I was visiting the Australian Synchrotron earlier today as part of a tour group; as the guide was going over the booster and storage rings I was reminded of something I learnt of quantum.

If I know my quantum well enough, every so often, there are spontaneous pairs of electrons and positrons created and destroyed everywhere. Now I know that they don't last very long, but I got to thinking, what would happen if a pair was created at just the right instant for the positron to collide with one of the super-high-energy electrons in the booster or storage ring of a synchrotron?

When I asked our guide he said that the probability was pretty low since there were only 1x10^6 electrons in there and naturally the pair would have to generate at just the right spot at just the right time. Ultimately he wasn't quite sure what would happen if the circumstances were correct for it though.

I have been doing some further thinking, if I remember correctly, these pairs form as an electron spiralling in towards a central positron. I think that therefore at ordinary energies any nearby electrons would be repelled by the electron around the positron (which is a little reminiscent of the atomic stuff of first year chemistry).

Question is, could an electron, at the sorts of energy levels of the order of giga-electron-volts, get past the spiralling electron and collide with the positron instead of its paired electron doing so? As this energy could not simply vanish, would you then get a lonely (now unpaired) electron and an explosion? or would the energy somehow transfer itself to the other electron?

My interest here is towards what would actually happen if all circumstances were just right; also appreciated would be anyone who might be able to tell me the theoretical frequency of electron-positron pairs being spontaneously generated, as well as helping me to think up a way of determining the probability of finding an electron within one of the rings (for instance, perhaps if electrons can be considered to have an effective 'volume' within which they are most likely to appear and comparing that to the volume of the ring?)

Thanks for your time, I really appreciate it.

  • $\begingroup$ Aside: The beams in synchrotrons spiral out. Here is how to figure it: the bending radius of a charged particle in a uniform magnetic field is linear in the particles momentum $r \propto p$, and the machine gives the particle a little boost on each half circle transit. $\endgroup$ Commented Jun 29, 2011 at 14:24
  • $\begingroup$ @scott: I know that the electron would think it were standing still but I was thinking that the spontaneously generated pair would probably not be generated within the same frame of reference as the high energy electron. In my mind, I was picturing the pair stationary relative to us and then the electron would see the pair as going relativistic. $\endgroup$
    – WizzPhiz
    Commented Jun 30, 2011 at 9:10
  • $\begingroup$ If the distribution of spontaneously generated pairs (assuming they exist) had a favored reference frame, then you would have a violation of Einstein's principle of equivalence. $\endgroup$ Commented Jul 3, 2011 at 15:12

2 Answers 2


The electron is annihilated by the positron of the pair, and the other electron stays.

The "normal" virtual creation/annihilation process is $$\text{nothing} \rightarrow e^- + e^+ \rightarrow \text{nothing}$$ If you add an electron nearby which does not interact with the pair, this becomes $$e^- \rightarrow 2e^- + e^+ \rightarrow e^-$$ and if the electron is annihilated by the virtual pair's, it becomes $$e^- \rightarrow 2e^- + e^+ \rightarrow e^-.$$ Since the electrons are undistinguishable, the two last equations are the same: there is no difference to which electron the positron "choses" to annihilate. And actually, they are the same as the boring equation where no virtual pair appears : $$e^-\rightarrow e^-$$

On a more technical side, the 3 processes correspond to 3 different Feymann diagrams, but the indistinguishability of their outputs make them interfere and and doesn't allow us to infer what has "really" happened. And the question of "what happened in reality" does not have a sense (that's why the pair is called "virtual"). We can only compute the contribution of the the various processes to movement of the electron.

  • $\begingroup$ I do not think that at the energies in the accelerators we have annihilation on virtual sea particles can happen. At most it has been theorized that the Unruh effect can be seen in that the electrons do not reach 100% polarization. $\endgroup$
    – anna v
    Commented Jun 29, 2011 at 20:00
  • $\begingroup$ If annihilation could energetically happen with a finite probability on a non virtual positron from the sea, there would be a lot of interactions creating pions etc and destroying the beam. It would be like the beam hitting a target. $\endgroup$
    – anna v
    Commented Jun 30, 2011 at 4:05

Welcome to the community WizzPhiz. You do not state your age in your profile, but I would think <20?

There are no spontaneously physical electron positron pairs created from the vacuum. The reason is called "conservation of energy" It would take energy to create such a pair. The vacuum sea consists of "virtual particles" and the electrons going around the accelerator cannot "see" them as in addition to being virtual they are created and annihilated in a small delta(time).

In this link, which is a festschrift for a scientist, in paragraphs 2 and later there are a lot of explorations of electrons scattering off radiation, even very low black body radiation that exists in a vacuum, but these are not the Dirac sea pairs you are asking about.

The vacuum pairs do have experimental signatures in the Casimir effect, and in the widening of the Lamb shift.

When one goes to General Relativity the vacuum particles theoretically may become physical in accelerated systems but there is no solid experimental evidence of this. The beam energies in the accelerators we have are not in that ball park of acceleration.

  • 1
    $\begingroup$ I thought that one of the ways that a black hole could evaporate was to 'swallow' a positron from a spontaneous creation and thereby lose some energy? $\endgroup$ Commented Jun 29, 2011 at 15:07
  • $\begingroup$ @anna v : You apparently didn't finish your answer. It ends quite abruptly. $\endgroup$ Commented Jun 29, 2011 at 17:26
  • $\begingroup$ @Frédéric Grosshans Sorry. it was left over from a previous writing. $\endgroup$
    – anna v
    Commented Jun 29, 2011 at 19:51
  • 1
    $\begingroup$ either it's the same as you started with as per frederic's answer because you can't distinguish the electrons from each other or is it that the virtual positron won't touch the real electron because it can't see it as per Anna's answer? How can we tell the difference? $\endgroup$
    – WizzPhiz
    Commented Jun 30, 2011 at 9:21
  • 1
    $\begingroup$ Frederic's answer is OK when one is writing diagrams to calculate crossections ignoring general relativity. Look for Unruh effect and accelerators when one includes it. That is the one place where the existence of the vacuum sea is measurably affecting the data, even though the accelerations are very low. The acceleration is necessary to provide the energy to get the pair, or part of it, out of the sea. $\endgroup$
    – anna v
    Commented Jun 30, 2011 at 10:59

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