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> There is a mathematical theorem that forbids you from writing down a discrete version of certain quantum field theories. > (...) > You know, if you take thi

by kbr- 4y ago

> There is a mathematical theorem that forbids you from writing down a discrete version of certain quantum field theories. > (...) > You know, if you take this theorem at face value, it’s telling us we’re not living in the Matrix. The way you simulate anything on a computer is by first discretizing it and then simulating. And yet there’s a fundamental obstacle seemingly to discretizing the laws of physics as we know it. So we can’t simulate the laws of physics, but it means no one else can either. So if you really buy this theorem, then we’re not living in the Matrix. I don't buy this reasoning. I can encode the continuous function f(x) = x^2 on a computer. Then I can calculate this function for any number up to any digit, if I allocate enough memory. I don't need to allocate a discrete domain up-front and then stick to it at all times. I can increase and decrease the accuracy as needed. In a similar fashion I could simulate a continuous universe encoded with continuous operators. I could simulate it on a discrete lattice with certain precision as long as nobody inside the simulation builds equipment that can measure things "in-between" the lattice points. And when somebody does, at that moment, I can simply pause the simulation, calculate the values of my operators using a locally-denser lattice, then unpause. The observer with their equipment wouldn't notice anything because the simulation was paused, they would just get the correct measurement.

blueprint 4y ago
What theorem is he talking about though?
- kbr- 4y ago
  > The theorem is called the Nielsen-Ninomiya theorem. Among the class of quantum field theories that you cannot discretize is the one that describes our universe, the Standard Model.
  - blueprint 4y ago
    ah. it talks about issues with putting fermions with spin on a lattice. there's more to fermions than a continuous exponential function though! :)
    - deleted 4y ago
      [deleted]
mxkopy 4y ago
I don't think continuous and discrete-with-arbitrary-precision are the same thing, maybe even to the degree where we can't use the latter as a universal approximation for the former. I think there are some cases where approximating a continuous function on a lattice produces an error that isn't a function of precision (I think the Weierstrass function is one).
- yarg 4y ago
  It really comes down not to the question of equivalency, but distinguishability. If the universe is lazy evaluating things discretely down to scales well below that of observation, there's probably no damned way to tell the difference.
  - mxkopy 4y ago
    I think it depends on what is doing the observing. If it's humans or conscious beings or something like that, then the admins could be keeping the precision perpetually out of our reach. If it's atoms or photons doing the observing, though, they're affected by the gravitational pull of every mass in their light cone. So an atom can distinguish between universes in which another atom 1000ly away moved by 0e or 1e (where e is the distance between two lattice points). But, it wouldn't have to know until 1000 years later. There does seem to be some computation-resource saving mechanic at play, but I don't think it has to do with discretization.
deleted 4y ago
[deleted]
blablabla123 4y ago
It's the current state of the art anyway. But also it's worth mentioning it's more complex than just a function x^2. When looking at a Feynman Path Integral you integrate an Operator (mapping a function to a function) over a an uncountable domain of functions. The other approach which is used to calculate probabilities goes with an approximation that cannot be used for simulations though. All this is not mathematically bullet-proof as Mechanics also speaking about Wightman axioms. (And even in Mechanics it's possible to construct infinities)
atty 4y ago
The point is that it can’t be simulated on any lattice of any density. It doesn’t matter how fine the lattice is compared to the sensitivity of the measurement. For your case you showed, it would be like if x^2 simply didn’t exist on a lattice, and couldn’t be calculated by a computer no matter how much memory you threw at the problem. Certain parts of QFT work fine in lattice based calculations (lattice-QCD, for instance, where it’s simulated on a certain lattice scale, and then extrapolated to the continuous limit), and some seemingly don’t work at all.
- kbr- 4y ago
  > The point is that it can’t be simulated on any lattice of any density. It doesn’t matter how fine the lattice is compared to the sensitivity of the measurement. It feels like you want to do sth like: first discretize (choose a lattice), then simulate (do calculations on this lattice). I want to do the opposite: first simulate (do calculations on a continuous domain), then discretize (restrict my results to a lattice). > For your case you showed, it would be like if x^2 simply didn’t exist on a lattice, and couldn’t be calculated by a computer no matter how much memory you threw at the problem. So the functions we're talking about do exist on continuous domains - but they don't have a corresponding definition on a lattice? Couldn't we embed a lattice in the continuous domain, then restrict the function along the embedding, thus getting a definition on a lattice? Unless it's not possible to embed the lattice in a continuous domain - then my reasoning breaks. (note: I know nothing about physics, I'm a programmer with math education, talk to me like I'm an idiot)
  - zmgsabst 4y ago
    If you have an oracle that can tell you the answer, you can subsample that at lattice points. But how do you compute the continuous answer in the first place?
    - kbr- 4y ago
      I guess the (crazy, I know) assumption that I made is that I have some analytical, symbolic expression for a function that describes the state of the universe at every point. This "state" describes some fundamental quantity (not necessarily a quantity we have a name for yet). Then we express the value of any particle field at every point as a (potentially very complex) symbolic expression that only uses the state function from my previous paragraph. All of these expressions need only finite memory to store. They describe functions with domain R^n to some co-domain of operators or whatever. Then I can calculate the value of this complex function at any point with any precision I like, with finite memory, although unbounded - I need to allocate more memory when I want more precision. Point is, I delay the process of "latticization" (calculating the numerical values at each point of a chosen lattice) to the very end - only then I have to choose how fine-grained my lattice is.
machina_ex_deus 4y ago
The theorem mentioned is Nielsen-Ninomiya. A good analogy is aliasing in signal processing. It's not that you can't define a theory and compute it. It's that under certain conditions, you get more than one electron, you will get electrons that behave like electron does but they are different, and they are an artifact of the discretization. This is similar to aliasing in signal processing. A high frequency signal behaving like low frequency. And there are solutions to this problem, but it's hard to reason which one is the "right" solution.
hyuijk 4y ago
But who says the computer running the simulation must be exact? We are using a lot of low precision computations in graphics, neural networks, .... Pi also can't be calculated exactly to infinite precision, yet that doesn't stop us to compute circles and so on.
x86x87 4y ago
You cannot have arbitrary small precision past a certain point no matter how much memory you allocate. You will run into this limit and it's a physical hard limit. Lookup why it's called quantum (mechanics, physics, field theory, etc). Before you simulate an entire universe try accurately simulating one atom. Let me know how it goes (spoiler alert: the computing power needed to do this is beyond the computing power we have today)

by kbr- 4y ago

blueprint 4y ago

What theorem is he talking about though?

kbr- 4y ago
> The theorem is called the Nielsen-Ninomiya theorem. Among the class of quantum field theories that you cannot discretize is the one that describes our universe, the Standard Model.
- blueprint 4y ago
  ah. it talks about issues with putting fermions with spin on a lattice. there's more to fermions than a continuous exponential function though! :)
  - deleted 4y ago
    [deleted]

mxkopy 4y ago

I don't think continuous and discrete-with-arbitrary-precision are the same thing, maybe even to the degree where we can't use the latter as a universal approximation for the former. I think there are some cases where approximating a continuous function on a lattice produces an error that isn't a function of precision (I think the Weierstrass function is one).

yarg 4y ago
It really comes down not to the question of equivalency, but distinguishability. If the universe is lazy evaluating things discretely down to scales well below that of observation, there's probably no damned way to tell the difference.
- mxkopy 4y ago
  I think it depends on what is doing the observing. If it's humans or conscious beings or something like that, then the admins could be keeping the precision perpetually out of our reach. If it's atoms or photons doing the observing, though, they're affected by the gravitational pull of every mass in their light cone. So an atom can distinguish between universes in which another atom 1000ly away moved by 0e or 1e (where e is the distance between two lattice points). But, it wouldn't have to know until 1000 years later. There does seem to be some computation-resource saving mechanic at play, but I don't think it has to do with discretization.

deleted 4y ago

[deleted]

blablabla123 4y ago

It's the current state of the art anyway. But also it's worth mentioning it's more complex than just a function x^2. When looking at a Feynman Path Integral you integrate an Operator (mapping a function to a function) over a an uncountable domain of functions. The other approach which is used to calculate probabilities goes with an approximation that cannot be used for simulations though. All this is not mathematically bullet-proof as Mechanics also speaking about Wightman axioms. (And even in Mechanics it's possible to construct infinities)

atty 4y ago

The point is that it can’t be simulated on any lattice of any density. It doesn’t matter how fine the lattice is compared to the sensitivity of the measurement. For your case you showed, it would be like if x^2 simply didn’t exist on a lattice, and couldn’t be calculated by a computer no matter how much memory you threw at the problem. Certain parts of QFT work fine in lattice based calculations (lattice-QCD, for instance, where it’s simulated on a certain lattice scale, and then extrapolated to the continuous limit), and some seemingly don’t work at all.

kbr- 4y ago
> The point is that it can’t be simulated on any lattice of any density. It doesn’t matter how fine the lattice is compared to the sensitivity of the measurement. It feels like you want to do sth like: first discretize (choose a lattice), then simulate (do calculations on this lattice). I want to do the opposite: first simulate (do calculations on a continuous domain), then discretize (restrict my results to a lattice). > For your case you showed, it would be like if x^2 simply didn’t exist on a lattice, and couldn’t be calculated by a computer no matter how much memory you threw at the problem. So the functions we're talking about do exist on continuous domains - but they don't have a corresponding definition on a lattice? Couldn't we embed a lattice in the continuous domain, then restrict the function along the embedding, thus getting a definition on a lattice? Unless it's not possible to embed the lattice in a continuous domain - then my reasoning breaks. (note: I know nothing about physics, I'm a programmer with math education, talk to me like I'm an idiot)
- zmgsabst 4y ago
  If you have an oracle that can tell you the answer, you can subsample that at lattice points. But how do you compute the continuous answer in the first place?
  - kbr- 4y ago
    I guess the (crazy, I know) assumption that I made is that I have some analytical, symbolic expression for a function that describes the state of the universe at every point. This "state" describes some fundamental quantity (not necessarily a quantity we have a name for yet). Then we express the value of any particle field at every point as a (potentially very complex) symbolic expression that only uses the state function from my previous paragraph. All of these expressions need only finite memory to store. They describe functions with domain R^n to some co-domain of operators or whatever. Then I can calculate the value of this complex function at any point with any precision I like, with finite memory, although unbounded - I need to allocate more memory when I want more precision. Point is, I delay the process of "latticization" (calculating the numerical values at each point of a chosen lattice) to the very end - only then I have to choose how fine-grained my lattice is.

machina_ex_deus 4y ago

The theorem mentioned is Nielsen-Ninomiya. A good analogy is aliasing in signal processing. It's not that you can't define a theory and compute it. It's that under certain conditions, you get more than one electron, you will get electrons that behave like electron does but they are different, and they are an artifact of the discretization. This is similar to aliasing in signal processing. A high frequency signal behaving like low frequency. And there are solutions to this problem, but it's hard to reason which one is the "right" solution.

hyuijk 4y ago

But who says the computer running the simulation must be exact? We are using a lot of low precision computations in graphics, neural networks, .... Pi also can't be calculated exactly to infinite precision, yet that doesn't stop us to compute circles and so on.

x86x87 4y ago

You cannot have arbitrary small precision past a certain point no matter how much memory you allocate. You will run into this limit and it's a physical hard limit. Lookup why it's called quantum (mechanics, physics, field theory, etc). Before you simulate an entire universe try accurately simulating one atom. Let me know how it goes (spoiler alert: the computing power needed to do this is beyond the computing power we have today)