Posted by: dangillis | April 14, 2011

## That’s Unpossible! Final Act: A Squishy Dimension

Click to embiggen this fractal. Prepare to have your mind blown.

Several days ago we went on an adventure exploring the Koch Snowflake.  During our exploration, I told you to trust me when I wrote that

$\displaystyle{\sum_{i=0}^{\infty}} \left(\frac{4}{9}\right)^{i}=\frac{9}{5}$.

A few days later, we learned that if $|p|<1$, we could write

$\displaystyle{\sum_{i=0}^{\infty}}p^{i}=\frac{1}{1-p}$.

And since in our example $p=\frac{4}{9}<1$, my original equation was correct.

Of course, I also stated that the snowflake somehow existed in a space that wasn’t quite 1 dimensional, but it also wasn’t quite 2 dimensional.  That is, it existed in a non-integer dimension.  What the what?  That’s crazy talk, you say.  Probably.  But let’s explore this further.

Image via Wikipedia

Most of you have probably heard of a Fractal.  The basic idea here is that you have an object that has a property known as self-similarity.  In a nutshell, if I were to zoom in on any part of the object, it would look like the original object.  If you think about it, one could easily get lost in a fractal because every region looks like every other region under some magnification.  It’s almost trippy if you think about it too long.

Regardless, the important point is that fractals are self-similar.  Can we say the same about other shapes?  And if so, what might this tell us about dimension?

Consider something in a 1 dimensional world: a line of length 1.  If we magnify the line 2 times, clearly we would end up with 2 copies of the original line.  That is, a 2x magnification gives us 2 copies of the original object.  If we magnified the line 3 times, we’d have 3 copies.  In fact, if we magnified the line $n$ times, we’d have $n$ copies.

What about something in 2 dimensions such as a square with side length of 1?  If we magnify the square by a factor of 2, you can imagine that we would end up with 4 copies of the original square.  In this case, a 2 times magnification gives us 4 copies of the original object.  A 3 times magnification would give us 9 copies, and in general, an $n$ times magnification would give us $n^{2}$ copies.

In a similar manner, a regular object in 3 dimensions would give us 8 copies if the original were magnified 2 times.  And in general, an $n$ times magnification would give us $n^{3}$ copies.

To recap:

• Scaling up something by a factor of $n$ in 1 dimension gives $n$ copies of the original,
• Scaling up something by a factor of $n$ in 2 dimensions gives $n^{2}$ copies of the original, and
• Scaling up something by  a factor of $n$ in 3 dimensions gives $n^{3}$ copies of the original.

In general, if we scale something up by a factor of $n$ in an $m$-dimensional space, we will end up with $n^{m}$ copies.  Cool!  That is, our scaling value $n$ raised to the dimension $m$ gives us the number of copies $s$.  In mathy terms

$s=n^{m}$.

We’re going to rearrange this in order to determine the dimension of a fractal.  That is, we need to solve for $m$.  Taking the $\log$ of both sides and isolating $m$ we get

$m=\frac{\log{(s)}}{\log{(n)}}$.

So, the dimension $m$ of an object is simply the ratio of the $\log$ of the number of copies $s$ one gets by scaling the object by a factor of $n$, to the $\log$ of the scaling factor $n$.  Sweet freaking awesome.

In the case of the Koch Snowflake, remember that we started with a line of length 1.  We then removed the middle third and inserted 2 lines of equal length (both being one-third the length of the original line).  If we were to magnify the line by a factor of 3 so that each of the line segments of length one-third looked like the original line segment of length 1, you’d see that we’d have 4 copies of the original line.  That is, $s=4$ while $n=3$.  This gives us a dimension of

$m=\frac{\log{(4)}}{\log{(3)}}\approx 1.262$.

Cripes on a cracker - an integer dimension.  That is, the Koch Snowflake isn’t quite 1 dimensional, and it isn’t quite 2 dimensional either.  Wild stuff!

Zooming in on the Koch Snowflake. No matter how much we zoom, the edge is always a bit fuzzy. Image via Wikipedia

For those that think this is all bunk, consider this: the Koch Snowflake starts with a line.  We then remove the middle third and add 2 line segments measuring one-third the original.  The result is a bent line.  If we continue this forever, we still end up with a bent line.  More specifically, an infinitely bent line.  So no matter how much we zoom in, it’s always going to look fuzzy (due to the infinite bending).   We won’t ever see crisp lines – things that exist in the 1 dimensional world.  But by the same token, we can’t say that the fuzziness that we see represents something that exists in the 2-dimensional world either.  Technically the lines have only length, but no width.  So clearly this object exists in some sort of nexus between these dimensions.  How mind-blowing-ly awesome is that?  Personally, I find the idea of a non-integer dimension completely satisfying.

Happy Mathematics Awareness Month all y’all!

## Responses

1. [...] Whilst perusing the Twitterverse, I stumbled on something most excellent.  Something that goes by the name of Hexaflake, which is a type of N-flake.  I won’t get into N-flakes, but trust me, they are cool.  Instead, I’m going to chat a little bit about this Hexaflake thing-y.  What it is, some of its very cool properties, and how it’s related to the Koch Snowflake that we talked about here, here, and here. [...]

2. [...] we’ve discussed the Koch Snowflake, Hexaflakes, infinite boundaries with finite areas, and non-integer dimensions.  You know, cuz we are nerdy like that.  Okay, maybe that’s just [...]

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