Anti-Matter

[Little Bang]

If we had a hydrogen atom and an anti-hydrogen atom located in close proximity would the repulsion between the electron and anti-proton, and the repulsion between the anti-electron and proton be enough to prevent annihilation?

[CraigD]

If we had a hydrogen atom and an anti-hydrogen atom located in close proximity would the repulsion between the electron and anti-proton, and the repulsion between the anti-electron and proton be enough to prevent annihilation?

No.

 

I think it’s OK to picture this semiclassically – that is, not worry about the probabilistic, wave-like nature of subatomic particles, but instead think of them as tiny classical balls with strong electrostatic charges.

 

When both the anti-proton + positron antihydrogen atom and the proton+electron hydrogen atom exist, they each have a 0 net charge, so are at neither attracted to repelled to one another.

 

Should they approach closely enough, the first particles to be very close to one another will be the electron and the positron.

 

The electron and positron have opposite charge, so will be attracted and annihilate, producing chargeless photons.

 

The remaining proton and antiproton will then have opposite charge, so will be attracted and annihilate.

 

This is very different than what happens when atoms of matter (or, in principle, atoms of antimatter) approach one another, and their electrons mutually repel. This effect is electron degeneracy pressure, and is what causes matter to behave in the usual way – solid objects not passing freely through one another, etc. – and also keeps matter from collapsing into very dense material due to gravitational attraction, except in very large bodies such as neutron stars.

 

Since changed particles have charge opposite their antiparticles, we could say their electron degeneracy pressure is negative, but rather than collapsing into very dense material, they annihilate into high-energy photons.

 

How such interactions would occur between large collections of matter and antimatter depends strongly on the assumption (soon to be or already experimentally tested, if the Athena collaborative is still on-track) that the mutual gravitational attraction of particles for antiparticles remains the same as particles for particles and antiparticles for antiparticles. Most physicists are nearly certain this is the case, but some of our members have put a lot of effort into imagining the large-scale physics of the universe if this is not the case.

[Little Bang]

So your saying the two atoms will be attracted to each other and annihilate?

[CraigD]

So your saying the two atoms will be attracted to each other and annihilate?

Yes – though a neutral atom (not an ion – having the same number of protons as electrons) and an anti-atom are attracted only by gravity, which is very, very weak (about [imath]10^{-38}[/imath] times as strong – see Fundamental interaction - Wikipedia, the free encyclopedia) compared to the electromagnetic attraction of ions, so for practical purposes, we could say they neither attract nor repel each other.

 

 

However, other than those that have been cooled to near absolute zero, atoms and move fast, and even in near vacuum, collide with one another frequently, so even with almost no net attractive force between them, antiatoms and atoms will collide and annihilate.

[Little Bang]

Craig, how far away from the proton is the orbital of an electron?

[CraigD]

Craig, how far away from the proton is the orbital of an electron?

Empirically, for hydrogen, about [imath]3.5 \times 10^{-11} \,\mbox{m}[/imath].

 

What “empirically” and various other terms to describe the radii of atoms means is complicated – see links from Atomic radii of the elements (data page) - Wikipedia, the free encyclopedia or other references for the details.

[Little Bang]

Thanks, I had already found it. I calculated the diameter of a wave in the shape of a torus using the wavelength related to the mass of an electron and found it to be ~1 X 10^-12 meters. Thought it interesting that it just fits inside your number.