Does a particle and its anti-particle always annihilate?

[Frank Castle]

When a particle and its corresponding anti-particle interact do they always annihilate or are there other possible interactions that can occur, such as them scattering off of one another?

If the former is true, why do they always annihilate? If the latter is true, is it the case that the most probable interaction is that they annihilate (hence why in pop-sci it is always said that matter and anti-matter annihilate if they come into contact)?

 

[Arman777]

When a particle and its corresponding anti-particle interact do they always annihilate or are there other possible interactions that can occur, such as them scattering off of one another?

If the former is true, why do they always annihilate? If the latter is true, is it the case that the most probable interaction is that they annihilate (hence why in pop-sci it is always said that matter and anti-matter annihilate if they come into contact)?

Well I think yes they should annihilate.
When they interect, energy,charge and angular momentum should conserved.Thats why I think the only possible solution is annihilation.

 

[Frank Castle]

When they interect, energy,charge and angular momentum should conserved.Thats why I think the only possible solution is annihilation.

These quantities can also be conserved if they don't annihilate though, for example, if they simply scatter off one another.

 

[mfb]

Scattering is possible, and it is the most likely outcome in most cases. Various other reactions are also possible if there is enough energy (e. g. proton plus antiproton -> hyperon plus kaon plus antiproton - changing the proton but not the antiproton).

If you bring macroscopic amounts of antimatter into contact with macroscopic amounts of matter, the scattering is irrelevant - the particles will annihilate and make a big explosion. If you shoot antimatter particles into matter, scattering is important to follow how the particle loses energy - but in the end the antimatter will annihilate and energy is released.

 

[Frank Castle]

If you bring macroscopic amounts of antimatter into contact with macroscopic amounts of matter, the scattering is irrelevant - the particles will annihilate and make a big explosion. If you shoot antimatter particles into matter, scattering is important to follow how the particle loses energy - but in the end the antimatter will annihilate and energy is released.

Why is this the case? Is it because on a macroscopic time-scale the probability that matter and antimatter (that initially scattered) to annihilate is large enough that it dominates over any probability for scattering?

 

[mfb]

Scattering process do happen, but you don't have to know about them to understand the outcome. It doesn't matter if the particles scatter x times before they annihilate.

 

[Frank Castle]

Scattering process do happen, but you don't have to know about them to understand the outcome. It doesn't matter if the particles scatter x times before they annihilate.

But why is it that macroscopically the matter and antimatter will completely annihilate? Couldn't there be small amounts of each left over that have scattered and then propagated away freely? Or is it that when there are macroscopic amounts of matter and antimatter in contact, particle number is so dense that it is impossible for any of the matter and antimatter not to be involved in a collision in which they annihilate?!

 

[mfb]

On Earth, matter is everywhere. Every antimatter particle you have on Earth will hit a matter particle quickly unless you keep the antimatter levitated in an extremely good vacuum.

 

[Frank Castle]

On Earth, matter is everywhere. Every antimatter particle you have on Earth will hit a matter particle quickly unless you keep the antimatter levitated in an extremely good vacuum.

So is my point in the previous post correct then? Even if there were only one antimatter particle, as there is so much matter (on Earth), although the antimatter particle my initially scatter with a few matter particles (of the same species) it will very quickly be involved in an interaction in which it annihilates with the corresponding matter particle.

 

[mfb]

Sure.

 

[Frank Castle]

Sure.

Ok cool, thanks for your help.

I'm assuming this is why it is so difficult to even maintain a single antimatter particle in a laboratory on Earth, since although it may initially scatter it will very quickly be involved in an interaction in which it is annihilated.

 

[Orodruin]

Note that, as mfb said, the typical antimatter particle will scatter a lot before annihilating. In fact, a positron will typically lose all its kinetic energy before annihilating. This is why you get two 511 keV photons from the annihilation. If the positron did not stop first, the photons would have a higher total energy (and generally the photon energies would be different).

 

[Frank Castle]

Note that, as mfb said, the typical antimatter particle will scatter a lot before annihilating. In fact, a positron will typically lose all its kinetic energy before annihilating. This is why you get two 511 keV photons from the annihilation. If the positron did not stop first, the photons would have a higher total energy (and generally the photon energies would be different).

So is the point that a typical antimatter particle will scatter, but if there is a macroscopic amount of matter present in the region that it is propagating then it will very quickly annihilate?

 

[Orodruin]

So is the point that a typical antimatter particle will scatter, but if there is a macroscopic amount of matter present in the region that it is propagating then it will very quickly annihilate?

No, it will eventually annihilate after scattering multiple times and losing most of its kinetic energy. On a human time scale, this eventually is a rather short time but the scatter time scale is even shorter.

 

[Frank Castle]

No, it will eventually annihilate after scattering multiple times and losing most of its kinetic energy. On a human time scale, this eventually is a rather short time but the scatter time scale is even shorter.

Ah ok. So when it is said that matter and antimatter annihilate one another this happens over some finite time scale with many scattering events before complete annihilation, although (as you said) this will be a relatively short amount of time on human timescales.

 

[mfb]

Right.
"Short amount of time" is here something shorter than a nanosecond.

 

[snorkack]

What can a neutrino and an antineutrino annihilate to, even if they meet each other?

 

[Frank Castle]

Short amount of time" is here something shorter than a nanosecond.

Is something that one can compute in principle?

 

[nikkkom]

When a particle and its corresponding anti-particle interact do they always annihilate

"Annihilate" is actually not a precise scientific term. They simply can react, and for heavy particles there are many possible end states, not at all limited to gammas. There could be pions, kaons, muons, electrons and positrons, etc.
Although in practice, daughter particles would themselves quickly decay or react with more matter, and almost all energy _eventually_ is converted to light and heat.

 

[mfb]

What can a neutrino and an antineutrino annihilate to, even if they meet each other?

Two other neutrinos (similar to scattering then)
Two photons, although that is an extremely unlikely process.
Other particle pairs, if their energy is sufficient
Single Z or Higgs, if their energy is sufficient

Is something that one can compute in principle?

Sure, it is routinely done in simulating particle physics experiments.

 

[Frank Castle]

Sure, it is routinely done in simulating particle physics experiments.

Is it simply the inverse of the decay rate for the process $e^{+}e^{-}\rightarrow\gamma\gamma$?

 

[mfb]

The scattering processes? No.

 

[Frank Castle]

The scattering processes? No.

No, I was meaning calculating the lifetime of an electron positron pair before annihilating.

 

[mfb]

If positronium forms, that is the inverse of the decay rate $e^{+}e^{-}\rightarrow\gamma\gamma$, sure. Otherwise we have to consider all possible processes with electrons bound to atoms.