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I'm interested in the meaning of the phrase "continuum limit" specifically as it is often used in expressions relating to the ability of a quantum gravity theory to recover GR in the continuum limit.

I believe I understand the meaning but want to make sure I am not missing some important part of the precise definition as my intuition may be off and I have not seen it defined anywhere.

Pointers to a place where it is defined in an online resource would be appreciated. A google search just turned up many references to it being used in papers and articles and such.

Thank you.

inflector
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    Odd, I found this link first in google after doing a search again: http://en.wikibooks.org/wiki/General_Mechanics/The_Continuum_Limit which seems to answer the question. I'll still take answers as this may not be correct. – inflector Dec 23 '10 at 18:56

1 Answers1

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Usually it relates to discrete models where it means "becoming less discrete".

To make this more formal and give a simple example, consider a set $$\Lambda_{\alpha}(M) := \{ x \in {\mathbb R}^d \,|\, x_i = k \alpha, \, 0 \leq k \leq \lfloor {M \over \alpha} \rfloor \}$$

and imagine there is an edge between nearest neighbors so that part of it looks like this

alt text

this can be thought of as discretization of a $d$-dimensional cube of side $M$ modelling some crystal. Now by continuum limit one understand the limit $\alpha \to 0$. In this case one would recover the original cube $[0, M]^d$. The reason this is important is that lattice is a much simpler object than a continuous space and we can compute lots of things there directly. In some cases we are then able to carry out limits and recover the original continuous theory.

As for your talk about GR, I think there is only one possible interpretation in the context of Loop Quantum Gravity which deals with discrete space-time. I don't know much about this theory but I'd assume that in order to recover GR one would need to perform the continuum limit of the space-time spacing.

But note that LQG doesn't mean quantum gravity in general. It is just one of proposed theories for quantum gravity and other theories (e.g. string theory) don't assume discrete space-time, so there is no continuum limit to carry out.

Update: David has made a very good point about the nature of continuum limit, so I decided to elaborate on the matter.

There are two semantically different concepts of continuum limit (although mathematically they are the same):

  1. One can start with a discrete microscopic theory that is already complete (e.g. lattice of crystals; or with space-time in LQG). In this case the semantics of continuum limit is getting rid of microscopic physics. Once you perform the limit, you obtain a simpler theory.

  2. But sometimes the complete theory can already be continuous (e.g. Quantum Field Theory) and one first wants to discretize it in order to be able to perform the calculations more easily. In this case when we are shrinking the scale we are actually making the discrete model a better approximation to the original theory until in continuum limit we recover the original theory completely. So here we also lose the microscopic degrees of freedom but we don't care because the discrete model wasn't physical anyway -- it was just a mathematical tool.

Marek
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    You're right that one would take the continuum limit of LQG to recover GR, but I think it's worth noting that LQG actually allows (if not assumes) some finite length scale for its spin networks. So if LQG is correct, that would mean that GR would fail to properly describe reality at sufficiently small scales. – David Z Dec 23 '10 at 21:58
  • @David: good point, see the update. – Marek Dec 23 '10 at 22:36
  • In this standard picture there is something which I never really understood. As far as I can see this limit leads to the set of rationals rather than the set of reals. In that sense you do not actually perform the continuum limit, only the "rational limit"! To get continuum, you'll need an extra "jump" at the end from Q to R, and there is no guarantee that everything valid for Q will be valid for R. – Igor Ivanov Dec 24 '10 at 19:16
  • @Igor: no, there is no problem with this and to see why, let me be a mathematician for a while :-) There is no notion of limit, sequence, or even continuity in rationals. Rationals are purely algebraic construction (a field of fractions of integers; this construction can be used to obtain a field from arbitrary integral domain, i.e. commutative ring without zero divisors). To talk about limits, one needs to introduce topology on rationals and this is usually done by introducing a norm. There are nonequivalent norms possible, giving either reals or else $p$-adic numbers as completions. (cont.) – Marek Dec 25 '10 at 00:25
  • @Igor: Anyway, the punchline is, that whenever you are talking about limits on rationals in standard physics, you are actually considering them as a subset of reals with a topology on them induced from the standard Euclidean topology on the real line (this coincides with order topology on rationals). So there really can't ever appear any sort of problem you are talking about. – Marek Dec 25 '10 at 00:28
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    @Marek — yes, I know how rational are introduced, and btw the reals are introduced in an even more non-physical way. But I am not considering a limit staying in Q. I am considering your procedure and what I see at the end is Q, not R. Take 1D lattice of fixed size (=1). You take small spacing a and take the limit $a \to 0$. These words are not enough, you need to give a specific prescription of what you exactly do and then prove the result is the same for all prescriptions. Let's adopt the procedure when we divide a by 2 at each step. (cont.) – Igor Ivanov Dec 25 '10 at 13:33
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    Then at first step we add a node at 1/2, then we add a node at 1/4, 3/4, etc. You see that you never add irrational points, and at the end you get a subset of Q. In order to claim that you have a true continuum limit, you need to devise an explicit procedure and show that every real appears at some stage. – Igor Ivanov Dec 25 '10 at 13:37
  • @Igor: no, read my comments again, because you missed the point completely. Anytime you perform a limit you are not working in $\mathbb Q$. You are working in some metric completion of it. In this case we are working in $\mathbb R$ because we usually work with the lattice embedded in standard Euclidean space, or something similar. Although it's certainly possible to talk about lattices as subset of $\mathbb Q$ considered as a subset of $p$-adic numbers, this is not the usual situation and I don't think there's any need to mentioning this explicitly. – Marek Dec 25 '10 at 18:37
  • @Igor: let me illustrate the misconception you are talking about on a simpler case. Let's consider rational approximations of $\sqrt 2 $ of the form $\alpha_n := {p_n \over 2^n}$ where $p_n$ is any integer that minimizes $|\sqrt 2 - \alpha_n|$ for a given $n$. Now, first few $\alpha_n$ are $1 \over 1$, $3 \over 2$, $6 \over 4$, $11 \over 8$, and so on. So you are saying that just because the set of these $\alpha_n$ is only a subset of $\mathbb Q$ for every finite $n$, we are not able to recover $\sqrt 2$. But it should be obvious that $lim_{n \to \infty} \alpha_n = \sqrt 2$ (cont). – Marek Dec 25 '10 at 18:45
  • @Igor: at least as long as it is understood that we are working in $\mathbb Q$ considered as a subset of $\mathbb R$. And this is naturally always the case both in real analysis and in physics, which constitute 99% of applications of rational numbers in standard math/physics. – Marek Dec 25 '10 at 18:47
  • @Marek — thanks for the effort, but still it does not sound persuasive enough to me. I don't really care about limit in the sense you mention: sure, Q is dense everywhere and its closure is R. I just wonder at which point we actually get continuity for the set of degrees of freedom, and as far as I see we still don't get it in the standard procedure. In my understanding continuity is introduced later: when we actually calculate something, we actually average whatever we get (Q or R, no matter) with some continuous weight functions. – Igor Ivanov Dec 25 '10 at 19:39
  • @Igor: I think it's time to move this over to the chat room: http://chat.stackexchange.com/rooms/71/physics. I like the discussion but comment format is quite unwieldy. – Marek Dec 25 '10 at 22:02