From: g_d_pusch_remove_underscores@xnet.com
Copy: root@mauve.demon.co.uk
Ian Stirling writes:
> Carey Sublette wrote:
>>
>> "Mike Miller" wrote in message
>> news:5dcb47db.0312300621.14d9d8de@posting.google.com...
>>> Jonathan Griffitts wrote in message
>> news:...
>>>
>>>> I recall reading that a matter/antimatter annihilation reaction would
>>>> NOT be a catastrophic explosion, because the reaction cross section is
>
>>> However, when larger quantities of anti-matter are released in a
>>> matter-rich environment (where the mass of matter is>> than the mass
>>> of anti-matter), the first matter/anti-matter reactions are going to
>>> heat up the anti-matter and "vigorously" disperse the remaining
>>> antimatter...
>>>
>>> Right into further (high velocity) contact with matter.
>
>> Nuclear weapons release most of their energy in a time scale of
>> under 100 nanoseconds, so this will be slower. But the outside world
>> does not interact with a nuclear explosion in this timescale, instead
>> the outside world doesn't "see" much of the energy until hundreds of
>> microseconds have passed, so based on the energy release rate their
>> would be little appaarent difference.
>
> Most of the energy isn't seen till tens or even hundreds of milliseconds
> due to the highly compressed air in front of the fireball obscuring it.
>
> I can't find (despite some half an hour websearch) how far 10.6GeV gammas
> will go in the atmosphere.
Where the heck are you getting "10.6 GeV gammas" from ?! The majority of
the annihilation energy comes out as _pions_ not gammas, with an average
multiplicity of about 3.5 (largest channel 3 pions, second largest 5 pions).
Annihilation energy is about 2 GeV per pair of nucleons, so the mean pion
energy will be around 570 MeV. About a third of those pions will be pi-zeros,
which _VERY_ promptly decay to two gammas most of the time, with mean
energies of around 290 MeV. The charged pions weak-decay to muons after
a mean time of ~70 nanoseconds, and the muons weak-decay to electrons and
positrons after a mean time of ~6.2 microseconds. The positrons annihilate
with electrons to produce two 0.511 MeV gammas most of the time. I see no
"10.6 GeV" _anythings_ here.
> I suspect that the initial fireball will be lots bigger.
Pion mean range is about 21 meters in vacuum; in air it's more on the
order of a few meters, since pions interact with nuclei with strong-
interaction cross-sections. Muon mean range is about 1.8 km in vacuum;
in most forms of matter (including air) muon range is around 1 gm/cm^2
at energies above 100 MeV, so dividing by the sea-level density of air,
the muon mean range is about 8.5 meters.
As for the gammas, the pi-zero decay gammas have a mean range on the order
of 20 gm/cm^2, while the positron annihilation gammas have a mean range on
the order of 40 gm/cm^2; at sea-level density, that works out to about
150 meters and 300 meters, respectively.
Conclusion: About 2/3 of the energy will be deposited as energetic charged
particles (pions and muons) in a volume less than ten meters in radius
over a period of about 10 microseconds. Since the energy is still
deposited in a volume much smaller than the fireball will expand to,
the fireball dynamics will be essentially indistinguishable from that
of a nuclear explosion having a comparable yield. However, since the energy
is deposited in a much larger volume over a much longer timescale than a
nuclear explosion, the shock wave will be much "softer" than for a nuclear
weapon of comparable yield. The ~1/3 of the yield that comes out as 290 MeV
gammas have much longer ranges than the initial fireball size, and will
pre-heat the air the shock-wave is propagating into weaking it still further.
Bottom line: Most of the yield will be released as heat and gamma radiation,
not as blast effect. An "antimatter bomb" will have therefore have an effect
somewhat similar to an "enhanced radiation weapon" (AKA "neutron bomb")
unless surrounded by a heavy, high-Z tamper so that most of the energy
will be deposited close to the bomb, instead of throughout a large volume
of atmosphere.
> If it's large enough, you may not get the normal "double flash" that
> normal bombs exhibit.
Possibly correct.
> I suspect EMP will be much, much worse.
Almost certainly incorrect, since the gamma-ray energy will be released
over a very long timescale compared to the cyclotron frequency of electrons
in the Earth's magnetic field. Since the gammas are not released as a
sharp pulse, they will not be able to excite coherent electron cyclotron
oscillations in the ionosphere; therefore, EMP effects will in fact be
quite weak compared to a nuclear weapon of comparable yield.
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
|
