Questions of the form:

An electron and a positron collide with E MeV of energy, what is the frequency of the photons released.

quite often come up in my A Level course (for often fairly arbitrary E). But this got me thinking. There is energy stored in the separation of an electron and a positron, which, as they get closer and closer together, should all be converted into kinetic energy. As the potential is of the form $\frac{1}{r}$, this implies that at arbitrarily small distances and arbitrarily high amount of energy is given off. Given that both electrons and positrons are typically regarded as point particles, in order for them to collide, they would have to be arbitrarily close together, which would imply that over the course of their collision they should have released arbitrarily high amounts of energy, in the form of kinetic energy. As this would imply photons of arbitrarily high frequency given off, I assume that I must have missed out some piece of physics somewhere, but I am uncertain where. Ideas I have had so far include:

  • Energy should be given off, anyway, by an accelerating electron, in the form of light, according to classical EM, although I don't know how this changes from classical to quantum ideas of EM - we certainly can't have all the energy given off in a continuous stream, because we need quantised photons, so does the electron itself experience quantised energy levels as it accelerates inwards (my only issue with treating the electron in such a quantised way is that, to my mind, it'd be equivalent of treating it mathematically as a hydrogen-like atom, where the probability of the electron colliding with the positron is still extremely low, and unlike electron capture, there'd be no weak force interaction to mediate this 'electron-positron atom').
  • The actual mechanism for the decay occurs at a non-zero separation distance, perhaps photons pass between the two particles to mediate the decay at non-infinitesimal distances.
  • At relativistic speeds our classical model of electrodynamics breaks down. Now, I know this to be true - considering the fact that magnetism is basically the relativistic component of electrodynamics. However, given the fact that magnetism is the only relativistic force which'd be involved, I don't see how it'd act to counteract this infinite release of energy - so is there another force which I'm forgetting?

These are just ideas I've come up with whilst thinking about the problem, and I don't know if any of them have any physical significance in this problem, so any advice is appreciated!

  • 2
    $\begingroup$ Thinking from the other side: if this would release infinite energy then you would have also needed an infinite amount to create the positrons. $\endgroup$
    – lalala
    Apr 8, 2019 at 6:48
  • $\begingroup$ Agreed - logically in order to create them, my misunderstanding would have stated that I would have needed to work against an infinitely high force to do so, thus precluding any matter/anti-matter creation in the universe... $\endgroup$ Apr 8, 2019 at 19:21

3 Answers 3


This is a great question! It can be answered on many different levels.

You are absolutely right that if we stick to the level of classical high school physics, something doesn't make sense here. However, we can get an approximately correct picture by "pasting" together a classical and a quantum description. To do this, let's think of when the classical picture breaks down.

In relativistic quantum field theory, the classical picture of point particles breaks down when we go below the Compton wavelength $$\lambda = \frac{\hbar}{mc}.$$ It isn't impossible to localize particles smaller than this length, but generically you will have a significant probability to start creating new particles instead. Now, the electric potential energy released by the time we get to this separation is $$E = \frac{e^2}{r} = \frac{e^2 m c}{\hbar}$$ in cgs units. Here one of the most important constants in physics has appeared, the fine structure constant which characterizes the strength of electromagnetism, $$\alpha = \frac{e^2}{\hbar c} \approx \frac{1}{137}.$$ The energy released up to this point is $$E \approx \alpha m c^2$$ which is not infinite, but rather only a small fraction of the total energy.

Past this radius we should use quantum mechanics, which renders the $1/r$ potential totally unapplicable -- not only do the electrons not have definite positions, but the electromagnetic field doesn't even have a definite value, and the number of particles isn't definite either. Actually thinking about the full quantum state of the system at this point is so hairy that not even graduate-level textbooks do it; they usually black-box the process and only think about the final results, just like your high school course is doing. Using the full theory of quantum electrodynamics, one can show the most probable final outcome is to have two energetic photons come out. In high school you just assume this happens and use conservation laws to describe the photons long after the process is over.

For separations much greater than $\lambda$, the classical picture should be applicable, and we can think of part of the energy as being released as classical electromagnetic radiation, which occurs generally when charges accelerate. (At the quantum level, the number of photons released is infinite, indicating a so-called infrared divergence, but they are individually very low in energy, and their total energy is perfectly finite.) As you expected, this energy is lost before the black-boxed quantum process starts, so the answers in your school books are actually off by around $1/137$. But this is a small enough number we don't worry much about it.

  • $\begingroup$ So, in some way, it would appear that we end up with some form of action at a distance. Would it be worthwhile to understand this in terms of some sort of photon exchange, or simply as some interaction between the probability fields (or not go down the route of using inaccurate intuitive understanding until I learn about QED?) $\endgroup$ Apr 6, 2019 at 13:50
  • $\begingroup$ @DoublyNegative Are you referring to the classical part of the process or the quantum part? The quantum process is definitely not a photon exchange, it's an annihilation along with the emission of two photons. Also, both the classical and quantum parts are 100% local, there is no action at a distance since everything is mediated by the field. $\endgroup$
    – knzhou
    Apr 6, 2019 at 13:57
  • $\begingroup$ I'm referring to the quantum part of the process. What counts as a collision when we only have two interacting wavefunctions? $\endgroup$ Apr 6, 2019 at 13:59
  • 6
    $\begingroup$ @DoublyNegative When the wavefunctions of the electron and positron overlap in space, they can decrease in magnitude, with the accompanying creation of a photon. Everything is perfectly local: if the wavefunctions don't overlap, they won't annihilate. Also, even if their wavefunctions collide head-on, they won't annihilate to photons all of the time -- there will also be some amplitude for them to pass right through each other. $\endgroup$
    – knzhou
    Apr 6, 2019 at 14:03

Positronium is what you're describing in your first idea.

Until/unless a believable model of the electron is developed, pretty much the best we can do is say that the rest mass of the electron-positron system is the only energy available to be converted to radiation in an annihilation event.

This experiment will reveal a lot more about what happens in very close encounters between electrons and positrons.

  • 1
    $\begingroup$ This would imply that the idea of using the kinetic energy of the particles is fundamentally flawed, then? $\endgroup$ Apr 6, 2019 at 13:57
  • 3
    $\begingroup$ Not exactly. The idea of rest mass is kind of messy. For example, the rest mass of a system of particles [think of them as being in a box] is the total energy of the particles: their rest masses, their kinetic energies, etc. If the box itself is massless but it can contain the particles, and if the mass of the box-plus-contents is measured, that might be called the mass of the system. The mass measured when the box is stationary would be its rest mass. $\endgroup$
    – S. McGrew
    Apr 6, 2019 at 14:03

I should point out that, even on a classical level, while the kinetic energy ramps up to infinity, the potential energy drops to negative infinity. The difference remains constant, equal to twice the electron mass (assuming they started from rest infinitely far away). When they annihilate, the kinetic energy disappears, but so does the potential energy. So the photons only carry off 2mc^2.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.