Attenuation correction for sample shape

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In cases of strongly scattering or absorbing samples, the detected scattering intensity is lower than the 'true' scattering. Moreover, for oddly-shaped samples, the extinction of scattering may be anisotropic: some parts of the detector image are more attenuated than others because of the differing path-lengths through the sample.

Formulation

In the following, we assume a transmission-scattering (TSAXS) experiment.

The measured scattering, S_m, at a particular scattering angle (\Theta_o, \chi_o) (where \Theta_o is the full (2 \theta) scattering angle between the scattered ray and the incident beam, and \chi_o is the azimuthal angle: \chi=0^{\circ} corresponds to the q_z axis, whereas \chi=90^{\circ} is along the q_r axis) can be computed by summing the scattering contributions for all the elements along the beam path through the sample.

We define a realspace coordinate system (x,y,z) where z points vertically, y points along the beam direction, and x points horizontally with respect to the sample. Let the sample size along the beam direction be L. Defining the point where the beam first enters the sample as (0,0,0) we write:


\begin{alignat}{2}
S_m(\Theta_o,\chi_o) = \int \limits_{l=0}^{l=L} S(l) \mathrm{d}l
\end{alignat}

The scattering from a particular location within the sample is affected by two attenuation effects:

  1. The beam flux within the sample decreases due to absorption/scattering, such that the flux at position l is not the incident flux, I_0 but attenuated to:
    I(l) = I_0 e^{- \alpha l }
    where \alpha is (Beer-Lambert like) extinction coefficient. If the 'true' scattering probability is given by \sigma (i.e. I_0 \sigma is the scattering observed in the absence of attenuation), then the scattering at l is:
    S(l) = I(l) \sigma = I_0 e^{-\alpha l} \sigma
  2. The scattered radiation is itself attenuated as it passes through the sample. Let this path-length (from scattering location (0,l,0) until it exits the sample along the direction (\Theta_o, \chi_o)) be denoted p(l). In such a case, the scattering that exits the sample is:
    
\begin{alignat}{2}
S(l) 
    & = I(l) \sigma \times \mathrm{attenuation}(l) \\
    & = I_0 e^{-\alpha l} \sigma e^{-\alpha p(l)} \\
    & = I_0 \sigma e^{-\alpha (l+p(l))}\\
\end{alignat}

The measured scattering is thus:


\begin{alignat}{2}
S_m(\Theta_o,\chi_o)
    & = \int \limits_{l=0}^{L} I_0 \sigma e^{-\alpha (l+p(l))} \mathrm{d}l \\
    & = I_0 \sigma \int \limits_{0}^{L} e^{-\alpha (l+p(l))} \mathrm{d}l \\
\end{alignat}

The integral of course depends on the form of p(l) which depends on the sample shape. Note that in the limiting case of weak attenuation (\alpha\approx0), we obtain the very simple result:


\begin{alignat}{2}
S_m(\Theta_o,\chi_o)
    & = I_0 \sigma \int \limits_{0}^{L} e^{0} \mathrm{d}l \\
    & = I_0 \sigma  L \\
\end{alignat}

As expected, scattering intensity scales with the scattering volume.

Normalization

To normalize-out the effect of attenuation, one can simply divide by the integral:


\begin{alignat}{2}
S_{\mathrm{norm}} (\Theta_o,\chi_o)
    & = \frac{S_m(\Theta_o,\chi_o)}{\int_{0}^{L} e^{-\alpha (l+p(l))} \mathrm{d}l } \\
    & = I_0 \sigma 
\end{alignat}

Of course in the case of weak attenuation the integral is simply L, and we are normalizing by the beam-path through the sample.

Coordinates

For a vector that starts at (0,0,0) and terminates at (x,y,z), pointing along direction (\Theta_o,\chi_o), the full length is:

|\mathbf{v}| = v = \sqrt{ x^2 + y^2 + z^2 }

We can consider triangles in various planes:

  1. xz plane (looking along beam):
     \tan(90^{\circ}-\chi_o) = \frac{z}{x}
     \cos(90^{\circ}-\chi_o) = \frac{x}{\sqrt{x^2+z^2}}
     \sin(90^{\circ}-\chi_o) = \frac{z}{\sqrt{x^2+z^2}}
     \tan(\chi_o) = \frac{x}{z}
     \cos(\chi_o) = \frac{z}{\sqrt{x^2+z^2}}
     \sin(\chi_o) = \frac{x}{\sqrt{x^2+z^2}}
  2. xy plane (looking from above):
     \tan(\omega_{xy}) = \frac{x}{y}
     \cos(\omega_{xy}) = \frac{y}{\sqrt{x^2+y^2}}
     \sin(\omega_{xy}) = \frac{x}{\sqrt{x^2+y^2}}
  3. yz plane (looking from side):
     \tan(\omega_{yz}) = \frac{z}{y}
     \cos(\omega_{yz}) = \frac{y}{\sqrt{y^2+z^2}}
     \sin(\omega_{yz}) = \frac{z}{\sqrt{y^2+z^2}}
  4. plane of beam elevation:
     \tan(\omega_{\mathrm{elevation}}) = \frac{z}{\sqrt{x^2+y^2}}
     \cos(\omega_{\mathrm{elevation}}) = \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}
     \sin(\omega_{\mathrm{elevation}}) = \frac{z}{\sqrt{x^2+y^2+z^2}}
  5. plane of full scattering angle:
     \tan(\Theta_o) = \frac{\sqrt{x^2+z^2}}{y}
     \cos(\Theta_o) = \frac{y}{\sqrt{x^2+y^2+z^2}}
     \sin(\Theta_o) = \frac{\sqrt{x^2+z^2}}{\sqrt{x^2+y^2+z^2}}

Height Z

If the vector's final point is at height z=Z:



\begin{alignat}{2}
v_Z
    & = \sqrt{ x^2 + y^2 +Z^2 } \\
    & = \frac{ \sqrt{x^2+Z^2} }{  \sin(\Theta_o)  } \\
    & = \frac{ 1 }{  \sin(\Theta_o)  } \sqrt{\left(  Z \tan(\chi_o)  \right)^2 + Z^2} \\
    & = \frac{ Z }{  \sin(\Theta_o)  } \sqrt{\tan^2(\chi_o)  + 1} \\
    & = \frac{ Z }{  \sin(\Theta_o)  } \sec(\chi_o) \\
    & = \frac{ Z }{  \sin(\Theta_o) \cos(\chi_o) }  \\
\end{alignat}

This has a minimum of v_z=Z when (\Theta_o,\chi_o)=(90^{\circ},0^{\circ}).

Depth L

If the vector's final position is at depth y=L:



\begin{alignat}{2}
v_Y
    & = \sqrt{ x^2 + L^2 +z^2 } \\
    & = \frac{ L }{  \cos(\Theta_o)  } \\
\end{alignat}

This has a minimum of v_Y=L when \Theta_o=0^{\circ}.

Width X

If the vector's final position is at width x=X:


\begin{alignat}{2}
v_X
    & = \sqrt{ X^2 + y^2 +z^2 } \\
    & = \frac{ \sqrt{X^2+z^2} }{  \sin(\Theta_o)  } \\
    & = \frac{ 1 }{  \sin(\Theta_o)  } \sqrt{X^2 + \left(  \frac{X}{ \tan(\chi_o) }  \right)^2} \\
    & = \frac{ |X| }{  \sin(\Theta_o)  } \sqrt{1 + \frac{1}{ \tan^2(\chi_o) } } \\
    & = \frac{ X }{  \sin(\Theta_o)  } \sqrt{\frac{\tan^2(\chi_o)+1}{ \tan^2(\chi_o) } } \\
    & = \frac{ X }{  \sin(\Theta_o)  } \frac{\sqrt{\tan^2(\chi_o)+1}}{ \sqrt{\tan^2(\chi_o) }} \\
    & = \frac{ X }{  \sin(\Theta_o)  } \frac{ \sec(\chi_o) }{ \tan(\chi_o) } \\
    & = \frac{ X \cos(\chi_o) }{  \sin(\Theta_o) \cos(\chi_o) \sin(\chi_o)  } \\
    & = \frac{ X }{  \sin(\Theta_o) \sin(\chi_o)  } \\
\end{alignat}

This has a minimum of v_X=X when (\Theta_o,\chi_o)=(90^{\circ},90^{\circ}).

Rectangular prism

If the sample is a rectangular prism with dimensions (2X, 2Y, 2Z) = (2X, L, 2Z) and the beam falls upon the center of the xz front-face, then the beam travels a distance L through the sample, and the scattered radiation in any quadrant passes through the rectangular prism of size (X,L,Z). The distance from (0,l,0) to the exit-point from the sample is the distance to the closest sample face:


\begin{alignat}{2}
p\left(l\right) 
    & = \mathrm{min}( d_{\mathrm{top}}(l) , d_{\mathrm{back}}(l) , d_{\mathrm{side}}(l) ) \\
\end{alignat}

The distances are:


\begin{alignat}{2}
d_{\mathrm{top}}(l)
    & = v_Z \\
    & = \frac{ Z }{  \sin(\Theta_o) \cos(\chi_o) }  \\
\end{alignat}

\begin{alignat}{2}
d_{\mathrm{back}}(l)
    & = v_{L-l} \\
    & = \frac{ L-l }{  \cos(\Theta_o)  } \\
\end{alignat}

\begin{alignat}{2}
d_{\mathrm{side}}(l)
    & = v_X \\
    & = \frac{ X }{  \sin(\Theta_o) \sin(\chi_o)  } \\
\end{alignat}