4 ms·

Wouldn't colliding with the walls of the chamber be extremely undesirable? That seems like it would create stress/damage in the chamber, and given how sensitive

by jxf 13y ago

Wouldn't colliding with the walls of the chamber be extremely undesirable? That seems like it would create stress/damage in the chamber, and given how sensitive and expensive ALPHA is I'd imagine that they don't want to do that, right?

InclinedPlane 13y ago
It's not a problem. It's important to keep an understanding of the scales involved here. A gram of Hydrogen contains 6e26 atoms. So a nanogram of material contains on the order of 10^17 atoms. That's ten million trillion atoms. A "beam of antimatter" which literally annihilates the walls of the vacuum chamber it is contained within may seem dramatic but in this case we are talking about mere dozens of anti-atoms. Against walls of solid metal the damage is so minute as to be undetectable with even the finest instruments. Even if the beam were far, far stronger it would still only cause erosion at the scale of picograms per trillion anti-atoms, which isn't much concern.
Sharlin 13y ago
We must remember we're talking about a few dozens of antiatoms here. The few gamma photons they turn into when annihilating are so utterly insignificant compared to natural background radiation it isn't even funny.
- InclinedPlane 13y ago
  Not to "actually" here, but actually... while positron/electron annihilation results in just photons most of the time the situation with anti-proton/proton annihilation is much more complex. Initially you end up with a bunch of high energy pi-mesons (you can think about this as the quarks and anti-quarks pairing off, which isn't really correct since typically more than just 3 pi-mesons result from an annihilation, the typical number is about twice that). These then decay into showers of other particles. Photons, certainly, but also leptons/anti-leptons (muons and electrons) and neutrinos. In terms of intensity you are absolutely correct though.
  - Sharlin 13y ago
    Thanks, yeah, for some reason I just considered the electron/positron case :)

by jxf 13y ago

InclinedPlane 13y ago

It's not a problem. It's important to keep an understanding of the scales involved here. A gram of Hydrogen contains 6e26 atoms. So a nanogram of material contains on the order of 10^17 atoms. That's ten million trillion atoms. A "beam of antimatter" which literally annihilates the walls of the vacuum chamber it is contained within may seem dramatic but in this case we are talking about mere dozens of anti-atoms. Against walls of solid metal the damage is so minute as to be undetectable with even the finest instruments. Even if the beam were far, far stronger it would still only cause erosion at the scale of picograms per trillion anti-atoms, which isn't much concern.

Sharlin 13y ago

We must remember we're talking about a few dozens of antiatoms here. The few gamma photons they turn into when annihilating are so utterly insignificant compared to natural background radiation it isn't even funny.

InclinedPlane 13y ago
Not to "actually" here, but actually... while positron/electron annihilation results in just photons most of the time the situation with anti-proton/proton annihilation is much more complex. Initially you end up with a bunch of high energy pi-mesons (you can think about this as the quarks and anti-quarks pairing off, which isn't really correct since typically more than just 3 pi-mesons result from an annihilation, the typical number is about twice that). These then decay into showers of other particles. Photons, certainly, but also leptons/anti-leptons (muons and electrons) and neutrinos. In terms of intensity you are absolutely correct though.
- Sharlin 13y ago
  Thanks, yeah, for some reason I just considered the electron/positron case :)