## Scattering, absorption and extinction Scattering is widely seen from photonics to material characterizations. Here we provide some intuitive ideas as well as mathematical relations about the concept scattering, absorption, and extinction. Some commonly applied parameters will be introduced. Consider the following experience: the plane wave impinges onto some particles, and a detector is placed on the other side to detect the transmission light along the incident direction. It is easily to imaginary that the intensity received by the detector is smaller than the incident beam ($I<I_{0}$). This is called **extinction** of light beam. And it includes two parts: - **Light scattering** by the particles "deviates" the transmission beam from the incident beam direction. - **Light absorption** in the particles converts the energy of photon to other forms. ![[Drawing 2024-03-31 22.17.28.excalidraw.svg]] ### Scattering, absorption and extinction cross-sections The concept of **cross-section** is introduced to quantify the efficiency of scattering, absorption and extinction. This concept is not limited to photonics or optics, but also widely applied for any scattering events, like for neutron or electrons (scattering has similar meaning but absorption for fermions could be different). The following definition is for EM (i.e., light under Maxwell eq, with wave properties). Consider a EM field, $(\mathbf{E}, \mathbf{M})$, the [Poynting vector](https://en.wikipedia.org/wiki/Poynting_vector) $\mathbf{S} = \mathbf{E}\times \mathbf{M}$ defines the energy flow in the direction of the propagation of the EM wave, with the dimension *power/unit area*. For a given surface $A$ with the normal vector $\mathbf{n}$, the electromagnetic power $W$ transferred across this surface is $\mathbf{S}\cdot \mathbf{n}A$. (Notice this is only valid for a **plane** $A$ and plane wave) If write this for a closed surface with all our particles enclosed, we should have $W=-\int_{A} \mathbf{S} \cdot \mathbf{n}\, \mathrm{d}A$ and $\sigma=\frac{W}{I_{0}}$ This W has the dimension of power, and intensity has the dimension of power per unit area, this gives $\sigma$ the dimension of an area. In similar way (perspective of energy), absorption and extinction cross-sections could be defined, and, $\sigma_{ext} = \sigma_{scat} +\sigma_{abs}$ It is worth noting that this area is not limited by the real space side of the particle, but could be much larger and much smaller. Some resonant particles could have extinction cross-sections for particular frequency that are much larger than the actual cross-sections areas. This means the incident wave is influenced beyond the physical boundaries of the particles. >[!Note]- >If not under EM wave definition, scattering cross-section could be defined in a quantum way, like in [[General introduction to scattering#Parameters in scattering experiments]]. ### Scattering, absorption and extinction cross-sections under Rayleigh approximation When the particle is much smaller than the incident wavelength, it can be considered as an electric dipole and use the scattering of electric dipole to compute the cross-sections. The induced dipole moment is proportional to the electric field with the coefficient $\alpha(\omega)$ as the polarizability, i.e., $\mathbf{p} = \alpha(\omega) \mathbf{E}(\mathbf{r}_{0})$ and cross-sections can be expressed by this polarizability. - Scattering cross-section: $\sigma_{\text{scat}} = \frac{k^4}{6\pi} |\alpha|^2$ - Absorption cross-section: $\sigma_{\text{abs}} = k \Im[\alpha(\omega)] - \frac{k^4}{6\pi} |\alpha|^2$ - Extinction cross-section: $\sigma_{\text{ext}} = \sigma_{\text{abs}} + \sigma_{\text{scat}} = k \Im[\alpha(\omega)]$ >[!Note] >One may wonder why $\sigma_{scat}$ has this form, here's the derivation: >The dipole is induced by the electric field, so we have >$\mathbf{p} = \alpha(\omega) \mathbf{E}(\mathbf{r}_{0})$ >For an oscillating electric dipole, the power radiated per solid angle $\Omega$ is >$\frac{\mathrm{d}P}{\mathrm{d}\Omega} = \frac{p_0^2 \omega^4 \sin^2(\theta)}{32\pi^2 \varepsilon_0 c^3}$ >where $p_{0}$ is the amplitude of the oscillating dipole moment, $\theta$ is the angle relative to the dipole axis. >Therefore, the total power would just be the integration over the solid angle, i.e., >$P = \int \frac{p_0^2 \omega^4 \sin^2(\theta)}{32\pi^2 \varepsilon_0 c^3} \mathrm{d}\Omega$ >The result is >$P = \frac{p_0^2 \omega^4}{12\pi \varepsilon_0 c^3}$ >The intensity of incident field $\mathbf{E}$ with the amplitude $E_{0}$ can be expressed as >$I_{0} = \frac{\varepsilon_{0}cE_{0}^{2}}{2}$ >this gives the scattering cross-section as >$\sigma_{scat} = \frac{P}{I_{0}}$ >$\sigma_{scat} = \frac{p_0^2 \omega^4}{12\pi \epsilon_0 c^3} \frac{2}{\epsilon_0 c E_0^2}$ >Substitute $\mathbf{p} = \alpha(\omega) \mathbf{E}(\mathbf{r}_{0})$, and only with the amplitude $p_{0} = \alpha E_{0}$, also the wavevector $k=\frac{\omega}{c}$, >$\sigma_{scat} = \frac{\alpha^2 k^4 E_0^2}{12\pi \epsilon_0 c^3} \frac{2}{\epsilon_0 c E_0^2}$ >$\sigma_{scat} = \frac{\alpha^2 k^4}{6\pi \epsilon_0^2 c^4}$ >$\sigma_{scat} = \frac{\alpha^2 k^4}{6\pi}$ >We finally get the expression $\sigma_{scat} = \frac{k^4}{6\pi} |\alpha|^2$. >Here the complex nature of dielectrics constant and orientation has been neglect for simplification. And the real case could be more complicated. >[!Info] >See more at: >https://en.wikipedia.org/wiki/Rayleigh_scattering >It is worth mention that, [[Dynamic light scattering]] is a technique using Rayleigh scattering.