## Scattering invariant >[!Notice] >This quantity does not rely on the *identical particles* assumption. Define a quantity $Q$ named **scattering invariant**, who has the expression, $Q = \int \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} (\mathbf{q}) \, \mathrm{d}^3q$ namely the integral of macroscopic differential scattering cross-section $\frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}$ over $d^{3}q$. (See [[Fraunhofer scattering#Scattering cross-section and scattering length density under Fraunhofer condition]] for the expression for cross-section) $\begin{aligned} Q &= \int \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}(\mathbf{q}) \, \mathrm{d}^3q \\ &= \frac{1}{V} \int \mathrm{d}^3q \int \mathrm{d}^3r \int \mathrm{d}^3r' \Delta \rho(r) \Delta \rho(r') e^{-i\mathbf{q} \cdot (\mathbf{r} - \mathbf{r}')} \\ &= \left(\frac{2\pi}{V}\right)^3 \int \mathrm{d}^3r \int \mathrm{d}^3r' \Delta \rho(r) \Delta \rho(r') \delta(\mathbf{r} - \mathbf{r}') \\ &= \left(\frac{2\pi}{V}\right)^3 \int \mathrm{d}^3r \Delta \rho^2(r) \\ &= \left(2\pi\right)^3 \left\langle \left(\rho(\mathbf{r}) - \langle\rho\rangle\right)^2 \right\rangle \end{aligned}$ We reach an arrangement independent term eventually. In the derivation we did a Fourier transform on $q$, and applied [Plancherel theorem](https://en.wikipedia.org/wiki/Plancherel_theorem), given as $(2\pi)^n \int \left| f(\mathbf{r}) \right|^2 d^n\mathbf{r} = \int \left| \mathcal{F}(\mathbf{q}) \right|^2 d^n\mathbf{q}$. This makes integration of $q$ becomes a delta function. Inside the bracket, we have the difference between SLDs at $\mathbf{r}$ and average. For any $\rho$, we have $\langle (\rho - \langle \rho \rangle)^2 \rangle = \langle \rho^2 \rangle - \langle \rho \rangle^2$ For a two phase system, we have phase 1 $\rho_{1}$, $\phi_{1}$ and phase 2 $\rho_{2}$, $\phi_2$, $\phi_{1}+\phi_{2}=1$. Then $\begin{aligned} \langle \rho^2 \rangle &= \phi_1 \rho_1^2 + \phi_2 \rho_2^2 \\ \langle \rho \rangle &= \phi_1 \rho_1 + \phi_2 \rho_2 \end{aligned} $ Insert it into $Q$, $\begin{aligned} Q &= \left(2\pi\right)^3 \left\langle \left(\rho(\mathbf{r}) - \langle\rho\rangle\right)^2 \right\rangle\\ &= (2\pi)^{3} (\phi_1 \rho_1^2 + \phi_2 \rho_2^2-(\phi_1 \rho_1 + \phi_2 \rho_2)^{2})\\ &= (2\pi)^{3}(\phi_{1}\phi_{2}\rho_{1}^{2} + \phi_{1}\phi_{2}\rho_{2}^{2}-2\phi_{1}\phi_{2}\rho_{1}\rho_{2})\\ &=(2\pi)^{3}\phi_{1}(1-\phi_{1})(\rho_{1}-\rho_{2})^{2} \end{aligned}$ This invariant gives volume fraction. Sometimes this scattering invariant is called Porod invariant. This Porod invariant is just $\frac{Q}{4\pi}$, though there is a pre-factor change but the idea is the same. For three-phase case, one have the Porod invariant written as $\text{Invariant} = 2\pi^2 (\rho_1 \varphi_1 \Delta_1^2 + \rho_2 \varphi_2 \Delta_2^2 + \rho_3 \varphi_3 \Delta_3^2)$ >[!Info] >See [[A New Insight into Growth Mechanism and Kinetics of Mesoporous Silica Nanoparticles by in Situ Small Angle X‑ray Scattering]] for 3 phase expression, and [[SANS_NR_Intro.pdf]]. The online version is https://www.ncnr.nist.gov/summerschool/ss10/pdf/SANS_NR_Intro.pdf. This is a very good introductory paper, answered a lot of questions that I'm not clear.