# Electron-Positron Annihilation Conservation Laws

1. Jun 12, 2015

### says

What ensures conservation of energy and conservation of momentum in Electron-Positron annihilation?

If e=mc2 and momentum = mass*velocity, couldn't the energy equivalence of the electron and the positron be converted to make one photon, or more that 2 photons? I've read that two photons are created and this is why conservation of energy is conserved, and they got in opposite directions, which conserves momentum.

2. Jun 12, 2015

### Staff: Mentor

The laws of physics.
That is true for particles at rest only.
That is an approximation for nonrelativistic particles only.

One photon would violate energy-momentum conservation. More than 2 photons are possible, but rare.

3. Jun 12, 2015

### says

Ahhhhhhh

Do particles not have energy-equivalence when they are in motion? Once the electron and positron collide couldn't we think of them at rest?

Can't one photon have the combined energy of the electron and positron?

4. Jun 12, 2015

### Orodruin

Staff Emeritus
If it did, it would break momentum conservation. It cannot simultaneously have the same momentum and energy as the electron-positron pair.

5. Jun 12, 2015

### ChrisVer

You can go to the reference frame where the electron-positron system is at rest. They (individually) will have some momentum $\vec{p}_-$ for electron and $\vec{p}_+$ for positron. The Center-of-Mass frame is given by $\vec{p}_- = - \vec{p}_+$ (or the particles have equal and opposite momenta,so that the total momentum is zero). In that frame you have that the initial 4-momentum is:
$(p_+ + p_-)^\mu = \begin{pmatrix} 2E_e \\ \vec{0} \end{pmatrix}$
since the energies of positron/electron are equal (they have the same mass and the same momentum magnitude).
What happens for the photon in that frame? Well the conservation of energy (in that frame) would imply that the photon has energy $2E_e$ (so far so good)... what about the momentum? The momentum of the photon will have to be zero.
That is impossible for a real photon, since the real photon should have equal momentum magnitude and energy -so that $E^2 - |\vec{p}|^2= m_\gamma^2 =0$.

That's why 1 photon alone cannot conserve energy/momentum.

If you have 2 photons you can assign their momenta/energy so that they cancel out ...that's why each photon would have a four momentum $p_\mu^{(\gamma1)}= \begin{pmatrix}E_e \\ \vec{p}_\gamma \end{pmatrix}$ and $p_\mu^{(\gamma2)}= \begin{pmatrix}E_e \\ -\vec{p}_\gamma \end{pmatrix}$ and so $p^{\mu~(\gamma1)} + p^{\mu~(\gamma2)}= p_+^\mu + p_-^\mu$. So in the center of mass frame of the electron/positron the photons go in opposite directions.

In a similar manner you can have more than 2 photons, however at each photon emission you get one factor of $\alpha_{em}$ (fine-structure constant for electromagnetism), which suppresses the interaction by approximately a factor of 100.

Last edited: Jun 12, 2015
6. Jun 12, 2015

### Staff: Mentor

They have, but you need the more general expression$$E^2 = m^2 c^4 + p^2 c^2$$
They can be, but the photon cannot.