Difference between revisions of "Talk:Geometry:TSAXS 3D"

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(Working results 1)
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====Working results 1====
 
:<math>
 
\begin{alignat}{2}
 
\mathbf{q} & = \frac{2 \pi}{\lambda} \begin{bmatrix} \sin \theta_f \cos \alpha_f  \\ \cos \theta_f \cos \alpha_f - 1 \\ \sin \alpha_f \end{bmatrix} \\
 
& = \frac{2 \pi}{\lambda} \begin{bmatrix} \sin \left( \arctan\left[ \frac{x}{d} \right] \right) \cos \left( \arctan \left[ \frac{z }{d / \cos \theta_f} \right] \right)  \\ \cos \left( \arctan\left[ \frac{x}{d} \right] \right) \cos \left( \arctan \left[ \frac{z }{d / \cos \theta_f} \right] \right) - 1 \\ \sin \left( \arctan \left[ \frac{z }{d / \cos \theta_f} \right] \right) \end{bmatrix} \\
 
  
& = \frac{2 \pi}{\lambda} \begin{bmatrix}
+
====Scratch/working (contains errors)====
\frac{x/d}{\sqrt{1+\left(x/d \right)^2}} \frac{d}{\sqrt{d^2+z^2\cos^2 \theta_f}}  \\
 
\frac{1}{\sqrt{1+\left(x/d \right)^2}} \frac{d}{\sqrt{d^2+z^2\cos^2 \theta_f}} - 1 \\
 
\frac{z \cos \theta_f}{\sqrt{d^2+z^2 \cos^2 \theta_f }} \end{bmatrix} \\
 
 
 
& = \frac{2 \pi}{\lambda} \begin{bmatrix}
 
\frac{x d}{\sqrt{d^2+x^2 }} \frac{1}{\sqrt{d^2+z^2\cos^2 \theta_f}}  \\
 
\frac{d}{\sqrt{d^2+x^2}} \frac{d}{\sqrt{d^2+z^2\cos^2 \theta_f}} - 1 \\
 
\frac{z \cos \theta_f}{\sqrt{d^2+z^2 \cos^2 \theta_f }} \end{bmatrix} \\
 
 
 
\end{alignat}
 
</math>
 
Note that <math>\cos \theta_f = d/\sqrt{d^2+x^2}</math>, and <math>\cos^2 \theta_f = d^2/(d^2+x^2)</math> so:
 
:<math>
 
\begin{alignat}{2}
 
\frac{1}{\sqrt{d^2+z^2 \cos^2 \theta_f }}
 
    & = \frac{1}{\sqrt{d^2+z^2 \left( d^2/(d^2+x^2) \right) }} \\
 
    & = \frac{1}{\sqrt{d^2} \sqrt{((d^2+x^2)+z^2)/(d^2+x^2)  }} \\
 
    & = \frac{\sqrt{d^2+x^2}}{d \sqrt{d^2 + x^2 + z^2  }} \\
 
 
 
\end{alignat}
 
</math>
 
And:
 
:<math>
 
\begin{alignat}{2}
 
\mathbf{q}
 
& = \frac{2 \pi}{\lambda} \begin{bmatrix}
 
\frac{x d}{\sqrt{d^2+x^2 }} \frac{\sqrt{d^2+x^2}}{d \sqrt{d^2 + x^2 + z^2  }}  \\
 
\frac{d}{\sqrt{d^2+x^2}} \frac{d \sqrt{d^2+x^2}}{d \sqrt{d^2 + x^2 + z^2  }} - 1 \\
 
\frac{z \left( d/\sqrt{d^2+x^2} \right) \sqrt{d^2+x^2}}{d \sqrt{d^2 + x^2 + z^2  }} \end{bmatrix} \\
 
 
 
& = \frac{2 \pi}{\lambda} \begin{bmatrix}
 
\frac{x}{ \sqrt{x^2 + d^2 + z^2  }}  \\
 
\frac{d }{\sqrt{x^2 + d^2 + z^2  }} - 1 \\
 
\frac{z }{\sqrt{x^2 + d^2 + z^2  }} \end{bmatrix} \\
 
 
 
 
 
\end{alignat}
 
</math>
 
 
 
As a check:
 
:<math>
 
\begin{alignat}{2}
 
\left( \frac{q}{k} \right)^2
 
    & = \left( \frac{x}{ \sqrt{x^2 + d^2 + z^2  }} \right)^2 + \left( \frac{d - \sqrt{x^2 + d^2 + z^2  } }{\sqrt{x^2 + d^2 + z^2  }} \right)^2 + \left( \frac{z }{\sqrt{x^2 + d^2 + z^2  }} \right)^2 \\
 
    & = \frac{x^2 + \left( d - \sqrt{x^2 + d^2 + z^2  }\right)^2 + z^2 }{x^2 + d^2 + z^2} \\
 
    & = \frac{x^2 + \left( d^2 - 2d \sqrt{x^2 + d^2 + z^2 } + x^2 + d^2 + z^2  \right) + z^2 }{x^2 + d^2 + z^2} \\
 
    & = \frac{2 x^2 + 2 d^2 + 2 z^2 - 2d \sqrt{x^2 + d^2 + z^2 } }{x^2 + d^2 + z^2} \\
 
    & = 2 \frac{( x^2 + d^2 + z^2 ) - d \sqrt{x^2 + d^2 + z^2 } }{x^2 + d^2 + z^2} \\
 
    & = 2 \left( 1  - \frac{d}{\sqrt{x^2 + d^2 + z^2}} \right)
 
\end{alignat}
 
</math>
 
 
 
====Working results 2 (contains errors)====
 
 
As a check of these results, consider:
 
As a check of these results, consider:
 
:<math>
 
:<math>

Revision as of 13:09, 13 January 2016

Scratch/working (contains errors)

As a check of these results, consider:

Where we used:

And, we further note that:

Continuing: