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Why not the $L_1$ or $L_3$ distances? Is there some deep reason why the universe (at least at human scales) looks pretty much Euclidean?

Could we imagine a different universe where a different $L_p$ metric would seem "natural"?

I know it's kind of a deep question, but the specialness of 2 here has always made me wonder.

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    $\begingroup$ Is the L2 norm the only one which is invariant under coordinate rotation? $\endgroup$
    – DanielSank
    Commented Sep 26, 2014 at 23:20
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    $\begingroup$ The L2 norm doesn't make much sense to a taxicab driver in Manhattan. The L1 norm, aka the taxicab norm, aka the Manhattan norm, makes a lot more sense to that taxicab driver. As for why the L2 norm locally appears special, the answer is "because the universe says so". The universe appears to locally conserve energy, linear momentum, and angular momentum. Why? Maybe Mach's principle, but even general relativity violates that to some extent. Ultimately, physicists don't quite know why. Observation says it is, and the descriptions comply with that mountain of observational data. $\endgroup$
    – David Hammen
    Commented Sep 26, 2014 at 23:25
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    $\begingroup$ Well, independent of the physical content, the L2 space is the only Lp space that is also an hilbert space, so, one can say that even mathematically it's a special norm $\endgroup$
    – Hydro Guy
    Commented Sep 26, 2014 at 23:34
  • $\begingroup$ @DavidHammen: That's a very good answer. One should add, that the L2 norm is physically only relevant for the free particle. Obviously, it will be trivially violated for any constrained problem (e.g. your lovely NYC cabbie!). $\endgroup$
    – CuriousOne
    Commented Sep 27, 2014 at 1:51

2 Answers 2

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If we want a sense of localness (or calculus to work), we'd like to be able to obtain the length by adding up the length from pieces of the path (for example using a ruler, or counting paces as we walk along the path between two points).

However, even considering just two dimensions we see something interesting for $L_p$.

$$\left(|x|^p + |y|^p\right)^{1/p} = \sum_{i=1}^N \left(\left|\frac{x}{N}\right|^p + \left|\frac{y}{N}\right|^p\right)^{1/p}$$

This trivially works with $p=1$, and due to a special symmetry at $p=2$ it works there as well. This will not work for other $p\neq 0$ (I am unsure of how to extend the definition to check $p=0$).

The special symmetry at $p=2$ is that the distance measurement becomes rotationally invariant. So the seemingly mundane reasons of

  • space has more than one dimension
  • locality
  • uniformity

seem to already select $L_2$ as special. Any other choice would give a preferred coordinate system, and possibly break locality.

So what would a different universe in which $L_1$ or something else is the natural choice? If you imagined an N dimensional Cartesian lattice world, so one with discrete lengths, and a clearly preferred coordinate basis, this would make $L_1$ a more natural choice.

I'm not sure of a good picture for a universe in which $L_p, p>2$ would be a natural choice. There would be preferred directions, and you could only consider an object as a whole (not in parts), which seems to suggest in such a hypothetical universe you couldn't even experience your life as a sequence of moments (which I guess would make sense if we have highly non-local physics and therefore causality is out the window).

Interesting question.

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    $\begingroup$ Yeah, I'd assumed rotational symmetry was important. Out of curiosity, are there other vector transformations that preserve the L3 norm, but not L2? $\endgroup$
    – hjfreyer
    Commented Sep 27, 2014 at 20:53
  • $\begingroup$ @hjfreyer I consider the inability to look at a path piecewise much more damning that the issue with rotational symmetry. Because of this, I'm not sure what would even be the appropriate way to discuss transformations of paths, but it would have to be very non-linear (for example we can't use linear algebra, with transformation matrices, if you'd hope to find a transformation that preserves L3). $\endgroup$
    – CuriousKev
    Commented Sep 27, 2014 at 21:21
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    $\begingroup$ This is wrong, that equation holds for every $p \neq 0$. $\endgroup$
    – user237040
    Commented May 4, 2020 at 4:35
  • $\begingroup$ The equation you wrote, $||{\bf v}|| = \sum^N ||\frac{\bf v}{N}|| = N\cdot||\frac{\bf v}{N}||$ follows from absolute homogeneity, $||\alpha{\bf v}|| = |\alpha|\cdot||{\bf v}||$. This is a property of all norms, not just $L_1,L_2$. $\endgroup$
    – Er Jio
    Commented Nov 7 at 15:02
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Comments to the question (v2):

  1. One nice property of the 2-norm (as compared to other norms, such as, e.g. the $p$-norm) is that it gives rise to an inner product via a so-called polarization trick. E.g. in the real case the polarization formula has 2 terms: $$ \langle u, v \rangle ~:=~\frac{1}{4} || u+v ||^2 -\frac{1}{4} || u-v ||^2 .$$ There is a similar 4-term polarization formula in the complex case. See also this related Phys.SE post.

  2. It seems appropriate to mention that there exist metric theories, which are not based on Riemannian manifolds with a metric tensor and its corresponding 2-norm. One class of such generalized metric theories is Finsler geometry, see e.g. arXiv.org.

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