## Differential scattering cross-section in SAS Finally we are able to answer the question posted at the beginning of [[Coherent and incoherent scattering]]. Recall the macroscopic differential cross-section given in [[Fraunhofer scattering#Scattering cross-section and scattering length density under Fraunhofer condition|Fraunhofer scattering]] $\frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}= \frac{1}{V} \iint \mathrm{d}\mathbf{r} \mathrm{d}\mathbf{r}' \, \rho(\mathbf{r}) \rho(\mathbf{r}') e^{-i \mathbf{q} \cdot (\mathbf{r} - \mathbf{r}')}$ We have multiple ways to interpret this result for small angle scattering. It is worth to re-emphasize that, this differential scattering cross-section is the **intensity** we measured. ### Cross-section signal as deviation from the average An intuitive idea is rewrite the cross-section as a deviation from the average, this does not requires separating solvent and particles. The SLD of the sample can be written as $\rho(\mathbf{r})=\langle \rho \rangle+\Delta\rho(\mathbf{r})$ Insert the expression in cross-section above, $\begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}&= \frac{1}{V} \left| \rho(\mathbf{q}) \right|^{2}\\ &=\frac{1}{V}\left|\int \mathrm{d}\mathbf{r} \rho(\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}} \right|^{2}\\ &=\frac{1}{V}\left|\int \mathrm{d}\mathbf{r} (\langle\rho \rangle +\Delta\rho(\mathbf{r}))e^{-i\mathbf{q}\cdot \mathbf{r}} \right|^{2}\\ &=\frac{1}{V}\left|\langle\rho\rangle\delta{\mathbf{q}}+ \int \mathrm{d}\mathbf{r} \Delta\rho(\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}} \right|^{2}\\ &=\frac{1}{V}\left| \int \mathrm{d}\mathbf{r} \Delta\rho(\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}} \right|^{2}\\ &=\frac{1}{V} \left|\Delta\rho(\mathbf{q})\right|^{2} \\ &=\frac{1}{V} \iint \mathrm{d}\mathbf{r}\mathrm{d}\mathbf{r'} \Delta\rho(\mathbf{r})\Delta\rho(\mathbf{r'})e^{-i \mathbf{q} \cdot (\mathbf{r} - \mathbf{r}')} \end{aligned} $ Forward scattering $\langle\rho\rangle\delta{\mathbf{q}}$ is neglected because we typically do not have it counted in our experiments. So the signal is from the contrast! In this part we have $\Delta\rho(\mathbf{r})=\rho(\mathbf{r}) -\langle \rho \rangle$, later we may have different definition (particles and solvents). Writing it this way gives [[Scattering invariant]]. ### Cross-section signal as particle-solvent contrast We now wanna convert the signal to something easier to analysis, like the pair distribution function and something describes the shape of the particle. That is, we wanna extract [[Structure factor S(q) and density autocorrelation functions]] and [[Form factor]] from the scattering signal. Consider mesoscopic particles like colloidal particles. ![[Drawing 2024-04-16 11.42.32.excalidraw.svg]] Since we are now dealing with mesoscopic particles, we use scattering length density of materials $\rho_{p}(\mathbf{r})$. Spatial dependence is still kept for uneven distribution of materials inside the particle. The solvent is $\rho_{0}$, which is a position independent constant. Then $\rho(\mathbf{r})$ can be expressed as: $\rho(\mathbf{r}) = \begin{cases} \rho_{p,i}(\mathbf{r}) & \text{inside the particle i} \\ \rho_{0} & \text{otherwise (i.e., in solvent)} \end{cases} $ Now the cross-section can be written as $\begin{aligned} V\frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}&= \Bigg|\int \mathrm{d}\mathbf{r} \, \rho(\mathbf{r}) e^{-i \mathbf{q} \cdot \mathbf{r}}\Bigg|^{2}\\ &= \Bigg|\int_{solvent} \mathrm{d}\mathbf{r} \, \rho(\mathbf{r}) e^{-i \mathbf{q} \cdot \mathbf{r}}+\int_{particles} \mathrm{d}\mathbf{r} \, \rho(\mathbf{r}) e^{-i \mathbf{q} \cdot \mathbf{r}}\Bigg|^{2}\\ &= \Bigg|\int_{solvent} \mathrm{d}\mathbf{r} \, \rho_0\ e^{-i \mathbf{q} \cdot \mathbf{r}}+\sum_{i}\int_{particle\ i} \mathrm{d}\mathbf{r} \, (\rho_{p,i}(\mathbf{r})+\rho_{0}-\rho_{0}) e^{-i \mathbf{q} \cdot \mathbf{r}}\Bigg|^{2}\\ &= \Bigg|\int_{entire\ sample} \mathrm{d}\mathbf{r} \, \rho_0\ e^{-i \mathbf{q} \cdot \mathbf{r}}+\sum_{i}\int_{particle\ i} \mathrm{d}\mathbf{r} \, (\rho_{p,i}(\mathbf{r}) -\rho_{0}) e^{-i \mathbf{q} \cdot \mathbf{r}}\Bigg|^{2}\\ &= \Bigg|\rho_{0}\ \delta(\mathbf{q})+\sum_{i}\int_{particle\ i} \mathrm{d}\mathbf{r} \, (\rho_{p,i}(\mathbf{r}) -\rho_{0}) e^{-i \mathbf{q} \cdot \mathbf{r}}\Bigg|^{2}\\ &=\left|\rho_{0}\ \delta(\mathbf{q})+ \sum_{i}\int_{particle\ i} \mathrm{d}\mathbf{r} \Delta \rho_{p,i} (\mathbf{r}) e^{-i \mathbf{q} \cdot \mathbf{r}}\ \right|^{2} \end{aligned}$ >[!Notice] >Special attention has to be made for vector $\mathbf{r}$. This is a vector covering the entire sample. So $\Delta \rho_{p,i}(\mathbf{r})$ is not a local property, but has the structure information (i.e., how particles are arranged) included. >To separation structure and form information, we need convert this delta to a localized property. That's why we introduce $\mathbf{r}_{i}$ later. Similarly, remove the forward scattering term $\rho_{0}\ \delta(\mathbf{q})$, we have the signal being $\frac{1}{V} \left|\sum_{i}\int \mathrm{d}\mathbf{r} \Delta \rho_{p,i} (\mathbf{r}) e^{-i \mathbf{q} \cdot \mathbf{r}}\ \right|^{2}$. Consider the shift vector $\mathbf{r}_{i}$, pointing form the initial point of $\mathbf{r}$ to a point inside particle $i$. Then we can write the original vector $\mathbf{r}$ as $\mathbf{r}_{i}+\mathbf{r}_{p}$, this $\mathbf{r}_{p}$ is a vector only inside the particle. Now insert $\mathbf{r}=\mathbf{r}_{i}+\mathbf{r}_{p}$ into the signal expression, we get $ \begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} &= \frac{1}{V} \left|\sum_{i}\int_{particle\ i} \mathrm{d}\mathbf{r} \Delta \rho_{p,i} (\mathbf{r}) e^{-i \mathbf{q} \cdot \mathbf{r}}\ \right|^{2}\\ &=\frac{1}{V}\left|\sum_{i}\int_{particle\ i} \mathrm{d}\mathbf{r} \Delta \rho_{p,i} (\mathbf{r}_{i}+\mathbf{r}_{p}) e^{-i \mathbf{q} \cdot (\mathbf{r}_{i}+\mathbf{r}_{p})}\ \right|^{2}\\ \end{aligned} $ Notice this scattering length density is still defined with respect to $\mathbf{r}$ and not a local term. Let $\varrho_{p,i}(\mathbf{r}_{p})=\rho_{p,i} (\mathbf{r}_{i}+\mathbf{r}_{p})$, in this way we define the term $\varrho$ a function of SLD inside a specific particle $i$ and only dependent on the local displacement vector $\mathbf{r}_{p}$. Therefore, we have $\begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} &=\frac{1}{V}\left|\sum_{i}\int_{particle\ i} d(\mathbf{r}_{i}+\mathbf{r}_{p}) \Delta \rho_{p,i} (\mathbf{r}_{i}+\mathbf{r}_{p}) e^{-i \mathbf{q} \cdot (\mathbf{r}_{i}+\mathbf{r}_{p})}\ \right|^{2}\\ &=\frac{1}{V}\left|\sum_{i}\int_{particle\ i} \mathrm{d}\mathbf{r}_{p} \Delta \varrho_{p,i} (\mathbf{r}_{p}) e^{-i \mathbf{q} \cdot (\mathbf{r}_{i}+\mathbf{r}_{p})}\ \right|^{2}\\ &=\frac{1}{V}\left|\sum_{i} e^{-i \mathbf{q} \cdot \mathbf{r}_{i}}\int_{particle\ i} \mathrm{d}\mathbf{r}_{p} \Delta \varrho_{p,i} (\mathbf{r}_{p}) e^{-i \mathbf{q} \cdot \mathbf{r}_{p}}\ \right|^{2}\\ \end{aligned}$ Now we separate $\mathbf{r}_{i}$ from the integration, and it's only a local integration for particle $i$ with respect to $\mathbf{r}_{p}$, and $\mathbf{q}$ is the only variable (SLD is something fixed if we have the system determined). Assign this integral a new name $F_{i}(\mathbf{q}) = \int_{particle\ i} \mathrm{d}\mathbf{r}_{p} \Delta \varrho_{p,i} (\mathbf{r}_{p}) e^{-i \mathbf{q} \cdot \mathbf{r}_{p}}$ Then the signal becomes $\frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}=\frac{1}{V}\left|\sum_{i} F_{i}(\mathbf{q})\, e^{-i \mathbf{q} \cdot \mathbf{r}_{i}} \right|^{2}$ >[!Notice] >It is always important to remember that $F_{i}$ is now a local property. Just like for atomic systems we can write individual scattering event (as scattering length) both in real space $b_{i}\delta(\mathbf{r}-\mathbf{r}_{i})$ and reciprocal space $b_{i} e^{-i \mathbf{q} \cdot \mathbf{r}_i}$, we can also represent such event for particle $i$ in real space as $\Delta \varrho_{p,i} (\mathbf{r}_{p})$ and reciprocal space as $F_{i}(\mathbf{q})$. **$F$ and $\Delta \rho$ is a Fourier pair.** >>[!Note]- >>As shown above, this $F_{i}$ is a function similar to scattering length. If one is interested, a *parallel definition* (which does not really exist) can be made for scattering length density. Use capital $\rho$, namely $\mathrm{P}$ for representation. A reasonable one would be >>$\mathrm{P}(\mathbf{q})=\sum_{i} F_{i}(\mathbf{q})\, e^{-i \mathbf{q} \cdot \mathbf{r}_{i}}$ >>$\mathrm{P}(\mathbf{r})=\sum_{i} \Delta \varrho_{p,i} (\mathbf{r}_{p})\chi_{i}(\mathbf{r})$ >>This $\chi_{i}$ is an indicator function, defined as >>$\chi_i(\vec{r}) = \begin{cases} 1 & \text{if } \vec{r} \in \text{particle } i, \\ 0 & \text{otherwise}. \end{cases}$ >> > >Some other aspects one should bear in mind is >- We may replace $\Delta \varrho$ with $\Delta \rho$ in later pages for simplicity. See the context to get the exact meaning. But this is typically not a problem, because >- If the material of particles (SLD) is well distributed (which is true in many of cases), $\varrho_{p,i}(\mathbf{r}_{p})=\rho_{p,i} (\mathbf{r}_{i}+\mathbf{r}_{p})=\rho_{p,i} (\mathbf{r}_{p})$. Translational vector does not affect the value of SLD. >- $\Delta \rho$ may be difference between particles and solvent, or deviation from average. Notice the context. >- It is safe here to integrate over particle $i$ because we did not and will not have Fourier transform on this part. In contrast, the previous integration range on $\rho_{0}$ was compensated to the entire sample (or say entire space) is because we performed Fourier transform. ^9b91ee $\begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}&=\frac{1}{V} \left| \sum_{i} F_i(\mathbf{q})e^{-i \mathbf{q} \cdot \mathbf{r}_i}\right|^2\\ &= \frac{1}{V}\sum_{i,j} F_i(\mathbf{q}) F_j^*(\mathbf{q}) e^{-i \mathbf{q} \cdot (\mathbf{r}_i - \mathbf{r}_j)} \\ &= \frac{1}{V} \left( \sum_{i} \left| F_i(\mathbf{q}) \right|^2 + \sum_{i \neq j} F_i(\mathbf{q}) F_j^*(\mathbf{q}) e^{-i \mathbf{q} \cdot (\mathbf{r}_i - \mathbf{r}_j)} \right) \end{aligned}$ Inside the parathesis, the first term $\sum_{i} \left| F_i(\mathbf{q}) \right|^2$ is the self part, having all the information of the properties of single particles that are scattered; $\sum_{i \neq j} F_i(\mathbf{q}) F_j^*(\mathbf{q}) e^{-i \mathbf{q} \cdot (\mathbf{r}_i - \mathbf{r}_j)}$ is the coherent part, containing all the information of the arrangement of all particles. ### Form and structure factor for identical particles Now assume we are having $N$ **identical particles** in the sample, namely removes the dependence of $i$ in $F_{i}$. Under such assumption we may extract the $F(\mathbf{q})$ out, $ \begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} &= \frac{N}{V} \colorbox{#a5d8ff66}{\(\left| F(\mathbf{q}) \right|^2\)} \colorbox{#ffa94d66}{\(\left( 1 + \frac{1}{N} \sum_{i \neq j} e^{-i \mathbf{q} \cdot (\mathbf{r}_i - \mathbf{r}_j)} \right)\)} \\ &= \frac{N}{V} \left| F(\mathbf{q} = \mathbf{0}) \right|^2 P(\mathbf{q}) {\left( 1 + \frac{1}{N} \sum_{i \neq j} e^{-i \mathbf{q} \cdot (\mathbf{r}_i - \mathbf{r}_j)} \right)} \\ &= \frac{N}{V} \left| F(\mathbf{q} = \mathbf{0}) \right|^2 P(\mathbf{q}) S(\mathbf{q}) \end{aligned} $ The $\colorbox{#a5d8ff66}{blue}$ and $\colorbox{#ffa94d66}{orange}$ parts gives the [[Form factor]] $P(\mathbf{q})$ and [[Structure factor S(q) and density autocorrelation functions]] $S(\mathbf{q})$. $\left| F(\mathbf{q} = \mathbf{0}) \right|^2$ are used to normalized the form factor, so we have $P(\mathbf{q}=\mathbf{0})=1$, and $ P(\mathbf{q}) =\frac{{\left| F(\mathbf{q}) \right|^2}}{\left| F(\mathbf{q} = \mathbf{0}) \right|^2} = \left| \frac{\int_{V_p} \mathrm{d}\mathbf{r} \Delta\rho_p(\mathbf{r}) e^{-i\mathbf{q} \cdot \mathbf{r}}}{\int_{V_p} \mathrm{d}\mathbf{r} \Delta\rho_p(\mathbf{r})} \right|^2 $ This is just inserting the definition of $F_{i}$ into the normalized expression. As for structure factor, we do not resolve the phase of each particle in actual system, so we have $S(\mathbf{q})$ being averaged also. $S(\mathbf{q}) = \left\langle 1 + \frac{1}{N} \sum_{i \neq j} e^{-i \mathbf{q} \cdot (\mathbf{r}_i - \mathbf{r}_j)} \right\rangle$ >[!Notice] >In the community some people like writing the phase term $e^{i \mathbf{q} \cdot \mathbf{r}}$. In previous derivation we used the Fourier transform $e^{-i \mathbf{q} \cdot \mathbf{r}}$, which is exactly the same. For structure factor, there is no mathematical or physical difference between these two. >To follow the convention, one just have to change the order of $\mathbf{r}_i$ and $\mathbf{r}_j$. Writing $\left| F(\mathbf{q} = \mathbf{0}) \right|^2$ as $\int_{V_p} \mathrm{d}\mathbf{r} \Delta\rho_p(\mathbf{r})$, $V_{p}$ is the volume of an individual particle, then put it back, we can have the differential cross-section being even simplified as $\begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}(\mathbf{q}) &= \frac{N}{V} \left| F(\mathbf{q} = \mathbf{0}) \right|^2 P(\mathbf{q})S(\mathbf{q}) \\ &= \frac{N}{V} \left[ \int_{V_p} \mathrm{d}\mathbf{r} \Delta\rho_p(\mathbf{r}) \right]^2 P(\mathbf{q})S(\mathbf{q}) \\ &= \frac{N}{V} {V_p^2}{\langle\Delta\rho_p\rangle^2} P(\mathbf{q})S(\mathbf{q}) \\ &= \phi V_p \langle\Delta\rho_p\rangle^2 P(\mathbf{q})S(\mathbf{q}) \end{aligned}$ Here the $\phi=\frac{NV_{p}}{V}$, the volume fraction of the particles. >[!Note] >It is reasonable to use $\langle \rho_{p}\rangle$, since we cannot ensure that inside the particle we still being homogeneous. Till now, we have turned the differential cross-section into an expression of volume fraction, particle volume, average particle SLD difference, and two factors describing the shape and arrangement of particles. It's the time to see how these parameters are reflected in the curve obtained from the scattering experiments. - Shape of the particles: [[Form factor]] - Arrangement of the particles: [[Structure factor S(q) and density autocorrelation functions]] - Volume fraction of each phase for two phase system: [[Scattering invariant]] - [[Radius of gyration]] could be extracted from form factor, under small $q$ - [[Fractal dimensions]] is related to $g(R)$, which can be extracted from structure factor $S$ - For high $q$, structure factor reflect [[Surface fractal]] - ... When extracting information from the scattering data, keep in mind that the case might not be ideal (monodisperse or has regular shape). With the level of complexity increases, the difficulty of gaining reasonable result would certainly be higher also. >[!Info] >See https://www.ncnr.nist.gov/programs/sans/pdf/sans_fund.pdf. This is a very good slide covering many questions I may considered. Local file: [[sans_fund.pdf]]. >Alternative version is https://www.ncnr.nist.gov/summerschool/ss10/pdf/SANS_NR_Intro.pdf. Many materials are provided under the link https://www.ncnr.nist.gov/summerschool/ss17. Change the last two numbers to access previous sessions. >