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The Uncertainty Principle is a very well established result which is quite generalized mathematically beyond Heisenberg's result. This has nothing to do with t

The Uncertainty Principle is a very well established result which is quite generalized mathematically beyond Heisenberg's result. This has nothing to do with that principle.

floatingatoll 9y ago
If you weaken the vector density until a point that you can generate multiple possible outputs from a given weakened input, then what level of weakening guarantees no unique outputs from the equations? These equations were designed for a system in which every 'atom' (vector) has a perfectly knowable 'spin', and they begin to produce unexpected results as the uncertainty is dialed up through weakening. It's just a shame that the considerations stop at "therefore we broke the equations" as opposed to "gee, that looks familiar". What's the Navier-Stokes equivalent of the diffraction experiment? What does the interference pattern of two vector fields even look like? Why aren't they trying to study the interference patterns of the one input, two outputs scenario? I get that this is all "obviously pointless" to others, but no one I've asked can actually explain why these comparisons are unacceptable. You completely dismiss it without any explanation other than "science is well-established", as if somehow that's meaningful. So, yeah, disappointment.
- yorwba 9y ago
  You seems to think that there is some kind of analogy between a weakened version of the Navier-Stokes equations having multiple solutions and the double-slit experiment generating an interference pattern. I don't think there is such an analogy. The equivalent of a spreading wave in quantum physics is the vector field describing the flow according to the Navier-Stokes equations. The equivalent of an interference pattern is the pattern of vortices in the fluid. When you do a double-slit experiment, you'll always see the same interference pattern, and it can be predicted exactly. There is only one solution to the equations. Having two different vector fields satisfy the Navier-Stokes equations with the same boundary conditions would be like seeing different interference patterns for no reason at all. There is no place for the Uncertainty Principle in this result, because that is a statement about the standard deviations of two complementary quantities, and there are no such quantities involved here.
- gjulian 9y ago
  Not the GP, but you have to take into account that these are studies done by mathematicians, not physicists, so there are to points to consider. First, weakening is not related to uncertainty, at least not in the normal sense that I think you refer to. It is not related to the physical solution itself but to our own rules of what we consider a solution. If instead of vector fields we were working with animals, the strong solution would be "we need this animal to be a duck" and the weakened one is "we need this animal to quack when we poke it with a stick". So it is not similar to the quantum uncertainty principle nor anything like it. Second, mathematicians are interested only in these specific equations. Breaking the equations means that they do not model correctly the real world, and finding those new equations is the job of physicists. Maybe there are other conditions on the solutions, or maybe the relaxation they did allows for non-physical solutions. In fact, they do not prove that those solutions satisfy the energy inequality, so it might be possible that all but one of those non-unique solutions are only possible if you allow fluids to magically gain energy out of nowhere (which obviously conflicts with thermodynamic laws).
  - im3w1l 9y ago
    Reminds me of the Banach-Tarski paradox, where non-smoothness breaks conservation.
    - snakeboy 9y ago
      Can you elaborate on what you mean here? Isn't B-T just a consequence of accepting the axiom of choice and performing some pathological decompositions of a ball? What do you mean by: >non-smoothness breaks conservation
      - im3w1l 9y ago
        Normally translations and rotations conserve volume. But not when you have these extremely non-smooth pieces.
  - claudius 9y ago
    > Breaking the equations means that they do not model correctly the real world This is wrong. We already know that these equations do not model correctly the real world. No model correctly models the real world and every model breaks down eventually at one point or another. Showing that a certain set of equations leads to non-unique results under certain conditions means nothing, unless you also show that those ‘certain conditions’ are true in cases where we previously assumed the model to hold. If you only show that in cases where we previously also assumed the equations not to hold, they actually output nonsense, this may be a mathematically curious and nice result but of no physical relevance. So far, this result looks more like realising that Newton’s gravitational field diverges for a point particle of nonzero mass, which is not really any indication whatsoever that it doesn’t correctly model the real world.
- mlvljr 9y ago
  "you have no idea how Sciance works" :)
bmcooley 9y ago
Could you possibly dive into this? The electron slit experiment shows that superposition exists in a fundamental way in our universe. You can even measure quantum effects with objects as large as buckyballs. Isn't it sort of obvious that a precise mathematical description of a macroscale, emergent system based on objects with quantum and probabilistic effects is always going to be an inaccurate abstraction?
- lmm 9y ago
  The Navier-Stokes equations are for a smooth fluid and don't capture quantum behaviour or even classical/macroscopic Brownian motion. This is known. It's also entirely beside the point.

The Uncertainty Principle is a very well established result which is quite generalized mathematically beyond Heisenberg's result. This has nothing to do with that principle.

floatingatoll 9y ago

If you weaken the vector density until a point that you can generate multiple possible outputs from a given weakened input, then what level of weakening guarantees no unique outputs from the equations? These equations were designed for a system in which every 'atom' (vector) has a perfectly knowable 'spin', and they begin to produce unexpected results as the uncertainty is dialed up through weakening. It's just a shame that the considerations stop at "therefore we broke the equations" as opposed to "gee, that looks familiar". What's the Navier-Stokes equivalent of the diffraction experiment? What does the interference pattern of two vector fields even look like? Why aren't they trying to study the interference patterns of the one input, two outputs scenario? I get that this is all "obviously pointless" to others, but no one I've asked can actually explain why these comparisons are unacceptable. You completely dismiss it without any explanation other than "science is well-established", as if somehow that's meaningful. So, yeah, disappointment.

yorwba 9y ago
You seems to think that there is some kind of analogy between a weakened version of the Navier-Stokes equations having multiple solutions and the double-slit experiment generating an interference pattern. I don't think there is such an analogy. The equivalent of a spreading wave in quantum physics is the vector field describing the flow according to the Navier-Stokes equations. The equivalent of an interference pattern is the pattern of vortices in the fluid. When you do a double-slit experiment, you'll always see the same interference pattern, and it can be predicted exactly. There is only one solution to the equations. Having two different vector fields satisfy the Navier-Stokes equations with the same boundary conditions would be like seeing different interference patterns for no reason at all. There is no place for the Uncertainty Principle in this result, because that is a statement about the standard deviations of two complementary quantities, and there are no such quantities involved here.
gjulian 9y ago
Not the GP, but you have to take into account that these are studies done by mathematicians, not physicists, so there are to points to consider. First, weakening is not related to uncertainty, at least not in the normal sense that I think you refer to. It is not related to the physical solution itself but to our own rules of what we consider a solution. If instead of vector fields we were working with animals, the strong solution would be "we need this animal to be a duck" and the weakened one is "we need this animal to quack when we poke it with a stick". So it is not similar to the quantum uncertainty principle nor anything like it. Second, mathematicians are interested only in these specific equations. Breaking the equations means that they do not model correctly the real world, and finding those new equations is the job of physicists. Maybe there are other conditions on the solutions, or maybe the relaxation they did allows for non-physical solutions. In fact, they do not prove that those solutions satisfy the energy inequality, so it might be possible that all but one of those non-unique solutions are only possible if you allow fluids to magically gain energy out of nowhere (which obviously conflicts with thermodynamic laws).
- im3w1l 9y ago
  Reminds me of the Banach-Tarski paradox, where non-smoothness breaks conservation.
  - snakeboy 9y ago
    Can you elaborate on what you mean here? Isn't B-T just a consequence of accepting the axiom of choice and performing some pathological decompositions of a ball? What do you mean by: >non-smoothness breaks conservation
    - im3w1l 9y ago
      Normally translations and rotations conserve volume. But not when you have these extremely non-smooth pieces.
- claudius 9y ago
  > Breaking the equations means that they do not model correctly the real world This is wrong. We already know that these equations do not model correctly the real world. No model correctly models the real world and every model breaks down eventually at one point or another. Showing that a certain set of equations leads to non-unique results under certain conditions means nothing, unless you also show that those ‘certain conditions’ are true in cases where we previously assumed the model to hold. If you only show that in cases where we previously also assumed the equations not to hold, they actually output nonsense, this may be a mathematically curious and nice result but of no physical relevance. So far, this result looks more like realising that Newton’s gravitational field diverges for a point particle of nonzero mass, which is not really any indication whatsoever that it doesn’t correctly model the real world.
mlvljr 9y ago
"you have no idea how Sciance works" :)

bmcooley 9y ago

Could you possibly dive into this? The electron slit experiment shows that superposition exists in a fundamental way in our universe. You can even measure quantum effects with objects as large as buckyballs. Isn't it sort of obvious that a precise mathematical description of a macroscale, emergent system based on objects with quantum and probabilistic effects is always going to be an inaccurate abstraction?

lmm 9y ago
The Navier-Stokes equations are for a smooth fluid and don't capture quantum behaviour or even classical/macroscopic Brownian motion. This is known. It's also entirely beside the point.