Can Black Holes Be Used as a Source of Antimatter?

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Discussion Overview

The discussion revolves around the energy/mass relationship in the context of nuclear weapons, particularly focusing on the materials used in fission and fusion reactions, and the potential future of antimatter as a weapon or energy source. Participants explore various aspects of nuclear physics, including binding energy, efficiency of mass conversion in explosions, and the implications of Einstein's equation, E=mc².

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants discuss the primary materials used in fission weapons, noting that U-235 and Pu-239 are commonly used due to their stability and availability.
  • Others mention that heavier elements are harder to produce and have shorter half-lives, which may limit their use in nuclear weapons.
  • There is a claim that most of the mass of plutonium or uranium is converted to energy in a nuclear explosion, contrasting with the low efficiency of energy conversion in chemical reactions like burning petroleum.
  • Some participants argue that while fission bombs use heavy elements, fusion bombs utilize lighter elements, which yield higher energy outputs due to the mass difference before and after the reaction.
  • A participant suggests that antimatter could be the next weapon of choice, proposing that even a small amount could cause massive destruction, but acknowledges the current challenges in its production and containment.
  • There is a debate about the role of electric force versus nuclear force in the functioning of nuclear bombs, with one participant asserting that the destructive force comes from electric energy rather than nuclear force.
  • Some participants challenge the accuracy of claims regarding the materials used in modern nuclear warheads, specifically regarding the use of U-238 versus Pu-239.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the materials used in nuclear weapons and the efficiency of mass conversion in explosions. The discussion remains unresolved, with differing opinions on the role of various forces in nuclear reactions and the future of antimatter as a weapon.

Contextual Notes

Participants reference various technical aspects of nuclear physics, including binding energy and the efficiency of mass conversion, but some claims are challenged or corrected without reaching a consensus on the accuracy of these statements.

denni89627
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Just curious about the energy/mass relationship and how it factors in when making nuclear weapons. (or hopefully a more peaceful use for such a reaction in the future.) I know manhattan era nukes used plutonium and polonium/ beryllium for fission reaction. Einstein's famous equation would have me believe the heaviest elements would yield much greater energy, fission or fusion. Do current nukes use the heaviest elements? Maybe they are too rare or too hard to isotope? Just curious. maybe its classified anyway.
 
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The two principle materials for fission weapons are U235 and Pu239. They both fission and can be made in quantity. Moreover, they are relatively stable in a stored environment. Half life of Pu239 is several hundred thousand years, while U235 is in millions. Heavier elements are much harder to produce in quantity and have much shorter half-lives.

The other materials you mentioned were used to generate neutrons to make sure the chain reaction started as soon as the bomb was triggered. They were not fission materials.
 
denni89627 said:
Just curious about the energy/mass relationship and how it factors in when making nuclear weapons. (or hopefully a more peaceful use for such a reaction in the future.) I know manhattan era nukes used plutonium and polonium/ beryllium for fission reaction. Einstein's famous equation would have me believe the heaviest elements would yield much greater energy, fission or fusion. Do current nukes use the heaviest elements? Maybe they are too rare or too hard to isotope? Just curious. maybe its classified anyway.

To help understand binding energy, fission and fusion, see -
Nuclear Binding Energy

Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together. This binding energy can be calculated from the Einstein relationship -
http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html

Current nuclear weapons are based on Pu-239, whereby spheres of Pu-239 are imploded. The technology is certainly classified. I believe all U-235 weapons have been deactivated and the U-235 recycled into commercial fuel stream.

Small weapons are primarily Pu-239, maybe with a DT booster. Thermonuclear weapons use a Pu-239 'trigger' to 'ignite' (initiate) the fusion reaction. That technology is also classified.
 
E=mc^2 and the nuclear bomb...the energy/mass relationship and how it factors in when making nuclear weapons.


The simplest factor is the amount of total mass that is converted into energy:

[tex]E = mc^2[/tex]

1 kiloton of TNT = 4.184 * 10^12 joules (j)
1 Megaton of TNT = 4.184 * 10^15 joules (j)

[tex]m(1 kt) = \frac{E_1}{c^2} = \frac{4.184 \cdot 10^{12} \; \text{j}}{c^2} = 4.655 \cdot 10^{-5} \; \text{kg}[/tex]

[tex]m(1 kt) = 4.655 \cdot 10^{-5} \; \text{kg}[/tex]

[tex]m(1 Mt) = \frac{E_1}{c^2} = \frac{4.184 \cdot 10^{15} \; \text{j}}{c^2} = 4.655 \cdot 10^{-2} \; \text{kg}[/tex]

[tex]m(1 Mt) = 4.655 \cdot 10^{-2} \; \text{kg}[/tex]
[/Color]
Reference:
http://en.wikipedia.org/wiki/Megaton
 
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Orion1 said:
The simplest factor is the amount of total mass that is converted into energy:

[tex]E = mc^2[/tex]

1 kiloton of TNT = 4.184 * 10^12 joules (j)
1 Megaton of TNT = 4.184 * 10^15 joules (j)

[tex]m(1 kt) = \frac{E_1}{c^2} = \frac{4.184 \cdot 10^{12} \; \text{j}}{c^2} = 4.655 \cdot 10^{-5} \; \text{kg}[/tex]

[tex]m(1 kt) = 4.655 \cdot 10^{-5} \; \text{kg}[/tex]

[tex]m(1 Mt) = \frac{E_1}{c^2} = \frac{4.184 \cdot 10^{15} \; \text{j}}{c^2} = 4.655 \cdot 10^{-2} \; \text{kg}[/tex]

[tex]m(1 Mt) = 4.655 \cdot 10^{-2} \; \text{kg}[/tex]
[/Color]
Reference:
http://en.wikipedia.org/wiki/Megaton

And a hundredth of a kilogram is ten grams, so the mass equivalent for a megaton explosion is 46.55 grams.
 
I believe that when something like petroleum is burned that only something like 1% is converted to energy, the rest just changing state to carbon monoxide and water. I assume that in a nuclear explosion that most of the mass of the plutonium or uranium is converted to energy. Thats what a teacher told me when I asked a similar question a while ago. Only a small amount of mass contains a huge amount of energy as the equation states so if the nuclear weapon is very efficient in converting all the mass to energy then you will have a big explosion.
 
there was actually a really good special on the National geographic channel about explosives a couple days ago. In the 50's scientists used a uranium or plutonium fission reaction as a trigger for hydrogen fusion. (as stated by astronuc also). The largest one ever detonated was "Tsar Bomba" by the russians. It yielded 38,000 times the energy of the atomic bomb dropped on Nagasaki. The video footage is unreal. I guess it's a bit rediculous to try and create something with a larger yield than that. But... scientists are doing just that! Experts agree that antimatter will be the next weapon of choice. It's too costly and difficult to make and contain currently, but may not be in the future. In theory a mass of antimatter equivalent to the mass of a paper clip would be enough to destroy an entire city. maybe they should work on an antimatter power plant instead. who's with me?
 
denni89627 said:
Einstein's famous equation would have me believe the heaviest elements would yield much greater energy, fission or fusion.
This is not the case. Fission bombs do use very heavy elements (and yet not the heaviest). However fusion bombs use the lightest ones and have higher yields. The issue is not the mass of the elements used. More important is the difference in mass between the nuclei before and after the reaction and the efficiency with which one reaction causes other reactions.

By the way, Einstein's famous equation is not limited to certain types of mass or energy. It applies universally to all energy flows such as conventional explosions and even the flap of a butterfly's wings.
 
dansydney said:
I believe that when something like petroleum is burned that only something like 1% is converted to energy, the rest just changing state to carbon monoxide and water. I assume that in a nuclear explosion that most of the mass of the plutonium or uranium is converted to energy.
The critical mass of plutonium is roughly 10 kg. As can be seen in Orion1's post #4, the amount of mass converted in a 1 kiloton nuclear explosion is on the order of one one-millionth of this, much less than 1%. The mass loss in the burning of petroleum or other chemical reactions is much smaller yet. It is so small that until Einstein published his famous equation, no one had noticed it. Had it been 1% as you suggest, the effect would have been easily measured and the famous equation would be somebody else's.
 
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  • #10
actualy,nuclear bombs don't use the nuclar force...they rather fight it...all that destructive force comes from the electric energy! you see,protons are all positive,so the electric force makes them get away from each other! luckely,there is the nuclear force that keeps them togeter,but the nuclear force has a very very short range...if you take a very hevavy element and bombard it with a neutron,the protons get out of the range of the enslaving nuclear force,and the electric force can do it's thing...to get an ideea, electric force is 1 bilion bilion bilion bilion times stronger then gravity!
so in conclusion,I don't think E=mc^2 really helps here...but you can aply it to an antimatter bomb,to see what it does(don't forget to add the mass of the normal matter,too! )
 
  • #11
Modern fission based nuclear warheads use U-238, for your information, not Pu-239.
U-238 is much more unstable in a reactive environment because of its number of electron shells.
 
  • #12
tormund said:
Modern fission based nuclear warheads use U-238, for your information, not Pu-239.

I find this very unlikely, as a) U-238 is difficult to fission, and b) it would allow the use on unenriched uranium, so why all the fuss about enrichment.
 
  • #13
tormund said:
Modern fission based nuclear warheads use U-238, for your information, not Pu-239.
U-238 is much more unstable in a reactive environment because of its number of electron shells.

That's totally erroneous. It is not possible to make a critical mass with U-238 and moreover the electrons have nothing to do with this.

What is correct, is that in a fission-fusion-fission weapon, the second "fission" part can be done with U-238 (which also serves as a tamper), but it is not a chain reaction. It just uses the fast neutrons created in the fusion stage to induce fission to boost the energetic yield (the neutrons themselves have 14 MeV, and if they induce a fission, they liberate 200 MeV, so there's in principle a factor of about 10 to be won with the second fission stage over the pure fusion although this is less of course as not every fusion-liberated neutron will give you a fission reaction).

The Tsar bomb test actually left out the second fission stage, which is why it "only" had 50 megatons, and not the 100 megatons or more.
 
  • #14
jimmysnyder said:
The critical mass of plutonium is roughly 10 kg. As can be seen in Orion1's post #4, the amount of mass converted in a 1 kiloton nuclear explosion is on the order of one one-millionth of this, much less than 1%. The mass loss in the burning of petroleum or other chemical reactions is much smaller yet. It is so small that until Einstein published his famous equation, no one had noticed it. Had it been 1% as you suggest, the effect would have been easily measured and the famous equation would be somebody else's.

Well, with a 10 kg plutonium core, you can have normally a few tens of kilotons of yield.
A 1 kiloton weapon is or a specially designed low-yield weapon, or a fizzle.
 
  • #15
...maybe they should work on an antimatter power plant instead. who's with me?

If we could compress some big object so much that it becomes a small black hole, we would have a source of anti-matter. As long as the Hawking temperature is higher than about
10^10 K, the black hole will emit positrons.
 

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