Talk:Scattering - Revision history

KevinYager: Blanked the page

2020-05-10T13:52:50Z

Blanked the page

KevinYager at 13:52, 10 May 2020

2020-05-10T13:52:43Z

KevinYager at 20:35, 9 May 2020

2020-05-09T20:35:17Z

KevinYager: Blanked the page

2015-12-30T15:31:42Z

Blanked the page

KevinYager: /* TSAXS 3D */

2015-12-30T15:30:28Z

‎TSAXS 3D

KevinYager: /* TSAXS 3D */

2015-12-30T15:25:06Z

‎TSAXS 3D

KevinYager: /* TSAXS 3D */

2015-12-30T15:20:26Z

‎TSAXS 3D

KevinYager: /* TSAXS 3D */

2015-12-30T15:18:26Z

‎TSAXS 3D

KevinYager: /* Check */

2015-12-30T15:03:38Z

‎Check

KevinYager: /* TSAXS 3D */

2015-12-30T14:37:19Z

‎TSAXS 3D

← Older revision		Revision as of 13:52, 10 May 2020
Line 1:		Line 1:

−	~~The scattering vector magnitude is:~~
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~q = \frac{4 \pi}{\lambda} \sin{ \theta_s } + \mathrm{constant}~~
−	~~\end{alignat}~~
−	~~</math>~~

@@ Line 3: / Line 3: @@
 :<math>
 \begin{alignat}{2}
-q = \frac{4 \pi}{\lambda} \sin{ \theta_s }
+q = \frac{4 \pi}{\lambda} \sin{ \theta_s } + \mathrm{constant}
 \end{alignat}
 </math>

← Older revision	Revision as of 20:35, 9 May 2020
Line 1:	Line 1:

	+	The scattering vector magnitude is:
	+	:<math>
	+	\begin{alignat}{2}
	+	q = \frac{4 \pi}{\lambda} \sin{ \theta_s }
	+	\end{alignat}
	+	</math>

← Older revision		Revision as of 15:31, 30 December 2015
Line 1:		Line 1:
−	~~===[[TSAXS]] 3D===~~
−	~~The ''q''-vector in fact has three components:~~
−	~~:<math>~~
−	~~\mathbf{q} = \begin{bmatrix} q_x & q_y & q_z \end{bmatrix}~~
−	~~</math>~~
−	Consider that the [[x-ray]] beam points along +''y'', so that on the [[detector]], the horizontal is ''x'', and the vertical is ''z''. We assume that the x-ray beam hits the flat 2D area detector at 90° at detector (pixel) position <math>\scriptstyle (x,z) </math>. The scattering angles are then:
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~\theta_f & = \arctan\left[ \frac{x}{d} \right] \\~~
−	~~\alpha_f ^\prime & = \arctan\left[ \frac{z}{d} \right] \\~~
−	~~\alpha_f & = \arctan \left[ \frac{z }{d / \cos \theta_f} \right]~~
−	~~\end{alignat}~~
−	~~</math>~~
−	where <math>\scriptstyle d</math> is the sample-detector distance, <math>\scriptstyle \alpha_f ^{\prime} </math> is the out-of-plane component (angle w.r.t. to ''y''-axis, rotation about x-axis), and <math>\scriptstyle \theta_f </math> is the in-plane component (rotation about ''z''-axis). The alternate angle, <math>\scriptstyle \alpha_f </math>, is the elevation angle in the plane defined by <math>\scriptstyle \theta_f </math>.
−	~~====Total scattering===~~
−	~~The full scattering angle is:~~
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~2 \theta_s = \Theta & = \arctan\left[ \frac{ \sqrt{x^2 + z^2}}{d} \right] \\~~
−	~~& = \arctan\left[ \frac{ \sqrt{(d \tan \theta_f)^2 + (d \tan \alpha_f^\prime )^2}}{d} \right] \\~~
−	~~& = \arctan\left[ \sqrt{\tan^2 \theta_f + \tan^2 \alpha_f^\prime } \right] \\~~
−	~~& = \arctan\left[ \sqrt{\tan^2 \theta_f + \frac{ \tan^2 \alpha_f }{ \cos^2 \theta_f } } \right] \\~~
−	~~\end{alignat}~~
−	~~</math>~~
−	~~The total [[momentum transfer]] is:~~
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~q & = \frac{4 \pi}{\lambda} \sin \left( \theta_s \right) \\~~
−	~~& = \frac{4 \pi}{\lambda} \sin \left( \frac{1}{2} \arctan\left [ \frac{\sqrt{x^2 + z^2}}{d} \right ] \right)~~
−	~~\end{alignat}~~
−	~~</math>~~

−	~~Given that:~~
−	~~::<math>~~
−	~~\begin{alignat}{2}~~
−	~~\cos( \arctan[u]) & = \frac{1}{\sqrt{1+u^2}} \\~~
−	~~\cos( 2 \theta_s ) & = \frac{1}{\sqrt{1 + (\sqrt{x^2+z^2}/d)^2}} \\~~
−	~~& = \frac{d}{\sqrt{d^2+x^2+z^2}}~~
−	~~\end{alignat}~~
−	~~</math>~~
−	~~We can also write:~~
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~q & = \frac{4 \pi}{\lambda} \sin \left( \theta_s \right) \\~~
−	~~& = \pm \frac{4 \pi}{\lambda} \sqrt{ \frac{1-\cos 2\theta_s }{2} } \\~~
−	~~& = \frac{4 \pi}{\lambda} \sqrt{ \frac{1}{2}\left(1 - \frac{d}{\sqrt{d^2+x^2+z^2}} \right) }~~
−	~~\end{alignat}~~
−	~~</math>~~
−	~~Where we take for granted that ''q'' must be positive.~~
−
−	~~====In-plane only====~~
−	~~If <math>\scriptstyle \alpha_f = 0 </math> (and <math>\scriptstyle \alpha_f ^{\prime} = 0</math>), then <math>\scriptstyle q_z = 0 </math>, <math>\scriptstyle 2 \theta_s = \theta_f </math>, and:~~
−	~~:<math>~~
−	~~q = k \sin \theta_f~~
−	~~</math>~~
−
−	~~====Components====~~
−	~~The [[momentum transfer]] components are:~~
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~q_x & = \frac{2 \pi}{\lambda} \sin \theta_f \cos \alpha_f \\~~
−	~~q_y & = \frac{2 \pi}{\lambda} \left ( \cos \theta_f \cos \alpha_f - 1 \right ) \\~~
−	~~q_z & = \frac{2 \pi}{\lambda} \sin \alpha_f~~
−	~~\end{alignat}~~
−	~~</math>~~
−
−	~~====Check====~~
−	~~As a check of these results, consider:~~
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~q & = \sqrt{ q_x^2 + q_y^2 + q_z^2 } \\~~
−	~~& = \frac{2 \pi}{\lambda} \sqrt{ \sin^2 \theta_f \cos^2 \alpha_f + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + \sin^2 \alpha_f } \\~~
−	~~\frac{q}{k}~~
−	~~& = \sqrt{ (\sin \theta_f)^2 (\cos \alpha_f)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + (\sin \alpha_f)^2 } \\~~
−	~~\frac{q^2}{k^2}~~
−	& = \left(\frac{x/d}{\sqrt{1+(x/d)^2}} \right)^2 \left(\cos \alpha_f \right)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + \left( \frac{z \cos \theta_f /d }{\sqrt{1+(z \cos \theta_f /d)^2}} \right)^2 \\
−	& = \left(\frac{x}{\sqrt{d^2+x^2}} \right)^2 \left(\cos \alpha_f \right)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + \left( \frac{z \cos \theta_f }{\sqrt{d^2+z^2 \cos^2 \theta_f }} \right)^2 \\
−	~~& = \frac{x^2}{d^2+x^2} \left(\cos \alpha_f \right)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + \frac{z^2 \cos^2 \theta_f }{d^2+z^2 \cos^2 \theta_f } \\~~
−	~~& = \frac{x^2}{d^2+x^2} \frac{d^4}{d^2+z^2 \cos^2 \theta_f} + \left ( \cos \theta_f \frac{d^2}{\sqrt{d^2+z^2 \cos^2 \theta_f}} - 1 \right )^2 + \frac{z^2 \cos^2 \theta_f }{d^2+z^2 \cos^2 \theta_f }~~
−	~~\end{alignat}~~
−	~~</math>~~
−	~~Where we used:~~
−	~~::<math>~~
−	~~\begin{alignat}{2}~~
−	~~\sin( \arctan[u]) & = \frac{u}{\sqrt{1+u^2}} \\~~
−	~~\sin \theta_f & = \sin( \arctan [x/d] ) \\~~
−	~~& = \frac{x/d}{\sqrt{1 + (x/d)^2}} \\~~
−	~~& = \frac{x}{\sqrt{d^2+x^2}}~~
−	~~\end{alignat}~~
−	~~</math>~~
−
−	~~And, we further note that:~~
−	~~::<math>~~
−	~~\begin{alignat}{2}~~
−	~~\cos( \arctan[u]) & = \frac{1}{\sqrt{1+u^2}} \\~~
−	~~\cos \theta_f & = \frac{1}{\sqrt{1 + (x/d)^2}} \\~~
−	~~& = \frac{d^2}{\sqrt{d^2+x^2}}~~
−	~~\end{alignat}~~
−	~~</math>~~
−
−	~~====cont====~~
−	~~Continuing:~~
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~\frac{q^2}{k^2}~~
−	& = \frac{x^2}{d^2+x^2} \frac{d^4}{d^2+z^2 \cos^2 \theta_f} + \left ( \frac{d^2}{\sqrt{d^2+x^2}} \frac{d^2}{\sqrt{d^2+z^2 \cos^2 \theta_f}} - 1 \right )^2 + \frac{z^2 }{d^2+z^2 \cos^2 \theta_f } \frac{d^4}{d^2+x^2} \\
−	~~& = d^4\frac{x^2+z^2}{(d^2+x^2)(d^2+z^2 \cos^2 \theta_f)} + \left ( \frac{d^4}{\sqrt{(d^2+x^2)(d^2+z^2 \cos^2 \theta_f)}} - 1 \right )^2 \\~~
−	~~& = \frac{d^4x^2+d^4z^2}{d^4+d^2x^2+d^4z^2} + \left ( \frac{d^4}{\sqrt{d^4+d^2x^2+d^4z^2}} - 1 \right )^2 \\~~
−	~~& = \frac{d^2x^2+d^2z^2}{d^2+x^2+d^2z^2} + \left ( \frac{d^8}{d^4+d^2x^2+d^4z^2} -2 \frac{d^4}{\sqrt{d^4+d^2x^2+d^4z^2}} + 1 \right ) \\~~
−	~~& = \frac{d^2x^2+d^2z^2}{d^2+x^2+d^2z^2} + \frac{d^6}{d^2+x^2+d^2z^2} -2 \frac{d^3}{\sqrt{d^2+x^2+d^2z^2}} + 1 \\~~
−	~~& = \frac{d^2x^2+d^2z^2 + d^6 -2d^3\sqrt{d^2+x^2+d^2z^2} + d^2+x^2+d^2z^2}{d^2+x^2+d^2z^2} \\~~
−	~~& = \frac{d^6 + d^2 + d^2x^2 + x^2 + 2d^2z^2 -2d^3\sqrt{d^2+x^2+d^2z^2}}{d^2+x^2+d^2z^2} \\~~
−	~~& = \frac{ (x^2 + z^2) } {(d^2 + x^2 + z^2)} \frac{(d^2 + x^2 + z^2)}{ (x^2 + z^2) } \frac{d^6 + d^2(1+x^2+2z^2) + x^2 -2d^3\sqrt{d^2(1+z^2)+x^2}}{d^2(1+z^2)+x^2} \\~~
−	~~& = ? \\~~
−	~~& = \frac{ x^2 + z^2 } {d^2 + x^2 + z^2} \\~~
−	~~\frac{q}{k} & = \sqrt{ \frac{ x^2 + z^2 } {d^2 + x^2 + z^2} } \\~~
−	~~& = \frac{ \sqrt{x^2 + z^2} } {\sqrt{d^2 + x^2 + z^2 }} \\~~
−	~~& = \frac{ \left[ \sqrt{x^2 + z^2}/d \right ] } {\sqrt{1 + \left[ \sqrt{x^2 + z^2}/d \right ]^2 }} \\~~
−	~~& = \sin \left( \arctan\left [ \frac{\sqrt{x^2 + z^2}}{d} \right ] \right) \\~~
−	~~q & = \frac{2 \pi}{\lambda} \sin \left( 2 \theta_s \right)~~
−	~~\end{alignat}~~
−	~~</math>~~

@@ Line 12: / Line 12: @@
 \end{alignat}
 </math>
-where <math>\scriptstyle d</math> is the sample-detector distance, <math>\scriptstyle \alpha_f ^{\prime} </math> is the out-of-plane component (angle w.r.t. to ''y''-axis, rotation about x-axis), and <math>\scriptstyle \theta_f </math> is the in-plane component (rotation about ''z''-axis). The alternate angle, <math>\scriptstyle \alpha_f </math>, is the elevation angle in the plane defined by <math>\scriptstyle \theta_f </math>. Also note that the full scattering angle is:
+where <math>\scriptstyle d</math> is the sample-detector distance, <math>\scriptstyle \alpha_f ^{\prime} </math> is the out-of-plane component (angle w.r.t. to ''y''-axis, rotation about x-axis), and <math>\scriptstyle \theta_f </math> is the in-plane component (rotation about ''z''-axis). The alternate angle, <math>\scriptstyle \alpha_f </math>, is the elevation angle in the plane defined by <math>\scriptstyle \theta_f </math>.
+====Total scattering===
 :<math>
 \begin{alignat}{2}
@@ Line 43: / Line 44: @@
 q & = \frac{4 \pi}{\lambda} \sin \left( \theta_s \right) \\
      & = \pm \frac{4 \pi}{\lambda} \sqrt{ \frac{1-\cos 2\theta_s }{2} } \\
-     & = \pm \frac{4 \pi}{\lambda} \sqrt{ \frac{1}{2}\left(1 - \frac{d}{\sqrt{d^2+x^2+z^2}} \right) }
+     & = \frac{4 \pi}{\lambda} \sqrt{ \frac{1}{2}\left(1 - \frac{d}{\sqrt{d^2+x^2+z^2}} \right) }
 \end{alignat}
 </math>
 The [[momentum transfer]] components are:
 :<math>
@@ Line 54: / Line 63: @@
 q_z & = \frac{2 \pi}{\lambda} \sin \alpha_f
 \end{alignat}
 </math>

@@ Line 34: / Line 34: @@
 \begin{alignat}{2}
 \cos( \arctan[u]) & = \frac{1}{\sqrt{1+u^2}} \\
-\cos( 2 \theta_s ) & = \frac{1}{\sqrt{1 + (x/d)^2}} \\
+\cos( 2 \theta_s ) & = \frac{1}{\sqrt{1 + (\sqrt{x^2+z^2}/d)^2}} \\
-     & = \frac{d^2}{\sqrt{d^2+x^2}}
+     & = \frac{d^2}{\sqrt{d^2+x^2+z^2}}
 \end{alignat}
 </math>
@@ Line 42: / Line 42: @@
 \begin{alignat}{2}
 q & = \frac{4 \pi}{\lambda} \sin \left( \theta_s \right) \\
      & = \pm \frac{4 \pi}{\lambda} \sqrt{ \frac{1-\cos 2\theta_s }{2} }
 \end{alignat}

@@ Line 38: / Line 38: @@
 q_z & = \frac{2 \pi}{\lambda} \sin \alpha_f
 \end{alignat}
 </math>
@@ Line 73: / Line 79: @@
 \end{alignat}
 </math>
 ====cont====
 Continuing: