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| 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: | | 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> | | :<math> |
− | \mathbf{q} = \begin{bmatrix} q_x \\ q_y \\ q_z \end{bmatrix} | + | \begin{alignat}{2} |
| + | \mathbf{q} & = \begin{bmatrix} q_x \\ q_y \\ q_z \end{bmatrix} \\ |
| + | & = \frac{2 \pi}{\lambda} \begin{bmatrix} \sin \theta_f \cos \alpha_f \\ \cos \theta_f \cos \alpha_f - 1 \\ \sin \alpha_f \end{bmatrix} |
| + | \end{alignat} |
| </math> | | </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: | | 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: |
Revision as of 13:03, 30 December 2015
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:
![{\displaystyle {\begin{alignedat}{2}\mathbf {q} &={\begin{bmatrix}q_{x}\\q_{y}\\q_{z}\end{bmatrix}}\\&={\frac {2\pi }{\lambda }}{\begin{bmatrix}\sin \theta _{f}\cos \alpha _{f}\\\cos \theta _{f}\cos \alpha _{f}-1\\\sin \alpha _{f}\end{bmatrix}}\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ecc685926083f1bc63d2930729d50de5c460ded)
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
. The scattering angles are then:
![{\displaystyle {\begin{alignedat}{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{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/facc9ad57cd58f15e7403d40dc08f08815d3662b)
where
is the sample-detector distance,
is the out-of-plane component (angle w.r.t. to y-axis, rotation about x-axis), and
is the in-plane component (rotation about z-axis). The alternate angle,
, is the elevation angle in the plane defined by
.
Total scattering
The full scattering angle is defined by a right-triangle with base d and height
:
![{\displaystyle {\begin{alignedat}{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{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce190aa4f7dd836349234c33033cb245c49d4f20)
The total momentum transfer is:
![{\displaystyle {\begin{alignedat}{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{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/978ccef9bfce6510674b68221b9f287f94e0bc79)
Given that:
![{\displaystyle {\begin{alignedat}{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{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07a031cc7bdbb545bcbf2e5d51b1899f74c01e74)
We can also write:
![{\displaystyle {\begin{alignedat}{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{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76394d813c7c0f44c04d51102fc176b8709ab6b0)
Where we take for granted that q must be positive.
In-plane only
If
(and
), then
,
, and:
![{\displaystyle {\begin{alignedat}{2}q&=2k\sin \left(\theta _{f}/2\right)\\&=2k{\sqrt {\frac {1-\cos(\theta _{f})}{2}}}\\&=2k{\sqrt {{\frac {1}{2}}\left(1-{\frac {1}{\sqrt {1+(x/d)^{2}}}}\right)}}\\&=2k{\sqrt {{\frac {1}{2}}\left(1-{\frac {d}{\sqrt {d^{2}+x^{2}}}}\right)}}\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85a03b5b0791e7c5fabe349d8abdcdec186d5533)
The other component can be thought of in terms of the sides of a right-triangle with angle
:
![{\displaystyle {\begin{alignedat}{2}q_{x}&=q\cos(\theta _{f}/2)\\&=2k\sin(\theta _{f}/2)\cos(\theta _{f}/2)\\&=k\sin(\theta _{f})\\q_{y}&=-q\sin(\theta _{f}/2)\\&=-2k\sin(\theta _{f}/2)\sin(\theta _{f}/2)\\&=-k\left(1-\cos \theta _{f}\right)\\&=k\left(\cos \theta _{f}-1\right)\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eefe88e0501da56939ef3a2eba477c2acec02a58)
Summarizing:
![{\displaystyle \mathbf {q} ={\frac {2\pi }{\lambda }}{\begin{bmatrix}\sin \theta _{f}\\\cos \theta _{f}-1\\0\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d001872554542314e9a6f6d754b61c68de634db)
Out-of-plane only
If
, then
,
, and:
![{\displaystyle {\begin{alignedat}{2}q&=2k\sin \left(\alpha _{f}/2\right)\\&=2k{\sqrt {\frac {1-\cos(\alpha _{f})}{2}}}\\&=2k{\sqrt {{\frac {1}{2}}\left(1-{\frac {d}{\sqrt {d^{2}+z^{2}}}}\right)}}\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c8bdb3fa7cd8b98f7e259e360f83a10287b1f4f)
The components are:
![{\displaystyle {\begin{alignedat}{2}q_{z}&=q\cos(\alpha _{f}/2)\\&=2k\sin(\theta _{f}/2)\cos(\theta _{f}/2)\\&=k\sin(\alpha _{f})\\q_{y}&=-q\sin(\alpha _{f}/2)\\&=k\left(\cos \alpha _{f}-1\right)\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cbdd0fdec873084db80ffca15d1c9970ee691f3)
Summarizing:
![{\displaystyle \mathbf {q} ={\frac {2\pi }{\lambda }}{\begin{bmatrix}0\\\cos \alpha _{f}-1\\\sin \alpha _{f}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eeca99475a3570261d9b0b8e36839d72f5e19343)
Components
For arbitrary 3D scattering vectors, the momentum transfer components are:
![{\displaystyle {\begin{alignedat}{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{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd41463d2553aa7060c90934ad0057df48f2fd08)
In vector form:
![{\displaystyle \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d7c5d5bc527e6ef01820ee67cd7a418cb3d5a03)
Total magnitude
Note that this provides a simple expression for q total:
![{\displaystyle {\begin{alignedat}{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 ^{2}\theta _{f}\cos ^{2}\alpha _{f}+\cos ^{2}\theta _{f}\cos ^{2}\alpha _{f}-2\cos \theta _{f}\cos \alpha _{f}+1+\sin ^{2}\alpha _{f}\\&=\cos ^{2}\alpha _{f}(\sin ^{2}\theta _{f}+\cos ^{2}\theta _{f})+\sin ^{2}\alpha _{f}-2\cos \theta _{f}\cos \alpha _{f}+1\\&=\cos ^{2}\alpha _{f}(1)+\sin ^{2}\alpha _{f}-2\cos \theta _{f}\cos \alpha _{f}+1\\&=2-2\cos \theta _{f}\cos \alpha _{f}\\q&={\sqrt {2}}k{\sqrt {1-\cos \theta _{f}\cos \alpha _{f}}}\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dede2da75fc32af533b44b889567367d4bbff4ce)
Check
As a check of these results, consider:
![{\displaystyle {\begin{alignedat}{2}q&={\frac {4\pi }{\lambda }}\sin \left(\theta _{s}\right)\\&={\frac {4\pi }{\lambda }}{\sqrt {\frac {1-\cos 2\theta _{s}}{2}}}\\\left({\frac {q}{k}}\right)^{2}&={\frac {4}{2}}\left(1-\cos 2\theta _{s}\right)\\&=2\left(1-{\frac {1}{\sqrt {1+\left({\sqrt {\tan ^{2}\theta _{f}+{\frac {\tan ^{2}\alpha _{f}}{\cos ^{2}\theta _{f}}}}}\right)^{2}}}}\right)\\&=2\left(1-{\frac {1}{\sqrt {1+\tan ^{2}\theta _{f}+{\frac {\tan ^{2}\alpha _{f}}{\cos ^{2}\theta _{f}}}}}}\right)\\&=2-{\frac {2}{\sqrt {1+{\frac {\sin ^{2}\theta _{f}}{\cos ^{2}\theta _{f}}}+{\frac {\sin ^{2}\alpha _{f}}{\cos ^{2}\alpha _{f}\cos ^{2}\theta _{f}}}}}}\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/314b83790a25af1f4266d4bae896e8e3027ea66f)
And:
![{\displaystyle {\begin{alignedat}{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{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87b8483858d87ba0bbb25e8a76b627b38e894211)