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The distance formula in one dimension is

$$D_1 = |x_2-x_1|$$

From the Pythagorean theorem, the distance formula in two dimensions is

$$D_2 = \sqrt{|x_2-x_1|^2 + |y_2-y_1|^2}$$

Now, in three dimensions, a student might innocently suppose that

$$D_3 = \sqrt[3]{|x_2-x_1|^3 + |y_2-y_1|^3 + |z_2-z_1|^3}$$

yet the Pythagorean theorem shows otherwise. Is there any physical intuition behind distances defined by $D_3$? What strange type of ruler would be needed to measure such a thing?

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  • $\begingroup$ Thanks Brian, I have clarified this. $\endgroup$ – Doubt Feb 14 '13 at 22:27
  • $\begingroup$ Actually in finite dimensional vector spaces any two norms are equivalent (mathematically speaking). $\endgroup$ – Daniel Robert-Nicoud Feb 14 '13 at 22:30
  • $\begingroup$ @DanielRobert-Nicoud: The OP isn’t interested in the topology generated by such norms. Rather, he’s interested in their geometrical properties. $\endgroup$ – Haskell Curry Feb 14 '13 at 23:03
  • $\begingroup$ How about using $D_1 = \sqrt{(x_2-x_1)^2}$ instead of $D_1 = |x_2-x_1|$? $\endgroup$ – steven gregory Nov 23 '18 at 22:34
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Instead of talking about distances $D(x,y)$ it is slightly easier to talk about norms $$N(x-y).$$

Your proposes $D_3$ is called the $L_3$ norm and is part of a family of norms $$\|x\|_p = \left(|x_1|^p + \ldots + |x_n|^p\right)^{1/p}$$ called the $L_p$ norms. For $p\geq1$ they obey all of the properties needed for a norm: positive for nonzero $x$, linear under multiplication by positive scalars, and satisfy the triangle inequality.

$p=2$ is the usual Euclidean norm and induces the usual Euclidean distance. In one dimension, all of the $L_p$ norms are equivalent, which is why you interpreted the $L_2$ norm as the $L_1$ norm.

So to answer your question, nothing goes wrong when you use the $L_3$ norm. You will get lengths that are somewhere between using the usual $L_2$ norm, and the $L_{\infty}$ or "max" norm which returns the magnitude of the largest component; as such, components of a vector that are small relative to the largest components will tend to get ignored by an $L_3$ ruler.

Here's what the unit circle would look like using such a norm:

enter image description here

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    $\begingroup$ I suspect that the OP knows this. The question is whether this norm has any natural physical interpretation, in the same way that the $L^2$ norm is ordinary Euclidean distance. $\endgroup$ – Brian M. Scott Feb 14 '13 at 22:36
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    $\begingroup$ As the question brings up the $L_3$ norm in the context of three dimensions, I think this image might be more appropriate. :) $\endgroup$ – Rahul Feb 15 '13 at 1:07

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