Is it possible for larger antimatter atoms to decay to matter and visa versa?

Following on from previous questions:

If you have antimatter-matter interactions where there is a larger antimatter particle (say carbon or Silicon), is there any reason to believe that the antimatter particle could decay to matter particles during an interaction, and visa versa?

Thanks

• I (a non particle physicist) believe the answer is no. The interaction has to conserve all quantum numbers, which are opposite between particles and their antiparticles. Thus, for example, anti-Silicon has 14 anti-protons, thus an electric charge of $-14\,e$ and carbon has $+6\,e$. It's the nucleusses that interact in any collision, so the nucleusses emerging from the reaction would need to have a total charge of $-8\,e$: there would need to be eight anti-protons in the products .... – Selene Routley Oct 9 '14 at 0:18
• ...A particle physicist will need to finish this answer, because there is the possibility that the system's electrons / positrons could be involved in the reaction, thus openning the possibility for what you propose. It would, I think, have a fantastically small amplitude. – Selene Routley Oct 9 '14 at 0:19
• It sounds like it should be tested to prove either way :D – sidewaiise Oct 9 '14 at 0:25

I (a non particle physicist) believe the answer is no. The interaction has to conserve all quantum numbers, which are opposite between particles and their antiparticles. Thus, for example, anti-Silicon has 14 anti-protons, thus an electric charge of $-14\,e$ and carbon has $+6\,e$. It's the nucleusses that interact in any collision, so the nucleusses emerging from the reaction would need to have a total charge of $-8\,e$: there would need to be eight anti-protons in the products.

Now you might think that this doesn't prevent the reaction if somehow the system's electrons and positrons are involved in the reaction to balance electrical charge. However, as I said, there are many conserved quantum numbers, amongst which is the baryon quantum number. The baryon quantum number $\frac{1}{3}\left(n_q-n_{\bar{q}}\right)$, where $n_q$ is the number of quarks and $n_{\bar{q}}$ the number of antiquarks, must be $28-12=+16$ if we're talking about anti-$^{28}Si$ reacting with $^{12}C$, for example. So there needs to be an excess of 48 quarks, i.e. 16 nucleons in the products.

In answer to the further question:

I guess the question should have been... what is the mechanism for a particle becoming anti-matter or matter?

The "mechanism" is simply any valid nuclear reaction with all the relevant quantum numbers conserved: matter and antimatter annihilate, yielding photons, which in theory can produce any new antimatter and matter in equal amounts and such that the total rest mass is less than the total rest mass of the reactants (the balance made up of photons). However, momentum needs to be conserved, which means that the photons go flying off such that their nett momentum is nought. If there are two photons (the commonnest number), they propagate in opposite directions, if the reactants begin with roughly balanced momentums. This means their paths do not classically intersect after having left the mutually annihilating "stuff", so the amplitude for them to meet again and beget new matter/antimatter is fantastically small. It is not nought, though, particularly if the reactants' momentum is highly nonzero, and such a reaction producing muon/ antimuon pairs is important in the interaction of cosmic rays with the biosphere. Cosmic rays are mostly high energy protons, which come crashing into the atmosphere producing all manner of pairs in this way. They are most likely to produce small particles, like electrons and positrons. However, being light, these are sharply accelerated by further interactions with the atmosphere and thus quickly radiate excess kinetic energy through bremstrahlung and get swallowed up by atmospheric matter. But there is a small amplitude to produce muon/ antimuon pairs: muons are long lived (2.2us lifetime) and moreover are 207 times as heavy as the electron. Being heavier, they are not as sharply accelerated by collisions, and so many manage to reach the Earth's surface. If you build a cloud chamber, such as described below, these muons are what you will see.

• I guess the question should have been... what is the mechanism for a particle becoming anti-matter or matter? – sidewaiise Oct 9 '14 at 1:28
• @sidewaiise See my updates above, particularly about cosmic ray begotten muons. – Selene Routley Oct 9 '14 at 2:43
• Thanks for that, conservation of energy is a very reasonable explanation. In beta-decay there are anti-particles - this is an example of an anti-particle coming from a regular particle. What goes on in that process to allow for it? – sidewaiise Oct 9 '14 at 2:51
• @sidewaiise Yes, that's a good example: here the $W^\pm$ boson is emitted and decays to $e^\pm$ and either an electron neutrino or antineutrino. See the wiki page for W and Z Bosons. – Selene Routley Oct 9 '14 at 3:05
• I see - so an electron pair (+-) are created and have opposite velocities on creation/exit? So energy is conserved. I'm still hung up on the proton though - it's a much larger particle. Just reading up on current production methods for anti-hydrogen. – sidewaiise Oct 9 '14 at 3:13

In some normal matter there is the phenomenon of positron decay. That is, an unstable atom decays by the emission of an anti-electron. Presumably there is a mirror form of this where antimatter decays by electron emission.

• But neither case violates the conservation of baryon or lepton numbers which means the total amount of "matterness" is the same at the end of each process as at the beginning. – dmckee --- ex-moderator kitten Oct 9 '14 at 12:24