Talk:Antimatter weapon

From Wikipedia, the free encyclopedia

	Portal This article is within the scope of WikiProject Physics, which collaborates on articles related to physics.
Start	This article has been rated as Start-Class on the assessment scale.
Low	This article is on a subject of Low importance within physics.
Help with this template This article has been rated but has no comments. If appropriate, please review the article and leave comments here to identify the strengths and weaknesses of the article and what work it will need.

Hmmm...the interesting thing about antimatter weapons is that they can be very small. A thimble full of antimatter is enough energy to blow up an entire city. If you could somehow hold the antimatter in a small "bottle", you could have a pocket sized nuke. Samboy 10:23, 30 Jan 2005 (UTC)

Perhaps, but producing the stuff seems to be prohibitively difficult, much less containing it in a small "bottle." I also don't think it would change the strategic/political situation of the world any more than nuclear weapons do (you can make fission weapon small enough to fit into a backpack). --Fastfission 22:41, 2 Feb 2005 (UTC)

A thimble full of antimatter? Hah! It costs billions of dollars just to produce a few atoms of it. --NoPetrol 02:00, 24 Feb 2005 (UTC)

Yes, it is expensive. For now. But so were computers, there is always a higher inital cost of development, and as the technology matures costs drop. How much? Who knows... --Zotel 11:32 EST July 24, 2005

Yes, but we are many, many orders of magnitude away from a practical "pure" antimatter device right now. There seems to be fundamental physical limits to how efficiently antimatter can be produced (giving an efficiency of maybe 0.1% in production), and there's not much available naturally (you might collect it with satellites from stuff captured in the Earth's magnetic field, but there's only miniscule quantities and it would be a massive undertaking). And energy costs, while tending to generally drop in real terms over time, drop at a fairly slow rate. So the energy required to make an antimatter arsenal seem unlikely to become practical, particularly when you can cook a nuclear weapon up in your back shed (well, not quite, but South Africa did it in secret for a few tens of millions of dollars), and as Fastfission pointed out the Americans made backpack-size nuclear weapons back in the 1950's.

--Robert Merkel

Well, I read the article and found couple grams of antimatter costs USD$25 mil.

P.S. Angels And Demons (By Dan Brown) has a part were antimatter was anilated.(Sorry for my spelling)

--Tehrussian5thgrader 15:28, 5 October 2006 (UTC)

Contents 1 Odd line 2 Mass needed? 3 Disputed 4 Prices? 5 buckyballs 6 Containment methods 7 Nuclear synthesis of Positrons 8 Citation needed 9 One gram of antimatter is NOT equal to 43 kilotons of TNT. 10 efficiency of nuclear fusion

[edit] Odd line

The article suggests that the production of "antimatter bombs" would be a foolish economic endeavor because the energy that would be required to produce enough antimatter for the bomb would be equal to the energy that the bomb would release.

This doesn't make any sense to me. This seems to imply that trying to use antimatter as fuel (as some sort of reactor) would be economically foolish (as you wouldn't net anything). But for a bomb, I don't think net energy gain is the point—it's about making something blow up. So even if it took a fission bomb's worth of energy to construct a fission bomb, it would still be worth it if it could instantly be delivered on another city, for example. I'm not saying I think that an antimatter bomb is practical but I'm not sure that this sentence really proves that point... perhaps others have thoughts? --Fastfission 22:55, 2 Feb 2005 (UTC)

Yeah - That's true - explosives are powerful not so much because they have a lot of stored energy, but because it's released quite rapidly, so there not being a net energy gain wouldn't be a problem. The problem (for would be antimatter weapons developers) is that they either have to produce the antimatter in advance and find a way to transport it without it reacting with any matter along the way, or they have to find a way to rapidly produce it right where they want it to detonate. And in the latter case, the lack of a net-energy-gain would be a problem. --Blackcats 06:13, 15 Apr 2005 (UTC)

[edit] Mass needed?

The Atomic bomb page states that just a gram of anti-matter would produce many times the energy of a standard nuclear weapon when destroyed. This page says it would take several kilograms -- who's right? One or the other needs to be corrected (I don't know which)

Actually, it just states that one gram of anti-matter would produce more energy than a 20KT nuclear device. You need half a gram to match Hiroshima bomb, 10-30 grams to match a typical modern nuclear warhead, slightly over 1 kg to match Tsar-Bomba.

[edit] Disputed

In the event of an antimatter detonation in the open atmosphere, most of the energy will ultimately be carried away by the neutrinos, and the remainder by 10-100 MeV gamma rays. The neutrinos would pass through the earth without being attenuated, while gamma rays are relatively weakly absorbed by matter: they lose roughly half of their energy per 500-1000 m of air, compared to only 20 cm of concrete. The explosion would not cause much physical damage because its energy would be evenly dispersed over large area, although the gamma rays may harm people standing nearby.

This seems to be quite incorrect - compare with Nuclear explosion, which states (correctly, as far as I am aware), that a large part of the energy of a regular nuclear explosion is released as gamma radiation, and it is quickly absorbed by the nearby air, superheating it and creating the fireball of the explosion. The same would most likely happen in an antimatter explosion, making it much the same as any nuclear explosion (minus the 60% lost to neutrinos).

--130.232.31.109 01:11, 10 July 2005 (UTC)

According to [1], kinetic energy of fission fragments accounts for 90% (170 out of each 190 MeV) of energy generated in a nuclear explosion, not counting neutrinos. Those gamma rays that do get produced have much lower energies and thus lower attenuation lengths. --Itinerant1 19:03, 10 July 2005 (UTC)

Well, that is no doubt a good point. However, you still can't dismiss the gamma energy like the article does - just the secondary visible-wavelength light from a normal nuclear explosion does a lot of burn damage even before the shockwave hits. Gamma rays might not get attentuated much by air, but they will get absorbed when they hit the ground, or buildings. And the energy radiated and absorbed will be huge, even at longer distances.

--130.232.31.109 17:52, 11 July 2005 (UTC)

Absolutely, a megaton is a megaton, even in gamma spectrum. The point is that, if there is not enough shielding material in the vicinity of the explosion, you won't have a usual shockwave. You'll have a huge ball of hot air ( and it won't be particularly hot: 1 kiloton of TNT can heat up 1 km³ of air by only 5 Kelvin ). Imagine all of the energy that goes into visible light from a conventional nuclear explosion, instantaneously absorbed within 1 km of epicenter. Then you'll have some interesting dynamics of this fireball - it will probably start expanding and then rising off the ground (due to Archimedean force), some fraction of energy will be reemitted as visible light; a mushroom cloud is not unlikely. Those gamma rays that do reach the ground will still be potent enough to kill exposed people within a few km. --Itinerant1 18:51, 11 July 2005 (UTC)

If the gamma rays produced by an antimatter explosion dissipating into the atmosphere and acting to heat a large area is a problem, it seems to me that it could be corrected by simply wrapping the weapon in lead (or some other material that shields well against gamma rays) and using that as a "sink" for the explosion to dump energy into. I imagine the secondary blast would be quite spectacular. The article is far too dismissive of the possibility of increasing desired blast effects, in any event.128.153.203.164 18:37, 29 January 2006 (UTC)

The problem here is that you seem to be viewing this in light of a rather small device. For larger devices, where the fireball of modern nukes would already be larger than 500-1000 meters the long distance asorbation ranges would likely be much less of a concern. Not only that, but the propogation of a powerful blast waves is not only possible, but likely. A 10 Megaton device, for instance, has the energy to heat a 1 km^3 volume by nearly 64,000 degrees Kelvin! (ignoring the change of specific heat...) Needless to say, that will create a rather potent shockwave. Further, given that the creation of an antimatter device is highly unlikely if the power can be upsurped by any other nuclear device (Given the other disadvantages of Antimatter), I have difficult seeing any real Antimatter 'bomb' reasonably being less than 10 Megatons. For such a bomb the propogation length of the gamma-rays is much less of a concern, and any material within that range will pretty much vaporize.

It is also worth noting that air become significantly more opaque to most wavelengths of light as the temperature raises -- especially if it is rendered into plasma. However, I don't know how this would react to such high-end gamma rays which tend to defy most standard rules. --Xylix 00:35, 25 February 2006 (UTC)

Seeing as the bomb would not simply be a ball of antimatter dropped into the air, the whole gamma ray thing is somewhat irrelevant. The most likely design (in my mind) for a antimatter bomb would be similar to an implosion type fission bomb, only rather than perfectly focused explosive lenses, a slightly less sophisticated system would be used to deliberately introduce turbulance between the antimatter and its surrounding matter tamper. The whole idea would be to mix the two as quickly and efficiently as possible. The tamper, the casing, and basically every other part of the bomb would be metal, and would interact with gamma rays and pions just fine, especially after it turned into a giant cloud of plasma--70.70.143.237 08:53, 1 June 2006 (UTC)

[edit] Prices?

The pricing of $25 billion per gramme conflicts with the $25 million per gramme on the Antimatter page, and the $2.5 million per gramme lower limit in this article. Clarification please! Was the billion figure meant to be for a Kg?--81.178.104.55 00:07, 20 October 2005 (UTC)

Originally Antimatter page quoted $25 billion per gram. The number in this article was copied from there verbatim. On July 13, 2005, someone replaced $25 billion with $25 million. I don't know which of the two figures is correct. --Itinerant1 03:01, 20 October 2005 (UTC)

All of the above estimations are incorrect. The actual cost TODAY of antiprotons which are needed to produce neutral antimatter is several million trillion US dollars per gram, as estimated below, 1 gram of antimatter contains 6x10^26 antiprotons + positrons. The Antiproton Decelerator facility of CERN can produce 4x10^12 antiprotons per year, at an operation cost of around $20 million including the 25-GeV proton synchrotron. Cost per gram is then around 1x10^19 dollars and it would take until the end of the earth to produce a gram. --User:137.138.16.5 17:48, 20 October 2005 (UTC)

CERN facility is research-oriented, not production-oriented, and may not be the cheapest possible antimatter factory. This article estimates production cost of antiprotons at CERN of $1.6*10^14 per gram, and it goes on to describe improvements that can lower it to $6.4*10^12 per gram. Robert L. Forward, apparently, predicted sometime during the '80's that antimatter could be produced at $10 billion per gram, but I can't find any good links. --Itinerant1 17:39, 21 October 2005 (UTC)

I think that all physicists would regard the contents of the paper you mention (i.e.producing antiprotons by simply putting two parallel plates together and using the Casimir effect) to be bad pseudo-science. The "production cost of antiprotons at CERN of $1.6*10^14 per gram" is a nonsensical value calculated by assuming that the accelerators can transfer 5% of the electrical power they consume into the beam used to produce the antiprotons, and that the other operational costs besides the electric bill are zero. The actual cost of running a multi-GeV accelerator is between several tens to several hundred million dollars per year. Also the antiproton production rates discussed in the introduction of the paper is far too optimistic and analyzed by non-experts who ignore the limits of accelerator technology. 10^7-10^8 antiprotons per proton bunch arriving every few seconds has been the limit at CERN and FNAL for many years, and even at the future facility at GSI. This is NOT because these facilities are "non-dedicated and research-oriented" etc. as these papers seem to claim. It is because of space-charge limitations in the accelerator which collect the antiprotons, efficiency of the stoachastic and electron cooling needed to cool down and decelerate the antiprotons, heat loading on the target used to produce the antiprotons. Now using various techniques (build an entirely new superconducting linear accelerator, liquid metal production targets, cascade of antiproton storage rings) using many billions of dollars it may be possible to increase that limit by maybe factor of 100, but there is NO way it would ever get NEARLY efficient enough to build weapons and rocketships and do these fantastic things. --User:137.138.16.5 17:37, 22 October 2005 (UTC)

Ok, I didn't read the article thoroughly, it appeared to have come from a reliable source. ( BTW, search in Compendex database yields 4 publications by Michael R. LaPointe, he does not seem to be involved with particle accelerators in any way ).

According to [2], running cost of CERN's Antiproton Decelerator is 600,000 Swiss francs ($350,000) a year. Presumably this number also includes electricity costs, so the maintenance cost is even lower. The number from NASA's article is for production of antiprotons only. 5% accelerator efficiency does seem a tad bit high ( although I'm not an expert in accelerator design ). Sure, it may be a lot more difficult and expensive to slow down and capture antiprotons than to produce them, no argument here. --Itinerant1 20:31, 22 October 2005 (UTC)

Three things are needed to produce antiprotons; a proton synchrotron PS (which accelerates protons to around 30-50 GeV), a target wherein the antiprotons are produced, and an antiproton decelerator to decelerate the antiprotons back to zero energy and trap them in a magnetic bottle. Now at CERN the PS (which costs several $10's million to run) will be primarily used to inject protons into the Large Hadron Collider (LHC) and only a small part of the beam will be used for antiprotons; that's why the cost of the antiproton programme is so low. But you also have to remember that it takes hundreds of people to run these huge accelerators and experiments and maintain them -- multiply that by the average salery including social security, taxes and insurance costs. If you built a dedicated antiproton factory from scratch my guess is that the total operational costs of the facility including R&D would easily be >$100 million per year --[[User:137.138.16.5] 13:59, 23 October 2005 (UTC).

[edit] buckyballs

The article stated: "Another, more hypothetical method is the storage of antimatter inside a buckyball. Because of the repulsion of all the carbon atoms, the antimatter would never combine with its opposite and no energy release will occur." This is nonsense. A neutral atom of antimatter would have antielectrons, which would annihilate with the electrons of the buckyball.--Bcrowell 06:16, 30 October 2005 (UTC)

You could put antiprotons inside a buckyball. —Keenan Pepper 03:37, 20 November 2005 (UTC)

Wait... I thought you can only put positrons inside a buckyball, correct me if I have mistaken the theory, but since the protons will be in the middle forming the ball, and the electrons on the outside forming the electron cloud. So the inside of the bucky ball would be positively charged. The antiprotons will very likely to be touching the protons inside because they would be attracted by the positively charged protons. MythSearcher 03:53, 9 December 2005 (UTC)

The electron cloud goes all around the nuclei, including the inside of the sphere. The nuclei are not naked on the inside. —Keenan Pepper 05:27, 9 December 2005 (UTC)

My intuition suggests to me that the antiproton, with its large mass and consequently very small Bohr radius, would rapidly enter the valence shell of one of the carbon atoms and from there proceed to drop down past the core electrons until it reached the nucleus. I couldn't begin to negotiate the math involved, however. --209.250.133.53 00:02, 18 February 2007 (UTC)

It was in John Ringo's "Legacy of the Alldenata," although I don't know if he came up with it. But if you put only antiprotons(-) in it, the electrons(-) might repell them --Dr.Moreau 23:38, 20 February 2006 (UTC)

I agree with Keenan, the antiprotons (which have a negative charge) would be suspended within the sphere of electrons on the inside of the carbon sphere. However, I can't see how you'd get them in there. Glooper 07:45, 8 July 2007 (UTC)

Do not complete the buckyball before you put the antiproton in there? MythSearcher^talk 16:06, 8 July 2007 (UTC)

[edit] Containment methods

A third hypothetical method is using cellular membranes. Since the membranes, like buckyballs, have one type of charges on one side and another on the outer side, it is possible to store a certain amount of anti-matter in them. However, both of the methods of storage using charged walls are not efficient because the storage and the extraction of the anti-matter will take months, if not years to re-obtain the anti-matter stored in them. There is, however, a fourth way proposed but not theorized. By trapping positrons inside highly compressed protons within a high pressure container. It will be able to store virtually unlimited amount of positron as long as a big (and strong) enough container is made. Although it will also be very difficult to separate the proton and positron afterwards. It is possible to use as an anti-matter weapon since all you would need to do is fire the container at your enemy and release the pressure. This proposed method has been criticized because of the virtual impossibility of containing the positrons inside the non-solid protons. Both the positrons and the protons would flow in the container and eventually some positrons would come in contact with the wall of the container, according to the critics.

The cell membrane thing sounds resonable enough, but it needs to be clarified. What, exactly, would take "months, if not years"?

The pressure thing, however, makes no sense to me. How can a positron be "inside" a proton? They have like charges, so they repel each other. I think instead of a "virtual impossibility" it's a logical impossibility and should not be mentioned. —Keenan Pepper 23:29, 8 December 2005 (UTC)

Well, the positrons are not trapped inside "a" proton, they are trapped inside "many" protons. It is kinda like jelly filled donuts, you have the positrons as jelly and protons like the crust. (with a harden solid container outside to maintain the pressure and the shape)

They do carry the same charge, that makes the positrons less likely to get through the barrier of protons, if enough pressure is applied(given the container itself can hold the force inside.) the protons will be able to separate the positrons from touching the container and thus storing it.

However, like it was said in the paragraph, the protons and positrons are not in a solid state and reasonable assumtion would be they are in a fluid state and eventually the positrons and protons would shift in position slowly(and randomly) and positrons would come into contact with the container.

The cell membrane method takes time because it would take time to put the anti-matter into the cell or buckyballs, and it would take time to take them out again, and process like that will take really long time with the current technology.MythSearcher 03:46, 9 December 2005 (UTC)

Do you mean the positrons are trapped inside a hollow nucleus of protons and neutrons? That might make some sense... I think... ?

Anyway, do you have a reference for this, or did you think it up yourself? —Keenan Pepper 05:37, 9 December 2005 (UTC)

Not exactly, I will try to make a picture for this...

I do not have a reference for this since I only remember I read it somewhere. I do not think it will work either (due to the fact I mentioned above in the paragraph and because the container might be penetrated by the protons if the electrons in the container wall did not start to flow towards the protons.)

However, since I have seen it, I might as well put it here. It is because even in modern physics, most of the time we'll assume we have rigid bodies when we don't. And there might be a container that can hold the protons inside and dense enough to prevent the positrons from getting through the protons.

BTW, the method you just proposed actually make some sense, too. If it is possible, I will try to ask a Physics professor I know and see what is his view point on it. Maybe I have mistaken what I have seen or memorized it wrong. You method seems to make more sense to me. MythSearcher 10:16, 9 December 2005 (UTC)

The ideal "containment" of "anti-matter" would be to hold the putative charge as ordinary matter and convert it into anti-matter in a short period of time with a mechanism within the bomb itself at the time of use. —The preceding unsigned comment was added by 74.130.179.209 (talk • contribs) .

How do you plan to achieve this violation of baryon number conservation? —Keenan Pepper 17:21, 9 September 2006 (UTC)

[edit] Nuclear synthesis of Positrons

Nobody seems to have mentioned that positrons can be made in another way. Some nuclei undergo positron emmission, which is like beta decay but the nucleus emmits a positron instead of an electron. Some nuclei that do it are carbon 11, nitrogen 13, and oxygen 15. A target could be bombared with some king of particle beam, and a postiron emmitter would be formed. This positron emmiter will emit positrons, which are captured in a trap and are somehow stored. There will be some energy involved in the nuclear synthesis, but there is no need to supply the whole mass-energy of an electron-postron pair. Polonium 20:51, 27 April 2006 (UTC)

[edit] Citation needed

The various numerical figures mentioned in this article should have inline citations to show that the numbers are authentic. Shawnc 22:19, 2 June 2006 (UTC)

[edit] One gram of antimatter is NOT equal to 43 kilotons of TNT.

According to one of the sources cited on this very article, http://sfgate.com/cgi-bin/article.cgi?file=/c/a/2004/10/04/MNGM393GPK1.DTL

It states, "One millionth of a gram of positrons contain as much energy as 37.8 kilograms". That directly contradicts a statement made in the article. From my quick calculations, 1 gram = 37.8 MEGAtons. I'm correcting this statement, and putting a cleanup notice at the top. This article is badly written. 69.69.73.94 03:44, 23 January 2007 (UTC)

No, multiplying 37.8 kilograms by one million gives 37.8 million kilograms. One ton is equal to 1000 kilograms. So 37.8 million kilograms is 37.8 thousand tons: that is, 37.8kT. Note that you did not correct the statement: you removed it. I have reverted that change. Could you be specific about what you think needs cleanup? --Dashpool 14:19, 25 January 2007 (UTC)

The 43 kiloton figure comes directly from high school physics: e=mc². One gram of antimatter is about two grams of annihilated material, and from e=mc², (.002 kg)(c²) = 1.8 x 10¹⁴ joules. If the TNT equivalent article is correct, 1 kiloton of TNT is 4.184 x 10¹² joules. That means 43 kilotons of TNT is approximately 1.8 x 10¹⁴ joules, which is the energy released by the destruction of a gram of antimatter and a gram of matter. --Benevolensaurus 13:30, 31 March 2007 (UTC)

[edit] efficiency of nuclear fusion

it is stated in this article the efficiency of nuclear fusion would be ~7%, I think this should be more like 7 promile.

Excuse my ignorance, but what is a promile? I cannot seem to find a definition for it. Anywhere. Glooper 07:42, 8 July 2007 (UTC)

A promille is a thousandth in german, I have to guess he means it is 7 thousandths, or 0.7% efficient —Preceding unsigned comment added by 70.70.136.240 (talk) 06:22, 18 October 2007 (UTC)