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In theory, I don't think there's such a thing as simulation without simplifications. The world seems to be continuous but our computers are discrete. There's a

• dekhn 5y ago
  I doubt it makes sense to assume the unverise is continuous (I'm glad you said "seems"). In particular, space could be spatially quantized (say, around the planck length) or any number of other details. People have done simulations with quad precision (very slow) but very few terms in molecular dynamics would benefit from that. In fact, most variables in MD can be single precision, exceptt for certain terms like the virial.
  • whatshisface 5y ago
    All of our current theories are set in continuous spacetime. At the present, there's no reason to assume anything else.
    • dekhn 5y ago
      the issue is that there are no theories based on experimental evidence at very small scales. I agree that in most situations, it would be silly to violate this assumption, unless you were working on advanced physics experiments.
    • jacquesm 5y ago
      True, but we do not actually know this for sure. There is a (small) possibility that we are simply looking at this at a scale where all we see is macro effects. It would require the quanta to be much smaller than the Planck distance though.
      • webmaven 5y ago
        > There is a (small) possibility that we are simply looking at this at a scale where all we see is macro effects. It would require the quanta to be much smaller than the Planck distance though. How much smaller?
        • jacquesm 5y ago
          Many orders of magnitude. How many? I do not know, I don't think anybody does. But photons resulting from the same event but with different energies arrive at detectors an appreciable distance away to all intents and purposes simultaneously, something that would not happen if spacetime were discrete at a level close to the Planck length. So it would have to be quite a big difference for an effect not to show up as a difference in time-of-flight.
          • aeternum 5y ago
            Is the idea that the two photons would traverse slightly different voxels due to the lower frequency wave being more spread out? What accounts for the expectation that they do not arrive simultaneously in a voxel-based universe?
    • feoren 5y ago
      I wouldn't say that "all our current theories" are set in continuous spacetime. For example, Quantum chromodynamics is set in SU(3), an 8-dimensional group of rotation-like matrices. Electric charge is discrete, spin is discrete, electron orbitals are discrete. In fact position and momentum would seem to be the outlier if they were not also discrete. I hardly call that "no reason".
      • whatshisface 5y ago
        SU(3) is a continuous group.
        • freemint 5y ago
          Yeah but it is very much not in space time.
          • whatshisface 5y ago
            But it is. SU(3) is the group for swapping colors around. It still has spacetime.
            • freemint 5y ago
              You can Cartesian product it with space time, yes. But that is possible for any system.
              • whatshisface 5y ago
                I'm having a hard time imagining quantum chromodynamics set in a single point. :)
      • drdeca 5y ago
        It is based on SU(3), but, does it really make sense to say that it isn't still set in spacetime? Like, quarks still have position operators, yes?
  • mensetmanusman 5y ago
    It’s definitely fun to think about. If the universe is discrete, how does one voxel communicate to the neighboring voxel what to update without passage through ‘stuff in between’ that doesn’t exist? Heh It seems physics is going the opposite way with infinite universes and multiple dimensions to smooth out this information transfer problem and make the discrete go away.
    • javajosh 5y ago
      As someone with keen interest in physics (and a bit of training) I find speculation about "discrete space" disquieting. It's the level of abstraction where intuition about space breaks down, and you have to be very careful. Remember that coordinate systems are short-hand for measurement. It's one thing to admit fundamental limits on measurement resolution, and quite another to say that space itself is quantized! Mostly I get around this by not thinking about it; most of these theories are only testable in atrocious and unattainable conditions, doing things like performing delicate QED experiments at the edge of a black hole. I don't think your "voxel" intuition can be right because it's a small jump from that to (re)introducing an absolute reference frame.
    • joshmarlow 5y ago
      > how does one voxel communicate to the neighboring voxel what to update without passage through ‘stuff in between’ that doesn’t exist? Heh That kind of reminds me of the 'aether' that was once hypothesized as a medium of transmission for light and radio waves [0]. Also, voxel's communicating sounds an awful lot like a higher-dimensioned cellular automata. [0] - https://en.wikipedia.org/wiki/Aether_theories https://en.wikipedia.org/wiki/Aether_theories
      • Yajirobe 5y ago
        Stephen Wolfram was right all along
• qboltz 5y ago
  All simulations have to make the Born-Oppenheimer approximation, nuclei have to be treated as frozen, otherwise electrons don't have a reference point. There will never be true knowledge of both a particle's location and momentum a la uncertainty principle, and will always have to be estimated.
  • aeternum 5y ago
    With a quantum computer could one theoretically input the super position of possible locations and momenta and run the simulation based on that?
  • phkahler 5y ago
    A simulation can have both.
    • Tagbert 5y ago
      Then is it an accurate simulation without the uncertainty?
      • contravariant 5y ago
        The pilot wave theory works perfectly fine with both exact position and momentum, but in other interpretations such particles simply don't exist.
  • drdeca 5y ago
    But, for a system of two quantum particles which interact according to a central potential, you can express this using two quantum non-interacting particles one of which corresponds to the center of mass of the two, and the other of which corresponds to the relative position, I think? And, like, there is still uncertainty about the position of the "center of mass" pretend particle, as well as for the position of the "displacement" pretend particle. (the operators describing these pretend particles can be constructed in terms of the operators describing the actual particles, and visa versa.) I don't know for sure if this works for many electrons around a nucleus, but I think it is rather likely that it should work as well. Main thing that seems unclear to me is what the mass of the pretend particles would be in the many electrons case. Oh, also, presumably the different pretend particles would be interacting in this case (though probably just the ones that don't correspond to the center of mass interacting with each-other, not interacting with the one that does represent the center of mass?) So, I'm not convinced of the "nuclei have to be treated as frozen, otherwise electrons don't have a reference point" claim.
• merely-unlikely 5y ago
  There's this concept that causation moves at the speed of light. When I first heard that, it sounded very much like a fixed refresh rate to me. Or maybe the "real world" is just another simulation
  • Filligree 5y ago
    It does if you put it that way, but another way of putting is that spacetime is hyperbolic (...well, lorentzian), and all (lightspeed) interactions are zero-ranged in 4D. As in, photons that leave the surface of the sun always strike those specific points in space-time which are at a zero spacetime interval from said surface. If you take the described geometry seriously, then "spacetime interval" is just the square of the physical distance between the events. (And any FTL path has a negative spacetime interval. If that's still the square of the distance, then I think we can confidently state that FTL is imaginary.)