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Energy of antimatter explosion

  1. Oct 3, 2012 #1
    Maximum is obviously 42 MT/kg.

    But how much of the energy is lost as neutrinos from pion and muon decay?

    Typical nucleon annihilation produces an average of 4,5 pions. Since the energy is 1880 MeV, each pion gets an average of 420 MeV, of which 140 MeV is rest mass. 1/3 of these are neutral pions which decay in a short distance to photons, which are captured by sea level air.

    How transparent are sea level air and common condensed substances to annihilation spectrum charged pions? Which proportion of these pions are captured by strong interaction with nuclei (resulting in fast nucleons, captured by air), and which proportion lives to decay to muons (where one half of the energy, and then 2/3 of the rest, is lost as neutrinos)?
     
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  3. Oct 3, 2012 #2

    mfb

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    Hmm, let's see:

    The dominant decay of neutral pions is to 2 photons within <1µm, with an average energy of 210 MeV in the frame of the bomb. As 420 MeV > 140 MeV, most annihilations will produce two high-energetic photons. According to PDG, mean free path length for a high-energetic photons should be something like ~38g/cm^2 or ~300m of air at sea level (~38cm for water, ~2cm for steel or similar bomb material), and the dominant process for initial photons is pair production. Therefore, we get electromagnetic showers within 1-2km (or less, if we have some bomb material in the path). That is probably small compared to the area of destruction - a smaller bomb can be built with cheaper methods ;). Most of the energy heats the atmosphere or is converted into Cherenkov photons.

    Charged pions have cτ=8m, with a boost of gamma~3 this corresponds to a typical decay length of ~25m.
    PDG (same as above) gives ~100m interaction length - most pions will decay before they hit anything, unless they are surrounded by metal or other bomb material (iron: 20cm).

    The decay to muons should keep most of their energy (~105/140=0,75 as rough estimate). Those muons have a typical decay length of ~3cτ=2km, and release electrons which can produce small showers as well (with ~1/4 of the original energy).
     
  4. Oct 4, 2012 #3
    And a bigger bomb can be built with simpler methods. There is no theoretical upper bound on fusion yield.
    Or non-bomb material - ground, walls etc.
    Which means that at a minimum (all pions escape and decay) 50 % of the energy goes to explosion (1/3 from neutral pions, and 1/4 of the remaining 2/3 from the decay electrons of charged pions).

    But... If an antinucleon annihilates with, say, a N-14 nucleus in air. Does this result in 4,5 pions escaping with all 1880 MeV of annihilation yield, and the remaining C-13 or N-13 nucleus receiving no energy? Or will the rest of the nucleus absorb any of the pions produced?
     
  5. Oct 4, 2012 #4

    mfb

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    Interesting question. As the annihilation process happens at O(GeV) and binding energies are some MeV/nucleon, I think the annihilation itself would look very similar - but the pions can react with the remaining nucleus afterwards. I would expect that this has been tested, and there is a publication somewhere.
     
  6. Oct 4, 2012 #5
    Do you mean O or 0?

    If a nucleus reacts with a pion (average energy 420 MeV, including its rest mass released on reaction) or even with a photon from neutral pion decay (average energy 210 MeV), this exceeds the total binding energy of nitrogen 14 nucleus (about 105 MeV). Would the photon scatter the nitrogen nucleus into 14 free protons and neutrons of 7,5 MeV kinetic energy each, or simply expel one proton with about 200 MeV kinetic energy and leave an intact carbon 13 nucleus?

    In case of fission explosion, less than 10 % of the energy goes to penetrating radiation - prompt hard gammas and fast neutrons. Over 90 % of the energy stays with fission fragments, whose range in air is mere 2...3 cm, and which are stopped by a few μm of metal, or by epidermis of skin.

    Whereas the range of 200 MeV protons in air has been quoted as 250 m.

    Does it mean that the energy of annihilation explosion can be spread out over a large volume of air, not concentrated near the original location of antimatter?

    If a large antinucleus annihilates with a single nucleon, can antinucleons be accelerated to high speeds?
     
  7. Oct 4, 2012 #6

    mfb

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    O, as a sloppy way to express "order of magnitude".

    Antiproton annihilation at rest in nitrogen and deuterium gas
    "The pion multiplicity, at least for the lighter nuclei, decreases only by 5 to 10%".
    Lambdas can be produced (rare), and protons can be ejected (0.70±0.03 per annihilation). Neutrons are hard to measure. The pion spectrum (figure 5) is similar to the p-antip-annihilation, with an average of 2.9±0.1 charged pions per annihilation (proton-antiproton annihilation: 3.2±0.1).

    Looks like that.

    Where do you get large antinuclei?
    That is symmetric to the antiproton / large nucleus situation, but I do not see numbers for the kinetic energy of the nucleus in the paper.
     
  8. Oct 4, 2012 #7
    Where do you get large amounts of antimatter?

    Most efficient fusion bomb is W41. 25 MT yield. Weight 4850 kg.

    The yield would represent complete fusion of about 400 kg lithium 6 deuteride.

    And could be replaced with annihilation of 600...1200 g antimatter, depending on how much pions are captured.

    For example, in Angels and Demons, 0,25 g antimatter was produced - and somehow stored up without exploding before triggered. Meaning 5 to 10 kT yield, depending on how much pions were captured by ground and surrounding walls.

    How is antimatter stored without exploding before getting triggered? And which antimatter substances can be so stored?

    Anyway, regarding small antimatter bombs - they do not seem to have an apparent lower yield limit.

    Fusion bombs have lower yield limit of 250 T. The necessary yield of fission primary. A fizzle yield below that will generate tritium via Li-6(n,α)t, but not heat the tritium enough to fuse.

    Fission bombs have no lower yield limit. But what they do have is lower size limit.

    Smallest fission arm is W53. Weight 23 kg. Yields at tests were 18 and 22 T.

    Obviously, even less powerful and smaller arms are possible - but not much smaller. The Demon Core was 6,2 kg - but that was the bare, subcritical metal core. It would be heavier counting the reflectors which did make it critical - tungsten carbide brick for Daghlian, beryllium hemispheres for Slotin.

    What was the Demon Core yield? Some μT or nT?

    It was the under 10 % of penetrating radiation that slew Daghlian, Slotin and over time others. Over 90% of fission fragment energy was absorbed by the core and only sufficed to warm it slightly - it could not melt the core nor expand it much.

    A very small yield, converted into moderately penetrating radiation (which is not stopped by a few metres of air or mm of skin, but which does deposit large fraction of its energy into a body) can be highly deadly.

    How small can an antimatter bomb be made?

    That depends on the unknown mechanism for storing antimatter....
     
  9. Oct 5, 2012 #8

    mfb

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    Produce many antiprotons and combine them with positrons. Not cheap, but at least possible.
    If you manage to freeze those antihydrogen atoms: Maybe with diamagnetic levitation.
    Another option might be a small net charge of the system, and something similar to a Penning trap.
    Hmm well, that is true, and they might be quite compact, too.
     
  10. Oct 10, 2012 #9
    So does anyone know the Demon Core yield?
     
  11. Oct 10, 2012 #10

    Drakkith

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  12. Oct 11, 2012 #11
  13. Oct 11, 2012 #12

    Drakkith

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    The yield? The core did not explode, so there was no yield. Are you wanting to know the amount of radiation emitted?
     
  14. Oct 12, 2012 #13
    Yes. The amount of fission that took place on the two occasions.

    The energy released was obviously small enough to cause only slight warming of Demon Core, and not enough to cause melting or explosive disassembly. But since yield is defined as energy released from nuclear reaction, it can obviously be measured, whether it is in μT, nT or other such units.
     
  15. Oct 16, 2012 #14

    QuantumPion

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    According to this report (pdf page 88-89), the yield of the 1946 accident was ~1016 fissions or ~300 kJ.
     
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