Scattering - Revision history

KevinYager: /* See Also */

2018-01-23T14:29:24Z

‎See Also

KevinYager: /* Geometry */

2015-12-30T15:31:16Z

‎Geometry

KevinYager: /* TSAXS 1D */

2015-12-30T15:03:34Z

‎TSAXS 1D

KevinYager: /* TSAXS 3D */

2015-12-29T21:45:03Z

‎TSAXS 3D

KevinYager: /* TSAXS 3D */

2015-12-29T21:30:02Z

‎TSAXS 3D

KevinYager: /* TSAXS 3D */

2015-12-29T21:21:54Z

‎TSAXS 3D

KevinYager: /* TSAXS 3D */

2015-12-29T21:08:22Z

‎TSAXS 3D

KevinYager: /* TSAXS 3D */

2015-12-29T21:05:28Z

‎TSAXS 3D

KevinYager: /* TSAXS 3D */

2015-12-29T21:01:51Z

‎TSAXS 3D

KevinYager: /* Geometry */

2015-12-29T20:56:10Z

‎Geometry

← Older revision		Revision as of 14:29, 23 January 2018
Line 54:		Line 54:
	* [[Diffraction]]		* [[Diffraction]]
	* [[Fourier transform]]		* [[Fourier transform]]
		+	* Geometry:
		+	** [[Geometry:TSAXS 3D\|TSAXS 3D]]
		+	** [[Geometry:WAXS 3D\|WSAXS 3D]]

@@ Line 4: / Line 4: @@
 ==Geometry==
 We define a vector in [[reciprocal-space]] as the difference between the incident and scattered x-ray beams. This new vector is the [[momentum transfer]], denoted by '''q''':
 :<math>
@@ Line 23: / Line 22: @@
 Where <math>\scriptstyle 2 \theta_s </math> is the scattering angle.
-===[[TSAXS]] 3D===
+See also [[Geometry:TSAXS 3D]].
 ==Theory==

@@ Line 17: / Line 17: @@
 :<math>
 \begin{alignat}{2}
-q = |\mathbf{q}| & = k \sin { ( \theta /2 ) } \\
+q = |\mathbf{q}| & = 2 k \sin { ( 2 \theta_s /2 ) } \\
-  & = \frac{2 \pi}{\lambda} \sin{ ( \theta /2 ) } \\
+   & = \frac{4 \pi}{\lambda} \sin{ \theta_s }
 \end{alignat}
 </math>
-Where <math>\scriptstyle 2 \theta </math> is the scattering angle.
+Where <math>\scriptstyle 2 \theta_s </math> is the scattering angle.
 ===[[TSAXS]] 3D===

@@ Line 25: / Line 25: @@
 ===[[TSAXS]] 3D===
-The ''q''-vector in fact has three components:
+TBD
 ==Theory==

@@ Line 40: / Line 40: @@
 \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_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}
@@ Line 50: / Line 50: @@
 \end{alignat}
 </math>
 ==Theory==

@@ Line 29: / Line 29: @@
 \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 pixel position <math>\scriptstyle (x,z) </math>. The scattering angles are then:
+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}
-\alpha_f & = 0 \\
+\theta_f & = \arctan( x/d ) \\
-\theta_f & = 0
+\alpha_f & = \arctan( z/d )
 \end{alignat}
 </math>
-where <math>\scriptstyle \alpha_f</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 total scattering angle <math>\scriptstyle 2 \theta = \Theta </math> is related via:
+where <math>\scriptstyle d</math> is the sample-detector distance, <math>\scriptstyle \alpha_f</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 total scattering angle <math>\scriptstyle 2 \theta = \Theta </math> is related via:
@@ Line 43: / Line 43: @@
 :<math>
 \begin{alignat}{2}
-q_x & = \frac{2 \pi}{\lamdbda} \\
+q_x & = \frac{2 \pi}{\lambda} \\
-q_y & = \frac{2 \pi}{\lamdbda} \\
+q_y & = \frac{2 \pi}{\lambda} \\
-q_z & = \frac{2 \pi}{\lamdbda}
+q_z & = \frac{2 \pi}{\lambda}
 \end{alignat}
 </math>