4 ms·

This is inconsistent with the claimed "specific force performance of 0.1N/kW". The ratio a photon's momentum to energy is fixed at 1/c, which is 3.3E-6 N/kW.

claudius 13y ago
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).
- throwaway_yy2Di 13y ago
  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)
  - scythe 13y ago
    See: https://en.wikipedia.org/wiki/Woodward_effect#Propellantless_propulsion 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... almost unheard of.
    - gus_massa 13y ago
      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.
- gus_massa 13y ago
  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 calculations.) So the best Moment/Energy ratio is for massless particles, like the photon. And the photon is the easiest particle to create and direct. But as throwaway_yy2Di said, the ratio with photons is 30000 smaller than the ratio they claim. So the only consistent explanation is that they are measuring something in a wrong way.
  - throwaway_yy2Di 13y ago
    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.
    - gus_massa 13y ago
      Yes, I should have stated that clearly.

claudius 13y ago

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).

throwaway_yy2Di 13y ago
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)
- scythe 13y ago
  See: https://en.wikipedia.org/wiki/Woodward_effect#Propellantless_propulsion 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... almost unheard of.
  - gus_massa 13y ago
    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.
gus_massa 13y ago
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 calculations.) So the best Moment/Energy ratio is for massless particles, like the photon. And the photon is the easiest particle to create and direct. But as throwaway_yy2Di said, the ratio with photons is 30000 smaller than the ratio they claim. So the only consistent explanation is that they are measuring something in a wrong way.
- throwaway_yy2Di 13y ago
  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.
  - gus_massa 13y ago
    Yes, I should have stated that clearly.