

When q_x=0:


\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}

Since sinc is an even function:


\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}

And:


\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}

qz

When q_z=0:


\begin{alignat}{2}
q_1 & = q_2 \\
q_3 & = q_4 \\
\end{alignat}

So:


\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}

q

When q=0:


\begin{alignat}{2}
q_1 & = q_2 = q_3  = q_4 = 0\\
\end{alignat}

So:


\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}

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{alignat}{3}
q_1 & = q_3 & = +\frac{q_z}{2} \\
q_2 & = q_4 & = -\frac{q_z}{2} \\
\end{alignat}

\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}

So:


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


\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 \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:


\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:

\mathbf{q}=(q_x,q_y,q_z)=(-q\sin\theta\cos\phi,q\sin\theta\sin\phi,q\cos\theta)

and use:


\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}

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{alignat}{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{ 3 R^2  - R^2 }\\
  & = \sqrt{ 2 } R \\
\end{alignat}

So that the pyramid face angle, \alpha is:


\begin{alignat}{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{alignat}

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:


\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:


\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:


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

Total:


\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 H=\sqrt{2}R and \tan(\alpha)=H/R = \sqrt{2}:


\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}

See Also

Form Factor:Octahedron