## all about: light-matter interaction

Or: “How photons and electrons say hello”

- Low energy —
__Photoelectric effect__
- This is the first one you learn: a photon knocks an electron out of its atomic orbit. It is most likely to occur at low energies… as you move up in energy it becomes more likely that the photon will be scattered rather than absorbed.

- Medium energy —
__Compton scattering__
- While in the photoelectric effect the energy of the incoming photons is absorbed completely by the electrons, at higher energies the photon will instead bounce off the electron, leaving some of its energy/momentum behind in the recoil.
Using relativistic energy/momentum formulas, you can derive the wavelength shift \(\lambda’-\lambda = \frac{h}{m_e c}\left(1-\cos\theta\right)\) (higher wavelength ⇒ lower energy).

- High energy —
__Pair production__
- γ → e
^{-} + e^{+} looks pretty reasonable, right? If the photon had enough energy, it could account for the mass of the created electron-positron pair, and charge is certainly conserved, so why not? Well, consider this: Given the conservation of momentum, output energy will be minimized by having the two electrons each take half the photon’s original momentum. But this gives \(E_\mathrm{out}=2\sqrt{\left(\frac{pc}{2}\right)^2+(m_e c^2)^2}>pc=E_\mathrm{in}\), so even in this best case we don’t have enough energy to support the electron’s/positron’s momentum.
All that means is that we need some other ingredient in the mix. One good option is an atomic nucleus… When the photon gets near, it can allow the nucleus to absorb some of its momentum, to make electron-positron pair production possible. This is why pair-production is a form of light-matter interaction, rather than just something light does on its own.

(And no matter what, this is definitely going to be a high-energy interaction: the incoming photon must have AT LEAST \(pc>2m_e c^2\).)