The Pauli Exclusion principle allows Bosons to be in the same energy state; but, because of the high energies required to overcome the coulomb barrier and allow the strong force to bind the nucleons together, your thermal randomness keeps the bosons from being in the same energy state anyway.
Deuterium is only a boson if it is ionized. And without the electrons, the whole mess has to be confined in some way, or it would expand due to the coulomb force between the atoms. With the electrons, the atoms cannot get together, because the electrons are individually exclusionary.
With He-4, a true neutral Boson, you can make a Bose-Einstein condensate, where any action on the condensate must affect the whole condensate. But I don't think that would help, because you'd have to somehow arrange for the energy to make the intermediate Be-8, and there goes your whole condensate idea.
Fusion Cross Sections
Perhaps irrelevant, but I think that the triple alpha process involved with helium burning in stars involves the formation of unstable Be8 which quickly breaks down into two alphas- reversing the process. But at the temperature and pressure in the cores of helium burning stars, the next step of adding a third alpha to the Be8 proceeds at even faster rates so the process does not stall at this step. The helium burning in stars accelerates rapidly once ~ 9,000 eV (~100,000,000 degrees C ) is surpassed. This suggests that the fusion crossection for this reaction has a steep slope around this temperature. I have wondered about the relative He4-He4 fusion rates compared to other fuels like D-D or P-B11.
Of course, the very brief window represented by the Be8 isotope half-life may mean that the process cannot proceed except in conditions where the density is very high (like stellar cores) since fusion rate scales as the square of the density, and these very high densities (much above atmospheric pressure) is needed to speed the rate to overcome the Be8 bottleneck.
Dan Tibbets
Of course, the very brief window represented by the Be8 isotope half-life may mean that the process cannot proceed except in conditions where the density is very high (like stellar cores) since fusion rate scales as the square of the density, and these very high densities (much above atmospheric pressure) is needed to speed the rate to overcome the Be8 bottleneck.
Dan Tibbets
To error is human... and I'm very human.
100 MK is less than 9 keV. Large tokamaks like ITER can easily hit this value.
Ivy Mike was a D-D device.
Stars are not all that powerful per unit volume; even reactors vastly outperform them, and bombs are light years beyond reactors because they can get the pressures as well as the temperatures. The Tsar Bomba - a 27-tonne aircraft-deliverable weapon - was let off at half yield and still had a power output of about 1.4% that of the entire sun. Didn't last long, but still...
Ivy Mike was a D-D device.
Stars are not all that powerful per unit volume; even reactors vastly outperform them, and bombs are light years beyond reactors because they can get the pressures as well as the temperatures. The Tsar Bomba - a 27-tonne aircraft-deliverable weapon - was let off at half yield and still had a power output of about 1.4% that of the entire sun. Didn't last long, but still...
The claim that stars are weak fusion reactors bugs me, because the comparison is between apples and oranges (or in this case heavy isotopes of hydrogen and normal hydrogen). Has anyone ever produced even miniscule levels of hydrogen (H1) fusion on Earth? Not to my knowledge. To compare stellar fusion to terrestrial fusion reactors you would have to pack the stellar cores with deuterium and and possibly tritium.
The fusion crossection for hydrogen is ~ 20 orders of magnitude less than Deuterium fusion with the P-P chain. Throw in a little carbon and increase the temperature some and the stellar fusion rate of hydrogen can reach within a few orders of magnitude of deuterium fusion rates in reactors (CNO cycle).
If the ratios hold, if the Sun was burning deuterium instead of hydrogen.the energy output would be ~ 10 ^20 times greater. The Earth would be vaporized within seconds. The Sun itself would quickly blow up
Even with hydrogen, under certain circumstances, stars can produce massively greater fusion- Type I Supernova (if I am remembering my classification right), and Novas.
Do stars burn deuterium- certainly. Early in their life cycle as they are condensing, the rare deuteriums will find each other and fuse, or perhaps more likely fuse with hydrogen (protons) as they are much more prevalent. I don't know how long this lasts and how much heat it contributes to the Stars output before it settles into the stable hydrogen burning portion of the Hertzsprung- Russel diagram, but it is significant.
In smaller red dwarf and larger brown dwarf stars the deuterium burning may actually provide most of the fusion heat as the hydrogen (P-P) fusion is very weak or virtually absent. I sometimes wonder if Jupiter's core is dense and hot enough to fuse the deuterium present- at relatively slow rates- not because of the temperature so much as the low concentration of deuterium in the hydrogen mix. Jupiter puts out more heat than it recieves from the Sun. I don't know the makeup of this heat- residual heat from gravitational collapse, heavy metal radioactive decay, and possible deuterium fusion.
[EDIT] PS: How good of a P-B11 reactor does the Sun make? The temperature is relatively low- but the density is somewhat higher than a Polywell- by a factor of ~ 10^8. Which means that the fusion rate would be ~ 10^16 times higher for that temperature. Assume that the P-B11 fusion crossection at ~ 1400 eV (~ 15,000,000 degrees K- estimated temperature of the Sun's core) is ~ 1 billionth of the peak crossection rate. Due to density, the net rate would still be ~ 10 million times greater per unit volume. Not exactly a feeble fusion output by terrestrial standards.
Dan Tibbets
The fusion crossection for hydrogen is ~ 20 orders of magnitude less than Deuterium fusion with the P-P chain. Throw in a little carbon and increase the temperature some and the stellar fusion rate of hydrogen can reach within a few orders of magnitude of deuterium fusion rates in reactors (CNO cycle).
If the ratios hold, if the Sun was burning deuterium instead of hydrogen.the energy output would be ~ 10 ^20 times greater. The Earth would be vaporized within seconds. The Sun itself would quickly blow up
Even with hydrogen, under certain circumstances, stars can produce massively greater fusion- Type I Supernova (if I am remembering my classification right), and Novas.
Do stars burn deuterium- certainly. Early in their life cycle as they are condensing, the rare deuteriums will find each other and fuse, or perhaps more likely fuse with hydrogen (protons) as they are much more prevalent. I don't know how long this lasts and how much heat it contributes to the Stars output before it settles into the stable hydrogen burning portion of the Hertzsprung- Russel diagram, but it is significant.
In smaller red dwarf and larger brown dwarf stars the deuterium burning may actually provide most of the fusion heat as the hydrogen (P-P) fusion is very weak or virtually absent. I sometimes wonder if Jupiter's core is dense and hot enough to fuse the deuterium present- at relatively slow rates- not because of the temperature so much as the low concentration of deuterium in the hydrogen mix. Jupiter puts out more heat than it recieves from the Sun. I don't know the makeup of this heat- residual heat from gravitational collapse, heavy metal radioactive decay, and possible deuterium fusion.
[EDIT] PS: How good of a P-B11 reactor does the Sun make? The temperature is relatively low- but the density is somewhat higher than a Polywell- by a factor of ~ 10^8. Which means that the fusion rate would be ~ 10^16 times higher for that temperature. Assume that the P-B11 fusion crossection at ~ 1400 eV (~ 15,000,000 degrees K- estimated temperature of the Sun's core) is ~ 1 billionth of the peak crossection rate. Due to density, the net rate would still be ~ 10 million times greater per unit volume. Not exactly a feeble fusion output by terrestrial standards.
Dan Tibbets
To error is human... and I'm very human.
@D Tibbets: You have a point. I was sleepy when I posted that. Though according to my calculations, bombs still have a substantial edge.
@Colonel_Korg: You do not have a point.
Wikipedia (I know - and it's tagged "citation needed" to boot) says 39 nanoseconds. Recompute and you find it lines up nicely.
@Colonel_Korg: You do not have a point.
Let's not.Let's say that the actual energy production in the Tsar bomb lasted 1 microsecond (1 millionth of a second).
Wikipedia (I know - and it's tagged "citation needed" to boot) says 39 nanoseconds. Recompute and you find it lines up nicely.
Last edited by 93143 on Thu Apr 28, 2011 6:15 pm, edited 4 times in total.
Dan,
True, but otoh here on Earth a compost heap isn't a very mass-efficient source of energy. But then I just like Sun-bashing.
The biggest advantage for the Sun seems to be in transport -- it takes millions of years for energy to get out, which is why it's hot even though energy generation per unit volume is so low -- which in turn is a function of its size and mass.
It looks like you're borrowing that trapped heat for the p-B11 calc, but I wonder if a chunk of B11 would very quickly create far more heat in the area, raising the local reaction rate in a runaway process till all the B11 was fused in something more resembling a bomb than a reactor. (BOOM! Take that, Sun!)
Which makes me wonder, if you started with a massive ball of B11 and hydrogen, under what conditions would it self-ignite, and could it reach equilibrium at some radius? There doesn't normally seem to be a major boron-burning phase in stellar evolution (I did find this and this interesting papers on stellar boron destruction), and I don't quite care enough to try to calculate that, but it's an amusing thought experiment.
True, but otoh here on Earth a compost heap isn't a very mass-efficient source of energy. But then I just like Sun-bashing.
The biggest advantage for the Sun seems to be in transport -- it takes millions of years for energy to get out, which is why it's hot even though energy generation per unit volume is so low -- which in turn is a function of its size and mass.
It looks like you're borrowing that trapped heat for the p-B11 calc, but I wonder if a chunk of B11 would very quickly create far more heat in the area, raising the local reaction rate in a runaway process till all the B11 was fused in something more resembling a bomb than a reactor. (BOOM! Take that, Sun!)
Which makes me wonder, if you started with a massive ball of B11 and hydrogen, under what conditions would it self-ignite, and could it reach equilibrium at some radius? There doesn't normally seem to be a major boron-burning phase in stellar evolution (I did find this and this interesting papers on stellar boron destruction), and I don't quite care enough to try to calculate that, but it's an amusing thought experiment.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...