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There has been a long discussion between me and Anna V on if the products of the annihilation will really cause a fireball to form and we haven't settled it yet.

Our point here is that gamma rays from the annihilation are too energetic to be absorbed by the atmosphere in a short time or a little space to actually create a fireball. Also the pions will pass without much probability of hitting a target nucleus and dissipating energy in the air so it won't probably contribute to the explosion.

In other words, if no fireball forms, then antimatter annihilation would not look like or have similar effects as nuclear detonations like fireball, blast wave and mushroom cloud but would rather be considered similar to something like a neutron bomb.

One last thing, this question is not duplicate, I have read the other questions and answers and none of them gave me a useful answer.

So what do you think ? is antimatter annihilation so different from nuclear detonations in the effects part ?

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A couple things: if your goal is really to collect opinions, you shouldn't be asking here. This is a site for questions that can be directly answered. So it would be much better if you could rephase the question so that it just asks what you want to know, don't ask for opinions. Also, to show that this is not a duplicate, link to the other questions you've read and say why they don't cover what you want to know. –  David Z Aug 12 '13 at 22:50
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up vote 9 down vote accepted

Although none of these questions is an exact duplicate, there is a lot of overlap, and I hope we can avoid stringing this kind of stuff out indefinitely. The good news is that you're apparently being very cautious about the safety hazards of your planned matter-antimatter spaceship -- hazards that science fiction authors typically blithely ignore. Please let me know the date and location of your launch, so I can make sure to be on another continent.

Amsler 1997 has a discussion of the decay products.

A lot of energy comes out as neutrinos, which escape without depositing any energy.

Some energy comes out in high-energy gammas such as $\gtrsim 67$ MeV gammas from decay of π0's. The ones that are emitted upwards probably penetrate most or all of the atmosphere, but the ones that go down probably create showers that only penetrate for ~1 meter into the ground. If the explosion is on or near the ground, then all of this energy is released within a very small volume. Definite fireball.

Some of the other products are charged particles. These include muons and charged pions. (The muons' half-life of 2 us, which gets increased by relativistic time dilation, implies that you won't see their decay products nearby. The neutral pion has a short half-life. The charged pions may or may not decay nearby -- one would need to estimate their energies and the time dilations involved.) All of these charged particles interact electromagnetically, and they will certainly dissipate energy in the surrounding matter at some rate $dE/dx$. They would be considered minimum ionizing, although at very high energies you do start getting bremsstrahlung, so dE/dx rises logarithmically with energy.

You will get 511 keV gammas from e+e- annihilation, and these will definitely stop fairly close to the explosion. These account for at least $\gtrsim 2\times 10^{-4}$ of the total mass-energy, so even for an explosion far above the ground (e.g., from lobbing a hunk of antimatter at our planet from outer space), that's a very solid lower limit on how much energy will go into a fireball.

So in summary, a great deal of the energy is either lost or dissipated far away, but there are quite a few mechanisms for releasing a pretty big percentage of the energy into heating of nearby matter. Yes, there will be an impressive fireball if the amount of antimatter is macroscopic. If you're really deeply interested in these issues, you should probably download and learn to use a sophisticated Monte Carlo package such as GEANT and start running actual simulations.

Amsler, http://arxiv.org/abs/hepex/9708025

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@AbanobEbrahim: ? - I would summarize the conclusion as the opposite of that -- an antimatter weapon would be extremely destructive. –  Ben Crowell Aug 12 '13 at 23:59
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Coming from a high energy background I would not characterize 67 MeV gammas as "extremely penetrating". They do go through a fair mass density, but the question to ask is "to what temperature do they heat the surrounding gas. One would expect a larger and cooler fireball than from a fission explosion of comparable energy, but the basic situation is still one of heating a lot of gas to very, very hot. –  dmckee Aug 13 '13 at 0:49
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@dmckee: I'd be interested in seeing a more quantitative analysis of what a 67 MeV gamma does in rock or whatever. It's outside my direct experience. I wonder if it matters whether the rock has already been turned into a plasma before the 67 MeV gamma gets there. I suppose it mainly interacts through pair production, which is basically a gamma-on-nuclei process? –  Ben Crowell Aug 13 '13 at 2:49
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@Ben The first place I look is the chapter of the Particle Data Book on the Passage of Radiation Through Matter. At that energy you can expect it to shower and the shower to extend through up to a hundred g/cm^2 of material which is a long way in air but not so long in rock. –  dmckee Aug 13 '13 at 2:54
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@BenCrowell,@dmckee,@anna, would a sphere or a case of Lead for example surrounding the annihilation process do any good at absorbing more energy ? I mean, if the annihilation near the ground will create "definite fireball", then some casing of a material good at absorbing gamma rays would be better, wouldn't it ? –  Abanob Ebrahim Aug 13 '13 at 16:37
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