Is the Compton scattering elastic or inelastic? Because the kinetic energy is conserved (in addition to the total energy conservation) so it should be elastic, but the energy of the photon is changed, so from that point of view is inelastic.
I have found both definitions online, which one is correct?
[Physics] Compton scattering: elastic/inelastic
particle-physicsphotonsscattering
Related Solutions
Raman scattering is inelastic scattering from molecules. The photon interacts with the molecule and changes the molecules vibrational, rotational or electron energy.
Rayleigh scattering is in the main elastic scattering from small particles whose size is less than that of the wavelength of the photon. The scattering can occur of atoms or molecules and for molecules the scattering can be inelastic with a change of rotational energy of the molecule.
Compton scattering is inelastic scattering of a photon from a free charged particle. If the charged particle is a bound electron then the energy of the photon must be much greater than the binding energy of the electron.
Side note: Rayleigh scattering is a particular case of Mie scattering. This theory explains in particular the white colour of objects which are made of particles of size greater than the typical wavelength : milk, clouds, chemical powders...
To add to the answer there is Thomson (no "p") scattering which is the elastic scattering of electromagnetic radiation by a free charged particle, as explained by classical electromagnetism.
It is just the low-energy limit of Compton scattering when the particle kinetic energy and photon frequency do not change as a result of the scattering.
This limit is valid as long as the photon energy is much smaller than the mass energy of the particle.
We can derive the Compton scattering formula using the 4-momentum. The 4-momentum before the collision is equal to the 4-momentum after the collision: $$\tag{1}P_{b}=P_{a}\label{1}$$ The 4-momentum before the collision is given by $P_{\gamma}+P_{e}$ and the 4-momentum after the collision is given by $P'_{\gamma}+P'_{e}$. In general case, the 4-momentum components are: $$\tag{2}p_{\mu}=\left(\frac{E}{c},\vec{p}\right)\label{2}$$ The dot product of 4-momentum with itself is: $$\tag{3}p_{\mu}\cdot p^{\mu}=\frac{E^2}{c^2}-p^2=m^2c^2\label{3}$$ Because photons are massless, the dot product of their 4-momentum is zero. We can rewrite equation (\ref{1}) in terms of photon and electron 4-momenta: $$\tag{4}P_{\gamma}+P_{e}=P'_{\gamma}+P'_{e}\label{4}$$ Rearranging equation (\ref{4}) and squaring: $$\tag{5}\left(P_{\gamma}+P_{e}-P'_{\gamma}\right)^2=\left(P'_{e}\right)^2\label{5}$$ Considering that the incoming photon is traveling along the $x$ axis, we can write its 4-momentum as $P_{\gamma}=\left(\frac{E_{\gamma}}{c},p_{\gamma},0,0\right)$. After the collision, the photon is scattered and will have a spatial momentum component in both $x$ and $y$ direction, so its 4-momentum can be written as $P'_{\gamma}=\left(\frac{E'_{\gamma}}{c},p'_{\gamma}\cdot \cos(\theta),p'_{\gamma}\cdot\sin(\theta),0\right)$. Also, considering the electron at rest before the collision, its 4-momentum is $P_{e}=\left(\frac{E_{e}}{c},0,0,0\right)$. Going back to the equation (\ref{5}) and using the 4-momentum expressions we obtain: $$\tag{6}0+m^2c^2+0+2\cdot\frac{E_{\gamma}E_{e}}{c^2}-2\left(\frac{E_{\gamma}E'_{\gamma}}{c^2}-p_{\gamma}\cdot p_{\gamma}'\cdot\cos(\theta)\right)-2\cdot\frac{E'_{\gamma}E_{e}}{c^2}=m^2c^2\label{6}$$ $$\tag{7}2\cdot\frac{E_{\gamma}E_{e}}{c^2}-2\cdot\frac{E'_{\gamma}E_{e}}{c^2}=2\left(\frac{E'_{\gamma}E_{\gamma}}{c^2}-p_{\gamma}\cdot p_{\gamma}'\cdot\cos(\theta)\right)\label{7}$$ $$\tag{8}E_{e}\left(E_{\gamma}-E'_{\gamma}\right)=E_{\gamma}E'_{\gamma}-p_{\gamma}c\cdot p_{\gamma}'c\cdot\cos(\theta)\label{8}$$ $$\tag{9}mc^2\left(\frac{hc}{\lambda}-\frac{hc}{\lambda'}\right)=\frac{hc}{\lambda}\frac{hc}{\lambda'}\left(1-\cos(\theta)\right)\label{9}$$ $$\tag{10}\left(\lambda'-\lambda\right)\frac{hc}{\lambda\lambda'}mc^2=\frac{hc}{\lambda}\frac{hc}{\lambda'}\left(1-\cos(\theta)\right)\label{10}$$ So we get the Compton scattering formula: $$\tag{11}\lambda'-\lambda=\frac{h}{mc}\left(1-\cos(\theta)\right)\label{11}$$
Best Answer
When you are considering scattering, you are looking to what happens to the incoming particle. In Compton scattering, the incoming particle, the photon, comes out after the scattering process with less energy than it started with, and so the scattering is called inelastic; whereas, with Thomson scattering, since the energy of the photon does not change, that would be termed as elastic scattering.
It might be that the confusion arises because of the term "elastic" alternatively being used for collisions when the total kinetic energy of the interacting particles is conserved?