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As can be seen in the David Morin's Classical Mechanics, there are some scaling strategies in order to calculate the moments of inertia of certain fractals, for example, the Cantor's set has a moment of inertia of $I_C = \displaystyle \frac{ml^2}{8}$, Sierpinski's Carpet has a moment of inertia of $I_S = \displaystyle \frac{3ml^2}{16}$ and Sierpinski's Triangle has a moment of inertia of $I_T=\displaystyle\frac{ml^2}{9}$ where m is the total mass of the set and l it's longitude of its sides.

However this doesn't involve any integrals, that's why I'm asking this. Is there any way to calculate this moments of inertia by meanings of the usual integral? For Cantor for example:

\begin{equation} I = \int_C x^2 dm = \frac{m}{l} \int_C x^2 dx \end{equation}

Where $dm=\rho dx = \frac{m}{l}dx$)

At first sight, this integral has a value of 0, as the cantor set has Lebesgue measure cero. For physical "reasons" a finite number is expected, that's why I thought that maybe I need to consider some kind of "fractional integral" over the Hausdorff dimension of the Cantor's set. This dimension is $d=\displaystyle \frac{log 2}{log 3}$. So maybe should I try to compute this?

\begin{equation} \int x^2 \chi_C dm_d \end{equation}

With $dm_d$ the d-dimensional Hausdorff measure, and $\chi_C$ the characteristic function over the Cantor's set.

I ask a similar question before, and an answer was given by John Dawkins involving a probability approach:

I'll take $m=1$ and $l=1$. The natural "uniform" mass distribution on the standard Cantor set is the distribution of the random variable $$ X:=\sum_{n=1}^\infty {\xi_n\over 3^n}, $$ where $\xi_1,\xi_2,\ldots$ are i.i.d. with $\Bbb P[\xi_n=0]=\Bbb P[\xi_n=2]=1/2$. (This distribution is also the normalized Hausdorff measure of dimension $d=\log 2/\log 3$.) Clearly $\Bbb E[X] =1/2$, while the variance of $X$ (a.k.a. the moment of inertia about the vertical axis throough the mean of $X$) is $$ \eqalign{ \Bbb E[(X-1/2)^2] &=\Bbb E[(\sum_n{\xi_n-1\over 3^n})^2]\cr &=\Bbb E\sum_{m,n}{\xi_m-1\over 3^m}{\xi_n-1\over 3^n}\cr &=\sum_{m,n} 3^{-m-n}\Bbb E[(\xi_m-1)(\xi_n-1)]}. $$ The expectations in this double sum are $0$ or $1$ according as $m\not=1$ or $m=n$; therefore $$ \eqalign{ \Bbb E[(X-1/2)^2] &=\sum_{n=1}^\infty 3^{-2n} = {1\over 8}.\cr } $$

So now I ask if this can be used to obtain the moment's of inertia of the other two fractals above or the integral approach is required or something else. Of course it would be extremely helpful if you could show me any bibliography where this is solved or if you can find any new methods in order to solve this. Maybe some fractional calculus is required or something similar to it.

Rafa
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  • The probability approach is pretty similar in spirit to integrating against the Frostman measure of the Cantor set. –  Mar 26 '16 at 16:35
  • @Rafu Try reading some more of the posts in this tag. Including this highly relevant post. – Zach466920 Apr 21 '16 at 13:06

1 Answers1

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The short answer is no. The mass distribution is mathematically modelled by a measure on the Cantor set. However, this measure, including the self similar one you considered, are not absolutely continuous with respect to the usual Lebesgue measure. You therefore cannot define a Radon-Nikodym derivative, i.e. you cannot write it as the integral of the Lebesgue measure with a suitable mass density.

Formally, you can therefore legitimately write: $$ I = \int x^2dm $$ but the passage: $$ \int x^2dm = \int x^2\rho dx $$ makes no sense since the mass density $\rho$ is not well defined. The next step of assuming $\rho$ constant and factoring out is even more problematic. This is why you can compute it with similarity arguments or approximation methods as you did.

Hope this helps.

LPZ
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