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Obviously experiment is the end-all-be-all of any science, but I'm curious if there's any a priori reason to model space as Euclidean three-space (from a pre-relativity viewpoint, of course; I'm ignoring the fact that space is only locally Euclidean). I'm comfortable with differential geometry if it's necessary.

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    $\begingroup$ No. In a sense you might say that Euclidean is the simplest possibility and any other space would require some explanation. But really our feeling that space should be Euclidean just comes from our experience. $\endgroup$ – Andrew Apr 8 '14 at 3:22
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I don't think there's an a priori reason, but there's certainly a good a posteriori empirical one: if you make two long, straight parallel things, they neither meet nor diverge away from one another. It was presumably this empirical fact that led Euclid to introduce his parallel postulate, although he probably wouldn't have seen it as empirical.

Moreover, if you send people up to the top of three nearby mountains and get each to precisely measure the angle between the other two, they add up to 180. There seems to be some debate about whether this experiment was performed (by Gauss) as an explicit test of Euclidean geometry (see http://www.springer.com/cda/content/document/cda_downloaddocument/9780387295541-c2.pdf?SGWID=0-0-45-301316-p86706747), but it's certainly plausible that it could have been.

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  • $\begingroup$ Evidently I completely missed the "aside from experimental evidence" in the title. OP, if this was remotely useful to you then let me know, otherwise I'll delete it. $\endgroup$ – Nathaniel Apr 8 '14 at 11:41
  • $\begingroup$ Oh, don't worry at all, the bulk of the answer may not have addressed the question, but it was still good information, both for me and for potential readers of the question! $\endgroup$ – EtaZetaTheta Apr 8 '14 at 19:52
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In the absence of evidence to the contrary we tend to assume the simplest possible description for physical systems.

Suppose the spatial scalar curvature had the non-zero value $S$. Immediately we have the question: why is it $S$ and not $S + 0.001$ or $S - 0.001$ or any other value. There must be some mechanism for making the curvature exactly $S$ and that suggests another layer of complexity that we would have to worry about.

If however $S = 0$ this seems simple because we assume it can't take any other value. If there is no physical mechanism that allows $S$ to change we don't have to lie awake at night wondering what that mechanism is and how it works.

You could argue that this view isn't terribly rational, and I would have to agree. Nevertheless there is a long and honourable history of not multiplying elements beyond necessity.

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The limit of non euclidean geometries, as the radius goes large, is euclidean. It's like relativity. Unless you have fancy equipment or fast things, the euclidean newtonian model does quite fine.

If you model space on hyperbolic geometry, with a curvature the same size of the earth, the observable universe would fit inside a sphere of radius 432000 miles. You can see from history, that even at this size, the model of being flat is good enough to build cities etc.

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