Watching one of Feynman's lectures, I came across something that puzzled me. What was Feynman referring to when he said the following?

What goes on in no matter how tiny a region of space and no matter how tiny a region of time, according to the laws as we understand them today, takes a computing machine an infinite number of logical operations to figure out. Now, how could all that be going on in that tiny space? Why should it take an infinite amount of logic to figure out what one stinky little bit of space-time is going to do?
Richard Feynman

To be clear, my question is: Why would Feynman say it would take an infinite amount of logic to simulate even a small region of space-time?

  • $\begingroup$ what do you think? $\endgroup$ – jinawee May 5 '14 at 22:43
  • $\begingroup$ @jinawee I have no idea what he might have been talking about. Continuity of space-time, maybe? But that can't be it, we already knew about Plank distance and time back then. $\endgroup$ – Paul Manta May 5 '14 at 22:50
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    $\begingroup$ He's talking about the fact that there are an infinity of feynman diagrams you can draw without legs. Technically, all of these contribute to the amplitude for (cube of spacetime) evolving to (cube of spacetime) $\endgroup$ – Jerry Schirmer May 5 '14 at 23:30
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    $\begingroup$ A less extreme example is the fact that just writing down the wavefunction for an iron atom in the position basis is impossible, because even if you discretize space into 1000 segments along each direction, it would still take more digital storage space than the universe could hold. $\endgroup$ – DumpsterDoofus May 5 '14 at 23:53

Feynman could be referring to several different things here, but I think most of them boil down to the following point: The currently accepted theories all assume some sort of spacetime continuum or otherwise continuously defined objects. Of course, a computer 'thinks' in terms of bits, and therefore does not deal with continuum variables or any other continuous objects. It therefore needs to 'chop things up', i.e. discretize.

However, this introduces some error, since it tries to model something continuous as something discrete. In order to perfectly simulate the universe cannot cut any corners, so we have to get a perfect approximation, which means the exact solution: We have to recover continuous objects. But this is not possible with any finite discretization. Note that your apparent impression that the Planck length & time preclude the existence of smaller time/space-intervals is incorrect. The Planck units are not the smallest or largest possible things. For instance, the Planck mass is about $10^{-5}$ grams!

One different interpretation of Feynman's statement was given by Jerry Schirmer in the comments: In quantum field theory, the vacuum is not what one intuitively might think it is: There are so-called vacuum fluctuations. These can be visualized using Feynman diagrams, without any 'external lines': Bubble diagrams! Since there are infinitely many such diagrams, every tiny little piece of the vacuum has infinitely much going on inside, making it impossible to simulate using any finite algorithm.

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    $\begingroup$ Irrespective of the Bekenstein Bound? $\endgroup$ – user56903 Jan 30 '16 at 22:23
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    $\begingroup$ @DirkBruere Note that the Bekenstein bound was not known back then. I'm just trying to interpret Feynman's statement, not assessing its correctness, per se. $\endgroup$ – Danu Jan 30 '16 at 22:26
  • $\begingroup$ But even within that bound, could one reasonably claims that an infinite amount of reversible computing is taking place? $\endgroup$ – user56903 Jan 30 '16 at 22:27
  • $\begingroup$ @user56903 - I believe that black hole entropy was derived using classical thermodynamics type arguments, and that there's no way to derive the entropy in the statistical mechanics sense of counting microstates without a theory of quantum gravity. So my guess would be that the question of whether it takes finite or infinite computing to predict how states will evolve "within that bound" would only be possible with a complete theory of quantum gravity. $\endgroup$ – Hypnosifl May 30 '17 at 22:02

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