# Allowed and Forbidden Electron-Photon Reactions

1. Aug 10, 2017

### Vrbic

1. The problem statement, all variables and given/known data
Show, using spacetime diagrams and also using frame-independent calculations, that the law of conservation of 4-momentum forbids a photon to be absorbed by an electron, e + γ → e.

2. Relevant equations
$\textbf{p}_{e1}+\textbf{p}_{\gamma}=\textbf{p}_{e2}$ : $(E_1,\vec{p_1})+(E_{\gamma},E_{\gamma}\vec{n})=(E,\vec{p_2})$,
where $\textbf{p}$ is 4-vector, $\vec{p}$ is ordinary 3-vector and $\vec{n}$ is unit vector in photon's direction.
$E=\sqrt{p^2+m^2}$

3. The attempt at a solution
I take only 3-vecotrs and I (I hope I may) suppose $\vec{p_1}=0$ i.e. is in rest.
$\vec{0}+E_{\gamma}\vec{n}=p_2\vec{n}=>\sqrt{E^2-m^2}=E_{\gamma}$
First time I put there $E=\frac{1}{2}mv^2$ but I feel it is not good probably. Is my begining right? What should I use or think about?

2. Aug 10, 2017

### Staff: Mentor

You used conservation of momentum so far.

The problem is easier in the reference frame of the electron in the final state.

3. Aug 11, 2017

### Vrbic

Ok, $(\gamma m, \vec{p}) + (E_{\gamma},E_{\gamma}\vec{n})=(m,\vec{0})$. Right? From energy part than we have $E_{\gamma}=m(1-\frac{1}{\sqrt{1-v^2}})<0$. And it is this problem, negative photon's energy?

4. Aug 11, 2017

### Staff: Mentor

Right.

Time-reversed, the process would be an electron at rest emitting a photon and gaining kinetic energy at the same time. Written like that, it should be clear that it cannot happen.

5. Aug 14, 2017

Thank you.