X-ray Scattering plots provide the intensity of X-ray signal defined as $I(\pmb k)$. The intensity can be related to the Fourier transform of the electron density denoted as $A$ (the electron density will be represented as $\rho(\pmb x)$).

$$A(\pmb{k})= \int \rho(\pmb x) e^{i\pmb{kx}} d\pmb x$$

$$ I(\pmb k) = |A(\pmb k)| = A(\pmb k) (A(\pmb k))^*$$

where $(A(\pmb k ))^*$ is the complex conjugate of $A(\pmb k)$. $I(\pmb k)$ and $\rho(\pmb x)$ have the following relationship.

$$ I(\pmb k) = \int \rho(\pmb x) e^{i\pmb{kx}} dx \int \rho(\pmb y) e^{-i\pmb {ky}} d\pmb y $$

$$ I(\pmb k) = \int \int \rho(\pmb x) \rho(\pmb y) e^{i\pmb k(\pmb x - \pmb y)} d\pmb x d\pmb y$$

Now lets make a change of variable and let

$$ \pmb z = \pmb x - \pmb y$$

Substituting this relationship back in the the equation we get.

$$ I(\pmb k) = \int \int \rho(\pmb z + \pmb y) \rho(\pmb y) e^{i\pmb k\pmb z} d\pmb z d\pmb y = \int \Gamma_{\rho}(\pmb z) e^{i\pmb k\pmb z} d\pmb z$$ $$ I(\pmb k) = \mathscr{F} \Big \{ \Gamma_{\rho}(\pmb z)\Big \} $$

where

$$ \Gamma_{\rho}(\pmb z) = \int \rho( \pmb x) \rho( \pmb x + \pmb z) d \pmb x $$

Therefore the autocorrelaiton of the electron density $\rho(\pmb x)$ equal to the inverse Fourier transform of the intensity $I(\pmb k)$.

$$ \Gamma_{\rho}(\pmb z) = \mathscr{F}^{-1} \{ I(\pmb k) \} $$

$$ \Gamma_{\rho}(\pmb z)= \int I(\pmb k) e^{-i\pmb k\pmb z} d\pmb k \space \space \space \space \pmb{(1)}$$

$$ \Gamma_{\rho}(\pmb z) = \int I(\pmb k) [cos(\pmb k \pmb z) - i sin(\pmb k \pmb z)] d\pmb k $$

$$ \Gamma_{\rho}(\pmb z) = \int I(\pmb k) cos(\pmb k\pmb z)d\pmb z - i \int I sin(\pmb k\pmb z)d\pmb k $$

In the case that $I(\pmb k)$ is symetric integral containing $sin(\pmb k \pmb z)$ will always be equal to zero.

$$ \Gamma_{\rho}(\pmb z) = \int I(\pmb k) cos(\pmb k\pmb z)d\pmb k $$

SAXS data provides the difference in electron density in the range of 1 to 1,000 nm. This difference can be represented as deviation from the mean electron density, $\bar{\rho}$ as shown below.

$$ \eta( \pmb z) = \rho( \pmb z) - \bar \rho$$

Let $\Gamma_{\eta} (\pmb x)$ represent the autocorrelation of $\eta ( \pmb x)$.

$$ \Gamma_{\eta} (\pmb x) = \int \eta ( \pmb y) \eta ( \pmb y + \pmb x) d \pmb y $$

Let $\gamma (\pmb x)$ represent the normalized autocorrelation of $\eta ( \pmb x)$ and is referred to as the Debye correlation function [1].

$$ \gamma (\pmb x) = \frac{\Gamma_{\eta} (\pmb x)}{\Gamma_{\eta} (0)} $$

The normalization constant is

$$\Gamma_{\eta}(0) = \int \eta ( \pmb y) \eta ( \pmb y + \pmb 0) d \pmb y = V \eta_o^2$$

where $V$ is the scattering volume and $\eta_o^2$ is the mean square perturbation of the scattering length density throughout the system [1].

The relationship between the Debye correlation function and the correlation function of $\eta( \pmb x)$ can be expressed as follows.

$$ \Gamma_{\eta} (\pmb x) = V \eta_o^2 \gamma (\pmb x) $$

This new definition allows us to write equation $(1)$ in terms of $\gamma (\pmb x)$.

$$ \gamma (\pmb x) = \frac{1}{V \eta_o^2} \int I ( \pmb k) e^{-i\pmb k\pmb z} d\pmb k \space \space \space \space (2)$$

The $\gamma (\pmb x)$ is related to the autocorrelation for local state one was shown below [2].

$$ \gamma (\pmb x) = f^{11}( \pmb x) - V_1^2 \space \space \space \space (3)$$

where $ f^{11}( \pmb x)$ is the autocorrelation of local state one and $V_1$ is the volume fraction of local state one.

By combining of equations $(2)$ and $(3)$ provides the relationship between the 2-point statistics, X-ray scattering intensity and volume fraction.

$$f^{11}( \pmb x) = \frac{1}{V \eta_o^2} \int I ( \pmb k) e^{-i\pmb k\pmb x} d\pmb k + V_1^2$$

[1] R. J. Roe, *Methods of X-Ray and Neutron Scattering in Polymer Science*, 174-176, Oxford University Press 2000.

[2] M. Baniassadi, F. Addiego, F. Hassouna, A. Laachachi, A. Makradi, S. Ahzi, V. Toniazzo, H. Garmestani, D. Ruch *Using SAXS approach to estimate thermal conductivity of polystyrene/zirconia nanocomposite by exploiting strong contrast technique*, Acta Mater (2011) doi:10.1016/j.actamat.2011.01.013.