## Fraunhofer diffraction of particles Fraunhofer diffraction, also known as far-field diffraction, refers to the pattern produced when light waves pass through a slit or around an obstacle and the resulting diffraction pattern is observed at a large distance from the slit or obstacle (or equivalently, at the focal plane of a converging lens). This type of diffraction occurs when the wavefronts can be considered planar because the source of light and the observation point are both at infinite distances from the diffracting aperture. Since we are talking about scattering, we may use Fraunhofer diffraction and Fraunhofer scattering interchangeably. ### Fraunhofer scattering condition and expression Now consider a simple case, which is valid for scattering of atoms (or particles with similar size): the scattering center is much smaller than the wavelength, namely $\text{size of scattering center} \ll \text{wavelength}$ >[!note]- >Since we known the wavelength of the incident wave, for x-ray and neutron are both at the range of $0.1\ nm$, this has the same size of an atom. And the approximation still holds is because the scattering involves interactions between electrons/nucleus, whose size is significantly smaller than the wavelength. In this case, if we have a incident plane wave, and the scattering center is considered as a point source, we will have the spherical wave as scattering waves. This is the solution of the point source Helmholtz equation. **Incident wave**: $E_i = \left| E_i \right| e^{i\mathbf{k}_i \cdot \mathbf{r}}$ $\mathbf{k}_i = \frac{2\pi}{\lambda} \hat{\mathbf{k}}_i$ **Scattered wave**: $E_f \propto \frac{-b}{r} e^{ikr}$ $k = \frac{2\pi}{\lambda}$ This $k$ is just the value of wavevector, not a vector, since this is a spherical wave. $b$ here describes how strongly the scattering is, and is a property of particles. ![[Drawing 2024-04-12 12.45.51.excalidraw.svg]] Consider the [[General introduction to scattering#Static and dynamic scattering|elastic scattering condition]], and assign the scattered wave a wavevector value of $\mathbf{k}_f$, we should have $|\mathbf{k}_i| = |\mathbf{k}_f|$. ![[Drawing 2024-04-12 12.58.24.excalidraw.svg]] Assume the vector between the sample center (i.e., selected scattering center of the entire sample) and the detector is $\mathbf{R}$, and the direction of $\mathbf{R}$ should be the same as $\mathbf{k}_f$. The individual scattering centers and the total scattering center have displacements $\mathbf{r}_{j}$. This gives the scattered field amplitude as a superposition of individual scattering amplitudes, namely $E_f = E_i \sum_j e^{i\mathbf{k}_i \cdot \mathbf{r}_j} \frac{-b_j}{|\mathbf{R} - \mathbf{r}_j|} e^{i\mathbf{k}_f |\mathbf{R} - \mathbf{r}_j|}$ Since for typical scattering setup, we have $|\mathbf{R}| \gg |\mathbf{r}_j|$ Use the estimation under this condition for any $|\mathbf{R}| \gg |\mathbf{x}|$, namely $\frac{1}{|\mathbf{R} - \mathbf{x}|} \approx \frac{1}{R}$ $|\mathbf{R} + \mathbf{x}| \approx R + \hat{\mathbf{R}} \cdot \mathbf{x}$ We have $E_{f}\approx E_i \sum_j e^{i\mathbf{k}_i \cdot \mathbf{r}_j} \frac{-b_j}{R} e^{i\mathbf{k}_f R} e^{-i\mathbf{k}_f \cdot \mathbf{r}_j}$ >[!Note]- >Here in the simplification $\hat{\mathbf{R}}$ is the unit vector, only indicating the direction. And at the final expression we have $e^{-i\mathbf{k}_f \cdot \mathbf{r}_j}$ because $\hat{\mathbf{R}}\parallel \mathbf{k}_{f}$. >This step is just changing the scattering centers from individual particles to an entire sample. Since the summation above is for $j$, we can extract the parts with no $j$, namely $\frac{e^{ik_{f}R}}{R}$. $\begin{aligned} E_f & \approx E_i \sum_j e^{i \mathbf{k}_i \cdot \mathbf{r}_j} \frac{-b_j}{R} e^{i k R} e^{-i \mathbf{k}_f \cdot \mathbf{r}_j} \\ & = -E_i \frac{e^{i k_f R}}{R} \sum_j b_j e^{i(\mathbf{k}_i - \mathbf{k}_f) \cdot \mathbf{r}_j} \\ & = -E_i \frac{e^{i k_f R}}{R} \sum_j b_j e^{i \mathbf{q} \cdot \mathbf{r}_j} \end{aligned} $ Here we replace the incoming and scattering wavevector difference as $\mathbf{q}$, namely $\mathbf{q} = \mathbf{k}_i - \mathbf{k}_f$. We may also call it scattering wavevector for simplicity. For a given light/neutron source, this value could be adjust by changing the scattering angle $\theta$ by adjusting the detector position (this $\theta$ also exist in Bragg condition, and the value of $|\mathbf{q}|$ is $q=\frac{4\pi}{\lambda} \sin(\theta)$, the angle between $\mathbf{k}_{i}$ and $\mathbf{k}_{f}$ is $2\theta$). And this is the scattering in the **far-field Fraunhofer approximation**. (in contrast, we have near field--the fast decay wave described in [[Diffraction limit and near field imaging]]) >[!Important] >For scattering in far field under Fraunhofer approximation, i.e., $\left| \mathbf{R} \right| \gg \left| \mathbf{r}_j \right| \quad \forall j$, an incident plane wave >$E_i = \left| E_i \right| e^{i \mathbf{k}_i \cdot \mathbf{r}}$ >will have a spherical scattering wave >$E_f = -E_i \frac{e^{i k R}}{R} \sum_j b_j e^{i \mathbf{q} \cdot \mathbf{r}_j}$ >with $\mathbf{q} = \mathbf{k}_i - \mathbf{k}_f$. >>[!Note] >>It is worth to recall some other terms inside this expression: >> >>Here $R$ is the distance from the sample scattering center to the detector. This center is *selected* by us, so $R$ is not constant term. This means, for actual scattering experiments, the result should be independent with $R$ selection. >> >>The wavevector $\mathbf{k}_{i}= \frac{2\pi}{\lambda}\hat{\mathbf{k}}_{i}$. >> >>$b_{j}$ has a **dimension of length**, although in the original spherical wave from Helmholtz eq solution it is just a term describing how strong the scattering is. >>Since we write $b$ as $b_{j}$, the result above indicates we could have different particles. >>This $b$ is called **scattering length**. ### Phase in the scattered beams Now consider any two particles with displacement $\mathbf{r}$, ![[Drawing 2024-04-13 15.35.16.excalidraw.svg]] If the upper scattering beam has the phase $e^{i\phi}$, then the lower one would have a phase $e^{i\phi}e^{i(\mathbf{k}_i - \mathbf{k}_f)\cdot r}=e^{i\phi}e^{i\mathbf{q}\cdot r}$. So here the phase difference is $e^{i\mathbf{q}\cdot r}$, compensates for the phase shift for displacement $\mathbf{r}$. If we have a crystal, then this displacement is uniform for all particles. And this is the Bragg condition for diffraction, which gives the relation between atomic spacing $d$ and wavelength $\lambda$ as $n \lambda = 2d \sin(\theta)$ $q = \frac{2\pi}{d}$ $q = \frac{4\pi}{\lambda} \sin(\theta)$ This is important for [[Scattering in crystals]]. Back to our samples, which is not necessarily to be a crystal (and typically not). The detector would only measure the amplitude, $|E_{f} (\mathbf{q})|^{2}$. And the phase parts are completely lost >[!Question] >Is it possible to obtain the phase information? Typically this is called tomographic imaging, and people like to apply for microscopic technique, using electron, x-ray and neutron. Some works are presented in past 20 years: >- [[Neutron Phase Imaging and Tomography]] >- [[Tomographic phase microscopy]] ### Scattering cross-section and scattering length density under Fraunhofer condition Recall the scattering cross-section definition in [[General introduction to scattering#Parameters in scattering experiments]]. The incident wave: $E_i = \left| E_i \right| e^{i \mathbf{k}_i \cdot \mathbf{r}}$ and the wavevector $\mathbf{k}_i = \frac{2\pi}{\lambda} \hat{\mathbf{k}}_i$ The area for differential: $\mathrm{d}A = R^2 \mathrm{d}\Omega$ The scattering wave: $E_f \propto \frac{-b}{r} e^{i k r}$ here $k = \frac{2\pi}{\lambda}$. This is for **one** scattering event, not for the entire sample. **Neutron beam** Flux: $\Phi=nv$ Neutron going through area $d A$, for detector: $\mathrm{d}A \left| E_f \right|^2 = R^2 \mathrm{d}\Omega \frac{b^2}{R^2} = b^2 \mathrm{d}\Omega$ Thus the differential cross-section is: $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{b^2 \mathrm{d}\Omega}{\mathrm{d}\Omega} = b^2$ and cross-section for scattered wave is: $\sigma = \int \mathrm{d}\Omega \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = 4\pi b^2$ >[!Notice]- >Here we calculate *Neutron going through area $\mathrm{d} A$, for detector*, and multiple a $|E_{f}|^{2}$, is because $\Phi$ is the flux is described for the incident beam. Although here we write it as $nv$, with a particle description, but if convert to wave it has the $|E_{i}|^{2}$ form. > >Although we are under the maxwell frame, we still should make $E_{f}$ being "normalized" in some extent to ensure energy conservation and match the incident wave. So the $|E_{f}|^{2}$ here can be considered as $\text{parameter}\times\frac{-b}{r} e^{i k r}|E_{i}|^{2}$. Although locally this could exceed the incident wave amplitude, but globally (integral on space and time) it should not be possible. > >The $|E_{f}|^{2}$ is a spatial decaying term. The same is for light scattering, just change the incident flux a little bit. **Light scattering** Flux: $\Phi = \frac{\left| E_i \right|^2}{\mu_0 c} \left[ \frac{J}{m^2 s} \right]$ Photon going through area $\mathrm{d} A$, for detector: $\mathrm{d}A \left| E_f \right|^2 = R^2 \mathrm{d}\Omega \frac{b^2}{R^2} = b^2 \mathrm{d}\Omega$ Thus the differential cross-section is: $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{b^2 \mathrm{d}\Omega}{\mathrm{d}\Omega} = b^2$ and cross-section for scattered wave is: $\sigma = \int \mathrm{d}\Omega \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = 4\pi b^2$ >[!Notice] >***ALL ABOVE*** in this section is for single scattering event, i.e., for an atom. >Also, we may, sometimes use *cross-section* and *differential cross-section* interchangeably. So check the actual meaning of the term would be necessary. Now let's consider one entire sample, with many point like scattering center. As shown before, we have the scattered wave $E_{f}$ $E_f = -E_i \frac{e^{i k R}}{R} \sum_j b_j e^{i \mathbf{q} \cdot \mathbf{r}_j}$ and the differential cross-section $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{ \mathrm{d}A \left| E_f \right|^2}{ \mathrm{d}\Omega} = \frac{R^2 \mathrm{d}\Omega}{\mathrm{d}\Omega} \left| \frac{1}{R} \sum_j b_j e^{i \mathbf{q} \cdot \mathbf{r}_j} \right|^2$ $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \sum_{j,k} b_j b_k e^{i \mathbf{q} \cdot (\mathbf{r}_j - \mathbf{r}_k)}$ Here $\mathbf{r}_{j}, \mathbf{r}_{k}$ are the vector from center defining $\mathbf{R}$ to particle $j, k$. The expression is just by writing above square separately. In the last step we remove the dependence of $R$, indicating this cross-section is not related with the selection of the scattering center, but with the relative positions of particles, as expected. For individual atoms, the scattering length $b$ is more like a constant and easy to describe. But for a bulk sample, the summation of $b_{j}$ multiply by the phase term for sure is not a good indicator, so another parameter should be consider. Notice $E_{f}$ has the form $E_f \propto \sum_j b_j e^{-i \mathbf{q} \cdot \mathbf{r}_j}$ This is the form of Fourier transform. If write the summation of b for all the particles for $\mathbf{r}$, we would have $\rho(\mathbf{r}) = \sum_j b_j \delta(\mathbf{r} - \mathbf{r}_j)$ This $\delta$ intrinsically fits the Fourier transform. And $\rho$ gives a real space distribution of $b$ on $\mathbf{r}$. This could be an indicator similar to scattering length, for bulk samples. To check this, we do the Fourier transform ($\mathbf{r} \rightarrow \mathbf{p}$) and see how this distribution be like in reciprocal space, which is more of our focus in scattering experiments. For wavevector difference $\mathbf{q}$, $ \begin{aligned} \rho(\mathbf{q}) &= \int \mathrm{d}\mathbf{r}\, \rho(\mathbf{r}) e^{-i \mathbf{q} \cdot \mathbf{r}}\\ &= \sum_j b_j \int \mathrm{d}\mathbf{r}\, \delta(\mathbf{r} - \mathbf{r}_j) e^{-i \mathbf{q} \cdot \mathbf{r}}\\ &= \sum_j b_j e^{-i \mathbf{q} \cdot \mathbf{r}_j}\\ &\propto E_{f} \end{aligned} $ This follows our expectation. >[!Notice] >The definition here is spatial dependence and describes atomic scattering, and not the common form (for materials) that would be applied later. With $\rho$ defined, we can write write the cross-section with respect to $\rho$, namely, $\begin{aligned} \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} &= \left| \rho(\mathbf{q}) \right|^2 \\&= \rho(\mathbf{q}) \rho^*(\mathbf{q})\\ & = \iint \mathrm{d}\mathbf{r} \mathrm{d}\mathbf{r}'\, \rho(\mathbf{r}) \rho(\mathbf{r}') e^{-i \mathbf{q} \cdot (\mathbf{r} - \mathbf{r}')} \end{aligned}$ This $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}$ is called microscopic (differential) cross-section, with the dimension $[\text{area}]$, the integration is made on the sample volume. and the macroscopic (differential) cross-section is $\begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} &= \frac{1}{V} \left| \rho(\mathbf{q}) \right|^2 \\ &= \frac{1}{V} \rho(\mathbf{q}) \rho^*(\mathbf{q}) \\ &= \frac{1}{V} \iint_{V} \mathrm{d}\mathbf{r} \mathrm{d}\mathbf{r}' \, \rho(\mathbf{r}) \rho(\mathbf{r}') e^{-i \mathbf{q} \cdot (\mathbf{r} - \mathbf{r}')} \end{aligned} $ This has the dimension $\left[ \frac{1}{\text{length}} \right]$, and the unit is typically $\frac{1}{\text{cm}}$. >[!Notice] >This $\rho$ is an interesting parameter. Under its definition it is possible to describe both real and reciprocal space quantities. It is even more critical that $\rho$ is directly related to differential scattering cross-sections, who is the **observables** in scattering experiments. It is worth to keep the expression of this $\rho$ in mind and see how it would connect to other quantities like radial distribution function, structural factors, etc. ### From cross-section to real space correlation This $\rho(\mathbf{r})$, here scattering length density, gives the real space distribution, and can be used to represent the [[Real space correlations, radial distribution function#Density autocorrelation function|real space distribution]]. It is worth noting that here we are still dealing with individual particles (or atomic scale). Check the scattering cross-section expression above. The integrations are made over the volume of the sample, $V$. If introduce a new variable $\mathbf{R}=\mathbf{r}-\mathbf{r}'$. We can write, $\begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} &= \frac{1}{V} \iint_{V} \mathrm{d}\mathbf{r} \mathrm{d}\mathbf{r}' \, \rho(\mathbf{r}) \rho(\mathbf{r}') e^{-i \mathbf{q} \cdot (\mathbf{r} - \mathbf{r}')}\\ &= \frac{1}{V}\iint_{V} \mathrm{d}\mathbf{r}' \mathrm{d}\mathbf{R} \, \rho(\mathbf{R}+\mathbf{r}') \rho(\mathbf{r}') e^{-i \mathbf{q} \cdot \mathbf{R}}\\ &= \frac{1}{V} \int_{V} \mathrm{d}\mathbf{r}' \int_{V} \rho(\mathbf{R}+\mathbf{r}') \rho(\mathbf{r}') e^{-i\mathbf{q} \cdot \mathbf{R}}\ \mathrm{d}\mathbf{R} \end{aligned} $ Here the first integration is the summation over the volume of the sample, **which effectively means an averaging of the second integration over the whole volume of the sample**. This gives the bracket outside. $\begin{aligned} \frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} &= \frac{1}{V} \int_{V} \mathrm{d}\mathbf{r}' \int_{V} \rho(\mathbf{R}+\mathbf{r}') \rho(\mathbf{r}') e^{-i\mathbf{q} \cdot \mathbf{R}}\ \mathrm{d}\mathbf{R} \\ & =\frac{1}{V} \times V \left\langle\int_{V} \rho(\mathbf{R}+\mathbf{r}') \rho(\mathbf{r}') e^{-i\mathbf{q} \cdot \mathbf{R}} \mathrm{d}\mathbf{R}\right\rangle\\ & =\left\langle\int_{V} \rho(\mathbf{R}+\mathbf{r}') \rho(\mathbf{r}') e^{-i\mathbf{q} \cdot \mathbf{R}} \mathrm{d}\mathbf{R}\right\rangle\\ &= \int_{V} \langle\rho(\mathbf{R}+\mathbf{r}') \rho(\mathbf{r}')\rangle e^{-i\mathbf{q} \cdot \mathbf{R}} \mathrm{d}\mathbf{R} \end{aligned}$ Write $\mathbf{r}'$ as $\mathbf{r}$, we get $\frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega}= \int_{V} \langle\rho(\mathbf{R}+\mathbf{r}) \rho(\mathbf{r})\rangle e^{-i\mathbf{q} \cdot \mathbf{R}} \mathrm{d}\mathbf{R}$ And microscopic cross-section is $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}= V\int_{V} \langle\rho(\mathbf{R}+\mathbf{r}) \rho(\mathbf{r})\rangle e^{-i\mathbf{q} \cdot \mathbf{R}} \mathrm{d}\mathbf{R}$ If in the volume we have homogeneity (i.e., no dependence on the position selection), then $G(\mathbf{R}) = \langle \rho(\mathbf{R})\rho(0) \rangle = \langle \rho(\mathbf{R} + \mathbf{r})\rho(\mathbf{r}) \rangle$ This $G(\mathbf{R})$ is an density autocorrelation function in real space, sometimes called the **pair correlation function**. This quantity is **orientation dependent**, unlike radial distribution function $g(r)$. With the definition of $G(\mathbf{R})$, we may write the differential scattering cross-section as $\frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} =\int \mathrm{d} \mathbf{R}\, e^{-i\mathbf{q}\cdot\mathbf{R}}G(\mathbf{R})= \iiint \mathrm{d}^3 R\, e^{-i\mathbf{q}\cdot\mathbf{R}}G(\mathbf{R})$ For isotropic material, like glass and liquid, $G(\mathbf{R})=G(R)$. This means one can writing above integration in spherical system as $\frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} = 4\pi \int_0^{\infty} \mathrm{d}R\, R^2 \frac{\sin(qR)}{qR} G(R) $ The left side is something related to our measurements, and its a function of $\mathbf{q}$, which is a controllable parameter. >[!Note]- >From here one can notice that the RHS has a $G(R)$, which is atomic arrangement originated. So for static sample, this should be a constant. And it has a pre-factor $\frac{\sin(qR)}{qR}$. It is this part changing with $\mathbf{q}$. >That's why we often see in many plots, the x axis is $qR$. By extracting the constant parts with number of particles, volume and scattering length, aka normalization, we get the spatial dependence parts written in $\delta(R)$ and $g(R)$, $G(R) = b^2 \frac{N}{V} \left[ \delta(R) + g(R) \right]$ Here $\delta(R)$ is given by the scattering center placed at the center of an atom, namely $R=0$, and the $g(R)$ is the radial distribution function, describing other atoms other than the $\delta(R)$. As shown in [[Real space correlations, radial distribution function]], $g(R)$ has the profile in the following figure, ![[Drawing 2023-07-27 11.54.48.excalidraw.svg]] For $q\rightarrow \infty$, $\frac{\mathrm{d}\Sigma}{\mathrm{d}\Omega} \rightarrow b^2 \frac{N}{V}$, cause when $q$ gets very large, the detectable distance in real space is very small, which means the scattering is mainly given by the atom at scattering center we selected ($\delta(R)$), and not by those neighboring atoms contribute to $g(R)$. More about the relationship among $g(r)$, $G(R)$ and $S(q)$ is discussed in [[Structure factor S(q) and density autocorrelation functions]]. >[!Notice] >***ALL ABOVE*** in this section is for scattering of particles, i.e., for multiple atoms or ions. >Do not tread nanoparticles/polymers or many particle ensembles in the same way! >>[!Tips] >>Those *particles*, considered not formed by many atoms but some consistent bulk matter could be treaded in a classic way, and the typical response to electromagnetic waves following [[Scattering, absorption and extinction#Scattering, absorption and extinction cross-sections under Rayleigh approximation|Rayleigh scattering]] or [[Mie resonances|Mie scattering]]. > >Also notice the meaning of $\mathbf{R}$ and $\mathbf{r}$, they sometimes are used interchangeably. $\mathbf{R}$ could be the vector form an assigned center to detector, or the vector to another particle by placing the center at one particle, even the vector from an assigned center to particle $i$. $\mathbf{r}$ could be the vector between any two particles, or the vector from an assigned center to particle $i$. **Their meanings are related to the expression of the equation so be careful!** >[!Info] >See https://www.fkf.mpg.de/5511283/11_Nuclear_Neutron_Scattering.pdf for more information on neutron scattering. Local file [[11_Nuclear_Neutron_Scattering.pdf]]. >The derivation of relationship between cross-section and autocorrelation function referred the book [[Neutrons in Soft Matter]], page 18. >For pair distribution function, some useful slides are given here: https://neutrons.ornl.gov/sites/default/files/NXS_2019_pdf_katepage_wide.pdf. Local file [[NXS_2019_pdf_katepage_wide.pdf]].