I thought that the annihilation process of positronium cannot take place without a third-party particle. This can be directly derived from energy & momentum conservation:

energy conservation: $$h\omega=E_e+E_p+E_{third}$$ momentum conservation: $$\frac{h\omega}{c}\hat{\vec{n}}=\frac{\vec{V}_e E_e}{c^2}+\frac{\vec{V}_p E_p}{c^2}+\frac{\vec{V}_{third} E_{third}}{c^2}$$

Short transformation of these two formulas lead to: $$|\frac{\vec{V}_e E_e}{c}+\frac{\vec{V}_p E_p}{c}+\frac{\vec{V}_{third} E_{third}}{c}|=E_e+E_p+E_{third}$$

Where $h\omega$ is a sum over all created photons, $\hat{\vec{n}}$ is a unit vector.

If there is no third particle, then this process cannot happen, because it would mean that electron and positron velocities had to be equal to speed of light. We cannot use vacuum fluctuations here, because vacumm balance must be preserved too (zero energy and zero momentum overall in vacuum fluctuations).

I start to wonder if that can become possible if only higher decay modes are used: three or five photons in case of parallel spin of $e^{+}$ and $e^{-}$, four or six photons in case of antiparallel spin. But it seems impossible for me, because the $E_{third}$ term has to be massive otherwise would just cancel out from both sides (because if the created particle is an extra photon it has a speed of light, and $E_{third}$ terms are equal). It seems to me that some massive particle must be created or participate along the way. Maybe for example a neutrino and antineutrino (created) or just some neutrino (participate)? Maybe this is another example showing that neutrinos have mass?

EDIT: well it appears that with your help we have found an answer. My original question ignored vectors. Now that the third formula includes a vector magnitude of momentum times c, the third particle can be a photon, and they cannot cancel out.

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    $\begingroup$ This iwhy you need at least 2 photons - your $E_{\rm third}$ can be another photon. $\endgroup$
    – Slaviks
    Jan 26, 2014 at 13:18
  • $\begingroup$ momentum is a vector and the second formula ignores the masses of the e an p too. $\endgroup$
    – anna v
    Jan 26, 2014 at 13:31
  • $\begingroup$ mass of electron and proton is $E/c^2$, the vector part is within velocity. $\endgroup$ Jan 26, 2014 at 14:04
  • $\begingroup$ $E_{third}$ cannot be another photon, because then $V_{third}=c$ and it just cancels out. $\endgroup$ Jan 26, 2014 at 14:05
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    $\begingroup$ It is pair production by Xrays that needs an interaction with a field in order to conserve momentum in the cms. hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html#c4 $\endgroup$
    – anna v
    Jan 26, 2014 at 15:10

1 Answer 1


The question is based on a misunderstanding.

We will do the work in the center of mass frame of the incident particles, and assume a two photon final-state (in practice it also occurs to a thre-photon state). We have a total incident a four-vector of $(\gamma m_e c^2,\vec{0})$ (this simple form is why I chose to work in the CoM frame). The two photons leave back-to-back in this frame (necessary to conserve three-momentum) and each having energy $\frac{1}{2} m_e \gamma c^2$. Without loss of of generality we pick a coordinate system such that each has momentum lying along the z-axis, giving total momentum $\frac{1}{2} m_e \gamma c - \frac{1}{2} m_e \gamma c = 0$.

Four-momentum is conserved.

It would be incorrect at this point to say that the problem is solved, because there never was a problem.

Boosting to other frames is trivial and is left as an exercise.


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