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It is a known fact that $\mathbb{R}^3$ does not model classical physical space accurately since it includes some non-invariant structure under translations, such as the origin $(0,0,0)$. However, I am unsure as to how one would show that.

My trouble is in understanding what exactly about $(0,0,0)$ is not invariant.

I fear this might belong to a more physics based exchange, but the kind of answer I am looking for is more mathematical than physical. I suppose this is trivial, since there is no one doing it explicitly, but I would appreciate it if someone could write a pseudo-formal proof of this.

Promethèus
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  • If you mean that $\mathbb R^3$ is a bunch of coordinates then note that physicists consider only things "physical" that do not depend on the choice of the coordinate system. This question is to broad and unspecific. You may want to read a decent book about special or general relativity. – Kurt G. Feb 08 '23 at 14:05
  • Well, I am reading a couple of mathematical physics books on classical mechanics, and they all more or less vaguely talk about this, but I haven't found much more information. – Promethèus Feb 08 '23 at 14:28
  • Start with this. – Kurt G. Feb 08 '23 at 16:39
  • There's not much to show, is there? Under a (nontrivial) translation $T(x)=x+a$, the zero vector $x=0$ is mapped to a nonzero vector $T(0)=a \neq 0$. – Hans Lundmark Feb 08 '23 at 20:35
  • KurtG. I've read what you suggested and I have found it of little use. The discussion does not really concern why $\mathbb{R}^n$ is not a good representation of our physical world as opposed to an affine space. If in your opinion it does, please say how. @HansLundmark. I like your comment a lot. So are saying that it is the special properties ($x + 0 = 0 + x = x$) of the zero vector that are not invariant under translations and that somehow hinder the whole of $\mathbb{R}^3$? is this reading correct? – Promethèus Feb 09 '23 at 14:52
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    Yes, it's the vector space structure of $\mathbb{R}^3$ that's incompatible with translations. If you consider $\mathbb{R}^3$ as an affine space (by “forgetting where the origin is”, or “viewing the elements as points rather than vectors”), then translations are fine, since in an affine space no point is different from any other point (in particular the origin isn't special), whereas in a vector space the zero vector is distinguished from the other vectors, for example by the properties you mention. – Hans Lundmark Feb 09 '23 at 16:54
  • @HansLundmark. Thanks, that makes sense to me. If you wanna copy that as an answer I'll happily accept it. – Promethèus Feb 09 '23 at 18:02

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The structure of $\mathbb{R}^3$ as a vector space is not preserved by (nontrivial) translations, since the zero vector is mapped to a nonzero vector.

But the structure of $\mathbb{R}^3$ as an affine space is preserved by translations; the origin is no different than any other point.

Hans Lundmark
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