Can Gamma Rays Alone Create Electron-Positron Pairs?

[Chaos' lil bro Order]

We know an electron colliding into a positron annihilate one another and 2 recoiling gamma rays are produced moving in opposite directions...
Is the process of pair-production, where an electron and positron are created, due to 2 colliding gamma rays annihilating each other? Furthermore, can a single gamma ray, of sufficient energy, spontaneously annihilate itself to create the same electron and positron pair? Thirdly, is there any basis in experiment or theory that allows for a single gamma ray to interact with the quantum field in a manner than could create an electron/ poistron pair? And finally, is there a version of string theory that would allow for a single gamma ray to interact with tiny dimensions whereupon an electron/ positron pair could be created?

 

[kvantti]

We know an electron colliding into a positron annihilate one another and 2 recoiling gamma rays are produced moving in opposite directions...
Is the process of pair-production, where an electron and positron are created, due to 2 colliding gamma rays annihilating each other?

Nope, it's due to a single > 1022 keV gamma ray photon decaying.

Furthermore, can a single gamma ray, of sufficient energy, spontaneously annihilate itself to create the same electron and positron pair?

Yup, see above.

Thirdly, is there any basis in experiment or theory that allows for a single gamma ray to interact with the quantum field in a manner than could create an electron/ poistron pair?

Yup, QED explains it theoretically and we have particle accelerators to show it in action. See http://cyclo.mit.edu/~bmonreal/photos/e-p1.jpg image for example.

 

[Meir Achuz]

We know an electron colliding into a positron annihilate one another and 2 recoiling gamma rays are produced moving in opposite directions...
Is the process of pair-production, where an electron and positron are created, due to 2 colliding gamma rays annihilating each other? Furthermore, can a single gamma ray, of sufficient energy, spontaneously annihilate itself to create the same electron and positron pair? Thirdly, is there any basis in experiment or theory that allows for a single gamma ray to interact with the quantum field in a manner than could create an electron/ poistron pair? And finally, is there a version of string theory that would allow for a single gamma ray to interact with tiny dimensions whereupon an electron/ positron pair could be created?

1. It is difficult to collide two gammas. It does occur vitually, but at such a high energy that just a single e-p (I'll use p for positron.)
pair is rare.
2. A single gamma cannot-->e-p. It is forbidden by E and p conservation.
Pair production usually occurs in the electric field of a nearby nucleus.
3. No, as in 2.
4. String theory also conserves E and p, if little else.

 

[Astronuc]

Pair production occurs with the interaction of a gamma-ray of at least 1.022 MeV (1022 keV), which is the threshold based on the rest masses of electron/positron, with the nucleus of an atom, or perhaps a heavy particle. Otherwise, photons scatter off electrons - Compton effect - or are absorbed - photoelectric effect.

 

[Chaos' lil bro Order]

2. A single gamma cannot-->e-p. It is forbidden by E and p conservation.
Pair production usually occurs in the electric field of a nearby nucleus.
QUOTE]


Your answer contradicts an earlier reply, who is right?
Also, I don't see why a single gamma -->e-p violates any law of conservation. The only conservation law I can think of is the conservation of lepton # which appears to be preserved by virtue of the e-p pair. Please explain your comment.

 

[Chaos' lil bro Order]

Nope, it's due to a single > 1022 keV gamma ray photon decaying.



Yup, see above.



Yup, QED explains it theoretically and we have particle accelerators to show it in action. See http://cyclo.mit.edu/~bmonreal/photos/e-p1.jpg image for example.

First off, very nice post. It was well-formatted, concise and insightful, so thank you. Secondly, your link to the bubble chamber picture puzzles me as an explanation for single gamma -->e-p. I'm far from an expert, but I can see the e-p diverge in opposite directions from a single S->North line, BUT this line continues even after the e-p pair is created. So IF this line is the gamma (which I believe it is?) how can it continue beyond the e-p divergence? Is the S->North line consisting of many gammas?

Thanks.

 

[CarlB]

2. A single gamma cannot-->e-p. It is forbidden by E and p conservation.
Pair production usually occurs in the electric field of a nearby nucleus.
QUOTE]


Your answer contradicts an earlier reply, who is right?
Also, I don't see why a single gamma -->e-p violates any law of conservation. The only conservation law I can think of is the conservation of lepton # which appears to be preserved by virtue of the e-p pair. Please explain your comment.

Meir Achuz is right. I think that Kvantti meant his comment is in the context of a bubble chamber where there is stuff available to carry off the excess energy. But then it wouldn't really be a single gamma ray turning into an electron and a positron, it would have something else involved.

A single gamma can't go to an e and p because it can never have enough energy to pay for the masses of the electron and proton (after balancing conservation of momentum). Another way of saying the same thing is that the photon always has too much momentum for the electron and positron to conserve momentum (assuming that the energies balance). Yet another way of saying this is that if you want to get the best rocket ship drive, use photons because they have the highest possible ratio of momentum to energy. Maybe one of these will help you remember it, if not, make up your own.

Here is the calculation, assume the photon travels in the +z direction:

Gamma ray energy-momentum 4-vector:
(E_{\gamma},0,0,P_{\gamma})

First electron 4-vector:
(E_1,Px,Py,Pz)

Second electron 4-vector (to get energy and momentum to add up):
(E_\gamma - E_1,-Px,-Py,P_\gamma-Pz)

Now if that were all there is to it, then sure, you could do the deed. (And in fact, you could if these were virtual particles that you will learn about much later in your career.) But that's not all there is to it. The particles have masses and the masses satisfy an equation like the following (forgive me, it's been 20 years so I could easily have a sign wrong):

E_\gamma^2 = P_\gamma^2c^2
E_1^2 = (Px^2 + Py^2 + Pz^2)c^2 - m^2c^4
(E_\gamma-E_1)^2 = (Px^2 + Py^2 + (P_\gamma-Pz)^2)c^2 - m^2c^4

where m is the mass of an electron = mass of a positron.

If you subtract the third equation from the second one you get:

E_\gamma^2 -2E_1\;E_\gamma = (P_\gamma^2 -2Pz\;P_\gamma)c^2

Taking into account the first equation gives you:

E_1 = Pz\;c

Squaring this and subtracting from the 2nd equation, you get three non negative terms on the right and zero on the left. Therefore the three non negative terms, Px\;c, Py\;c, m^2c^4 must all be zero. But m^2c^4 cannot be zero.

From this, you see that if one massless particle decays into two particles, the two created particles have to be both massless and they both have to have the same momentum. In other words, they have to travel on a path just like the original massless particle. And if all the interesting quantum numbers are conserved, how could you know that a decay took place?

Carl

 

[Chaos' lil bro Order]

A single gamma can't go to an e and p because it can never have enough energy to pay for the masses of the electron and proton (after balancing conservation of momentum). Another way of saying the same thing is that the photon always has too much momentum for the electron and positron to conserve momentum (assuming that the energies balance).


Carl

Forgive my daftness but let me set up a scenario for you to illustrate my question.

Consider a single gamma ray traveling Northwards. This gamma ray spontaneously annihilates itself by some process unknown to us, that we shall reserve comment on for the time being. This annihilation produces an e-p pair. The e travels Northwards at an arbitrary velocity equal to 6. The p travels Southwards at an arbitrary velocity equal to 2. IF the gamma ray was initially traveling Northwards at a velocity equal to 4, can we say that conservation of momentum was conserved? Is this even possible for the e traveling Northwards to have been imparted with a such a large velocity relative to the p traveling Southwards?

Finally, and this may be a very dumb question, are there any known laws of nature that forbid a photon from decaying into smaller particles, other than the fact that we presently believe photons to be indivisible fundamental particles?

 

[kvantti]

I think that Kvantti meant his comment is in the context of a bubble chamber where there is stuff available to carry off the excess energy.

Yes, something like that. I meant that you don't actually need two photons to have an e-p pair produced; a single gamma photon can "kick off" from an atomic nucleus, for example, to produce a electron-positron pair.

I'm far from an expert, but I can see the e-p diverge in opposite directions from a single S->North line, BUT this line continues even after the e-p pair is created.

I bet it is a neutron, because it sure doesn't seem to have a charge (it doesn't steer to any specific direction in the magnetic field of the bubble chamber). So I think the image is actually showing a single gamma photon "kicking off" from a neutron to produce the e-p pair. Maybe, maybe not. It could be a high-energy atomic nucleus also.

 

[simic4]

Just a comment on CarlB's proof. There is a much easier and more general way to prove that a photon cannot decay into electron anti electron pair.

Let P be the fourmomentum of the photon. let P1 and P2 be the 4momenta of the resulting electron and anti electron respectively. the photon deacying is the statement:

P = P1 + P2 => P^2 = 0 = m^2 + m^2 + 2P1*P2

now go to the rest frame of the electron, P1 = (m,0,0,0)

and you get the contradiction that 0 > 2m^2 >0.

thus a photon (or any other massless particle) can never split into two massive particles, atleast in the absence of external fields.

 

[Chaos' lil bro Order]

What is the heaviest known pair of particles produced from gamma annihilation?