Antimatter/nuclear megaton equivalency?
Cyde Weys
antimatter (measured in grams) and equivalence in megatons of TNT
(which seems to be what nuclear explosions are measured in). Also,
what would be the equivalence between grams of antimatter and the
casualty radius of the explosion? We have some data for such as
explosions as Hiroshima and Nagasaki, which were in the kiloton range,
but I don't know how to go from there to larger amounts and to grams of
antimatter versus nuclear weapon weight.
By the way, when figuring out antimatter explosive potential according
to E=mc^2, don't forget to multiply by two ... the full mass of the
antimatter is turning into energy as well as an equivalent mass of
normal matter. The normal matter is irrelevant when stating the
strength of the weapon though; it can either be part of the antimatter
containment system, the ground that the antimatter bomb happens to hit,
whatever. When specifying the strength of an antimatter weapon I think
you'd just measure it in something like "3 grams of antimatter"; the 3
grams of normal matter it is to react with is assumed, since everything
around us (and us) is made of normal matter.
rap...@netscape.net
Cyde Weys wrote:
TNT has an explosive yield of 4.184 MJ/kg
1g of matter has a yeild of 90,000,000MJ giving 21.5 kT of TNT (approx
a fission bomb)
1kg of matter/antimatter has a yield of 180,000,000,000MJ or 43MT of
TNT
My understanding is that the radius scales with the cube root of the
energy. 1kg of anti matter would therefore of around 10 times the
radius of Hiroshima or Nagasaki.
However, antimatter can be reduced below a fisson bombs critical mass.
1,000,000 mg explosions would do more damage than one 1kg explosion.
Kinda nuclear blanket bombing.
Wayne Throop
: I'm looking for a simple equation that relates explosion energy of
: antimatter (measured in grams) and equivalence in megatons of TNT
: (which seems to be what nuclear explosions are measured in).
Well, a gram of antimatter (or two grams of mass, as you point out)
is about 1.8e14 joules, and a ton of tnt is about 4.2e9 joules. Divide
joules per gram-o-antimatter by joules per tons-o-tnt to get tons-o-tnt
per gram-o-antimatter, and I get about 43 kilotons tnt per gram antimatter.
I suppose a number of kilotons per gram counts as an "equation"? I could
have dropped a decimal, but you can check my arithmetic.
The 1.8e14 you can get by E=mc^2, or (2/1000)*3e8^2 kilogram meter^2 / sec^2,
and the 4.2e9 you can get by using the unix "units" program, saying you
have a ton of tnt, and ask for how many joules that is. (Sadly, some
versions of the units program don't have "tnt" as a conversion factor
from mass to energy, in which case, you could have googled it.)
From there, you can google an expression for kill radius per kiloton tnt.
But roughly, it's two hiroshima's per gram, iirc.
Wayne Throop thr...@sheol.org http://sheol.org/throopw
Nyrath
> Well, a gram of antimatter (or two grams of mass, as you point out)
> is about 1.8e14 joules, and a ton of tnt is about 4.2e9 joules. Divide
> joules per gram-o-antimatter by joules per tons-o-tnt to get tons-o-tnt
> per gram-o-antimatter, and I get about 43 kilotons tnt per gram antimatter.
> I suppose a number of kilotons per gram counts as an "equation"? I could
> have dropped a decimal, but you can check my arithmetic.
Nah, all decimal present or accounted for!
I check you.
James Nicoll
cube of the yield but heat and light effects will go with the inverse
square. This means that for large explosions, you get incinerated
farther away than you have buildings toppled on top of you.
--
http://www.cic.gc.ca/english/immigrate/
http://www.livejournal.com/users/james_nicoll
Wayne Throop
: One minor correction: the blast effects scale with the inverse
: cube of the yield but heat and light effects will go with the inverse
: square. This means that for large explosions, you get incinerated
: farther away than you have buildings toppled on top of you.
Ah. Yes, and that makes a big difference if the application is
space warfare rather than, um, urban renewal.
Nyrath
in a matter-antimatter reaction will be available as
destructive energy.
http://www.stardestroyer.net/Empire/Tech/Torpedoes/Torpedo1.html
It is non-trivial to create a weapon that will ensure all the
matter will reaction with all the antimatter. This will
reduce the percentage of available energy.
The electrons will annihilate with the positrons to produce
100% deadly gamma rays, which is good news for the weapon
designers.
However, the protons will annihilate with the antiprotons
and produce only 30% gamma rays. The other 70% will
be in the form of charged pions.
I do not know what percent of the blast will be due
to electron-positron annihilation and what percent will
be due to proton-antiproton annihilation.
Charged pions will have a destructive effect, but will
quickly decay into harmless neutrinos. The link gives
a figure of several meters to several hundred meters,
I do not know how accurate that figure is.
Anyway once you factor in the inefficiency in mixing
the ingredients and the pion problem, the amount of
energy in the blast will be drastically less than
the theoretical maximum.
Mike Van Pelt
Nyrath <nyr...@projectrho.com.INVALID> wrote:
>However, it is important to note that not all the energy
>in a matter-antimatter reaction will be available as
>destructive energy.
>
>http://www.stardestroyer.net/Empire/Tech/Torpedoes/Torpedo1.html
>
>It is non-trivial to create a weapon that will ensure all the
>matter will reaction with all the antimatter. This will
>reduce the percentage of available energy.
>
>The electrons will annihilate with the positrons to produce
>100% deadly gamma rays, which is good news for the weapon
>designers.
And are more or less quickly absorbed by air, becoming heat.
>However, the protons will annihilate with the antiprotons
>and produce only 30% gamma rays. The other 70% will
>be in the form of charged pions.
Pions decay very quickly (microseconds?) to mesons, neutrinos,
and high-energy gamma rays. The mesons also decay
(milliseconds?) to electrons and positrons. The neutrinos
escape, of course, but not a lot of the energy goes into them.
Mostly, it's gammas on the order of 200 MeV. The gammas become
heat, ultimately, when they're absorbed by matter.
So, the energy release will be spread out a bit in time (the
decay time of the anihilation products) and spread out in space
(how far the products travel before decay/anihilation/absorption).
Wild guess, I think there's going to be a size limit below
which antimatter will produce not a bang, but a general
irradiation of the surroundings.
I suspect a hand grenade energy release worth of antimatter is
going to give little or no blast effect, but heat up the
surroundings a bit, while toasting any unfortunate nearby
biologicals to a crackly crunch with all the nasty gammas.
--
Tagon: "Where's your sense of adventure?" | Mike Van Pelt
Kevyn: "It died under mysterious circumstances. | mvp at calweb.com
My sense of self-preservation found the body, | KE6BVH
but assures me it has an airtight alibi." (schlockmercenary.com)
Erik Max Francis
> One minor correction: the blast effects scale with the inverse
> cube of the yield but heat and light effects will go with the inverse
> square. This means that for large explosions, you get incinerated
> farther away than you have buildings toppled on top of you.
There's also the effect that as the yields get _really_ large, more and
more of the energy in the explosion is simply radiated into space
instead of causing more widespread damage. I'm not sure how the effect
scales, but it should make modifications to these terms that make them
somewhat less than inverse cube and inverse square, respectively.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Golf is a good walk spoiled.
-- Mark Twain
Erik Max Francis
> Charged pions will have a destructive effect, but will
> quickly decay into harmless neutrinos. The link gives
> a figure of several meters to several hundred meters,
> I do not know how accurate that figure is.
Charged pions can't just decay into neutrinos (which are uncharged).
They decay into muons, (muon) neutrinos and photons, all of which are
stable particles. Neutrinos interact so weakly that they can be
completely ignored. Muons can also induce certain other decays but they
also rapidly lose their energy on interaction with matter and won't
cause any significant damage to my knowledge. But you're certainly
correct that the energy that goes into creating pions is pretty much
lost in terms of causing damage.
(Note that proposed antimatter _drives_ can actually get some benefit
out of these pions, since they're charged.)
> Anyway once you factor in the inefficiency in mixing
> the ingredients and the pion problem, the amount of
> energy in the blast will be drastically less than
> the theoretical maximum.
Right. Still, the effect is a lot more than fission or fusion, even
with the inefficiencies worked in. The practical yields
Erik Max Francis
> Pions decay very quickly (microseconds?) to mesons, neutrinos,
> and high-energy gamma rays. The mesons also decay
> (milliseconds?) to electrons and positrons. The neutrinos
> escape, of course, but not a lot of the energy goes into them.
I don't think this is right. Charged pions decay into muons and
muon-neutrinos. Neutral pions decay into an electron, positron, and a
photon. (Pions are themselves mesons; none of their decay products are.)
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Who, my friend, can scale heaven?
-- _Gilgamesh_, ca. 3rd C. BC
Erik Max Francis
> Charged pions can't just decay into neutrinos (which are uncharged).
> They decay into muons, (muon) neutrinos and photons, all of which are
> stable particles.
Err, scratch that. Charged pion decay only results in two-particles, a
muon and a muon-neutrino. Photons are only produced (in addition to an
electron and a positron) in neutral pion decays.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Andy Shepard
> Charged pions can't just decay into neutrinos (which are uncharged).
> They decay into muons, (muon) neutrinos and photons, all of which are
> stable particles. Neutrinos interact so weakly that they can be
> completely ignored. Muons can also induce certain other decays but they
> also rapidly lose their energy on interaction with matter and won't
> cause any significant damage to my knowledge. But you're certainly
> correct that the energy that goes into creating pions is pretty much
> lost in terms of causing damage.
Muons are not stable. They decay to electrons or positrons, and neutrinos
with a half-life of 2.2 microseconds.
--
Andy Shepard |
andy+...@andyshepard.org |
http://www.andyshepard.org/ | http://www.livejournal.com/users/vox_soli/
Mike Van Pelt
>
>> Pions decay very quickly (microseconds?) to mesons, neutrinos,
>> and high-energy gamma rays. The mesons also decay
>> (milliseconds?) to electrons and positrons. The neutrinos
>> escape, of course, but not a lot of the energy goes into them.
>
>I don't think this is right. Charged pions decay into muons and
>muon-neutrinos. Neutral pions decay into an electron, positron, and a
>photon. (Pions are themselves mesons; none of their decay products are.)
ACK! You're right, muons is what I meant.
Pions and muons are short for pi mesons and mu mesons, respectively.
Erik Max Francis
> ACK! You're right, muons is what I meant.
>
> Pions and muons are short for pi mesons and mu mesons, respectively.
No. Pions are mesons, but muons aren't mesons, they're leptons (like
electrons or neutrions). Mesons are hadrons consisting of two quarks;
leptons aren't composed of quarks at all. No decay products of pions
(or decay products of those decay products) are mesons.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
There is nothing so subject to the inconstancy of fortune as war.
-- Miguel de Cervantes
Nyrath
annihilating an average chunk of matter, what percentage
of the resulting energy will be in dangerous form
(i.e., gamma rays and particles which will strongly interact
with matter) and what percentage will be harmless?
I would assume that on average the number of protons
and electrons would be equal, but what about neutrons?
uray
news:1139254601.3...@g14g2000cwa.googlegroups.com...
> I'm looking for a simple equation that relates explosion energy of
> antimatter (measured in grams) and equivalence in megatons of TNT
> (which seems to be what nuclear explosions are measured in). Also,
> what would be the equivalence between grams of antimatter and the
> casualty radius of the explosion? We have some data for such as
> explosions as Hiroshima and Nagasaki, which were in the kiloton range,
> but I don't know how to go from there to larger amounts and to grams of
> antimatter versus nuclear weapon weight.
Others have answered the energy issue, but I've always wondered about
another point. Exactly what is it going to *look* like, both in space and on
the Earth.
There have been a number of threads in various groups concerning this
subject, but I never did form a clear picture of the resulting explosion.
Points I recall are:
1. It will be slower than fission or fusion.
2. Some of the energy will appear to materialize at a distance from the
center due to secondary decay.
A big question from what I recall is exactly how a chunk of antimatter will
annihilate when exposed in a room environment. In other words, drop it on
the floor. The outer electrons are going to go first, but after that I
don't think there's clear consensus on what's going to happen after that.
IIRC some have predicted a expanding shell, pushing the matter away and
slowing the reaction considerably.
In space it may be even more difficult, how do you ensure maximum
annihilation? Is the impact velocity of a "gun" style U-235 bomb sufficient
to annihilate an A/M pellet implosion?
Idle questions, I feel like the guy in Larry Niven's _Lucifer's Hammer_ who
wanted to know what it would "sound like" when the comet struck :-)
Erik Max Francis
> A big question from what I recall is exactly how a chunk of antimatter will
> annihilate when exposed in a room environment. In other words, drop it on
> the floor. The outer electrons are going to go first, but after that I
> don't think there's clear consensus on what's going to happen after that.
> IIRC some have predicted a expanding shell, pushing the matter away and
> slowing the reaction considerably.
With a physical chunk of antimatter in a matter atmosphere, you'll get
an ambiplasma surrounding the chunk, which is a very high-energy plasma
composed of matter and antimatter. Due to the energy of the reaction,
the ambiplasma will get created and basically moderate the reaction,
slowing it down. Needless to say, this isn't what you want for a bomb.
I don't have specific knowledge of any real plans for an antimatter
bomb, but I'd imagine what you'd want is your antimatter in _gas_ form
(which practically speaking is how you'd be storing it anyway) and then
disperse it into the atmosphere very quickly so that an ambiplasma
doesn't have time to develop. In other words, some kind of primary
explosion like you have in thermonuclear devices, although the goal here
isn't to compress the fusion fuel to create fusion reactions, but rather
do disperse the antimatter so violently that it can't help but
annihilate rapidly before any kind of moderating ambiplasma can develop.
That's just a guess on my part.
> In space it may be even more difficult, how do you ensure maximum
> annihilation? Is the impact velocity of a "gun" style U-235 bomb sufficient
> to annihilate an A/M pellet implosion?
That I really don't know. Just like with criticality studies that were
done for the development of nuclear weapons, I'm sure there are
geometries, whether they're gun-type or implosion-type, that can
maximize the amount of annihilation done in the device before it blows
itself apart (because then, just like with nuclear weapons, it really is
too late to get anymore energy out of it, whereas that's not the case
with an antimatter bomb detonated in a matter atmosphere).
I'm sure some studies have the proper geometries have been done,
although whether anybody would be willing to tell us about it is another
question :-).
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Maybe this world is another planet's Hell.
-- Aldous Huxley
Erik Max Francis
> Can anybody offer a WAG given an average chunk of matter
> annihilating an average chunk of matter, what percentage
> of the resulting energy will be in dangerous form
> (i.e., gamma rays and particles which will strongly interact
> with matter) and what percentage will be harmless?
Nucleons and antinucleons interacting is a messy business, as is now
doubt clear from this thread. The reason it's messy is because nucleons
aren't fundamental particles, they're baryons, composed of three quarks
each. So, in some sense, how those quarks happen to interact is how you
get the shower of secondary particles.
My understanding (which may well be incorrect) is that a
nucleon-antinucleon annihilation typically liberates 2 or 3 pions (and
occasionally a really wacky baryon, as well). Neutral pions decay into
photons directly, or electron-positron pairs and photons. Charged pions
decay to muons and neutrinos; muons decay to electrons or positrons and
more neutrinos. Pions have very short half-lives and don't get very far
at all before they decay (the neutral pions have _really_ short
half-lives). Muons have a half-life on the order of about a
microsecond, but my understanding is that they're slowed down quite a
bit on interaction with matter (particular the superdense hell that
would in the periphery of a serious antimatter explosion) so I'm not
sure whether they'd get very far before decaying themselves.
So when all is said and done, what you're left with is electrons,
positrons, photons, and neutrinos. If you're in an atmosphere, then
obviously those positrons will also eventually annihilate with a random
electron and be turned into photons as well.
Doing some simple Google research, I'm seeing conflicting numbers, but
it looks like something like 30-60% of the total mass-energy of the
nucleons and antinucleons ends up in the form of neutrinos when
everything has settled down. Depending on the circumstances, the rest
of that energy should end up either as heat liberated (think a nuclear
fireball) or a high-energy electrons (beta radiation).
> I would assume that on average the number of protons
> and electrons would be equal, but what about neutrons?
Do you mean the ratio of neutrons to protons in normal matter? Elements
with higher atomic number tend to have more neutrons to keep them
stable, so heavy metals and transuranics are going to have a lot more
neutrons involved in the reaction than, say, gases or elements like
carbon and oxygen.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Nyrath
> Doing some simple Google research, I'm seeing conflicting numbers, but
> it looks like something like 30-60% of the total mass-energy of the
> nucleons and antinucleons ends up in the form of neutrinos when
> everything has settled down. Depending on the circumstances, the rest
> of that energy should end up either as heat liberated (think a nuclear
> fireball) or a high-energy electrons (beta radiation).
Simplistic elementary school math to follow, please check
assumptions (and the math as well):
So for a WAG, if one assumes that there are approximately
the same number of electrons as nucleons, then approximately
50% of the annihilation energy will come from electron-positron
annihilation, and 50% from nucleon-antinucleon annihilation.
Of the electron-positron energy, 100% is "bomb-like" energy,
that is, it is in the form of gamma rays that will strongly
interact with matter, like that enemy spacecraft you are
aiming at.
Of the nucleon-antinucleon energy, 40% to 70% will be
bomb-like energy.
So the theoretical maximum of bomb-like energy produced
will be 70% to 85% of the total annihilation energy.
I take this to mean that in order to get a WAG of the
destructive force of an antimatter warhead
[1] total the amount of matter and antimatter
[2] multiply by the percentage of matter-antimatter that
your bomb design actually allows to react (an ancient
post on r.a.s.s suggested you might be lucky to get 10%)
[3] run the result through E=Mc^2
[4] multiply by 70% to 85%, depending upon whether one
is an optimist or pessimist.
You are going to yell at me for using unit specified
variables, but:
E = 21480.8MBR
where
E = blast energy in kilotons of TNT
M = total mass of matter and antimatter in kilograms
B = percentage of total energy that is blast energy
= 0.70 to 0.85
R = percentage of total mass that actually participates
in the annihilation reaction
21480.8 = approximately c^2 / joules-per-kiloton-TNT
Aaron Denney
> Erik Max Francis wrote:
>> Doing some simple Google research, I'm seeing conflicting numbers, but
>> it looks like something like 30-60% of the total mass-energy of the
>> nucleons and antinucleons ends up in the form of neutrinos when
>> everything has settled down. Depending on the circumstances, the rest
>> of that energy should end up either as heat liberated (think a nuclear
>> fireball) or a high-energy electrons (beta radiation).
>
> Simplistic elementary school math to follow, please check
> assumptions (and the math as well):
>
> So for a WAG, if one assumes that there are approximately
> the same number of electrons as nucleons, then approximately
Not too horrible, gets hydrogen right, but off by about 50 % for
everything else.
> 50% of the annihilation energy will come from electron-positron
> annihilation, and 50% from nucleon-antinucleon annihilation.
Except, of course that each nucleon is 1800 times as massive.
--
Aaron Denney
-><-
chorned...@hushmail.com
Nyrath wrote:
> Erik Max Francis wrote:
> > Doing some simple Google research, I'm seeing conflicting numbers, but
> > it looks like something like 30-60% of the total mass-energy of the
> > nucleons and antinucleons ends up in the form of neutrinos when
> > everything has settled down. Depending on the circumstances, the rest
> > of that energy should end up either as heat liberated (think a nuclear
> > fireball) or a high-energy electrons (beta radiation).
>
> Simplistic elementary school math to follow, please check
> assumptions (and the math as well):
>
> So for a WAG, if one assumes that there are approximately
> the same number of electrons as nucleons, then approximately
> 50% of the annihilation energy will come from electron-positron
> annihilation, and 50% from nucleon-antinucleon annihilation.
>
times more mass, and there are about 2,5 times more of them.
> Of the electron-positron energy, 100% is "bomb-like" energy,
> that is, it is in the form of gamma rays that will strongly
> interact with matter, like that enemy spacecraft you are
> aiming at.
>
Let us follow the partitioning of nucleon-nucleon annihilation energy.
If one proton and one antiproton annihilate, they generate on average
4,5 pions total, at slow speed, more if they approach fast.
Proton and antineutron, I believe, have similar average numbers. This
is average total; minimum is 2 (from momentum conservation), maximum
would be about 13 if their approach is slow and more if they approach
fast (from the pion rest mass). I suppose that spins might interfere
with that, too.
Now if an antiproton hits a large nucleus, then the pions are very
strongly absorbed by nucleons. About half are absorbed by target
nucleus and go to fast nucleons; the other half escape. The matter
could get thornier if the annihilation involves small nuclei - or a
bunch of very close by nuclear fragments, as in a nucleus already
disrupted.
On average, 1/3 of the pions are neutral (The total number of charged
pions in individual annihilation is constrained by charge
conservation). The neutral pions, if escaping the nucleus, decay to 70
MeV photons - if stationary. As they are not (they wind up with average
of about 400 MeV per pion IIRC), the photons should have a broad peak
at around 200 MeV, extending below 70 MeV and also above. And carry a
total of 1/3 annihilation energy (if proton-proton) or 1/6 (if
proton-nucleus).
Next, the charged pions. Their behaviour depends interestingly on where
exactly the annihilation takes place.
I have seen numbers like, free fast pions are half absorbed by 5 cm of
tungsten or 3 m of nitrogen or 7 m of hydrogen - and they half decay in
12 m of vacuum.
This means that if the spot of annihilation is in dense matter,
practically all charged pions are stopped by nuclei and go to excite
them - average 400 MeV ending up in each. In vacuum, they decay to
muons and neutrinos, and in air of near-ground density, a significant
par is absorbed while a significant part decays.
If they decay, about slightly over half the free energy should go to
neutrinos and slightly under half to muons.
Muons are charged particles. They could be destroyed by hitting
nucleons, but this is a weak interaction. Alternatively, they can shed
some energy through electric resistance, and finally decay - whereupon
2/3 of their energy goes to neutrinos.
So, at a maximum, in dense matter, all annihilation energy is
destructive.
At a minimum, hydrogen-hydrogen annihilation in vacuum, about 4/9 will
(1/3 from neutral pions, 1/9 from electrons through muon decay).
Nyrath
> Except, of course that each nucleon is 1800 times as massive.
D'oh!!! <smacks forehead>
Nyrath
> At a minimum, hydrogen-hydrogen annihilation in vacuum, about 4/9 will
> (1/3 from neutral pions, 1/9 from electrons through muon decay).
Thank you! It is good to get some hard numbers.
So since electron-positron annihilation's contribution to
the blast is about 1/5000th that of nucleon-antinucleon,
it can be safely ignored. Correct?
This implies that the blast equation is
E = (21480.8 x 4/9)MR
or
E = 9547.0 MR
where
E = blast energy in kilotons of TNT
M = total mass of matter+antimatter in kilograms
R = percentage of total mass that actually participates
in the annihilation reaction
For megatons it would be
E = 9.6 MR
which I believe is a WAG of the equation
originally requested by Mr. Weys at the start
of the thread.
Mr. Weys, when antimatter warheads are initially
invented, R will likely be rather low (say 0.01),
and will gradually grow larger as the technological
state of the art advances. The theoretical
maximum is of course 1.0
Mike Van Pelt
>
>> ACK! You're right, muons is what I meant.
>>
>> Pions and muons are short for pi mesons and mu mesons, respectively.
>
>No. Pions are mesons, but muons aren't mesons, they're leptons (like
>electrons or neutrions). Mesons are hadrons consisting of two quarks;
>leptons aren't composed of quarks at all. No decay products of pions
>(or decay products of those decay products) are mesons.
When I was in college, they were, now they aren't. I did a bit
of googling to check on your point, and found that you're right,
I somehow missed that terminology change.
From Wikipedia: "For historical reasons, muons are sometimes
referred to as mu mesons, even though they are not classified
as mesons by modern particle physicists." (Yeah, I know, but
this not being a controversial political issue, I'd give
Wikipedia the benefit of the doubt here, and it is an easy
place to check.)
Still, *originally*, muon was short for mu meson, though it
isn't any longer.
Mike Van Pelt
>numbers, but it looks like something like 30-60% of the total
>mass-energy of the nucleons and antinucleons ends up in the
>form of neutrinos when everything has settled down.
Wow, that's a *lot* more than I would have expected.
Those must be some really, *really* high-energy neutrinos.
Fascinating... I'll have to look into this.
Mike Van Pelt
>you'll get an ambiplasma surrounding the chunk, which is a very
>high-energy plasma composed of matter and antimatter. Due to
>the energy of the reaction, the ambiplasma will get created and
>basically moderate the reaction, slowing it down.
So, it might look kind of like E. E. "Doc" Smith's loose atomic
vortices, as in "The Vortex Blasters"? Interesting.
Erik Max Francis
> When I was in college, they were, now they aren't. I did a bit
> of googling to check on your point, and found that you're right,
> I somehow missed that terminology change.
>
> From Wikipedia: "For historical reasons, muons are sometimes
> referred to as mu mesons, even though they are not classified
> as mesons by modern particle physicists." (Yeah, I know, but
> this not being a controversial political issue, I'd give
> Wikipedia the benefit of the doubt here, and it is an easy
> place to check.)
>
> Still, *originally*, muon was short for mu meson, though it
> isn't any longer.
I didn't know that, that's interesting. I wasn't even aware of the
terminology change. Turns out that the use of _meso-_ (meaning
"intermediate" in mass between protons and electrons) was much more
literal there and that was before the decay products of muons and the
other later discovered mesons really were different enough to warrant a
further distinction as muons not being like the other mesons (though it
appears _meso-_ was first used to identify muons, an interesting twist
of evolving terminology). Then, of course, quark theory and the
Standard Model came along to explain what was really going on.
Interesting history lesson; learn something new every day.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
We are circumstances of our birth.
-- Sade Adu
Erik Max Francis
> So for a WAG, if one assumes that there are approximately
> the same number of electrons as nucleons, then approximately
> 50% of the annihilation energy will come from electron-positron
> annihilation, and 50% from nucleon-antinucleon annihilation.
As others have pointed out, this is by far the biggest error. Almost
all the energy comes from the nucleon-antinucleon annihilation; in fact,
the electron-positron annihilation can be safely ignored.
Probably the best watered-down version way of looking at it is that, in
space, roughly half of the energy of the blast is lost in neutrinos. In
an atmosphere, somewhat less than that will be lost.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
I got you in my sight / Gun in my palm / Surprisingly calm
-- Sole
chorned...@hushmail.com
Nyrath wrote:
> chorned...@hushmail.com wrote:
> > At a minimum, hydrogen-hydrogen annihilation in vacuum, about 4/9 will
> > (1/3 from neutral pions, 1/9 from electrons through muon decay).
>
> Thank you! It is good to get some hard numbers.
>
> So since electron-positron annihilation's contribution to
> the blast is about 1/5000th that of nucleon-antinucleon,
> it can be safely ignored. Correct?
>
> This implies that the blast equation is
> E = (21480.8 x 4/9)MR
No. You forgot the factor of 2 from the rest energy of the matter
annihilated.
> or
> E = 9547.0 MR
> where
> E = blast energy in kilotons of TNT
> M = total mass of matter+antimatter in kilograms
> R = percentage of total mass that actually participates
> in the annihilation reaction
>
> For megatons it would be
> E = 9.6 MR
>
Usually much more - minimum 19,2, maximum 42,9, normally over 30.
> which I believe is a WAG of the equation
> originally requested by Mr. Weys at the start
> of the thread.
>
> Mr. Weys, when antimatter warheads are initially
> invented, R will likely be rather low (say 0.01),
Unlikely.
Antimatter is simply not easy to preserve anywhere near ground, once it
is not contained. R would exceed 0,99 from start.
Look at it this way: cosmic radiation is generated by high-energy
particles hitting atmosphere.
Some reaches ground - but this is a small proportion. There is much
more radiation in space.
Since atmosphere blocks most radiation from space from reaching ground,
for a near-ground annihilation explosion it would also block most
radiation from escaping into space.
Since half of the atmosphere is in the lower 5-6 km, it follows that
most radiation from an annihilation explosion would be stopped by the
first hundreds of metres to kilometres of air
chorned...@hushmail.com
Mike Van Pelt wrote:
> In article <NLednYBdnJiXX3re...@io.com>,
> Nyrath <nyr...@projectrho.com.INVALID> wrote:
> >However, it is important to note that not all the energy
> >in a matter-antimatter reaction will be available as
> >destructive energy.
> >
> >http://www.stardestroyer.net/Empire/Tech/Torpedoes/Torpedo1.html
> >
> >It is non-trivial to create a weapon that will ensure all the
> >matter will reaction with all the antimatter.
In space, perhaps. In atmosphere, it is trivial.
I doubt it. Mainly because of the heat per weight is highly
concentrated.
If you discharge a big condenser by shortcircuiting it with a fat,
conductive copper bar, the energy is dissipated all over the massive
bar, so it is only slightly warmed - no light or sound generated. Nor
gammas.
If you were to discharge it through a cathode-ray tube, you would get a
lot of nasty x-rays for nearby biologicals, and heat the anode a bit,
but you would not get blast sound.
But if you discharged the condenser by a spark through air, the energy
would be spent on a narrow spark channel, which contains only a small
weight of air having little heat capacity. It heats to very high
temperature, radiating visible flash and creating a loud blast. Even
very small amounts of energy, if concentrated in the form of a spark,
cause relatively audible sound and visible flash, compared to total
energy expended.
Nyrath
> Nyrath wrote:
>> This implies that the blast equation is
>> E = (21480.8 x 4/9)MR
>
> No. You forgot the factor of 2 from the rest energy of the matter
> annihilated.
>
>> M = total mass of matter+antimatter in kilograms
I'm confused. In my equation I am
(perhaps in a non-standard manner)
using the term "M" to symbolize
the combined mass of the matter and antimatter.
> Usually much more - minimum 19,2, maximum 42,9, normally over 30.
Check me on this, so you are saying that the
percentage of nucleon-antinucleon annihilation
energy that goes into "dangerous" forms
varies from minimum 0.4444 (i.e., 4/9),
maximum 1.0, normally over 0.7?
(again, "dangerous" meaning "interacts
strongly with matter", or "acts like a
bomb blast")
>> Mr. Weys, when antimatter warheads are initially
>> invented, R will likely be rather low (say 0.01),
>
> Unlikely.
>
> Antimatter is simply not easy to preserve anywhere near ground, once it
> is not contained. R would exceed 0,99 from start.
Arrrgh. I'm sorry.
I was talking about an antimatter warhead on
a interplanetary warhead that would be detonated
in a deep space environment, NOT a lump of
antimatter dropped into a planetary target
with an atmosphere. I did not make that clear.
In space, when a quantity of matter and antimatter
come into contact, the atoms at the point of contact
will annihilate, creating a source of radiation
pressure which will tend to push the masses
*out* of contact.
Actually it will probably be powerful enough
to vaporize the masses of matter and antimatter,
blowing them apart with great force, preventing
them from participating in the reaction. They
will just disperse into the vacuum of space.
Creating a warhead geometry to counteract this
is non-trivial (though I suspect that some
research into binary chemical explosives
could provide insights)
Thomas Womack
Nyrath <nyr...@projectrho.com.INVALID> wrote:
> In space, when a quantity of matter and antimatter
> come into contact, the atoms at the point of contact
> will annihilate, creating a source of radiation
> pressure which will tend to push the masses
> *out* of contact.
>
> Actually it will probably be powerful enough
> to vaporize the masses of matter and antimatter,
> blowing them apart with great force, preventing
> them from participating in the reaction. They
> will just disperse into the vacuum of space.
>
> Creating a warhead geometry to counteract this
> is non-trivial (though I suspect that some
> research into binary chemical explosives
> could provide insights)
I don't think binary chemical explosives are the place to look: my
mental model of an antimatter bomb looks like a standard implosion
nuke, with a lot of chemical explosive imploding a thick shell of
depleted uranium (I assume high-Z is at least slightly helpful in
absorbing the gammas; 511keV is a good deal lower than the energy of
most fission gammas, though obviously the 2GeV nucleon-nucleon
annihilations are hard to absorb) onto a speck of antimatter in the
centre.
And I suspect you want to do that in atmosphere as well, since I suspect
you get a quicker annihilation (and so a brighter fireball) if you use
a dense metal annihilator rather than plasma-state air.
Tom
JimboCat
>in space, roughly half of the energy of the
>blast is lost in neutrinos
Neutrinos are generally considered harmless: "they go right through a
light-year of lead", after all. Thus "lost". But a few of them do
interact with matter, and if there are *lots* of neutrinos there will
be lots of interactions. And we're talking *lots* of neutrinos, here!
I'm wondering if those neutrino interactions could be sufficiently
destructive to do damage even to people/facilities that are adequately
shielded from other effects of a matter-antimatter annihilation.
A quick google indicates that neutrino interactions with matter are of
two major types:
neutrino-electron scattering - if enough energy is exchanged, this will
be an ionizing event.
neutrino absorption - this can raise the atomic number of an atom: I
think the new electron (to balance charge) is usually ejected, so it's
also an ionizing event at the same time. This can alternately lower the
atomic number, with a positron emitted: that gives a nasty gamma-ray
photon generated inside the target.
Cross-sections for these interactions are on the order of 10^-43 S_n^2
cm^2 MeV^-2. Do we have any clue how energetic the neutrinos emitted
from an antimatter explosion might be?
Jim Deutch (JimboCat)
--
"We must go forth and crush every world view that doesn't believe
in tolerance and free speech," - David Brin
Mike Van Pelt
JimboCat <10313...@compuserve.com> wrote:
>Erik Max Francis wrote:
>
>>in space, roughly half of the energy of the
>>blast is lost in neutrinos
>
>Neutrinos are generally considered harmless: "they go right through a
>light-year of lead", after all. Thus "lost". But a few of them do
>interact with matter, and if there are *lots* of neutrinos there will
>be lots of interactions. And we're talking *lots* of neutrinos, here!
>
>I'm wondering if those neutrino interactions could be sufficiently
>destructive to do damage even to people/facilities that are adequately
>shielded from other effects of a matter-antimatter annihilation.
My calculations from Supernova 1987A were that you'd get 500
roentgens of neutrino exposure at about 10 AU from the
supernova. (I left it as roentgens, as I have no idea what the
RBE of neutrinos might be.) Later, I saw a similar figure in
National Georgraphic, so I don't guess I slipped a decimal
point anywhere.
The nifty (if it isn't you) feature of this is, it doesn't
matter which side of the planet you are from the supernova;
you get virtually the same exposure.
Of course, this hardly matters, because lacking a warp drive,
you going to get morally vaporized in the near future anyway.
But, turning to the antimatter issue... Yeah, a lot of neutrinos,
but I (no calculations, just SWAG) don't think there's that much
of an overwhelming abundance of them. On the other hand... half
of the energy goes off as neutrinos? Wow. Those must be
exceedingly high-energy neutrinos. And, IIRC, really, really
high energy neutrinos are more likely to interact with less
ephemeral matter, so maybe there is enough of them after all.
Erik Max Francis
> Neutrinos are generally considered harmless: "they go right through a
> light-year of lead", after all. Thus "lost". But a few of them do
> interact with matter, and if there are *lots* of neutrinos there will
> be lots of interactions. And we're talking *lots* of neutrinos, here!
>
> I'm wondering if those neutrino interactions could be sufficiently
> destructive to do damage even to people/facilities that are adequately
> shielded from other effects of a matter-antimatter annihilation.
Even if they're much more energetic than those created in normal fusion,
they're still effectively harmless. The cross-section is tiny.
Consider that the LD-50 dose from the neutrino burst of a _supernova_
is ~1 au.
Neutrinos generated in any kind of nuclear interaction can be utterly
ignored in terms of doing damage.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Strange is our situation here upon earth.
-- Albert Einstein
Erik Max Francis
> But, turning to the antimatter issue... Yeah, a lot of neutrinos,
> but I (no calculations, just SWAG) don't think there's that much
> of an overwhelming abundance of them. On the other hand... half
> of the energy goes off as neutrinos? Wow. Those must be
> exceedingly high-energy neutrinos. And, IIRC, really, really
> high energy neutrinos are more likely to interact with less
> ephemeral matter, so maybe there is enough of them after all.
In fusion and antimatter interactions (at least antimatter interactions
in space), neutrinos are the real show. Remember, the vast majority of
the energy released in a supernova is released in _neutrinos_. By
comparison, the piddly little electromagnetic event is a tiny burp in
comparison.
But even for the ridiculously enormous fluxes involved in a _supernova_,
you have to be really, really close to worry about neutrino radiation
exposure. For any kind of non-stellar event, neutrino release is really
a matter of energy escaping completely from the reaction. If you want
to extract any energy from them, you're going to need light-years of lead.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
John Schilling
says...
>
>"Cyde Weys" <cyde...@gmail.com> wrote in message
>news:1139254601.3...@g14g2000cwa.googlegroups.com...
>> I'm looking for a simple equation that relates explosion energy of
>> antimatter (measured in grams) and equivalence in megatons of TNT
>> (which seems to be what nuclear explosions are measured in). Also,
>> what would be the equivalence between grams of antimatter and the
>> casualty radius of the explosion? We have some data for such as
>> explosions as Hiroshima and Nagasaki, which were in the kiloton range,
>> but I don't know how to go from there to larger amounts and to grams of
>> antimatter versus nuclear weapon weight.
><snip>
>
>Others have answered the energy issue, but I've always wondered about
>another point. Exactly what is it going to *look* like, both in space and on
>the Earth.
>
>There have been a number of threads in various groups concerning this
>subject, but I never did form a clear picture of the resulting explosion.
>Points I recall are:
>1. It will be slower than fission or fusion.
Yes. But note that "slower than fission or fusion" means "takes whole
microseconds!"
>2. Some of the energy will appear to materialize at a distance from the
>center due to secondary decay.
Where the distance is on the order of ten meters, yes.
>A big question from what I recall is exactly how a chunk of antimatter will
>annihilate when exposed in a room environment. In other words, drop it on
>the floor. The outer electrons are going to go first, but after that I
>don't think there's clear consensus on what's going to happen after that.
>IIRC some have predicted a expanding shell, pushing the matter away and
>slowing the reaction considerably.
For macroscopic quantities of antimatter, none of this really matters.
Nobody cares whether the initial manifestation of a 20-kiloton explosion
is 8.4E13 joules released over five microseconds in a ten-meter sphere
or 8.4E13 joules released over fifty nanoseconds in a ten-centimeter
sphere. Hiroshima is going to look just the same either way.
>In space it may be even more difficult, how do you ensure maximum
>annihilation? Is the impact velocity of a "gun" style U-235 bomb
>sufficient to annihilate an A/M pellet implosion?
As others have noted, this is good for maybe a factor of two difference
in destructively-"useful" energy release.
Even if you somehow arrange for some of your antimatter to not annihilate
at all in the bomb, that just means some antimatter gets sprayed around
the landscape to impinge on and annihilate with any targets in the area.
So, 8.4E13 joules of charged particles and gamma rays irradiating everything
in the vicinity, or a gram of antimatter impacting on everything in the
vicinity and generating 8.4E13 joules of charged particles in the process?
--
*John Schilling * "Anything worth doing, *
*Member:AIAA,NRA,ACLU,SAS,LP * is worth doing for money" *
*Chief Scientist & General Partner * -13th Rule of Acquisition *
*White Elephant Research, LLC * "There is no substitute *
*schi...@spock.usc.edu * for success" *
*661-951-9107 or 661-275-6795 * -58th Rule of Acquisition *
chorned...@hushmail.com
When a nucleon and antinucleon annihilate, roughly 1900 MeV + their
kinetic energy is distributed between an average of 4,5 pions. Thus
about 400 MeV total energy per each.
A charged pion decays into a muon and a mu neutrino. Muon has a rest
mass of about 100 MeV. So, BOTE the average energy of neutrino should
be around 200 MeV. Conservation of energy and monmentum should give a
more precise value.
By contrast, neutrinos generated from beta decay, including fission
reactor operation, OR a supernova (which is essentially inverse beta
decay to a neutron star) have average energies in few MeV. It follows
that neutrinos from annihilation explosion interact with matter far
more strongly and cause far more damage per energy passing through.
Mike Van Pelt
<chorned...@hushmail.com> wrote:
>> I suspect a hand grenade energy release worth of antimatter is
>> going to give little or no blast effect, but heat up the
>> surroundings a bit, while toasting any unfortunate nearby
>> biologicals to a crackly crunch with all the nasty gammas.
>>
>I doubt it. Mainly because of the heat per weight is highly
>concentrated.
But a whole lot of the energy from antimatter anihilation is
in the form of particles and gamma rays that travel quite a
ways, on the average, before interacting with anything. So
the energy being released is spread over some area. For a
large enough quantity of antimatter, it may not make much
difference, but for a hand grenade level of energy release?
Erik Max Francis
> A charged pion decays into a muon and a mu neutrino. Muon has a rest
> mass of about 100 MeV. So, BOTE the average energy of neutrino should
> be around 200 MeV. Conservation of energy and monmentum should give a
> more precise value.
>
> By contrast, neutrinos generated from beta decay, including fission
> reactor operation, OR a supernova (which is essentially inverse beta
> decay to a neutron star) have average energies in few MeV. It follows
> that neutrinos from annihilation explosion interact with matter far
> more strongly and cause far more damage per energy passing through.
Sure, and neutrino cross-section scales roughly as the square of the
neutrion energy [1], at least at neutrino energies are much less than
quarks they're scattering off of [2]). But for neutrino emissions from
traditional nuclear reactions have cross sections on the order of 10^-47
m^2. Increasing the neutrino energies by a factor of 10^2 only
increases their cross-sections by a factor of 10^4, which sounds like a
lot, but now you're only up to 10^-43 m^2. By comparison, light
interacts with atoms with cross-sections of 10^-20 m^2 and the
interaction of nuclei in the nucleus is on the order of 10^-28 m^2.
The mean free path D of a particle of (nuclear) cross-section sigma
through matter of density rho is
D = u/(sigma rho)
For a neutrino with a normal fusion/beta decay cross-section of ~10^-47
m^2, that's a mean free path through lead (rho = 11 400 kg/m^3) of
~10^16 m, or ~1 ly. Increase the energy such that cross-section as
increased by 10^4, and the mean path drops to ~10^12 m, or ~10 au.
That's a huge difference, but that's the _mean free path_. Each
neutrino would have to travel through ten times the distance between the
Sun and the Earth of _solid lead_ in order to have a 50% chance of an
interaction.
Even for neutrinos with energies measured in the hundreds of GeV (the
region where the response becomes linear to energy), the cross-section
is ~10^40 m^2, which gives you a mean free path of ~1 Gm.
So sure, even if the neutrinos are vastly more energetic, you're
changing the cross-section from "so ridiculously small it's utterly
negligible" to "bigger but still negligibly small -- at least, not large
enough that you're going to have to seriously consider it as a danger
relative to plenty of other forms of radiation that an antimatter bomb
will give off."
.
1. _Principles of Stellar Evolution and Nucleosynthesis_, Clayton, p. 261.
2. From
http://hyperphysics.phy-astr.gsu.edu/hbase/particles/neutrino3.html, it
looks like neutrinos in this energy realm have roughly linear
cross-sections; this reference is also consistent (give or take an order
of magnitude) with the given estimate of the cross-section for neutrinos
from beta decay (and other typical nuclear reactions).
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Melancholy men, of all others, are the most witty.
-- Aristotle
Erik Max Francis
> But a whole lot of the energy from antimatter anihilation is
> in the form of particles and gamma rays that travel quite a
> ways, on the average, before interacting with anything. So
> the energy being released is spread over some area. For a
> large enough quantity of antimatter, it may not make much
> difference, but for a hand grenade level of energy release?
Actually, the atmosphere is rather opaque to very high energy gamma
rays. That's why a nuclear explosions generates the huge fireball --
that's the energy of those gamma rays being absorbed by the air
molecules, being reemitted, absorbed by another, etc., etc. ...
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
When you talk to her / Talk to her
-- India Arie
Carey Sublette
"Erik Max Francis" <m...@alcyone.com> wrote in message
news:zoKdnbq9RYd8UHbe...@speakeasy.net...
> Mike Van Pelt wrote:
>
>> But a whole lot of the energy from antimatter anihilation is
>> in the form of particles and gamma rays that travel quite a
>> ways, on the average, before interacting with anything. So
>> the energy being released is spread over some area. For a
>> large enough quantity of antimatter, it may not make much
>> difference, but for a hand grenade level of energy release?
>
> Actually, the atmosphere is rather opaque to very high energy gamma rays.
> That's why a nuclear explosions generates the huge fireball --
> that's the energy of those gamma rays being absorbed by the air molecules,
> being reemitted, absorbed by another, etc., etc. ...
Any small nuclear device (e.g. a few hundrd tons or less) is in essence a
radiation weapon (the lethal range for radiation considerably exceeds the
lethal range of other weapon effects). This is true even for pure fission
devices, where most of the energy is initially in the form of fission
fragment kinetic energy and only several percent in the form of gammas or
neutrons. The 10-20 ton yield Davy Crockett recoilless rifle warhead
depended on this.
MeV range gammas have a half-thickness value in the air of several hundred
meters. This value is based on medium density and is almost independent on
composition. Energetic gamma can penetrate a few inches of lead, which is
10,000 times denser than air so they can penetrate few thousand feet of air.
Carey Sublette
JimboCat
>A charged pion decays into a muon and a mu neutrino. Muon has a rest
>mass of about 100 MeV. So, BOTE the average energy of neutrino should
>be around 200 MeV. Conservation of energy and monmentum should give a
>more precise value.
>
>By contrast, neutrinos generated from beta decay, including fission
>reactor operation, OR a supernova (which is essentially inverse beta
>decay to a neutron star) have average energies in few MeV. It follows
>that neutrinos from annihilation explosion interact with matter far
>more strongly and cause far more damage per energy passing through.
If the neutrinos from an antimatter explosion are about two orders of
magnitude more energetic than those from a supernova, then I think the
interaction cross-sections for those neutrinos are about four orders of
magnitude bigger.
We've had two estimates of the "neutrino-fatality" distance from a
supernova: LD50 at ~1 AU (Erik Max Francis) and 500 Roentgens at 10 AU
(Mike Van Pelt). BOTE conclusion: if you have a supernova-sized
antimatter explosion, the neutrinos alone could give a fatal dose of
radiation at something close to a light-year distance. Yowza!
So how big does an antimatter explosion have to be to sterilize a
planet? You need a fatal dose at the distance of the explosion from the
planet plus the planet's diameter... There's (at least) two problems
with this plan, however: half the energy of your explosion is in forms
that won't pass through the planet, so the (prompt) destruction is
going to be much worse on the near side, and a whole lotta neutrino
energy is going to be liberated _within_ the planet: I suspect we're
talking major volcanism and earthquakes just due to that heating.
I don't think this is a good way to sterilize a planet for immediate
re-occupation.
Jim Deutch (JimboCat)
--
"When ideas fail, words come in very handy." - Goethe
chorned...@hushmail.com
Actually, comparing a small annihilation explosion with a small fission
event...
What precisely happens to fission fragments in air?
If you are near a critical assembly of solid fissionable material, then
it is about a few cm or more across (an unreflected sphere of metal
U-235 is about 50 kg in about 2,5 litres, meaning 17 cm across; I think
Pu-239 and U-233, and some other isotopes would be less, but still over
or around 10 cm; reflectors would mean a smaller critical mass and
volume of metal, but the thickness of the reflector would make the
whole assembly as big or bigger than unreflected assembly...). As long
as the assembly is not having a fission chain reaction, they generate
alpha particles, with energy of a few MeV.
Alphas with a few MeV are stopped by a few cm of air. Or a much thinner
layer of heavy metal (how much?).
Now, the moment fission chain starts...
Those 10-15 cm of heavy metal stop, by definition of criticality, about
40 % of all fast neutrons generated through fission capture. Another
few % are captured in non-fission ways, generating U-236, Pu-240 etc.
So about half neutrons escape into air and once there, can travel
hundreds of metres unless they hit something dense like your body.
Almost all the neutrons that escape the metal are available to
irradiate anything in the first few metres, though they naturally
spread with distance.
The prompt gammas are slightly more penetrating in air, so it stands to
reason that most of them escape the metal, but a significant part is
absorbed.
What about fragments?
They have energies much more than the few MeV of alphas. As they have
around 180 MeV to share, the conservation of momentum requires that the
lighter fragment should have about 110 MeV and the heavier one about 70
MeV.
How far do fission fragments generated by spontaneous or induced
fission near metal surface travel in air? And how far do they travel in
metal?
In any case, a large majority of fission takes place inside the
critical assembly and the fragments only heat up the surroundings. If
there is so much fission that the whole assembly reaches millions of
degrees, well then the fireball radiates soft x-rays which travel a bit
before they are captured and heat up the air. If the fission only heats
the core to a few thousands of degrees so it evaporates and is blown
apart, stopping the chain, there would be a flash and bang, but no soft
x-rays from thermal heating, since the heat would be enough only for
near IR and visual photons. If the chain reaction stops after heating
the assembly by only a few degrees, there should be no bang and no
flash other than slight increase in the far thermal IR from the
critical assembly. But still a lot of prompt gammas and neutrons.
Now imagine antimatter explosion...
Suppose that an antiproton were to meet a nitrogen 14 nucleus - the
most common in air.
One nucleon from the target is annihilated. Could be proton or neutron,
the charges of the remaining few charged pion would sort out the
charge.
About half the pions with about half the energy - around 1000 MeV -
escape and have free path of a few m in air.
The other half with another 1000 MeV are captured by the remaining 13
nucleons. What will they do?
The total binding energy of the nucleus would be around 100 MeV. There
would be plenty of energy to destroy the nucleus totally, and leave 900
MeV to be distributed between 13 nucleons, kinetic energy 13 MeV each.
Alternatively the nucleus might emit just a single fast neutron, with
energy of about 900 MeV, and leave a stable C-12 nucleus recoiling at
75 MeV.
My guess is that neither happens - there would be a few very fast
nucleons, both protons and neutrons, carrying away the bulk of energy
in a few hundreds of MeV per each, then several others having merely a
few tens of MeV, then a few having only a few MeV, and some nucleons
would be left over sticking to each other as fast or slow fragment
nuclei. Basically anything that is remotely capable of sticking
together - something perfectly stable as a high-energy alpha, or
something totally odd, as helium-6, or beryllium 8, or boron-8...
How far would high-energy, heavily charged nuclear fragments travel in
air?
sh...@cadence.com
produced a flash, though I might be remembering the Hollywood version.
This would not have been created by heat, given the limited temperature
increase. It would have been something like Cerenkov radiation, or
ionization of the air.
A bang can also be created without enough energy to vaporize the core.
An accident with the Godiva assembly did not quite melt any of the
core, but still produced a thermal shockwave. The energy release was
equivalent to 1.7 pounds of high explosive, but only 1.4% of the energy
(or the equivalent of 0.024 pounds of HE) was converted to kinetic
energy.
rap...@netscape.net
JimboCat wrote:
> Erik Max Francis wrote:
>
> >in space, roughly half of the energy of the
> >blast is lost in neutrinos
>
> Neutrinos are generally considered harmless: "they go right through a
> light-year of lead", after all. Thus "lost".
Assuming a 50% chance of making it through 1 LY of lead. That means
that the chances of it passing through 0.01LY of lead is only 0.007.
The odds of a specific neutrino interacting is virtually 0. This means
that the percentage of energy emitted is almost entirely lost.
Carey Sublette
<chorned...@hushmail.com> wrote in message
news:1139592608.0...@f14g2000cwb.googlegroups.com...
They plow through about a centimeter or two, making a heck of a lot ions,
then they stop.
Not very far (several microns). A thin foil will block them completely.
The world managed to not discover fission between 1935 and late 1938 partly
due to this fact. It was common practice when bombarding uranium and thorium
targets with neutrons to place a foil over them to stop alphas that would
confound the measurement of neutrons. This of course stopped the fission
fragments which would have created enormous ionization signals.
Erik Max Francis
> Any small nuclear device (e.g. a few hundrd tons or less) is in essence a
> radiation weapon (the lethal range for radiation considerably exceeds the
> lethal range of other weapon effects). This is true even for pure fission
> devices, where most of the energy is initially in the form of fission
> fragment kinetic energy and only several percent in the form of gammas or
> neutrons. The 10-20 ton yield Davy Crockett recoilless rifle warhead
> depended on this.
That's a very good point. We've been talking generally about nuclear
weapons vs. antimatter weapons but we haven't really been talking
specific yields, so (in both cases) there are going to be different
effects in either weapon and also what the purpose of the weapon is
(tactical vs. strategic; destructive vs. enhanced radiation; etc.).
With a "hand grenade" type weapon of antimatter I'm not really sure
which effect wins out. Since there's no fission fragment kinetic
energy, and I'm not sure what gamma ray flux density you need to get the
fireball effects, I definitely miscalled that.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
He who has a _why_ to live can bear with almost any _how_.
-- Friedrich Nietzsche
Erik Max Francis
> If the neutrinos from an antimatter explosion are about two orders of
> magnitude more energetic than those from a supernova, then I think the
> interaction cross-sections for those neutrinos are about four orders of
> magnitude bigger.
I think that's right.
> We've had two estimates of the "neutrino-fatality" distance from a
> supernova: LD50 at ~1 AU (Erik Max Francis) and 500 Roentgens at 10 AU
> (Mike Van Pelt). BOTE conclusion: if you have a supernova-sized
> antimatter explosion, the neutrinos alone could give a fatal dose of
> radiation at something close to a light-year distance. Yowza!
500 R is about 5 Gy, so if Q = 1 for neutrino-induced beta decay, which
seems about right, then that's about 5 Sv, which is a similar short-term
lethal dose.
I've seen both ~1 au and ~10 au figures for the estimate; it really
depends on a lot of assumptions and arbitrary choices so the answer is
probably right to within only a few orders of magnitude anyway (i.e.,
both answers are correct). Here's the details of a calculation that
gets ~1 au:
http://www.tass-survey.org/richmond/answers/snrisks.txt
(I just think that the 1 au figure is interesting because if you're
talking about a type II supernova, the supergiant progenitor is already
well bigger than 1 au!)
There are details that are glossed over. He chose a minimum lethal dose
of 10 Gy (not 10 Sv, he didn't consider the quality factor Q), which is
probably pretty high. If we're talking about neutrinos (equal numbers
of neutrinos and antineutrinos) interacting with nuclei, then you're
going to get get an electron (beta radiation) or a positron, which in
matter will relatively quickly annihilate with an electron and create a
few gamma rays (only a few of which will deposit their energy in the
target). Plus, those nucleon transmutations should cause some
additional radioactivity which should settle down with the emission of
more alphas, betas, or gammas. The snrisks.txt URL above also mentions
issues of elastic scattering which aren't taken into account, either.
So there's pluses and minuses there, I'm not sure which ones win out. Q
= 1 is probably a good estimate all things considered.
> So how big does an antimatter explosion have to be to sterilize a
> planet? You need a fatal dose at the distance of the explosion from the
> planet plus the planet's diameter... There's (at least) two problems
> with this plan, however: half the energy of your explosion is in forms
> that won't pass through the planet, so the (prompt) destruction is
> going to be much worse on the near side, and a whole lotta neutrino
> energy is going to be liberated _within_ the planet: I suspect we're
> talking major volcanism and earthquakes just due to that heating.
>
> I don't think this is a good way to sterilize a planet for immediate
> re-occupation.
If we're talking about an antimatter explosion big enough that the
neutrino flux density is lethal on the other side, then obviously we
don't have to worry about calculating what happens with the other types
of radiation. Assuming you're talking about making fleshy bags of water
(density 1000 kg/m^3) on the other side absorb a lethal LD50 short term
whole-body dose of about 4 Sv = 4 Gy (Q = 1 as above), then the Earth
(average density about 5000 kg/m^3) is going to absorb about five times
that much, or about 20 Gy = 20 J/kg. Assuming the antimatter bomb is
detonated relatively far from the planet (just so we can treat
everything as linear), that would liberate ~10^26 J in heat alone in the
Earth itself. Solar insolation over the whole Earth's dayside is 1.8 x
10^17 W, so you're talking something like 20 yr of solar insolation
liberated throughout the Earth basically instantaneously.
And that's the amount of energy that's actually absorbed in terms of
neutrinos; the vast majority of it passes through unscathed. With a
cross-section of ~10^-43 m^2 and a density of 5000 kg/m^3, the mean free
path is ~1 Tm. Taking a characteristic length of the Earth as ~10^7 m,
your absorption efficiency should be something like 10^-6 -- so the half
of the bomb's energy has to go into neutrino production needs to be
10^32 J, or a week or so of the Sun's total output. (Note this is a
seriously back-of-the-envelope calculation since I'm basically computing
the problem exactly backwards, but you get the idea.)
Needless to say, I don't think you're going to have much of a planet
left when you're done with this.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Get married, but never to a man who is home all day.
-- George Bernard Shaw
Carey Sublette
The quality factor issue applies only to chronic hazards of radiation
exposure (i.e. genetic damage leading to cancer of heritable defects). For
the effects of acute radiation exposure injury appears to depend only on the
energy deposited, not its form thus grays are appropriate (the mechanisms of
acute radiation syndrome, and chronic effects are different). If you
assuming Q=1 though then the analysis is unaffected.
Carey Sublette
chorned...@hushmail.com
Erik Max Francis wrote:
> Carey Sublette wrote:
>
> > Any small nuclear device (e.g. a few hundrd tons or less) is in essence a
> > radiation weapon (the lethal range for radiation considerably exceeds the
> > lethal range of other weapon effects). This is true even for pure fission
> > devices, where most of the energy is initially in the form of fission
> > fragment kinetic energy and only several percent in the form of gammas or
> > neutrons. The 10-20 ton yield Davy Crockett recoilless rifle warhead
> > depended on this.
>
> That's a very good point. We've been talking generally about nuclear
> weapons vs. antimatter weapons but we haven't really been talking
> specific yields, so (in both cases) there are going to be different
> effects in either weapon and also what the purpose of the weapon is
> (tactical vs. strategic; destructive vs. enhanced radiation; etc.).
>
> With a "hand grenade" type weapon of antimatter I'm not really sure
> which effect wins out. Since there's no fission fragment kinetic
> energy, and I'm not sure what gamma ray flux density you need to get the
> fireball effects, I definitely miscalled that.
>
kinetic energy. Even if pions escape into air, most of them are
absorbed by nitrogen (and oxygen) nuclei in the next few metres, with
spectacular effects.
The Davy Crockett warheads had design yield around 10-20 tons. Which
meant that they generated powerful and deadly prompt radiation. They
weighed 34 kg, though, and therefore needed a recoilless rifle for
launch.
It should be possible to design an assembly to undergo fission with
design yields in kg or g of TNT. Even the latter would be lethal -
think of deliberately replicating Slotin or Daghlian type excursions.
However, the device would need to have almost the mass and volume of
the Davy Crockett warhead - it cannot be significantly smaller, or it
will not fission at all.
What would be the TNT fission yield at which a deliberate fission is as
deadly, at short ranges through air, as a pure high explosive infernal
machine of similar size could be?
Now, you cannot have small fusion explosions at all, not with fission
primary - it is only with fission yields above hundreds of tons TNT
that there is enough heat for thermonuclear fusion.
What about annihilation explosion?
Well, that depends on the antimatter technology. How precisely is
antimatter contained before annihilation?
For a prompt radiation weapon, hand grenades are problematic. 20 tons
fission explosion is lethal at 400 metres - too far to throw a hand
grenade. 20 kg would still have appreciable lethal range - how far? I
suspect that annihilation explosion has even greater penetrating power.
Whereas infernal machines... How far would be the lethal range of
radiation from criticality excursion equivalent to 10 grams of TNT?
>From annihilation generating as much energy? And for comparison, from
an explosion of 10 g conventional explosive - assuming it is
spherically directed, not targeted to propel a bullet to perforate a
specific target?
Now, 10 grams TNT is 10 000 milligrams. Using 42 milliards antimatter
to TNT means 10 grams TNT equivalent can be generated from 240
picograms of antimatter. Further assuming 20 grams per millilitre would
mean 12 femtolitres of antimatter would be enough. Which you can have
inside a sphere 3 micrograms across. Everything now depends on what the
antimatter containment requires. If you can levitate 240 picograms of
antimatter with a vacuum gap of under a micrometer wide, and ordinary
matter vacuum vessel and levitation equipment less than another
micrometer across, and get the trigger for the infernal machine inside
the shell...
With antimatter, you might design rather lethal infernal machines of
volume such that they cannot be detected by naked eye, and once they
explode, the violence of explosion evaporates the device so completely
nothing is left to reverse engineer the infernal machine.
Erik Max Francis
> The quality factor issue applies only to chronic hazards of radiation
> exposure (i.e. genetic damage leading to cancer of heritable defects). For
> the effects of acute radiation exposure injury appears to depend only on the
> energy deposited, not its form thus grays are appropriate (the mechanisms of
> acute radiation syndrome, and chronic effects are different). If you
> assuming Q=1 though then the analysis is unaffected.
You know, that never clicked before, but on looking it up you're
absolutely right. In the past I had seen short-term whole-body dosages
measured in grays (or rads) and had always thought this was a minor
oversight (since the type of radiation was always clear). Thanks, learn
something new every day.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Society attacks early when the individual is helpless.
-- B.F. Skinner