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I stumbled upon this wikipedia article on antimatter weaponry.

Being greatly appalled by the sad fact that large sums of money are being wasted on this, I could not stop myself from thinking for a moment about the physics behind it.

If you somehow put together a gram of antimatter and a gram of matter all at the same instant and at high density (so the "annihilation efficiency" would be 100%), would there actually be any explosion?

AFAIK, this would just produce large amounts of gamma photons, neutrino's etc., but there's be very little (anti)matter left to absorb the energy in the form of kinetic energy. In other words -- it would be a radiation bomb. There wouldn't even a flash of light to warn anyone.

Would this indeed be the case? Or am I overlooking something here?

Rody Oldenhuis
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    antimatter is the ultimate weapon, yes, sad kittens.. but don't forget that it is also the ultimate fuel: there will be no relativistic interstellar travel for us without it – user56771 Aug 28 '12 at 15:16
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    @user56771 There is no guarantee we can get interstellar travel using antimatter. Carrying a lot of antimatter fuel with you for years seems like a particularly chancy way to travel.

    But also, by the time we're ready to try interstellar travel we will have several hundred years more physics study under our belt. Our current understanding of physics will be to them like physics several hundred years ago is to us. Newton could speculate about what kind of cannon it would take to reach the moon....

     – J Thomas Jul 27 '22 at 14:33

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Have a look at these cross section plots of proton proton scattering and anti-proton proton, where the anti proton has an order of magnitude higher probability of interacting.

It is not true that most of the energy goes into radiation, it goes into creating particles, with an average multiplicity for annihilation at rest of about five charged particles. The interaction is strong and gluon mediated, the photons produced come from pi0 decays, direct photons are a higher order effect.

Thus one gets an "explosion" as this image shows, once one obtains large numbers of such annihilations in a small space

enter image description here

The charged pions will eventually end up as electrons and muons carrying a lot of the kinetic energy of the reaction and a destructive power. The piOs, about 1/3 of the charged number, will give two photons which on average will have enough energy to be destructive when hitting nuclei.

Glorfindel
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anna v
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  • Would that make any practical difference? Either way (pions or gamma) it is all going to be converted into heat within a few cm and in any nuclear weapon it's the heating and subsequent expansion of the air that does the damage. – Martin Beckett Aug 28 '12 at 13:29
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    @MartinBeckett Well, neutrinos are weakly interacting, charged pions, because they are relatively long lived will manage to hit nucleons and by the strong interaction multiply the effective dispersion of the energy. Photons and electrons are interacting electromagnetically which is still orders weaker than the strong force. It is the pions that will concentrate the energy in an explosive format, i.e. a lot of energy in a small volume. – anna v Aug 28 '12 at 16:09
  • I think this is the best answer I'm going to get on this matter (pun intended); thank you anna. – Rody Oldenhuis Aug 28 '12 at 18:06
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Electron + positron = 2 * gamma going opposite directions and having nearly the same energy of electron + positron mass-energies. Indeed this is a polar-explosion(1 gamma to one side and the other to opposite side).(considering low energy state) http://en.wikipedia.org/wiki/Electron%E2%80%93positron_annihilation

huseyin tugrul buyukisik
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One Gram of matter and one gram of antimatter, equivalent mass conversion to energy is roughly 180 Terajoules. In a warhead, equivalent to 43 kilotons of TNT or twice yield of the Atomic bomb destroying Hiroshima in 1945.

LazyReader
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