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 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 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

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 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

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} 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} }

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 = \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 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 \alpha} , truncated at H (along z axis), we note that the in-plane size of the pyramid at height z is:

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 R_z = R - \frac{ z }{ \tan \alpha }}

Integrating with Cartesian coordinates:

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}) & = \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):

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_{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:

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}) & = \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:

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{ 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 $q_{y}=0$:

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

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:

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

Isotropic Form Factor Intensity

To average over all possible orientations, we note:

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

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:

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

Surface Area

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

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

The base is simply:

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} }