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Really annoying to see false expertise pop up again and again. > conservation of momentum, which is just a subset of conservation of energy This is false. Mom

Really annoying to see false expertise pop up again and again. > conservation of momentum, which is just a subset of conservation of energy This is false. Momentum and energy are two different things, even with mass-energy equivalence. In particular, momentum is a vector while energy is a scalar, and their mass-energy equivalent units (impulse vs energy) are also different. Also: pair production doesn't necessarily require some amazing quantum device to keep the pair separate. > There is even some research into using a vibrating "quantum mirror" that forces virtual photons to become real using only energy from quantum fluctuations but that's even more complicated and difficult to engineer. This is true if you consider conservation of energy to be an engineering problem. Under no scientific theory can you pluck energy from the vacuum.

tprice7 10y ago
Did you see my response to the parent, where I give a sketch of how conservation of momentum follows directly from conservation of energy in all frames of reference? What is your objection to that argument?
- cynicalkane 10y ago
  Remove either conservation law and work out the equations for an elastic collision, and you'll see a new degree of freedom pop up. In classical mechanics they are logically independent, and since the space is affine, frames of reference don't help you. Your sketch takes as a premise, something that is actually a consequence of conservation laws: the way quantities change under frame-of-reference changes. You can get both by looking at the least action principle under frame of reference changes and using Noether's theorem, but that's not what your sketch is about.
  - tprice7 10y ago
    Is there some objective sense in which conservation of momentum is more fundamental than conservation of energy in a different reference frame? Since either can be derived from the other, how do you determine which is "actually a consequence" of which? (I ask out of genuine curiosity)
    - cynicalkane 10y ago
      Well, what is fundamental isn't the point. If you imagine a world where different quantities are conserved, there would no doubt be different change-of-frame equations describing how they are. The point is you can imagine a world where energy is conserved but momentum is not. Actually it's possible to imagine physical systems, with affine frames of reference, where nothing is conserved at all. Fundementality is a different thing, but it's more natural to assume conservation laws and take frame-of-reference equations are a consequence. You can do it the other way (I'm pretty sure, but not 100% sure because my physics math is rusty) but you're still making assumptions, and you still don't have one conservation law following from the other.
  - akiselev 10y ago
    You've got it backwards: in an elastic collision, kinetic energy is conserved by definition. Removing one conservation law makes the whole exercise physically meaningless because elastic collisions require as a given that kinetic energy will not be converted to other forms of energy. This premise guarantees that the momentum is treated as a quantity separate from the total energy of the system (which, as a reply to your first post pointed out, when expressed in relativistic units, includes not just the energy-mass equivalency but energy-momentum as well). In reality, momentum in almost every collision will be converted into other forms of energy such as when intramolecular bonds are broken and materials are deformed in a car accident or when momentum is converted into mass in particle colliders. The elastic collision equations are only useful as a rough approximation precisely because they go out of their way to simplify the energy-momentum equivalence to conservation of kinetic energy just like Newtonian mechanics is a simpler but useful rough approximation of relativity at low energies. You can't use elastic collisions to reason about energy-momentum equivalence because their most basic assumption is that there is no energy-momentum equivalence.
deleted 10y ago
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pdonis 10y ago
> Momentum and energy are two different things, even with mass-energy equivalence. In particular, momentum is a vector while energy is a scalar, and their mass-energy equivalent units (impulse vs energy) are also different. None of this is correct when you take relativity into account. In relativity, momentum and energy are both parts of a 4-vector, and that 4-vector is conserved. And in natural units, in which c = 1, energy and momentum have the same units. Particle physicists use these units all the time.
- akiselev 10y ago
  Right, most people think of mass energy equivalence which is E=mc^2 but this only describes energy for a mass at rest, with a momentum of zero. The full equation relating energy to intrinsic mass and momentum is E^2=(pc)^2+(mc^2)^2 which is why smashing particles in a collider works. The momentum of two particles colliding at relativistic speeds is "conserved" in the classical sense but it is really converted to photons and massive particles, among other things. What I meant in my original post was that intuition regarding conservation of momentum in classical mechanics may not apply here because as long as the total energy of the system is conserved, the EM drive is free to convert energy to particles and motion because momentum will still be conserved between the drive and the new particles (which is how regular chemical propulsion works). The system experiences a net change in momentum of zero but what matters is that the total energy of the EM drive slowly falls as energy is converted to particles that provide propulsion.
  - pdonis 10y ago
    > the EM drive is free to convert energy to particles and motion because momentum will still be conserved between the drive and the new particles I agree that this would conserve momentum if if were what was happening in the EmDrive. But all of the papers I have read insist that it is not what is happening; no particles are being ejected from the test device; there is no "rocket exhaust".
    - akiselev 10y ago
      How do they measure the exhaust? With the EM field in the cavity I would imagine precisely measuring the photon emissions would be very difficult and the SNR might not be high enough to draw a conclusion. If the force is generated by new massive particles it would be headline news in experimental physics as the most efficient energy-mass converter ever made but if it's somehow generated by high energy photons, what are the chances they have adequate imaging to measure UV, X, and gamma rays, especially if the emissions have some sort of complex interference pattern? (although the theoretical maximum for force/power input is much lower with photon pressure than the results in this paper so it's still unlikely)
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raattgift 10y ago
"Under no scientific theory can you pluck energy from the vacuum." You can if you're very patient, have a lot of material, and lucky aim, thanks to the metric expansion of space. From within a galaxy cluster shoot one end a very very long wire into the central black hole of a galaxy in a distant galaxy to anchor the far parts of the wire, and wind some near part of the wire around the pole of an electrical generator in your galactic cluster. You won't get much power for a long time by human standards, but you'll get an awful lot of energy (and eventually substantial power give or take divergence in the stress-strain tangent modulus and related quantities) even after the gravitationally anchored far end has receded outside your generator's Hubble volume. Mining \Lambda dark energy for fun and profit, brought to you by $nabla_\mu T^{\mu\nu} = 0$, where $nabla_mu$ is the covariant derivative. It only takes a large up-front investment that you'll wish you had made billions of years ago.

tprice7 10y ago

Did you see my response to the parent, where I give a sketch of how conservation of momentum follows directly from conservation of energy in all frames of reference? What is your objection to that argument?

cynicalkane 10y ago
Remove either conservation law and work out the equations for an elastic collision, and you'll see a new degree of freedom pop up. In classical mechanics they are logically independent, and since the space is affine, frames of reference don't help you. Your sketch takes as a premise, something that is actually a consequence of conservation laws: the way quantities change under frame-of-reference changes. You can get both by looking at the least action principle under frame of reference changes and using Noether's theorem, but that's not what your sketch is about.
- tprice7 10y ago
  Is there some objective sense in which conservation of momentum is more fundamental than conservation of energy in a different reference frame? Since either can be derived from the other, how do you determine which is "actually a consequence" of which? (I ask out of genuine curiosity)
  - cynicalkane 10y ago
    Well, what is fundamental isn't the point. If you imagine a world where different quantities are conserved, there would no doubt be different change-of-frame equations describing how they are. The point is you can imagine a world where energy is conserved but momentum is not. Actually it's possible to imagine physical systems, with affine frames of reference, where nothing is conserved at all. Fundementality is a different thing, but it's more natural to assume conservation laws and take frame-of-reference equations are a consequence. You can do it the other way (I'm pretty sure, but not 100% sure because my physics math is rusty) but you're still making assumptions, and you still don't have one conservation law following from the other.
- akiselev 10y ago
  You've got it backwards: in an elastic collision, kinetic energy is conserved by definition. Removing one conservation law makes the whole exercise physically meaningless because elastic collisions require as a given that kinetic energy will not be converted to other forms of energy. This premise guarantees that the momentum is treated as a quantity separate from the total energy of the system (which, as a reply to your first post pointed out, when expressed in relativistic units, includes not just the energy-mass equivalency but energy-momentum as well). In reality, momentum in almost every collision will be converted into other forms of energy such as when intramolecular bonds are broken and materials are deformed in a car accident or when momentum is converted into mass in particle colliders. The elastic collision equations are only useful as a rough approximation precisely because they go out of their way to simplify the energy-momentum equivalence to conservation of kinetic energy just like Newtonian mechanics is a simpler but useful rough approximation of relativity at low energies. You can't use elastic collisions to reason about energy-momentum equivalence because their most basic assumption is that there is no energy-momentum equivalence.

deleted 10y ago

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pdonis 10y ago

> Momentum and energy are two different things, even with mass-energy equivalence. In particular, momentum is a vector while energy is a scalar, and their mass-energy equivalent units (impulse vs energy) are also different. None of this is correct when you take relativity into account. In relativity, momentum and energy are both parts of a 4-vector, and that 4-vector is conserved. And in natural units, in which c = 1, energy and momentum have the same units. Particle physicists use these units all the time.

akiselev 10y ago
Right, most people think of mass energy equivalence which is E=mc^2 but this only describes energy for a mass at rest, with a momentum of zero. The full equation relating energy to intrinsic mass and momentum is E^2=(pc)^2+(mc^2)^2 which is why smashing particles in a collider works. The momentum of two particles colliding at relativistic speeds is "conserved" in the classical sense but it is really converted to photons and massive particles, among other things. What I meant in my original post was that intuition regarding conservation of momentum in classical mechanics may not apply here because as long as the total energy of the system is conserved, the EM drive is free to convert energy to particles and motion because momentum will still be conserved between the drive and the new particles (which is how regular chemical propulsion works). The system experiences a net change in momentum of zero but what matters is that the total energy of the EM drive slowly falls as energy is converted to particles that provide propulsion.
- pdonis 10y ago
  > the EM drive is free to convert energy to particles and motion because momentum will still be conserved between the drive and the new particles I agree that this would conserve momentum if if were what was happening in the EmDrive. But all of the papers I have read insist that it is not what is happening; no particles are being ejected from the test device; there is no "rocket exhaust".
  - akiselev 10y ago
    How do they measure the exhaust? With the EM field in the cavity I would imagine precisely measuring the photon emissions would be very difficult and the SNR might not be high enough to draw a conclusion. If the force is generated by new massive particles it would be headline news in experimental physics as the most efficient energy-mass converter ever made but if it's somehow generated by high energy photons, what are the chances they have adequate imaging to measure UV, X, and gamma rays, especially if the emissions have some sort of complex interference pattern? (although the theoretical maximum for force/power input is much lower with photon pressure than the results in this paper so it's still unlikely)

deleted 10y ago

[deleted]

raattgift 10y ago

"Under no scientific theory can you pluck energy from the vacuum." You can if you're very patient, have a lot of material, and lucky aim, thanks to the metric expansion of space. From within a galaxy cluster shoot one end a very very long wire into the central black hole of a galaxy in a distant galaxy to anchor the far parts of the wire, and wind some near part of the wire around the pole of an electrical generator in your galactic cluster. You won't get much power for a long time by human standards, but you'll get an awful lot of energy (and eventually substantial power give or take divergence in the stress-strain tangent modulus and related quantities) even after the gravitationally anchored far end has receded outside your generator's Hubble volume. Mining \Lambda dark energy for fun and profit, brought to you by $nabla_\mu T^{\mu\nu} = 0$, where $nabla_mu$ is the covariant derivative. It only takes a large up-front investment that you'll wish you had made billions of years ago.