# Geometrical objects whose volumes are fractional powers of their sizes

While studying properties of foams (imagine bubbly soap or microscopical grids/networks), I started wondering on the relationship between the volume occupied by the matter $V_s$ itself and the overall volume $V_a$ of a given foam sample (matter+voids). Two lengths characterize these objects: the size of each pore $l_a$ and the thickness $l_s$ of the matter that encloses such a pore.

You can picture a foam like a stack of cubes (the pores). When you suppose that each pore is fully closed by walls of matter, you obtain $V_s/V_a\propto l_s/l_a$. When you suppose that each pore is fully open, such that filaments are joined to each other and forming a sort of network, you obtain $V_s/V_a\propto (l_s/l_a)^2$. This is easy to picture, and comes naturally from the fact that basic geometrical shapes have a volume proportional to the powers of typical distances.

Now, many foams exhibit $V_s/V_a\propto (l_s/l_a)^k$, where $k$ is fractional, between 1 and 2. This means that the volumes are not proportional to integer powers of the characteristic sizes in the problem.

How is that possible? Is there a visual example of this situation? It seems related to fractals however I cannot pinpoint how.

I am looking for a non-technical answer as I am not a mathematician. I have basic knowledge on fractals.

UPDATE Following some answers that have already been proposed, I find it difficult to imagine a fractal object that does not introduce another length (than $l_a$ and $l_s$). Iterating the fractal keeps producing more lenghts somewhat related to each other. Then, how is it possible to define only one $l_s$ in these conditions?

• What do you mean with the overall enclosing volume $V_\alpha$?
– Newb
Oct 29, 2013 at 1:19
• @Newb, Imagine that you have a cube containing many filaments like a neural network. Then, $V_a$ is the volume of the whole cube and $V_s$ is the volume of the filaments themselves. Oct 29, 2013 at 2:24
• what do you mean by fully closed and fully open ? can you describe more ? fully closed : I picture a sphere (void) and a bigger sphere enclosing it (matter). fully open : no idea Nov 3, 2013 at 9:56
• @Thomas, the foam is constituted of many many pores, which are little void volumes next to each other. Imagine for example many cubes (the pores) stacked together filling entirely a bigger volume. If you make each of the cubes having thick walls (facets), made of matter, the pores are closed. If the cubes have no thick walls but only "strings" of matter joining their vertices, then they are open. Nov 4, 2013 at 18:07

I like this problem and the way you got to fractals as a non mathematician. You are almost there.

There can be many models for how the matter surrounds a spore.

Model 1 (flat surface): a big cube (size ls+la) encloses a (just a bit) smaller cube (size la), the smaller cube being empty. The volume of the matter is $Vs=(ls+la)^3 - la^3$, and $Vs/Va = ls/la$ (in first approximation + removing constants), you are right.

Model 2 (strings): just strings of diameter ls and length la, joining vertices of the cube (size la). $Vs=la * ls^2$ and $Va=la^3$. This time $Vs/Va=(ls/la)^2$ (again, not including constants and this is an approximation when $ls/la -> 0$)

Model 3 (soap): the matter is a bubble of soap. It is thick at the boundaries (the vertices) and not thick at the center of the cube, like a 2d parabole. The equation of the thickness is $1 - 16 * x *(1-x) * y*(1-y)$ on cube of size 1 (just rescale x -> x*la and y -> y*la). I am just inventing here, I haven't read any papers, but it could also be the solution of a partial differential equation with fixed boundaries like a laplacian=0 equation, or whatever that SCALES, i.e the shape of matter will be the same whether the cube is big or small. Then you will still end up having $Vs/Va=(ls/la)^2$. For a (simple) demonstration of this, consider that the equation of the thickness is f(x, y) with $x in [0, 1] and y in [0, 1]$ and the volume is $integral(integral(f(x, y) dxdy))$. When you apply this to the foam you have to rescale, which means x -> la*x etc etc and the volume is ls * la * la * C, C=integral over [0,1]x[0,1]. Roughly saying here that you choose the shape of your matter, then when you reduce it, the ratio stays (of course) the same.

Model 4 (fractal) : instead of a soap parabole you decide to have a constant thickness distributed evenly on a fractal drawn on the cube's face (if you don't picture it I will edit something but it should be easy to imagine. Just think of a fractal (like sierpinsky's triangle) and color it on the face of a cube, and imagine there is 1inch of thickness where it is coloured and nothing on the face elsewhere. Vs=ls * FractalArea(la) and Va=la^3. The thing with fractals is that their length/area/volume doesn't follow usual geometry rules. If you have a square and multiply its length by x, its area is multiplied by x^2 (because dimension 2). If it is a cube, the volume is multiplied by x^3. With fractal of dimension d, the area is multiplied by x^d. So when the size of the spore and $ls/la -> 0$, FractalVolume(la) goes like ls * la^d with d between 2 and 3 (because this is a fractal drawn on the 2d surface of the cube). So $Vs/Va = ls*la^d / la^3 = ls / la^(3-d)$. This is not exactly your formula but Im not a fractal specialist nor a foam specialist and the model is not good here, what I have described is impossible in reality (constant thickness). If the thickness was "fractalian" too we could have another formula. But still we see that fractals can bend the relation and have their dimension in the equation.

If this is too far stretched, just imagine this : Because models 1 and 2 are the extremes (constant surface = matter on 2 dimensions = power of 2 in the approx vs no surface = matter along 1 dimension = power of 1 in approx) you can imagine that when the distribution of the matter is in between this (a lot on the edges and not much on the center of the faces -> string -> power of 1+smthg, a lot everywhere -> power of 2-smthg) you get a constant power in between 1 and 2 (if the distribution is fractalian). If the distribution is anything on the surface that follows some equation, even complicated like cos(x)*sin(y)^2, you will have the (ls/la)^2. The difference between fractals and 2d equations is that the "f(x, y)" changes when you reduce the size of the foam and of the cube (making ls and la -> 0) so at every step it is of the form Vs/Va=C(ls, la)(ls/la)^2 but the constant C(ls, la) will change over time, increasingly if the fractal takes more and more space (bigger dimensions), decreasingly if it is shorter and shorted. And it is this effect that gets factorized into the power

Hope it is clear, don't hesitate to comment on that

http://en.wikipedia.org/wiki/Fractal_dimension

• This answer points to the right direction and I have similar pictures in mind to explain how fractals can exist in foams. But still it does not answer the question. How come the fractals have a volume that is not an integer power? You give the example of sierpinsky's triangle, but its area at infinite iterations is zero, and at finite iterations, it is an integer power of it's size ! Where does the area, or volume, become a non integer power ? Nov 9, 2013 at 7:35
• Yes and no, it depends on the space available to the fractal. Like atoms can't be divided. Assume for instance that you can go 1 iteration further in the fractal only if the minimum size is greater than some threshold. Then the limit of 0 is not reached. It's like a drop of water. Maybe it can't be smaller that 0.1 ml because of interaction forces, or a snowflake can't have edges smaller than something because of the weight of the snow. Because of this physical limit, the number of iterations can't be infinite, and you have the non integer power relation Nov 9, 2013 at 7:56
• For Sierpinskiy : you start with area0 = (roughly) la^2 and area(n)=la^2 * (3/4)^n. edge0=la and edge(n)=la/(2^n) You will stop when edge(N) ~= ls. So this gives N=log(la/ls)/log(2) and if you put this back into area(N) you have area = la^2 * (la/ls)^(log(3/4)/log(2)) which is of the form we mentioned Nov 9, 2013 at 8:04
• I have a hard time seeing how you can identify edges(N) with ls. It is not clear to me. Mybe this triangle is a bas example. Nov 9, 2013 at 23:49
• edge(n) is the size of the edge of a sub sub sub (n times) triangle. edge(0) = la. edge(n) is divided by 2 at each iteration. You stop dividing by 2 when the size of the edge is small enough (ls) Nov 10, 2013 at 15:26

Vihart has a couple of recent videos on fractals and things like size and length related to them. Apart from being really entertaining videos themselves I believe you can gain some insight into your problem.

Doodling in Math Class: Dragon Dungeons

Doodling in Math Class: Dragon Scales

The later part especially talks about how fractals' lengths, areas and volumes scale up when you expand them.

• The second video was very helpful. Now, how could this be applied to a foam? Is there an example which identifies ls and la? And it seems that it requires infinite iterations to obtain fractional power: how does that apply in physics, where you don't do infinite iterations? Nov 12, 2013 at 3:09

This is going to be more of a pictorial answer, and I will leave out the math. Hopefully it is more enlightening this way. There are two ways you can consider expanding your fractal, in 2 dimensions and in 3. If we only look at the case where our "bubbles" are actually cubes, which makes this argument simpler to see, we have these basic pictures for both types. Let an iteration be, for 2 dimensions, dividing each area in 4 equal parts. For 3 dimensions, let an iteration be dividing our cubes into 8 equal parts. Below are the first four iterations of both 2 and 3 dimensions.

Both of the 2 dimensional and 3 dimensional pictures are of the form you mentioned where we are "open," and therefore give a proportion of $V_s/V_a \leftrightarrow (l_s/l_a)^2$. The other option is to use cubes with solid sides, which gives the $V_s/V_a \leftrightarrow (l_s/l_a)$ (Not pictured.) One does not really have to stretch their imagination to see that if we take the limit the 2 dimensional squares as we iterate the fractal, we would end up with a solid square. So, to increase the power of our proportion, we want to make our 2 dimensional squares a higher iteration, BUT, we want to keep our 3 dimensional iteration lower. This will give us a higher value of $V_s$ and $l_s$, but keep our $V_a$ and $l_a$ the same. The difference is that we know this must get us a lower power because if we let the sides, or 2 dimensional pieces tend toward the solid squares, we must get our $V_s/V_a \leftrightarrow (l_s/l_a)$ relation. As this is, in some sense, continuous and well behaved, we see that the in between steps will give $V_s/V_a \leftrightarrow (l_s/l_a)^k$, for $1\leq k \leq 2$. (Here is a cut away picture to show what I mean. This would be the 3 dimensional iteration done once and the 2 dimensional iteration done twice after that.)

This tells us that for a random foam, we will get a fractional power proportion relation when we increase the number of divisions of the surfaces area of the bubbles and not divide the volume contained in them as much. One more picture, which might help. Consider every colored section of the sphere to be a subdivision, or "2 dimensional" iteration while keeping the volume contained in them the same.

• Your explanation seems to be different from fractals. There are actually more lengths than just la and ls. The fractional power looks like it appears like an artefact of one of these lengths being somewhat related to the others. Nov 9, 2013 at 23:55
• Actually, i don't see how your example with squares is a fractionnal power. I can calculate the occupied volume Vs by summing small volumes which sizes are fractions of ls or la. Thus the total volume Vs is still an integer power. Nov 10, 2013 at 1:52
• @fffred Have you tried it for more of the 2 dimensional iteration while keeping the 3 dimensional iterations constant, but more than 1? I find my values of $k$ are changing when I plug through the calculations. Nov 10, 2013 at 4:14
• I think we have different calculations. Doing an iteration in 3d is meaningless because it changes the size of the pores: one 3d iteration multiplies the number of pores by 8. So, the repeated pattern is the same but smaller. Only the number of 2d iterations above the number of 3d ones matters. 3d iterations just displace the problem. Now, you're left with a box having holes in its walls. Its size is la and wall thickness ls. I get Vs=3*lalals*(3/4) approximately. So, no fractional power. Nov 10, 2013 at 5:20
• Since it is a fractional power, it is harder to see where it comes up. Instead think about the limiting behavior of doing the 2D iterations. I hope you agree that this will result in solid walls, which has the linear relation, as opposed to the quadratic. Well, this process will not have quadratic relation for all iterations and then suddenly at infinity jump to a linear relation. It does this in some vaguely continuous pattern. So, as we iterate our 2D, we must start getting this fractional relation showing up, even if it is hard to see in our calculations. Nov 10, 2013 at 16:07