High energy density fuels

[Fingolfin_Noldor]

How about spinning black holes?

In relativistic terms, energy is calculated as thus: E = Sqrt( (m c^2)^2 + (p c)^2) where p is moment, m is mass, c is speed of light.

Spinning black holes are a special case, because (assuming the singularity model is correct) nothing is actually moving. Angular momentum in normal matter is just the net effect of conventional linear momentum in the object's constituent particles. I only have a very limited understanding of this, but AFAIK spin on a black hole is modeled the same way as particle spin, i.e. as an intrinsic property of a zero-dimensional object.

Even with Quantum spins, while there are degenerate spin levels, there are also different angular momentum levels. The number of spin levels is defined by the angular momentum level. The energies between angular momentum levels are different.

In the famous Rubidium-87 D-2 line, the two F=1 and F=2 ground states have a separation of about 6.8GHz.

[Darth Wong]

How about spinning black holes? You can store a huge amount of energy (up to G(M^2)/c) in the angular momentum of a rotating black hole. Does this energy have mass equivalence?

Inside a black hole, the distinctions between mass and energy cease to have meaning anyway. It's all just mass/energy.

[Darth Wong]

If we accelerate a mass m of normal iron filings or something to 0.99c then its going to have kinetic energy equal to gamma*mc^2 = 7mc^2.

So if we stuff that m of iron into a wanktacular magnetic field of obscene power and keep our iron going in a circle then we can extract energy from it by... errr... putting something near it and letting the resulting friction massively heat it and heat water and drive some turbines or something.

I think you could just convert the kinetic energy into electricity by making the slug conductive and slowing it down magnetically; PERMANENT mentions such a scheme for recovering some energy from a lunar mass driver by slowing down the launch buckets and "recycling" their KE back into electricity (with some inefficiency, of course). This would be similar, but using vastly more energy. I believe this also the principle that regenerative braking works on.

It doesn't matter; the kinetic energy itself has mass equivalence. Or, to put it another way, at high relativistic speeds, the faster it goes, the heavier it is. You still can't get past E=mc².

[Cykeisme]

Actually this is something I was wondering about, with regard to matter-antimatter annihilation. If I have anti-deuterium and fuse it before allowing the resulting anti-helium to annihilate with matter, or if I have anti-uranium and fission it before annihilating the fission products, is the total energy output identical to simply annihilating the antimatter without first having a nuclear reaction? I would think so, since those reactions release energy by allowing the nuclei to fall into a more stable state, but annihilation of heavy atoms should effectively lose energy (relative to annihilation of free particles) proportional to the atom's binding energy in the process of tearing it apart.

When you fuse deuterium hydrogen into helium, the post-fusion end result has less mass than the initial pre-fusion nuclei.

So assuming you perform fusion with your anti-deuterium, I think the anti-helium will have less anti-mass than the anti-deuterium you began with.. therefore you'll have less to annihilate.

Fusing your anti-matter simply released a bit of the energy, before you released all the rest in matter-antimatter annihilation; there's no change to the total amount released.

Actually, I'm talking out my ass. Can someone tell me if that's right?

[Steel]

Actually this is something I was wondering about, with regard to matter-antimatter annihilation. If I have anti-deuterium and fuse it before allowing the resulting anti-helium to annihilate with matter, or if I have anti-uranium and fission it before annihilating the fission products, is the total energy output identical to simply annihilating the antimatter without first having a nuclear reaction? I would think so, since those reactions release energy by allowing the nuclei to fall into a more stable state, but annihilation of heavy atoms should effectively lose energy (relative to annihilation of free particles) proportional to the atom's binding energy in the process of tearing it apart.

When you fuse deuterium hydrogen into helium, the post-fusion end result has less mass than the initial pre-fusion nuclei.

So assuming you perform fusion with your anti-deuterium, I think the anti-helium will have less anti-mass than the anti-deuterium you began with.. therefore you'll have less to annihilate.

Fusing your anti-matter simply released a bit of the energy, before you released all the rest in matter-antimatter annihilation; there's no change to the total amount released.

Actually, I'm talking out my ass. Can someone tell me if that's right?

That is indeed correct.

You can look up the exact masses of each element in 'amu' s (atomic mass units or just 'u's i think is the modernised measure, defined as 1/12 the mass of carbon 12) then if you take the difference in amu*c^2 between the reactants and the products you'll get the binding energies. (Minus of course any chemical energy, but thats typically 5 oom smaller so doesnt matter really.)

[Ariphaos]

It is probably worth noting that, since an antimatter reaction gives off a minimum of 56% (or so) of its mass off as neutrinos, it is certainly theoretically possible to be more -efficient- than antimatter.

The only way you can get more energy than mass from something is if that something is not the true source - a dimensional tap, harvesting device, whatever.

[Fingolfin_Noldor]

It will be quite interesting to find a way to tap the neutrino output, but currently, the physics says it's either impossible or very hard.