# Can we have non continuous models of reality? Why don't we have them?

This question is about Godel's theorem, continuity of reality and the Luvenheim-Skolem theorem.

I know that all leading physical theories assume reality is continuous. These are my questions:

1) Is reality still continuous according to string theory? what is the meaning of continuous if there is a finite size for a string? (My understanding of string theory is very limited :)

2) According to the Luvenheim-Skolem theorem, any enumerable first order logic theory (physics?) has an enumerable model. This means we must be able to have a discrete theory of physics which is as good as the continuous ones. Is this accurate?

3) Why haven't we found one? I've heard some people say that Heisenberg's formulation of quantum mechanics doesn't speak of anything continuous... is this accurate?

4) If we can find one / have found one, does this mean that our understanding of physical laws can always be reduced into a Turing machine?

5) If we can find one / have found one, how can we explain the use of real constants in physics? (such as Pi)

6) Does Godel's theorem mean that we could always "notice" that some things are true but not be able to prove them from within the universe?

Thanks.

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–  Qmechanic May 26 '11 at 9:48
there are probably several senses in which one could axiomize the phrase 'is reality continuous in [[theory X]]?'. Certainly, conventional QM happens on a $\mathbb{R}^{3}$ background, but with a lot of quantum mechanical discreteness happening on top of that background. It's difficult to answer a question that can be formulated several ways mathematically. –  Jerry Schirmer Jun 1 '11 at 0:14
@Jerry : Part of answering this question certainly involves defining it more accurately, as I'm not an expert in these areas... However, IMHO, if we use the computer program analogy, storing wave functions with infinite accuracy requires more than enumerable storage, which makes it "a continuous model". –  Uri Jun 1 '11 at 13:08

Physics isn't first order logic!

What you want to do is write down an axiomatization of physics using first order logic. If you do that, even the real numbers will need to be encoded using first order logic, say ZFC. The nonstandard model you are looking for requires the axiom of choice. Now, we are really far out from the territory of physics. with the axiom of choice, don't expect computational decidability.

Your question has nothing to do with physics.

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The Axiom of Choice isn't required to define the reals or to define continuous functions. –  Dan Brumleve Jun 1 '11 at 2:10
Could you specify why physics isn't first order logic, why we need the axiom of choice for real numbers, and why do we even need real numbers to begin with, and not simply rational numbers? Furthermore - how does the axiom of choice contradicts decidability? What if you allow randomness for the TM? (overall this doesn't answer the question and naturally I don't agree with the statement that this question has nothing to do with physics...) –  Uri Jun 2 '11 at 15:54