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Quantum vacuum plasma thruster

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Re: Quantum vacuum plasma thruster

#21

Post by zackmorris »

The reason this is important (if it works) is that currently there is no known propellantless rocket that produces more thrust than light pressure. Light pressure is incredibly weak:

https://en.wikipedia.org/wiki/Light#Light_pressure

The formula is F = P/c so 1 watt of power (say, from a laser) only results in a force of 3.33 nano newtons or 0.75 nano pounds. In other words 1 kW (a little over 1 horsepower) results in about 1 micro newton or 1 micro pound of force. A craft that used light pressure for propulsion would use up most of its mass (even with fusion or antimatter fuel) to get to some fraction of the speed of light.

So if this effect is real and has a performance higher than 3.33 nano newtons per watt, to me it's one of the greatest breakthroughs in the history of propulsion. It would have all kinds of ramifications for science because even though it doesn't break any laws of physics, it means that we can no longer assume that motion only comes from reactions with other matter or light. It would basically open the door to modifying the mass and energy of empty space. So I will remain highly skeptical until someone reproduces the claimed results, but I would love it if they did!

Re: Quantum vacuum plasma thruster

#23

Post by wcoenen »

valarauca1 wrote: post #10
baq wrote: post #7

> equivalent specific impulse of ~1x10^12 seconds this is so much more than pretty much anything else it's hard to comprehend why there hasn't been a billion dollars put into it just to rule out the possibility that it could work.

Most likely because of the 0.1N/kW. The best current made RTG's generate about 2.8 watts per kilogram. [source Wikipedia] Which means based on these current numbers well need 3500 kilograms per Newton of thrusts. This would generate a acceleration of 0.00002 m/s^2 of acceleration. This would take about an hour, and a half to make a 1m/s delta-v course change. And that's without the drive, communication, heating, or s…

With a TOPAZ reactor you can get 15 watts per kilogram.

Re: Quantum vacuum plasma thruster

#24

Post by throwaway_yy2Di »

gus_massa wrote: post #19
claudius wrote: post #15

Earlier quoted context omitted.

Which is why using the vacuum fluctuations makes sense – these are particle/antiparticle pairs created randomly in the vacuum. Since these do have finite masses, you can accelerate them to larger momenta than would be possible with photons (which can also carry arbitrarily large momenta ℏω, but creating these high-frequency photons is difficult).

The generalization of the well known E=mc^2 equation is E^2 = p^2 c^2 + m^2 c^4 (where p is the momentum and m is the “rest mass”). So: p = Sqrt( (E/c)^2 - (mc)^2 ) Therefore, for massive particles, you get less moment for the same energy. (An easy way to get a general idea this without calculation is that part of the energy goes to create the mass, and the other part goes to create the moment. But I prefer calculati…

To clarify an important point: this is including rest mass as part of the energy budget, as E=mc^2. In most discussions the conclusion is the opposite: you get vastly higher moment/energy for massive particles, ordinary molecules like hydrogen, because engineers aren't counting the energy required to create hydrogen from nothing -- only the energy to accelerate it! The moment/energy of light is actually terribly low for practical purposes.

Re: Quantum vacuum plasma thruster

#25

Post by gus_massa »

throwaway_yy2Di wrote: post #24
gus_massa wrote: post #19

Earlier quoted context omitted.

The generalization of the well known E=mc^2 equation is E^2 = p^2 c^2 + m^2 c^4 (where p is the momentum and m is the “rest mass”). So: p = Sqrt( (E/c)^2 - (mc)^2 ) Therefore, for massive particles, you get less moment for the same energy. (An easy way to get a general idea this without calculation is that part of the energy goes to create the mass, and the other part goes to create the moment. But I prefer calculati…

To clarify an important point: this is including rest mass as part of the energy budget, as E=mc^2. In most discussions the conclusion is the opposite: you get vastly higher moment/energy for massive particles, ordinary molecules like hydrogen, because engineers aren't counting the energy required to create hydrogen from nothing -- only the energy to accelerate it! The moment/energy of light is actually terribly low…

Yes, I should have stated that clearly.

Re: Quantum vacuum plasma thruster

#26

Post by gus_massa »

scythe wrote: post #20
throwaway_yy2Di wrote: post #16

Earlier quoted context omitted.

That doesn't solve anything, it only shifts the violation somewhere else? Of course you can't keep the virtual particles, that's a net creation of mass-energy. " which can also carry arbitrarily large momenta ℏω, " With proportionally large energy. That's the concern here, not photon count. (For anyone confused, it's ℏω/c in conventional units. Some unit systems define c=1 for convenience)

See: https://en.wikipedia.org/wiki/Woodward_effect#Propellantless... The claim appears to be that the momentum is transferred first to the local stress-energy tensor, and then to celestial bodies. It almost sounds feasible at a glance, but, well, it's quantum gravity. I would say it certainly seems like there's something to be learnt by this, though. A quantum effect with measurable gravitational consequences is... a…

There is not a complete Quantum Gravity Theory, so all what I will say are only educated guess, supposing that the final theory will mix gracefully with the current theories for the other forces.

> [...] thereby producing a propulsive force thereon without having to expel propellant from the object. [...] Local momentum conservation is preserved by the flux of momentum in the gravity field that is chiefly exchanged with the distant matter in the universe.

The gravity field should be quantized, and the associated particles are the gravitons. So in this effect they are using a jet of gravitons instead of a jet of photons (like in my example) or a jet of atoms/ions (like in a conventional rocket).

The exact quote can be used to describe how the “laser motor” works:

(modified quote) [...] thereby producing a propulsive force thereon “without” having to expel “propellant” from the object. [...] Local momentum conservation is preserved by the flux of momentum in the electromagnetic field that is chiefly exchanged with the distant matter in the universe.

The correctness of this phrase depends on if you count the photons/gravitons as propellant or not.

The graviton shold be massless, because the gravity force has unlimited range. The relation of the energy-momentum in a graviton should be a cuadrivector, or in more simple terms, the formula in my other comment should be valid, with a “rest mass” m = 0.

So the formula can be simplified and you get the same p = E/c relation of the photons, with is very small as throwaway_yy2Di noted.

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