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It is clear that we need at least the real numbers to measure lengths as it is very easy to construct a length of $\sqrt{2}$ geometrically. However, consider every physics situation where we want to measure some thing (eg: length , mass time etc ) are real numbers enough to measure any physical system?

tryst with freedom
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  • Many things are certainly easier using complex numbers (RLC circuits, quantum mechanics), but one could laboriously use (magnitude,phase) pairs instead of them I suppose. – Jon Custer Feb 28 '22 at 14:19
  • Those I feel are auxillary i.e: Only for the theory. When you actually measure something, I don't think you get a complex number measurement – tryst with freedom Feb 28 '22 at 14:20
  • Well, you asked if real numbers are enough "to mathematically model" the physical system, which pretty much implies the theory, right? – Jon Custer Feb 28 '22 at 14:23
  • I mean in the sense of strictly measuring things @JonCuster – tryst with freedom Feb 28 '22 at 14:25
  • Sure, with ideal instruments you can construct some irrationals, but you need infinite precision to properly measure reals. And even when you're apparently measuring a rational number an argument can be made that you're actually measuring an integer (or two). – PM 2Ring Feb 28 '22 at 14:32
  • What you mean and what you asked are different then. But, as I said, one can always use only real numbers if you choose, it just makes many analyses more complex (pun intended). – Jon Custer Feb 28 '22 at 14:33
  • The Feynman Lectures talk about this. https://www.feynmanlectures.caltech.edu/I_22.html – mmesser314 Feb 28 '22 at 15:39
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    @Buraian, your question says you are interested in whether "real numbers [are] enough to mathematically model the physical system?", but in comments you imply what you're really interested in how we measure a system. Please edit your question to make clear which one you are interested in. – The Photon Feb 28 '22 at 16:22

3 Answers3

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Are real numbers enough to mathematically model physical system ?

Not really. You need complex numbers to model the behaviour of quantum wave functions. Of course, you could replace complex numbers by pairs of real numbers with appropriate rules for addition and multiplication. But then you could replace real numbers with limits of sequences of rational numbers with appropriate rules for addition and multiplication. And you could replace rational numbers with pairs of integers with appropriate rules for addition and multiplication. So if you go down this route, you could model every physical system using only integers - plus a lot of increasingly complicated rules for modelling other number systems using only integers. Just like you could work out the cost of your weekly shopping trip using only Roman numerals - but why would you want to ?

Of course, we don't actually know for certain that space and time are continuous. If they are discrete then maybe real numbers are unnecessary - you could model the whole of reality using only (very large) integers (and, in the complex realm, Gaussian integers).

gandalf61
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consider every physics situation where we want to measure some thing

Since you are talking about measurements there is no need to go to the reals. All measurements have some uncertainty. And regardless of how small that uncertainty is, there are an infinite number of rational numbers that are consistent with the measurement to within that uncertainty. So you need rationals to represent measurements, but not reals.

real numbers enough to mathematically model the physical system?

Now, this is a different question. Just because the measurements only need rationals does not mean that the models will work with only the rationals. However, it is difficult to say that there are no alternative models that might work with a different set of numbers. I think that all we can do is say what types of numbers our current models do actually use.

Our current models use real numbers, complex numbers, and tensors formed with real and complex numbers. Our current models also would work well or even better with the hyperreals. And some of the statements that we make about our models are actually incorrect with the reals but correct with hyperreals. So I think that there are good arguments that hyperreals may be even more natural than reals in our existing models.

But again, I think it is wise to recognize that measurements don’t require anything beyond the rationals.

Dale
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  • I don't agree with your argument here. Measurements with uncertainty are characterized not merely by endpoints, but a probability distribution over the reals. – J.G. Feb 28 '22 at 18:24
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    @JG actually, all measurements with uncertainty are characterized by discrete random variables with some discrete probability mass function. They are ubiquitously modeled as a probability distribution over the reals, but that is a model. An actual measurement always has a discrete set of outcomes. – Dale Feb 28 '22 at 20:29
  • You're talking about values the measurement can return; I'm talking about our Bayesian posterior after such measurements for the physical state. For the former, $N$ measurements give an exponential number of possible results, each with its own e.g. mean. Statistics is the art of how we learn about states from these, not of "believing in" this exponentially large finite set of possible values. – J.G. Feb 28 '22 at 20:32
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    @JG the Bayesian posterior is not a measurement. It is a model, so it would be covered in the second part of my answer. Also, in principle you can do Bayesian inference with discrete distributions. So again the use of a probability distribution over the reals belongs to the model, not the measurement, as I said in the answer – Dale Feb 28 '22 at 21:14
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It is clear that we need at least the real numbers to measure lengths as it is very easy to construct a length of $\sqrt2$ geometrically.

Is it, now? As I understand it, the traditional, geometric way to construct a length of $\sqrt2$, from a length of $1$, is to draw a certain sequence of perfectly straight lines and perfect circles. While this is nice and well-defined in the mathematical world of geometry, it is also unphysical. In physics, all measurements have nonzero uncertainties, and since the rationals are dense in the reals, it suffices to represent measured quantities with rational numbers. Of course, since $\mathbb{R}$ is an extension field of $\mathbb{Q}$, rationals can be treated as elements of $\mathbb{R}$, but the key property of $\mathbb{R}$ that distinguishes it from $\mathbb{Q}$ (completeness) is not that important in this context, so it's fine to treat measured quantities as elements of $\mathbb{Q}$.

Sandejo
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