# Lasers so powerful they could rip apart empty space

An article in Science says:

According to the theory of quantum electrodynamics (QED), which describes how electromagnetic fields interact with matter, the vacuum is not as empty as classical physics would have us believe. Over extremely short time scales, pairs of electrons and positrons, their antimatter counterparts, flicker into existence, born of quantum mechanical uncertainty. Because of their mutual attraction, they annihilate each another almost as soon as they form.

But a very intense laser could, in principle, separate the particles before they collide. Like any electromagnetic wave, a laser beam contains an electric field that whips back and forth. As the beam's intensity rises, so, too, does the strength of its electric field.At intensities around $10^{24}$ W/cm2, the field would be strong enough to start to break the mutual attraction between some of the electron-positron pairs, ... . The laser field would then shake the particles, causing them to emit electromagnetic waves—in this case, gamma rays. The gamma rays would, in turn, generate new electron-positron pairs, and so on, resulting in an avalanche of particles and radiation that could be detected.

Ethan Siegel writes about this in his blog "Starts With A Bang":

...you can — through physics — use these electromagnetic properties of empty space to generate real particle/antiparticle pairs.

You ought to notice, right away, that even the dream scenario of the science article gives intensities that are still 100,000 times too small to reach this threshold, and whenever you’re below that threshold, your ability to produce particle/antiparticle pairs is exponentially suppressed. The mechanism at play is quite different than simply the reverse of pair production, where instead of an electron and positron annihilating to create two photons, two photons interact to produce an electron/positron pair. (That process was first experimentally demonstrated way back in 1997.) In the laser setup, no individual photons have enough energy to produce new particles, but rather their combined effects on the vacuum of space causes particle/antiparticle pairs to pop into existence with a particular probability. Unless, however, those intensities approach that critical $10^{29}$ W/cm² threshold, that probability might as well be zero.

But in a reference frame moving in the direction of the beam, the power density would be lower. If it is below the threshold, the effect should not be seen by an observer in that frame. Can you explain this apparent issue?

• anna v's answer to Pair production at high laser intensity? suggests that this can only occur in the presence of electrons or atomic nuclei. – Keith McClary Feb 13 '18 at 1:48
• Anna is correct in the case of one laser beam, but two counterpropagating beams can in theory pair-produce without any third actor to conserve momentums: I don't know how one would arrange this practically, though, the self-vaporizing apparatus is quite comical to imagine. One might use quarter waveplates and polarizers to prevent each beam reaching the other source, but I suspect you'd run into practical difficulties at such stupendous intensities. – WetSavannaAnimal Feb 13 '18 at 2:03
• @WetSavannaAnimalakaRodVance As Ethan says, the photons have much too low energy for pair production. This (theoretically predicted) effect depends on the electric field intensity and is supposed to happen in a single beam. – Keith McClary Feb 13 '18 at 2:27
• @WetSavannaAnimalakaRodVance they are contemplating gamma gamma colliders slac.stanford.edu/pubs/beamline/26/1/26-1-kim.pdf – anna v Feb 13 '18 at 8:05
• @annav Wow! And the powers and other parameters don't look too unreal either. Amazing – WetSavannaAnimal Feb 13 '18 at 11:14

## 1 Answer

But in a reference frame moving in the direction of the beam, the power density would be lower. If it is below the threshold, the effect should not be seen by an observer in that frame. Can you explain this apparent issue?

The issue is clarified in the center of mass of the real e+e- pair created. In that center of mass of the pair, a single photon will still be going with velocity c and zero mass, whereas the pair has the invariant mass of two electrons and cannot be going with velocity c, reductio ad absurdum.

A third field is necessary in order that energy and momentum are conserved.

If two photons in the beam have invariant mass equal or larger to twice the mass of the electron, so as to create a pair, at their center of mass they have to be back to back for momentum conservation, thus cannot be in the beam which will be going through with velocity c in the beam direction even at the center of mass of these two putative parent gammas. So it cannot be a two photon effect, and if a pair is created, there should be a third real photon /gamma coming out of the interaction.

Hand waving for the possibility of multiphoton effects:

It is true that the individual photons building up the beam in synergy are not all pointing in the same optical ray direction, which means that at some density of energy a "bunch" of photons might interact with one photon through QED diagrams and if the angular divergence is large enough it would be similar to the two photon argument discussed in the previous paragraph. There would still be the need for a third real particle coming out of the fray, with an analogous argument.

• To clarify, are you saying this is kinematically allowed for >2 particles created?(BTW, that is the same paper Ethan linked. ) – Keith McClary Feb 14 '18 at 4:27
• @KeithMcClary yes. There is a mathematical contradiction of only two particles from one when the rest mass is zero. (decays of massive ones happen all the time) – anna v Feb 14 '18 at 5:01
• This is not quite what I ask, though. Regardless of the details of the interaction or the types of particles produced, in a given rest frame the number of events/volume should be a function $W(I, \nu)$ where $I$ is the energy intensity and $\nu$ the frequency. I argued that if there is a perfectly sharp threshold (meaning $W(I, \nu)$ drops to $0$ below some values of the arguments) then observers in different frames would calculate different results for the same light beam (one observer would calculate $0$ events). So there must be some constraint on the sharpness of the function $W(.,.)$. – Keith McClary Mar 1 '18 at 5:05
• You are thinking classical electrodynamics. Particle creation is quantum electrodynamics, different mathematics fit the data. It is not the power density that determines the creation of particles but the quantum mechanical probability of interaction which is lorenz invariant so one does not have to think of volumes etc. – anna v Mar 1 '18 at 5:16
• Yes, but when you've done the calculation you come up with a formula predicting events/volume observed (by experimentalists) in a beam of given $I$ and $\nu$. Your formula should work for observers in any frame looking at the same beam (using their respective values of $I$ and $\nu$). So any old function \$W(.,.) can't be right. – Keith McClary Mar 1 '18 at 5:35