A electron-positron annihilation creates two photons. What if the particles were accelerated - three photons?
I think the result is just going to be a little blueshifting of the photons - because there is a reference frame in which you could ignore the kinetic energy of one or other of the particles, therefore in a reference frame where you can’t ignore it, the number of photons is the same, but their energies are different.
Blueshifting - should be relatively easy to test??? Alredy done?
Yes. Already done. It’s a basic part of the conservation of energy. The photons created contain all of the energy of the initial particles, kinetic energy included.
Although now I want someone more knowledgeable than me to explain what happens with gravitational potential energy in annihilation!
The annihilation products appear at the same place in the same gravitational field as the original particles were when they met, so the gravitational potential energy is automatically conserved. If the resulting photons should happen to leave that gravity well, they’ll be redshifted as they do so.
I don’t understand the bit about gravitational potential energy, photons don’t have mass. Or is this some other concept that I’m not understanding? I’m taking a physics course right now and we’ve talked a little about this, the threshold energy is 1.022Mev, and the two particles have one half of that energy .511Mev opposite in a straight line 180 degrees.
A photon doesn’t have mass, but it’s still subject to gravity.
Otherwise, why would black holes be black? The basic definition of a black hole includes its event horizon, the radius from the center of mass at which the escape velocity equals the speed of light. This wouldn’t be a factor if light weren’t affected by gravity.
See also gravitational lensing.
What about angular momentum? The positron and electron each have spin +/-1/2 so together they either have spin +/-1 or 0. If they have spin 1 together how can that be achieved with two photons each having spin +/- 1?
n/m
Neutrinos take care of the conserved spin.
Two particles with spins m and n can have a total spin of anywhere between m + n and |m - n|, in integer increments. In particular, two particles of spin 1 can also have a total spin of anywhere between 0 to 2 in integer increments, which of course includes 1. Very roughly speaking, this is due to the fact that angular momentum is a vector, and the magnitude of the sum of two vectors can be anywhere between sum of their magnitudes and the difference of their magnitudes, depending on the directions in which they point (the triangle inequality.)
That said, an electron and a positron can form a bound state called positronium; and if the electron and positron have a total spin of 1, it turns out that the particles can’t produce two photons when they annihilate but must instead produce three. This has more to do with charge conjugation symmetry than the conservation of angular momentum, though. If you’re interested, you could read this paper for the gory details.
You’re thinking of beta-decay of protons/neutrons in a nucleus. For electrons & positrons annihilating, no neutrinos are produced.
Are you sure about that. It was my understanding that neutrinos are produced in a low energy electron positron annihilation only around one time in thousands. But the spins should add match and to +/- 1 half the time.
I thought that photons had to have spin +1 or -1 on a vector which pointed in the direction they were traveling? In the frame of reference in which the electron and positron have zero total momentum, if two photons produced don’t they have to leave in opposite directions? If all this is correct, they’d have to have +2, 0 or -2 spin.
Your other answer does clear things up though.
Thought experiment: You have a hollow spherical shell of unobtanium. Inside the sphere you have two equal masses of matter and antimatter. From outside of the shell, you can measure the combined masses of shell, matter, and antimatter.
Now the matter and antimatter annihilate and are converted into photons, which bounce endlessly off the inside of the perfectly reflective unobtanium shell. Does the gravitational mass you measure from outside the shell change?
Looks like I was completely wrong about the neutrinos. Apparently this sort of interaction almost never produces neutrinos.
No, it does not change (if it did, you could make a perpetual motion machine that way). The key is that, even though an individual photon does not have mass, a group of two or more photons collectively can and usually does have mass. Mass is not simply additive, as most people assume: That’s just an approximation that works pretty well in everyday life.
I one has no mass, but several do - is mass an attribute of interaction?
It can be.
Basically, for any given system, you can measure the total energy. Now, observers in different reference frames will come up with different values for the total energy: For instance, if I’m sitting in an airplane eating a peanut, an observer on the ground will say that the peanut has some kinetic energy, and will include that in the total, but relative to me, it’s still, and so doesn’t have any kinetic energy. And different observers might also disagree about which part of a system the energy is in. But for any given system, there’s some minimum value of total energy, such that all observers, regardless of reference frame, will always measure at least that much energy. And that minimum amount of energy, the “invariant portion” of the energy, is what is defined to be the mass.
So, for example, with just one single photon (or a set of photons all going in exactly the same direction), I can redshift them or blueshift them by moving in the same or opposite direction as them. The faster I go, the more redshifted they’ll be, and I can make their energy as close to zero as I like. So an individual photon (or a set of them all going exactly the same direction) has zero mass.
But now suppose that I have two photons going in opposite directions. I can redshift one of them by moving very quickly in the same direction as it, but in the process, I’m also blueshifting the other one. And if I redshift the second photon instead, then I’m blueshifting the first. If you work out the math, it turns out that the absolute lowest possible energy I can have for the two of them combined is what I observe in the frame where they’re each the same energy, evenly balanced. So that’s the mass of that pair of photons.
And for a pair of photons moving in different directions, there’s always some reference frame where they’re moving in opposite directions, so this applies to that case, too.
Before the thread gets diverted down the well-worn path of “invariant mass of a gas of photons” (search the board for many essentially identical threads on that topic), one mustn’t bury the lead with regard to OP’s question.
You’re on the right path. Adding kinetic energy to the system can change what gets produced. The exact answer depends on how the particles are moving relative to one another.
The cleanest reference frame to address this question is the “center of mass” or “center of momentum” reference frame, which for your example is the reference frame where the two particles are moving directly toward each other at equal speed. You can pick a different reference frame, in which case the electron and positron will have more total kinetic energy, but that extra kinetic energy doesn’t matter when asking “what does the annihilation produce?”
If your electron and positron (in the center of mass reference frame) are moving very slowly toward one another, then they will annihilate to two photons. As you increase their energy, the annihilation can produce ever heavier particles. Thus, the annihilation products can be pretty much anything you want, within conservation laws. When multiple products are possible, it’s a random process as to which one happens, with the probabilities governed by the energy available and the physics of the fundamental interactions (electromagnetic, weak, and strong interactions).
What you’ve described is precisely how electron-positron colliders work. The highest energy electron-positron collider ever was the machine that lived in the same tunnel now occupied by the LHC. This was LEP. LEP brought electrons and positrons together at high energy to let them annihilate with each other – exactly your hypothetical situation – to produce new particles and study the properties of said particles (and associated interactions).
So, the products of electron-positron annihilation can include photons, Z bosons, W bosons, Higgs, muons, quarks, neutrinos, … anything, as long as there’s enough energy and along as certain conservation laws are respected (like angular momentum and charge conservation).