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If spacetime is continuous you effectively get infinite precision -> information density - at every point (of which there are an infinite number). This seems u

If spacetime is continuous you effectively get infinite precision -> information density - at every point (of which there are an infinite number). This seems unlikely for a number of reasons. This doesn't mean spacetime is a nice even grid, but it does suggest it comes in discrete lumps of something, even if that something is actually some kind of substrate that holds the information which defines relationships between lumps.

amelius 5y ago
> If spacetime is continuous you effectively get infinite precision -> information density - at every point (of which there are an infinite number). Even if space is continuous, that doesn't mean we can get information in and out of it in infinite precision. Look at quantum physics. Maxwell's equations don't suggest existence of photons (quantized information). But atoms being atoms, they can only emit and absorb in quanta.
scythe 5y ago
People complain about infinite information, but the theory of real fields is complete and consistent, while arithmetic is not: https://en.wikipedia.org/wiki/Decidability_of_first-order_theories_of_the_real_numbers https://en.wikipedia.org/wiki/Decidability_of_first-order_th...
- j-pb 5y ago
  I often wonder if we simply constructed math the wrong way around. People tend to mentally construct the natural numbers from set theory, wholes from naturals, rationals from wholes, reals from rationals and so on. But what if there is some universe (in the math sense), which is actually complete and decidable, it's just that the moment you take discrete subsets of it, you also remove the connections that make it consistent or complete. The very act of formalising mathematical concepts into words and paper is a quantisation step after all, because both are symbols. Maybe there are proofs that can be inuitioned about (assuming brains are continuuous in some sense) but neither verbalised nor formalised.
  - burrows 5y ago
    > But what if there is some universe (in the math sense), which is actually complete and decidable, it's just that the moment you take discrete subsets of it, you also remove the connections that make it consistent or complete. Presumably this hypothetical universe implements arithmetic, so it’s not complete.
    - j-pb 5y ago
      You missed my point. Incompleteness means that there are true statments for which there are no proofs. But that doesn't preclude proofs that are beyond the proof system that you proofed incompleteness for. A non discete/symbolic proof might exist after all. It would therefore not be the existence of the natural number subset that causes undecidability, but the missing parts of the non-natural superset required to talk about the proofs that cause undecidability.
      - burrows 5y ago
        Doesn’t the incompleteness theorem say that any consistent proof system which implements arithmetic must be incomplete? Then the system you are proposing, does it implement arithmetic or not?
        j-pb 5y ago
        The incompleteness proof comes to a russel paradox like contradiction, caused by a sentence of the form "I am not proovable." encoded via goedel numbering onto peano atrithmetic. But proof by contradiction itself is problematic, because it relies on the law of the excluded middle, which only holds in classical two valued logic. If you construct math from the top down rather than from the bottom up, then it is natural that it also has a multi valued logic. In fact, it also would have infinitely valued logic. Infinite sentences, infinite theorems. Such math is non expressible for us, because we rely on discrete descriptions. But that doesn't preclude its existence. Why should it though? What evidence do we have, if we can't express it or fit it into our existing mathematical framework. To illustrate, think of Goedelsz approach and turn it backwards for a second. Instead of taking predicate logic and assigning each sentence a natural number, imagine that predicate/classical logic is a different view on the natural numbers. Now that means that there might also be logical interpretation of the reals, the hyperreals and so on. (We could do the same with Alephs, in fact they might be the more fundamental objects.)
- syki 5y ago
  How do you know arithmetic is not complete or consistent? In ZFC arithmetic is complete and arithmetic is consistent. This assumes that ZFC is consistent. The second order Peano axioms are categorical so I assume you mean only the first order theory. At any rate, what does any of this have to do with information capacity in the universe? Is the information capacity of the universe related to the consistency/completeness of arithmetic?
  - burrows 5y ago
    > How do you know arithmetic is not complete or consistent? In ZFC arithmetic is complete and arithmetic is consistent. This assumes that ZFC is consistent. Aren’t you just begging the question by assuming ZFC is consistent to demonstrate that arithmetic is consistent?
    - syki 5y ago
      I don’t think so. The point is, saying “arithmetic is inconsistent” doesn’t mean anything without talking about where this theory resides. The larger point is that this has no relationship to whether or not the universe is infinite so it shouldn’t be talked about at all within that context. Suppose for a moment that it makes sense to say arithmetic is part of the fabric of the universe (whatever that is supposed to mean). How would one know if arithmetic is consistent or complete within the context of being part of the universe? Suppose ZFC is part of the universe. Then arithmetic (the model of it as being part of the universe) is complete and consistent. Now what? I claim nothing of consequence follows from this in relation to whether or not space is continuous.
      - burrows 5y ago
        > How would one know if arithmetic is consistent or complete within the context of being part of the universe? The same way I know that it’s incomplete when formulated in ZFC (assuming ZFC is consistent), from the first incompleteness theorem. > Suppose ZFC is part of the universe. Then arithmetic (the model of it as being part of the universe) is complete and consistent. Why does this conclusion follow? Are you raising the question of whether mathematical propositions can be justified a posteriori? Or are you arguing that the incompleteness theorems don’t necessarily apply to theories formulated ‘as part of the universe’?
rndphs 5y ago
> If spacetime is continuous you effectively get infinite precision -> information density - at every point (of which there are an infinite number). This would be true if objects existed at perfectly local points. However we know that a perfectly localised wavefunction has spatial frequency components that add up to infinite energy. Any wavefunction with finite energy is band-limited. At non-zero temperature the Shannon-Hartley theorem will give a finite bit rate density over frequencies, and since the wavefunction is band limited it will therefore only have the ability to carry a finite amount of information.
- TheOtherHobbes 5y ago
  This and the comment under it are making my point for me. Relativity assumes spacetime is continuous. Quantum theory implies that quantum phenomena are bandlimited and therefore the information spacetime can hold is limited. The difference is we know what the quantised components of field theory are. We don't have any idea what the quantised components of spacetime are supposed to be, or how they operate. The various causal propagation theories (like causal dynamical triangulation) may be the first attempts at this, but it's going to be hard to get further without experiments that can probe that level - which is very difficult given the energies involved. Without that, we're just guessing.
elcritch 5y ago
Obviously that "something" is the float type used to calculate the simulation. Probably some ultra dimensional IEEE style spec that some CPU vendor intern booked anyways. ;)
- dtgriscom 5y ago
  Your mistake is that you didn't pony up for the "double" feature in your personal universe.
  - throwaway17_17 5y ago
    Im pretty sure my math is right here, so… It would be more appropriate to say they failed to shell out for the 256 bit processor. Because at 256 bits per int a vector of 3 ints can easily encodes any location in the observable universe as Planck length coordinates.
    - pixl97 5y ago
      Ya, but the observable universe only exists for cache locality reasons. You dont want to have to transfer information across too many processors as your bus bandwidth limited. The actual universe is much larger.
      - elcritch 5y ago
        We should be careful then not to overload the 256 range. If we probe too closely at the QM level or use too many quantum computers it might overload the local processor node and crash it. Be a bad day for everyone.
echelon 5y ago
> If spacetime is continuous you effectively get infinite precision -> information density - at every point (of which there are an infinite number). Without some smallest resolution, we'd need infinite amounts of information to track point particle displacement in one dimension.
somewhereoutth 5y ago
Care to to enlighten us these number of reasons?
deleted 5y ago
[deleted]

amelius 5y ago

> If spacetime is continuous you effectively get infinite precision -> information density - at every point (of which there are an infinite number). Even if space is continuous, that doesn't mean we can get information in and out of it in infinite precision. Look at quantum physics. Maxwell's equations don't suggest existence of photons (quantized information). But atoms being atoms, they can only emit and absorb in quanta.

scythe 5y ago

People complain about infinite information, but the theory of real fields is complete and consistent, while arithmetic is not: https://en.wikipedia.org/wiki/Decidability_of_first-order_theories_of_the_real_numbers https://en.wikipedia.org/wiki/Decidability_of_first-order_th...

j-pb 5y ago
I often wonder if we simply constructed math the wrong way around. People tend to mentally construct the natural numbers from set theory, wholes from naturals, rationals from wholes, reals from rationals and so on. But what if there is some universe (in the math sense), which is actually complete and decidable, it's just that the moment you take discrete subsets of it, you also remove the connections that make it consistent or complete. The very act of formalising mathematical concepts into words and paper is a quantisation step after all, because both are symbols. Maybe there are proofs that can be inuitioned about (assuming brains are continuuous in some sense) but neither verbalised nor formalised.
- burrows 5y ago
  > But what if there is some universe (in the math sense), which is actually complete and decidable, it's just that the moment you take discrete subsets of it, you also remove the connections that make it consistent or complete. Presumably this hypothetical universe implements arithmetic, so it’s not complete.
  - j-pb 5y ago
    You missed my point. Incompleteness means that there are true statments for which there are no proofs. But that doesn't preclude proofs that are beyond the proof system that you proofed incompleteness for. A non discete/symbolic proof might exist after all. It would therefore not be the existence of the natural number subset that causes undecidability, but the missing parts of the non-natural superset required to talk about the proofs that cause undecidability.
    - burrows 5y ago
      Doesn’t the incompleteness theorem say that any consistent proof system which implements arithmetic must be incomplete? Then the system you are proposing, does it implement arithmetic or not?
      - j-pb 5y ago
        The incompleteness proof comes to a russel paradox like contradiction, caused by a sentence of the form "I am not proovable." encoded via goedel numbering onto peano atrithmetic. But proof by contradiction itself is problematic, because it relies on the law of the excluded middle, which only holds in classical two valued logic. If you construct math from the top down rather than from the bottom up, then it is natural that it also has a multi valued logic. In fact, it also would have infinitely valued logic. Infinite sentences, infinite theorems. Such math is non expressible for us, because we rely on discrete descriptions. But that doesn't preclude its existence. Why should it though? What evidence do we have, if we can't express it or fit it into our existing mathematical framework. To illustrate, think of Goedelsz approach and turn it backwards for a second. Instead of taking predicate logic and assigning each sentence a natural number, imagine that predicate/classical logic is a different view on the natural numbers. Now that means that there might also be logical interpretation of the reals, the hyperreals and so on. (We could do the same with Alephs, in fact they might be the more fundamental objects.)
syki 5y ago
How do you know arithmetic is not complete or consistent? In ZFC arithmetic is complete and arithmetic is consistent. This assumes that ZFC is consistent. The second order Peano axioms are categorical so I assume you mean only the first order theory. At any rate, what does any of this have to do with information capacity in the universe? Is the information capacity of the universe related to the consistency/completeness of arithmetic?
- burrows 5y ago
  > How do you know arithmetic is not complete or consistent? In ZFC arithmetic is complete and arithmetic is consistent. This assumes that ZFC is consistent. Aren’t you just begging the question by assuming ZFC is consistent to demonstrate that arithmetic is consistent?
  - syki 5y ago
    I don’t think so. The point is, saying “arithmetic is inconsistent” doesn’t mean anything without talking about where this theory resides. The larger point is that this has no relationship to whether or not the universe is infinite so it shouldn’t be talked about at all within that context. Suppose for a moment that it makes sense to say arithmetic is part of the fabric of the universe (whatever that is supposed to mean). How would one know if arithmetic is consistent or complete within the context of being part of the universe? Suppose ZFC is part of the universe. Then arithmetic (the model of it as being part of the universe) is complete and consistent. Now what? I claim nothing of consequence follows from this in relation to whether or not space is continuous.
    - burrows 5y ago
      > How would one know if arithmetic is consistent or complete within the context of being part of the universe? The same way I know that it’s incomplete when formulated in ZFC (assuming ZFC is consistent), from the first incompleteness theorem. > Suppose ZFC is part of the universe. Then arithmetic (the model of it as being part of the universe) is complete and consistent. Why does this conclusion follow? Are you raising the question of whether mathematical propositions can be justified a posteriori? Or are you arguing that the incompleteness theorems don’t necessarily apply to theories formulated ‘as part of the universe’?

rndphs 5y ago

> If spacetime is continuous you effectively get infinite precision -> information density - at every point (of which there are an infinite number). This would be true if objects existed at perfectly local points. However we know that a perfectly localised wavefunction has spatial frequency components that add up to infinite energy. Any wavefunction with finite energy is band-limited. At non-zero temperature the Shannon-Hartley theorem will give a finite bit rate density over frequencies, and since the wavefunction is band limited it will therefore only have the ability to carry a finite amount of information.

TheOtherHobbes 5y ago
This and the comment under it are making my point for me. Relativity assumes spacetime is continuous. Quantum theory implies that quantum phenomena are bandlimited and therefore the information spacetime can hold is limited. The difference is we know what the quantised components of field theory are. We don't have any idea what the quantised components of spacetime are supposed to be, or how they operate. The various causal propagation theories (like causal dynamical triangulation) may be the first attempts at this, but it's going to be hard to get further without experiments that can probe that level - which is very difficult given the energies involved. Without that, we're just guessing.

elcritch 5y ago

Obviously that "something" is the float type used to calculate the simulation. Probably some ultra dimensional IEEE style spec that some CPU vendor intern booked anyways. ;)

dtgriscom 5y ago
Your mistake is that you didn't pony up for the "double" feature in your personal universe.
- throwaway17_17 5y ago
  Im pretty sure my math is right here, so… It would be more appropriate to say they failed to shell out for the 256 bit processor. Because at 256 bits per int a vector of 3 ints can easily encodes any location in the observable universe as Planck length coordinates.
  - pixl97 5y ago
    Ya, but the observable universe only exists for cache locality reasons. You dont want to have to transfer information across too many processors as your bus bandwidth limited. The actual universe is much larger.
    - elcritch 5y ago
      We should be careful then not to overload the 256 range. If we probe too closely at the QM level or use too many quantum computers it might overload the local processor node and crash it. Be a bad day for everyone.

echelon 5y ago

> If spacetime is continuous you effectively get infinite precision -> information density - at every point (of which there are an infinite number). Without some smallest resolution, we'd need infinite amounts of information to track point particle displacement in one dimension.

somewhereoutth 5y ago

Care to to enlighten us these number of reasons?

deleted 5y ago

[deleted]