21
$\begingroup$

According to wikipedia, all antimatter annihilation produces gamma rays (along with potentially other elements).

Why specifically Gamma rays? Why not electromagnetic waves of other wavelength?

$\endgroup$
4
  • 12
    $\begingroup$ ...because the energy of two annihilating objects is such that if you divide it among two photons you mostly get gamma rays? $\endgroup$ – ACuriousMind Jul 17 '16 at 18:11
  • 3
    $\begingroup$ One has to distinguish between the annihilation process and secondary thermalization, which could produce a very wide range of spectral emissions. We usually only look at the primary products, but what would happen in a scenario that involves other matter is complicated. $\endgroup$ – CuriousOne Jul 17 '16 at 18:38
  • 1
    $\begingroup$ Basically because in order for the photons to have energy equivalent to the mass of the electron annihilated, it needs to be very high frequency. i.e., gamma-rays. If we could observe the annihilation of smaller particles (and that pretty much only leaves the lighter neutrinos) then the frequency could be lower than gamma rays. But could luck trying to do that, the hurdles are pretty big, I'm not sure if we even know whether neutrinos have a separate anti-particle or not. $\endgroup$ – RBarryYoung Jul 18 '16 at 18:08
  • $\begingroup$ Because matter corresponds to so much energy $E=mc^2$ that the likeliest energy levels for (just a few) photons created would be so large that the quanta with enough energy would be gamma rays. Then why just a pair or a few photons would be created and not, say a billion small rays are created simultaneously from one annihilation? I don't know that. But presumably there is something limiting the number of quanta that can be produced from a single annihilation (?) $\endgroup$ – mathreadler Jul 18 '16 at 20:57
4
$\begingroup$

To expand on a @ACuriousMind's comment, consider an electron-positron pair such that their center of mass is at rest. Since energy and momentum are conserved in the annihilation process, the resulting photons will be emitted in opposite directions with an energy $E = m_e = 511\ \text{keV}$ each (units such that $c = 1$). If the pair is instead in motion, one photon will be blueshifted and the other redshifted.

However, to redshift a $511\ \text{keV}$ photon to even $100\ \text{keV}$ requires strongly relativistic velocities $v > 0.95c$; to get to $10\ \text{keV}$ you need $v > 0.999c$. Now if the electron is part of "the environment", it's approximately at rest with respect to the lab system, so the positron must be moving at close to $c$ relative to the electron if we are to observe anything other than $\approx 511\ \text{keV}$.

But as the relative velocity increases, the cross section for annihilation decreases rather rapidly. For the velocities mentioned above, $\gamma \approx 3, 22$ so you can see that annihilation of relativistic positrons is suppressed. Thus we don't expect to see many photons that don't have energies close to $511\ \text{keV}$.

$\endgroup$
21
$\begingroup$

At tree level, a matter-antimatter annihilation reaction doesn't just produce gamma rays, nor can you exclude neutrinos in the final state.

Even the simplest such reaction can—given enough energy—produce a variety of particle-pairs. However, those pairs are subject to two subsequent processes:

  1. If the particles are not stable, they will decay towards lighter and more stable particles eventually reaching electron-positrons, nucleons/anti-nucleons, and neutrino/antineutrinos.

  2. The anti-particles in the final state (excepting some of the anti-neutrinos) that don't decay will eventually find some matching particle to annihilate with.

So if we wait a while, the final state of the annihilation will be photons (mostly in the gamma band, but possible some low energy ones as well) and neutrinos (both matter and anti-matter (or both right and left handed if neutrinos are Majorana in nature)).

The neutrinos are often ignored because for practical purposes they have no effect other than carrying energy and momentum away from the interaction.

$\endgroup$
5
  • 1
    $\begingroup$ the final state of the annihilation will be photons (mostly in the gamma band, but possible some low energy ones as well) . Why mostly in the gamma band? Why not in the visible or IR or UV band? $\endgroup$ – Charles Shiller Jul 17 '16 at 21:57
  • 1
    $\begingroup$ @CharlesShiller The amount of energy in the final state is the same as that in the initial state which means at a minimum there will be $E = 2mc^2$ where $m$ is the mass of each of the initial particles. The mass energy of even an electron is already in the gamma band. The number of photons in the simplest case is 2 photons, though 3 can be required depending on the angular momentum of the initial state. Low energy photons are allowed, but they appearance is governed by a probability calculation involving a factor of $\alpha^2$ where $\alpha \approx 1/137$ is the fine structure constant. $\endgroup$ – dmckee --- ex-moderator kitten Jul 17 '16 at 22:03
  • 1
    $\begingroup$ "A tree level a mater" - I don't understand this. $\endgroup$ – Jens Jul 17 '16 at 22:05
  • $\begingroup$ @Jens Aside from the typo you can read that as "To a first approximation". More exactly it means "considering only the simplest Feynman diagrams that contribute to the process", but that doesn't tell you much unless you know roughly how quantum field theories work. $\endgroup$ – dmckee --- ex-moderator kitten Jul 17 '16 at 22:07
  • $\begingroup$ @CharlesShiller In addition, one could also observe low-energy photons if the annihilating system is ultrarelativistic relative to the lab as then one photon would be redshifted. However, it is unlikely for such annihilations to happen in the first case: physics.stackexchange.com/a/268607/38551 $\endgroup$ – Robin Ekman Jul 17 '16 at 23:47
12
$\begingroup$

In the field of PET (positron emission tomography) people tend NOT to call the product of the annihilation "gamma rays", but rather "annihiliation photons". While that may be a subtle distinction, the view is that a "gamma ray" is emitted by a nucleus, while a "photon" is a more general term used for a quantum of electromagnetic energy. But according to Wikipedia, the definition / use of the term seems to depend on the field:

Electromagnetic radiation from radioactive decay of atomic nuclei is referred to as "gamma rays" no matter its energy, so that there is no lower limit to gamma energy derived from radioactive decay. This radiation commonly has energy of a few hundred keV, and almost always less than 10 MeV. In astronomy, gamma rays are defined by their energy, and no production process needs to be specified. The energies of gamma rays from astronomical sources range to over 10 TeV, an energy far too large to result from radioactive decay.[1]

(My emphasis).

So depending on the field - a photon is called a gamma ray if it has energy greater than 100 keV, or it is produced by nuclear decay. I think that no annihilation process will produce quanta less than 100 keV - but as @dmckee pointed out in his answer, there are annihilations where other particles (eg neutrinos) may be created.

You may ask "why not less than 100 keV?". That has to do with the mass of the particles. In the case of electron/positron pairs, their combined mass is equivalent to 1022 keV (2 x 511 keV). And if all matter disappears, it all has to turn to energy. It is possible though unlikely that an annihilation produces more than two photons - see this answer and links therein. This mentions that the electron-positron pair will annihilate into three photons in about 1/370th of all annihilations. I cannot find a description of the energy distribution in that case - but I imagine that the probability of one of these photons having energy less than 100 keV would be very small - especially in the c.o.m. frame of the annihilation (of course when your two particles are traveling near the speed of light relative to the observer, crazy things might happen to the energies of the observed particles.) This was measured in "Gamma-ray energy spectrum from orthopositronium three-gamma decay", Chang, Tang and Li, Phys Lett B, Volume 157, Issues 5–6, 25 July 1985, Pages 357-360 to which I unfortunately don't have access without handing over $39.95...

$\endgroup$
2
  • $\begingroup$ The habit of distinguishing types of photons by origin has a long history, but it only makes sense when you set up a system where you can know the origin, which is not generally the case. As a result there is a pretty strong contingent out there that argues fiercely against it and prefers to label photons only by band. Personally I'm wishy-washy: the origin based terminology makes it easier to discuss some topics in radioactivity and nuclear physics so I want to have the convention available from time to time. $\endgroup$ – dmckee --- ex-moderator kitten Jul 17 '16 at 20:44
  • 1
    $\begingroup$ @dmckee in the case of PET imaging the origin of the 511 keV radiation is pretty clear... but I agree that "no hard and fast rules" is probably the right approach. $\endgroup$ – Floris Jul 17 '16 at 21:09
5
$\begingroup$

The amount of energy to emit is enormous ($2mc^2$ plus kinetic energy), so the emitted energies per photon will be at least the rest mass of the electron, which is half GeV (not counting some extreme and unlikely redshifts due to the collision point moving with respect to the observer). And that's well in the gamma spectrum.

$\endgroup$
1
$\begingroup$

It is possible.

The rule is simple: the total energy (which is at least mc^2, m the sum of both masses) will be converted into other particles (photons or particle-anti-particle pair). So it is possible for a photon to pick up only a very few energy (for example by having a lot of photons, which is possible, even dough very unlikely).

I think, it is written as this, as it mostly, which means probably around 99.9% of the time, decays into high-energetic photons.

If you are interested, do the math by yourself: the total energy at the LHC is currently around 13 TeV. How many photons are required so that the mean energy is lower then gamma-energy? (and btw, for each photon added, the probability decreases by a factor of 1/100 roughly)

$\endgroup$
0
$\begingroup$

The main reason is that lightest common particle/anti-particle pair is the electron/positron. When they collide, the overwhelmingly most likely result is emitting photons. The electron and positron weigh 511 kev, so when they annihilate, there is 1.02 MeV of energy. So each photon has 511 kev. Photons with this energy are gamma rays. Other photons like visible light, X-rays, radio-waves, etc are lower energy photons.

::: The question had some simple assumptions so the above complete but synoptic answer had some simple assumptions as well. Below are some more nuanced discussions of related issues for those bothered by details omitted or glossed over above. (apologies beforehand if I neglect below to mention a nuance of particular interest to you.)

1) Why two photons and not one? The electron and positron have opposite charge, opposite spin, etc. At the point of annihilation, the system has 0 for most quantum variables (except mass-energy) So we are left with a small (instaneous) energy bundle with baryon number 0, lepton number 0 charge 0, ... If a single photon was emitted, the photon would have spin of 1 thus not conerserving spin. So (at least) two photons are emitted and they will have opposite spins. I.e. emitting a single photon is prohibited by the combined conserfcations laws of spin and momentum.

2) Why not more than two photons? Well you could have 3 photons. In fact if the electron and positron have the same spin (or more precisely are in a spin 1 state) then at least three photons are required. The outgoing set of photons must have a net spin of 1. This can only occur with an odd number of photons. However, emitting 1 photon is impossible from a consideration of momentum conservation. This can be seen by looking at the center of mass system. In that frame, the total momentum is zero, by construction. However, a photon always has a momentum of Energy/c. So the before and after conditions cannot conserve momentum if only a single photon is emitted - i.e. it is a forbidden channel. So spin 1 pre-annihilation system must have (at least) 3 photons. Now we could in principle have more than 3 emitted photons but they are much less likely. Because to emit 4 photons (for the spin 0 e-/e+ state) or 5 photons (for the spin 1 e-/e+ state) the effective vertex operator is much higher order and thus is suppressed by another 2 orders of the coupling parameter. So they are much, much less likely (but still possible)

3) Many things can be created, not just photons Well yes. But the energy available for e-/e+ system is only 1 MeV so only particles with energy < 511 KeV could be created on shell (i.e. real particles, not virtual particles) photons and all zero mass particles fit that bill. But not many are available. For example, creation of a neutrino/anti-neutrino pair is possible but the probability is very small given that relative strength of the weak force versus the EM force a that energy scale. That is almost all e-/e+ annihlations lead to photons. A very, very small number lead to neutrinos. So in practice, good luck doing that experiment. The cross section might be too small to have any hope of measuring it.

4) But there are other matter/antimatter annihilation possiblities beyond e-/e+ Yes. good point. proton/anti-proton for example or "other things". In those cases, the standard analysis is to consider creation out of the annihilation of particles of less mass. So for example, proton/anti-proton annihlation can lead to the creation of e-/e+ pairs. Moreover, one can accelerate "thingies" for annihilation where the center of mass energy is much higher than rest mass. A e-/e+ pseudo-atom in an hydrogen-like bound state will only have about 1 MeV of energy available after annihilation. However, if the e-/e+ were brought together in a high-energy accelerator (such as the B-factory for example) then the center of mass energy could be high enough to generate a zoo of post-annihilation products (hence the name B-factory) e-/e+ are nice for these experiments because so many quantum numbers are 0: baryon number, charge, ... Proton colliders have baryon number of 2 and so the zoo of products is much messier.

5) Many products are subject to various forces and often recombine So for outgoing photons this is very clean. You annihilate matter and antimatter, emit photons which do not interact and they travel outward - nice and simple. Suppose you emit a quark and anti-quark? Well they will continue to pull on each other and even interact with their force mediating gluons and thus may have much more complex post annihilation behavior leading to things like jets of particles.

Now if you have gotten this far, you deserve a nice lemon-drop treat for persevering. Consider the gravitational case. What happens when a matter and anti-matter blackhole collide? For exmaple two planck-mass sized mini-black holes - An interesting brain teaser that I will not (yet) spoil with an answer. enjoy.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.