# Fractal dimensional analysis?

I know how to use a ruler to approximate a length of an object (like a wire or a stick) in meters. I could also use the ruler to approximate a two dimensional area (like a table top or a parking lot) in $$\text{meters}^2$$ by dividing it into a grid and counting squares.

I read that we can estimate the length of the coast of Britain to have fractal dimension 1.25. Is there a value in $$\text{meters}^{1.25}$$ giving the 1.25-dimensional fractal measure of that coast? Can I calculate it using my ruler or rulers of different sizes/precisions? Or if I drew a Koch snowflake whose largest triangle had side length $$1\text{ meter}$$, could I find it's $$\ln(4)/\ln(3)$$-dimensional measure in some analogous way?

If there is such a thing as fractional-dimensional measure for dimension $$d$$, can we give a fractional unit like $$\text{meters}^d$$ physical meaning?

• Don't have time for a full answer, but the idea of Hausdorff Measure plays a role. Aug 28, 2017 at 21:32
• If a full answer is too long to fit nicely here at least describe some ideas and scenic points along the route. Jan 9, 2020 at 5:56
• Zhu, Zhi Wei; Zhou, Zuo Ling; Jia, Bao Guo (October 2003). "On the Lower Bound of the Hausdorff Measure of the Koch Curve". Acta Mathematica Sinica. 19 (4): 715–728. doi:10.1007/s10114-003-0310-2 claims the $\log_3(4)$-dimensional Hausdorff measure of the Koch curve is between $0.5$ and $0.589$ (the upper bound is a result from a different paper) Jan 9, 2020 at 14:59
• @Claude See the edited the bounty notice. An explanation of the difficulties in handling the Koch snowflake will help the community. Or a study of some other fractal :-) Jan 10, 2020 at 14:53
• For what it's worth I've learned a simple, non-rigorous way to understand Hausdorff dimension. Looking closely at any "edge" of the Koch snowflake, you can see it is self similar to 4 copies of itself, and it is 3 times bigger than any of these copies. Hence $\log(4)/\log(3) = log_3(4)$ is the Hausdorff dimension. This agrees with the dimension of Euclidean space in that if you look at a cube of space in $\mathbb{R}^n$ you can see it's broken up into $m^n$ cubes of side-length $1/m$ the original. That gives Hausdorff dimension of $\log(m^n)/\log(m) = n$. Can Hausdorff meausure be related? Jan 10, 2020 at 20:39

Definition: A metric space is a set $$X$$ together with a distance function $$d(x,y)$$ that assigns a nonnegative number to any pair of points $$x$$ and $$y$$ from $$X$$, and $$d$$ satisfies:

• $$d(x,y) = 0 \iff x=y$$.
• $$d(x,y)=d(y,x)$$, and
• $$d(x,y) \leq d(x,z) + d(z,y)$$ ("the triangle inequality").

Definition: for a subset $$A \subset X$$, $$\operatorname{diam} A := \sup \{d(x,y): x \in A, y \in A \} .$$

Definition: for a point $$x \in X$$, and $$r>0$$, the ball of radius $$r$$ centered at is $$B(x,r): = \{ y\in X: d(x,y) < r \}.$$

The story of the Hausdorff measure/dimension begins by extending our Euclidean lengths, areas, and volumes to arbitrary dimensions, in a rather crude way: in Euclidean $$\mathbb{R}^n$$, the $$n$$-dimensional volume of a ball is $$\omega_n r^n,$$ where $$\omega_n$$ is a constant independent of radius. Notice that if radius is scaled by $$c>0$$ the volume scales by $$c^n$$. The numbers $$\omega_n$$ are the all famous ones in dimensions 1, 2, and 3 and for the higher dimensions $$\omega_n$$ is obtained from $$\omega_{n-1}$$, recursively, by applying Fubini's theorem: a slice of $$n$$-sphere is an $$(n-1)$$-sphere with radius depending on the height at which we slice. Think about cutting a potato, a 3-ball, and getting (2D) disks!

It follows also that $$\omega_n$$ has an explicit formula via the Gamma function. Therefore, one can plug in noninteger values in place of $$n$$'s and (formally) define $$\omega_\alpha$$.

(In what follows $$\alpha \in [0,\infty)$$.)

Assume you have a metric space and you suspect that it is $$\alpha$$-dimensional. Then you should assign an $$\alpha$$-dimensional volume of $$\omega_\alpha r^\alpha ,$$ to the balls of radius $$r$$, i.e $$B(x,r)$$.

And a subset $$A$$ compared to a ball of same diameter has volume "comparable" (no precise argument here) to $$\omega_\alpha (\frac{\text{diam} \ A}{2})^\alpha.$$

Now, given a subset $$E\subset X$$, to define its $$\alpha$$-dimensional Hausdorff measure, $$H^\alpha(E)$$, we first find for a $$\delta >0$$ the quantity $$H^\alpha_\delta (E) = \inf \sum_i \omega_\alpha (\frac{\text{diam} \ A_i}{2})^\alpha,$$ where infimum is over all countable coverings $$\cup_{i=1}^\infty A_i \supset E$$ with no restrictions on $$A_i$$'s other than $$\operatorname{diam} A_i < \delta$$ for each $$i$$. That is, we look for the tightest coverings with sets of diameter less than $$\delta$$.

(In the definitions of $$H^\alpha_\delta$$ and $$H^\alpha$$ we allow infinities.)

Now, $$H^\alpha(E) = \lim_{\delta \to 0} H^\alpha_\delta(E).$$ Notice that smaller $$\delta$$ makes $$H^\alpha_\delta$$ larger, since we do not allow some of the coverings that previously worked, hence the infimum goes up. So, this limit exists since $$H^\alpha_\delta$$ increases as $$\delta \to 0$$.

With all this buildup we are ready to harvest the fruits!

Lemma: Given any metric space $$(X,d)$$, any subset $$A \subset X$$, there exists an $$\alpha_0 \in [0,\infty)$$ such that \begin{align} H^\alpha(A) &= 0 &\forall \alpha > \alpha_0, \\ H^\alpha(A) &= \infty &\forall \alpha < \alpha_0. \end{align}

We call this $$\alpha_0$$ the Hausdorff dimension of the (sub)set $$A$$. The theorem tells us that too large measuring cups will not detect $$A$$ and smaller measuring cups won't contain $$A$$, which is quite intuitive and geometric: the area of a line is zero, and length of a (solid/filled) square is infinite.

This is how and why of a dimension of $$1.25$$ for the coast of Britain—this is the right $$\alpha_0$$ for it. Now, you ask then how much is this measure? This is the same as asking what happens right at $$\alpha = \alpha_0$$, which the lemma above leaves out?

There are three possibilities: $$H^\alpha(E) = 0$$, or $$=\infty$$, or it is a finite positive number. In each case the value deserves to be called the $$\alpha$$-dimensional measure/volume of $$E$$. Easy examples can be given for all cases. For instance, although a plane, extended indefinitely, is 2D, its 2D area is infinity nevertheless.

With specific and explicit metric spaces, specially those constructed through self-similar processes, it is usually straightforward to find the dimension—using the steps of the construction to find coverings that apply to the definition of $$H^\alpha_\delta$$. But finding the actual value $$H^\alpha(E)$$, may prove more difficult.

I do not know if this counts as "physical meaning" to non-integer Hausdorff dimensions and measures, but to me it is so natural and geometric!

• This is a helpful explanation of the Hausdorff dimension. Can you give an example of a set of a fractal Hausdorff dimension AND calculate its Hausdorff measure (of the prescribed dimension), non-zero and finite. Jan 12, 2020 at 5:52
• I know for the standard 1/3-Cantor set its appropriate dimensional measure is nonzero and positive. I am looking for a reference and if I find i will share it here... Jan 12, 2020 at 14:25
• The book "Geometric Measure Theory, a beginner's guide" by Frank Morgan, 4th edition, in exercise 2.6 of chapter 2 we are asked to show that the usual Cantor set, $C$, its H-dimension is $\alpha=\frac{\ln 2}{\ln 3}$. The exercise asks us to "try to prove that $H^{\alpha}(C)=\frac{\omega_\alpha}{2^\alpha} \ ,$ or at least that $H^\alpha(C) > 0.$" Jan 12, 2020 at 14:41
• That book, BTW, has many rather concrete examples of fractals and exotic spaces. Jan 12, 2020 at 14:43
• It is a worthy exercise! Jan 12, 2020 at 15:02