Antiproton reactions with water

[pxe2]

This started out with interest on antiproton radiotherapy but as it turns out there's little to no publication on the reaction equations that's happening. So this is just an attempt to have some equations at hand.

The reactions are for antiproton with water. For simplicity everything is considered at rest.
Water is just ordinary H2O but since it randomly dissociate into hydronium (H₃O⁺)and hydroxide (OH⁻) ions those will be included for completeness.
The basic reactions are antiproton with antiproton and neutrons. Everything else is just a longer chain of the same with varying larger molecules.
Feel free to correct, amend, and expand etc.

(1) p + p⁻ → 3 π± + 2 π⁰ → 3 π± + 4 γ → 3 μ± + 4 γ → 3 e± + 4 γ releases about 3.76 GeV
(2) p⁻ + n → 2.5 π⁻ + 1.5 π⁺ + 2 π⁰ → 4 e± + 4 γ releases about the same energy
-----------------
(3a) p⁻ + H₃O⁺ → (1) + H₂O
(3b) → (1) + H₃? + ¹⁵N
(3c) → (2) + H₃? + ¹⁵O → (2) + H₃? + ¹⁵N + e⁺
(4a) p⁻ + H₂O → (1) + OH⁻
(4b) → (1) + H₂ + ¹⁵N
(4c) → (2) + H₂ + ¹⁵O → (2) + H₂ + ¹⁵N + e⁺
-----------------
(5a) p⁻ + OH⁻ → (1) + O
(5b) → (1) + H⁺ + ¹⁵N
(5c) → (2) + H⁺ + ¹⁵O → (2) + H⁺ + ¹⁵N + e⁺
(6a) p⁻ + O → (1) + ¹⁵N
(6b) → (2) + ¹⁵O → (2) + ¹⁵N + e⁺
-----------------
(7) p⁻ + H₂ → (1) + H⁺

The fragments could react with water and form more reaction chains so there's plenty left to cover.
It would be nice to include all the energies and kinetic energies involved but without experimental data that's not possible. There's virtually nothing available to lookup.

 

[Vanadium 50]

If you annihilate an antiproton on a nucleus, the nucleus almost never remains intact.

 

[mfb]

If the antiproton is slow then the total energy released will just be twice the mass of the proton, 2*0.94 GeV = 1.88 GeV, with a smaller correction from the binding energy. The same applies to an annihilation with a neutron.

As mentioned the nucleus is unlikely to stay in one piece, and the chemical bonds are even less likely to survive. You have three completely different energy scales.

 

[Astronuc]

π⁰

One needs to consider the branching ratios (in annihilation and decay) and disposition of the neutral pions and their decay products, which are highly energetic gamma rays (~67.5 MeV). Those will travel a long distance, and may cause photoneutron, pair production, or photodisintegration reactions in light elements, as well as Compton scattering as the photons scatter toward lower energies.

Also, consider what it takes to produce antiprotons in the first place. It is more practical, and a lot less energy intensive, to use protons in the first place, or neutron, as in boron-neutron capture therapy.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5296588/

 

[fresh_42]

This bugs me since I saw the thread title.

If the injected anti-proton is slow, doesn't it simply "swim" on top of the water due to electromagnetic repelling? Or does the asymmetry of the water molecule force annihilation anyway?

 

[malawi_glenn]

This bugs me since I saw the thread title.

If the injected anti-proton is slow, doesn't it simply "swim" on top of the water due to electromagnetic repelling? Or does the asymmetry of the water molecule force annihilation anyway?

Is there vacuum "above" the water?

 

[fresh_42]

Is there vacuum "above" the water?

How about a closed tank with 100% water in it? In the case of a vacuum, we probably run into thermodynamic problems.

 

[Vanadium 50]

doesn't it simply "swim" on top of the water due to electromagnetic repelling?

From what? Water is a polar molecule. so there is an attractive side and a repulsive side.

 

[fresh_42]

From what? Water is a polar molecule. so there is an attractive side and a repulsive side.

That was my question. Does the asymmetry cause annihilation?

 

[pxe2]

If the antiproton is slow then the total energy released will just be twice the mass of the proton, 2*0.94 GeV = 1.88 GeV, with a smaller correction from the binding energy. The same applies to an annihilation with a neutron.

As mentioned the nucleus is unlikely to stay in one piece, and the chemical bonds are even less likely to survive. You have three completely different energy scales.

Right. I was doubling the energy and for some reason did it twice.

If you annihilate an antiproton on a nucleus, the nucleus almost never remains intact.

Alright.

One needs to consider the branching ratios (in annihilation and decay) and disposition of the neutral pions and their decay products, which are highly energetic gamma rays (~67.5 MeV). Those will travel a long distance, and may cause photoneutron, pair production, or photodisintegration reactions in light elements, as well as Compton scattering as the photons scatter toward lower energies.

Also, consider what it takes to produce antiprotons in the first place. It is more practical, and a lot less energy intensive, to use protons in the first place, or neutron, as in boron-neutron capture therapy.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5296588/

Yes, even though they last between 22-70ns (up to 250MeV) that's enough to reach as far as 21m.
Anyhow, about 30% of the reaction energy is deposite close by. And that's supposed to be twice as much as with protons, hence, more effective.

 

[pxe2]

This bugs me since I saw the thread title.

If the injected anti-proton is slow, doesn't it simply "swim" on top of the water due to electromagnetic repelling? Or does the asymmetry of the water molecule force annihilation anyway?

It's in an enclosed tank but other than a simple sketch without anything I couldn't find any detailed design description.
see DOI:10.3389/fphy.2013.00037

 

[Vanadium 50]

That was my question. Does the asymmetry cause annihilation?

What asymmetry? A proton sees the same effect, although of course opposite "poles".

Maybe the way to think about it is like this: atomic physics occurs at the eV scale. Above this scale, the particle doesn't see atoms - just their constituents. If I get the antiproton moving slowly enough (and this is far from trivial) that a few eV makes a difference, its wavelength is so large that it sees the nucleus and can annihilate with it.

 

[Astronuc]

(1) p + p⁻ → 3 π± + 2 π⁰

Charge conservation? Shouldn't this be (assuming the reactants are relatively low energy) should produce 2 π± , or rather π⁺ + π^- + 2 π⁰?
One may wish to look into the Crystal Barrel experiment at CERN's LEAR.

The Crystal Barrel Experiment ran at the Low Energy Antiproton Ring, (LEAR), from 1989 through the end of 1996 and studied antiproton proton and antiproton deuterium annihilations both at rest and in flight.

http://www-meg.phys.cmu.edu/cb/

Proton-Antiproton Annihilation and Meson Spectroscopy with the Crystal Barrel
https://arxiv.org/pdf/hep-ex/9708025.pdf

https://home.cern/news/news/experiments/crystal-clear-30-years

Anyhow, about 30% of the reaction energy is deposite close by. And that's supposed to be twice as much as with protons, hence, more effective.

Where or how did one obtain this information?

 

[pxe2]

What asymmetry? A proton sees the same effect, although of course opposite "poles".

Maybe the way to think about it is like this: atomic physics occurs at the eV scale. Above this scale, the particle doesn't see atoms - just their constituents. If I get the antiproton moving slowly enough (and this is far from trivial) that a few eV makes a difference, its wavelength is so large that it sees the nucleus and can annihilate with it.

Likewise they prefer higher Z numbers whenever available.

Charge conservation? Shouldn't this be (assuming the reactants are relatively low energy) should produce 2 π± , or rather π⁺ + π^- + 2 π⁰?
One may wish to look into the Crystal Barrel experiment at CERN's LEAR.

http://www-meg.phys.cmu.edu/cb/

Proton-Antiproton Annihilation and Meson Spectroscopy with the Crystal Barrel
https://arxiv.org/pdf/hep-ex/9708025.pdf

https://home.cern/news/news/experiments/crystal-clear-30-yearsWhere or how did one obtain this information?

It's actually
(1) p + p⁻ → 1.5 π⁻ + 1.5 π⁺ + 2 π⁰
and the main difference to p⁻ + n reaction.

Thanks, I will look them up.

That's my summary of these papers:
DOI:10.3389/fphy.2013.00037
https://iopscience.iop.org/article/10.1088/0031-9155/53/3/017

 

[Vanadium 50]

Likewise they prefer higher Z numbers whenever available.

Who prefers what? How do you know this? (Which is really asking 'what are you saying?)

 

[pxe2]

Who prefers what? How do you know this? (Which is really asking 'what are you saying?)

The antiprotons would react more with a higher-Z material. In this case it would be the rare H₃O⁺ so it doesn't real matter here..
Based on:
Ponomarev L I 1973 "Molecular structure effects on atomic and nuclear capture of mesons", Annu. Rev. Nucl.
Sci. 395–430 or Annu. Rev. Nucl. Sci. 1973.23:395-430
It's a paper re-examing the violation of the Z-law. An investigation on meson capture by the nucleus.
Also referenced above.

 

[Vanadium 50]

You probably didn't expect people to look up any of those references, as Ponomarev says exactly the opposite of what you claim.

Antiprotons are mentioned exactly once, on page 423, There it says Equation 24 has a of about 11. Equation 24 says the probability W is

$$W = \frac{an}{Z^2(n+mZ)} $$

As Z goes up W goes down.

 

[pxe2]

You probably didn't expect people to lookj up any of those references, as Ponomar4ev says exactly the opposite of what you claim.

Antiprotons are mentioned exactly once, on page 423, There it says Equation 24 has a of about 11. Equation 24 says the probability W is

$$W = \frac{an}{Z^2(n+mZ)}$$

As Z goes up W goes down.

Yes, but on the following page 424

Thus, if the direct nuclear
capture of π⁻ mesons by protons occurs with a noticeable but still smaller
probability than the cascade transition in the Zπ atom, then in going from π⁻
and antiprotons the probability W₂ and, consequently, the a coefficients in
formula 24 must increase.

As the mass m is extremely small and n also small while a is getting larger, the result is frankly reversed.
And to quote the other paper:

When antiprotons enter a chemical
compound consisting of several materials, the majority of the annihilations will take place
on high-Z materials (Ponomarev 1973).
For example, in the case of polystyrene only 1% of
the antiprotons will annihilate on hydrogen with the remaining 99% annihilating on carbon.