Question: Can a single photon have enough energy to create two pairs?

[Gonzolo]

Can 1 photon become 2 pairs? (Edited 17 Aug 2004)

Hi, what happens if the energy of a photon is more than that of an e- and an e+, can it create two pairs instead or will the 1st pair simply have more kinetic energy? Or are other types of particles involved?

Added August 17 2004 :

Yeah, yeah I won't forget now, you need 2 photons to make an e+ e- pair because of C. of Momentum. A stupid drawing in an otherwise good book shows one only becoming a pair and has remained in my head like the Smurf's theme song. I thus translate the question to :

Can 2 photons become 2 pairs?

Or more generally :
Is <number of photons> / <number of pairs> necessarily 1?

Or even more generally, what happens if the energy of a single photon is much more than (1/2)mc2, where m is the mass of an electron?

 

[jcsd]

A photon can't decay by itself, otherwise four momentum will be violated, but it can decay if it inetracts with other particles.

 

[Nenad]

Hi, what happens if the energy of a photon is more than that of an e- and an e+, can it create two pairs instead or will the 1st pair simply have more kinetic energy? Or are other types of particles involved?

Im preety shure it can only create one pair.

 

[jcsd]

Im preety shure it can only create one pair.

It can create just about anything as long as the fundmantal symmetries (such as the conservation of four momentum) are adhered to.

 

[meteor]

It seems to me that a photon can decays spontaneously in a process called Parametric down-conversion, but this decay produces two photons, not a pair electron-positron

 

[jcsd]

It seems to me that a photon can decays spontaneously in a process called Parametric down-conversion, but this decay produces two photons, not a pair electron-positron

AFAIK this process involves putting a photn throuh a crystal to produce two daugter photons.

 

[selfAdjoint]

It can create just about anything as long as the fundmantal symmetries (such as the conservation of four momentum) are adhered to.

And since the magnitude of four-momentum is mc^2, and the photon is massless, its four-momentum has magnitude zero, while any massive particle will have a four-momentum greater than zero. Therefore conservation of four-momentum means the photon can't decay into massive particles.

Once learned, never forgotten.

 

[Gonzolo]

It seems to me that a photon can decays spontaneously in a process called Parametric down-conversion, but this decay produces two photons, not a pair electron-positron

This is non-linear optics, it relates to what happens when many photons of "regular" (visible, IR etc.) energy and crystals are involved.

What I'm questioning about is what happens at the upper "end" (?) of the energy scale of a single photon.

It can create just about anything as long as the fundmantal symmetries (such as the conservation of four momentum) are adhered to.

What it creates depends on what particle it interacted with to change?

...Therefore conservation of four-momentum means the photon can't decay into massive particles.

Assuming your are serious, I'm confused. It is well known that a photon can decay into an electron and a positron, which are massive.

 

[jtolliver]

Assuming your are serious, I'm confused. It is well known that a photon can decay into an electron and a positron, which are massive.

Actually that takes 2 photons.

 

[jcsd]

As I've been trying to say:

1) an isolated photon cannot decay, if you notice I said this in the second post, and it was again re-iterated by self-adjoint.

2) it may decay if it interacts with other particles, provided that the fundmantal symmetries of nature are adhered to, for example the total chrage of the daughter particles must be zero, etc. these fundamenatal symmetries are the only limits on what the daughter particles will be.

 

[ZapperZ]

Maybe, at this point, it requires an elementary illustration.

Let's say you have a photon with energy exactly equal to the rest mass energy of an electron and a positron. So in principle, this photon has enough energy for this pair production. But while conservation of energy is maintained, there's problem with conservation of linear momentum. The photon orginally had a well-defined momentum. But after the pair production, the there's enough energy to only created the pair, but with no kinetic energy. Thus, the net momentum of the pair system is zero. Violation of conservation of momentum.

That why it has been said a few times on here that pair production cannot occur in isolation (or in vacuum). It can only occur in the vacinity of a "massive" object, such as a solid crystal. This object provides the necessary momentum conservation for the pair production to occur. The typical scenario for pair production is by passing high energy photons through something like Be crystal (or something else, I don't quite remember this).

Zz.

 

[Nereid]

Ditto with high energy gammas in space - whether interstellar space or intergalactic space. 0.51 MeV 'line emission' is observed by satellite observatories (e.g. Compton), from objects many Mpc distant.

What happens to gammas in the TeV to PeV range? There are several astrophysical processes which can generate copious quantities of them - can they travel through 100 Mpc between superclusters?

 

[ArmoSkater87]

Hi, what happens if the energy of a photon is more than that of an e- and an e+, can it create two pairs instead or will the 1st pair simply have more kinetic energy? Or are other types of particles involved?

You would need for two energetic photons (gamma) to collide head on (to conserve momentum) in order to create an electron/positron pair. The opposite happens when the pair anihhilate...two gamma rays are emitted in opposite directions.

 

[ArmoSkater87]

Decays must always conserve momentum, in fact it led to the discovery of the neutrino. In beta decay of elements with unstable proton/neutron ratios, depending on whether the ratio is higher or lower than the stable ratio, the nucleus of the element will either decay its protons or neutrons.
Proton => neutron + positron + neutrino
Neutron => proton + electron + antineutrino.
When this decay was observed, it was noticed than the neutron and positron (or proton and electron) were emitted at an angle, therefore violating conservation of momentum, which lead people to believe than there was something else there that they were not detecting.

 

[Gonzolo]

See corrected question (original post).

 

[Nereid]

Or even more generally, what happens if the energy of a single photon is much more than (1/2)mc2, where m is the mass of an electron?

If the photon's energy is > 2 x m_e (i.e. >1 MeV), pair production is possible. Interesting questions: under what conditions? How does the (photon->pair) rate vary with photon energy (all other things being equal)?

 

[selfAdjoint]

Here we go again. Energy conservation isn't enough for pair production; you also need conservation of momentum. That's relativistic 4-momentum of course since we're talking about photons here.

Now the magnitude of a particle's 4-momentum is mc^2. The photon's mass is zero so its 4-momentum has zero magnitude; that's where we get all those null trajectories and so on for the photon.

But the mc^2 value for the electron is greater than zero, because it has a finite mass.

So we have impossible physics; zero magnitude momentum on one side and non-zero on the other side. Conclusion: the photon cannot spontaneously decay into massive particles. No go.

If it makes you feel better I made this same mistake about a year ago and got corrected then. Pass it on.

 

[Nereid]

Clearly, photons of energy > 1MeV aren't like neutrinos - they can't go through a light-year of lead without being absorbed

If such a gamma enters a region with matter - a fully ionised plasma, a gas, a liquid, a solid, degenerate matter - can it undergo an interaction from which an electron/positron pair emerges? In what way does the target affect the pair production rate?

 

[what_are_electrons]

Clearly, photons of energy > 1MeV aren't like neutrinos - they can't go through a light-year of lead without being absorbed

If such a gamma enters a region with matter - a fully ionised plasma, a gas, a liquid, a solid, degenerate matter - can it undergo an interaction from which an electron/positron pair emerges? In what way does the target affect the pair production rate?

Based on my understanding, Yes, That's it. The target is the source of the pair.

 

[selfAdjoint]

Clearly, photons of energy > 1MeV aren't like neutrinos - they can't go through a light-year of lead without being absorbed

If such a gamma enters a region with matter - a fully ionised plasma, a gas, a liquid, a solid, degenerate matter - can it undergo an interaction from which an electron/positron pair emerges? In what way does the target affect the pair production rate?

Yes. The photon is absorbed, the other particle (massive) goes into a raised energy state and subsequently decays into other particles.

 

[Nereid]

http://besch2.physik.uni-siegen.de/~depac/DePAC/DePAC_tutorial_database/grupen_istanbul/node20.html describes pair production "in the Coulomb field of a nucleus" and "in the Coulomb-field of an electron", and gives a cross section for the former in terms of various target parameters.

Clearly energy and momentum are conserved, but does pair production require 'a Coulomb field'? If the photon were to enter a neutron gas, would there be pair production?

 

[LarryHerene]

Yes quarks carry charge

The charge of the constituent ( and virtual ) up and down quarks would be able to interact with the photon.

 

[kurious]

Gravitons are massless.So if a photon decayed into gravitons four momentum could be conserved.Yes or no?

 

[Nereid]

Gravitons are massless.So if a photon decayed into gravitons four momentum could be conserved.Yes or no?

Gravitons are hypothetical (photons are not), so their properties are dependent on the theory which 'creates' them. I rather doubt there would be any theories in which a photon could decay into gravitons ... but if you come across one, please let us know

 

[kurious]

If photons decayed into gravitons this might explain where energy of redshifted photons from distant galaxies goes - there is a debate in the relativity section on physics forums about the conservation of energy in general relativity.

 

[Nereid]

If photons decayed into gravitons this might explain where energy of redshifted photons from distant galaxies goes - there is a debate in the relativity section on physics forums about the conservation of energy in general relativity.

Well I guess it *might*, but it would also certainly require a pretty dramatic re-writing of most areas of astrophysics ... and QM!

Why? Well for one thing, we can observe galaxies that are certainly redshifted (in terms of their observed redshift being partly/largely due to the expansion of the universe), and we can see that the stars which comprise these galaxies are just like those here in the Milky Way ... if photons -> gravitons, then the appearance of those stars in distance galaxies would be an illusion; they would 'actually' be quite different from nearby stars ... so back to the drawing board re stellar evolution models, inc the pp process, CNO cycle, neutrinos, He burning, white dwarfs, neutron stars, supernovae ...

IIRC, someone (meteor?) posted a preprint of paper (in Astronomy & Cosmology?) which proposed an alternative to universal expansion as the reason for the Hubble relation; it was >60 pages long, and barely scratched the surface on all the areas of astrophysics that needed varying degrees of re-interpretation! (Needless to say, it seemed not to mention some of the areas that I would have guessed would be hardest to 'unexplain').

 

[kurious]

I wasn't questioning expansion - just saying that the longer a photon takes to reach Earth the more redshifted it will be (if it emitted gravitons continuously).
Anyway, how does conservation of four momentum work out for a radioactive nucleus that yields a gamma ray.Is this analagous to a time-reversal of a gamma ray colliding with a lead atom and producing positron-electron pairs?

 

[Nereid]

Whoa! 'emitting' is a whole different kettle of fish from 'decaying into'! :grumpy:

Photons 'emitt[ing] gravitons continuously' would require a rather strange theory of gravitons for this kind of thing to not break all manner of conservation rules ... I'm not saying such a theory wouldn't be possible, just that from my tiny understanding of these things, it would be a far more radical departure from any serious theories currently on the drawing board are from well established ones.