Geometry:TSAXS 3D - Revision history

KevinYager: /* Components (distances) */

2016-01-13T17:11:38Z

‎Components (distances)

KevinYager: /* Total magnitude */

2016-01-13T17:08:52Z

‎Total magnitude

KevinYager: /* Components */

2016-01-13T17:06:58Z

‎Components

KevinYager: /* Total scattering */

2016-01-13T16:42:07Z

‎Total scattering

KevinYager at 17:03, 30 December 2015

2015-12-30T17:03:21Z

KevinYager: /* Components */

2015-12-30T17:00:12Z

‎Components

KevinYager: /* Check */

2015-12-30T16:48:22Z

‎Check

KevinYager: /* Alternate check */

2015-12-30T16:37:43Z

‎Alternate check

KevinYager: /* Check 2 */

2015-12-30T16:37:19Z

‎Check 2

KevinYager: /* Check */

2015-12-30T16:36:49Z

‎Check

@@ Line 176: / Line 176: @@
 </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>
+::<math>
 \begin{alignat}{2}
 \frac{1}{\sqrt{d^2+z^2 \cos^2 \theta_f }}

@@ Line 205: / Line 205: @@
 ===Total magnitude===
 :<math>
-  - \frac{d}{\sqrt{x^2 + d^2 + z^2}}\begin{alignat}{2}
+\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 \\

@@ Line 103: / Line 103: @@
 </math>
-==Components==
+==Components (angular)==
 For arbitrary 3D scattering vectors, the [[momentum transfer]] components are:
 :<math>
@@ Line 154: / Line 154: @@
      = & \frac{\cos \theta_f \cos \alpha_f }{ \sqrt{ \cos^2 \alpha_f (\cos^2 \theta_f + \sin^2 \theta_f) + \sin^2 \alpha_f }}   \\
      = & \cos \theta_f \cos \alpha_f
 \end{alignat}
 </math>

@@ Line 46: / Line 46: @@
 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) }
+     & = \frac{4 \pi}{\lambda} \sqrt{ \frac{1}{2}\left(1 - \frac{d}{\sqrt{d^2+x^2+z^2}} \right) } \\
 \end{alignat}
 </math>

@@ Line 1: / Line 1: @@
 In transmission-[[SAXS]] ([[TSAXS]]), the x-ray beam hits the sample at normal incidence, and passes directly through without [[refraction]]. TSAXS is normally considered in terms of the one-dimensional [[momentum transfer]] (''q''); however the full 3D form of the ''q''-vector is necessary when considering [[scattering]] from anisotropic materials. The ''q''-vector in fact has three components:
 :<math>
-\mathbf{q} = \begin{bmatrix} q_x \\ q_y \\ q_z \end{bmatrix}
+\begin{alignat}{2}
 </math>
 This vector is always on the surface of the [[Ewald sphere]]. 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:

← Older revision		Revision as of 16:48, 30 December 2015
Line 142:		Line 142:
	</math>		</math>
	And:		And:
		+	:<math>
		+	\begin{alignat}{2}
		+	& \left( 1+\frac{\sin^2 \theta_f}{\cos^2 \theta_f} + \frac{ \sin^2 \alpha_f }{ \cos^2 \alpha_f \cos^2 \theta_f } \right) ^{-1/2} \\
		+	= & \left( \frac{\cos^2 \alpha_f \cos^2 \theta_f}{\cos^2 \alpha_f \cos^2 \theta_f}+\frac{\cos^2 \alpha_f \sin^2 \theta_f}{\cos^2 \alpha_f \cos^2 \theta_f} + \frac{ \sin^2 \alpha_f }{ \cos^2 \alpha_f \cos^2 \theta_f } \right) ^{-1/2} \\
		+	= & \left( \frac{\cos^2 \alpha_f \cos^2 \theta_f + \cos^2 \alpha_f \sin^2 \theta_f + \sin^2 \alpha_f}{\cos^2 \alpha_f \cos^2 \theta_f} \right) ^{-1/2} \\
		+	= & \left( \frac{\cos^2 \theta_f \cos^2 \alpha_f }{\cos^2 \alpha_f \cos^2 \theta_f + \cos^2 \alpha_f \sin^2 \theta_f + \sin^2 \alpha_f} \right) ^{+1/2} \\
		+	= & \frac{\cos \theta_f \cos \alpha_f }{ \sqrt{ \cos^2 \alpha_f (\cos^2 \theta_f + \sin^2 \theta_f) + \sin^2 \alpha_f }} \\
		+	= & \cos \theta_f \cos \alpha_f
		+	\end{alignat}
		+	</math>

← Older revision		Revision as of 16:37, 30 December 2015
Line 142:		Line 142:
	</math>		</math>
	And:		And:
−
−	~~====Alternate 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 } \\~~
−	~~\left( \frac{q}{k} \right)^2~~
−	~~& = (\sin \theta_f)^2 (\cos \alpha_f)^2 + \left ( \cos \theta_f \cos \alpha_f - 1 \right )^2 + (\sin \alpha_f)^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>~~
−	~~Continuing:~~
−	~~:<math>~~
−	~~\begin{alignat}{2}~~
−	~~\left( \frac{q}{k} \right)^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{\sqrt{d^2+x^2+z^2} - d}{2 \sqrt{d^2+x^2+z^2}} \\~~
−	~~& = \frac{1}{2}\left(1 - \frac{d}{\sqrt{d^2+x^2+z^2}} \right) \\~~
−	~~q & = \frac{4 \pi}{\lambda} \sin \left( \theta_s \right)~~
−	~~\end{alignat}~~
−	~~</math>~~

@@ Line 143: / Line 143: @@
 And:
-===Check 2===
+====Alternate check====
 As a check of these results, consider:
 :<math>
@@ Line 175: / Line 175: @@
 \end{alignat}
 </math>
 Continuing:
 :<math>