Infinitizing Discrete Annulus

Historically, the development of calculus was accompanied with introducing the concept of infinitesimals. Nowadays, discrete structures are often approximated by contiguous models, assuming that – if the discrete structure grows siffuciently large – from a bird’s eye view, the difference between the actual discrete reality and the contiguous approximation is neglectable. Motivation for modelling discrete structures with real-valued contiguous models is that many mathematical operators are applicable in a real-valued context only.

Approximation of discrete structures by real-valued models works surprisingly well in many cases, but fails or requires unexpected correction factors when switching between the models. Such correction factors, in fact, may be highly relevant in physics when modelling quantized properties by real-valued models on a large scale. In the end, such factors even may be responsible for still unexplained phenomena such as dark matter (which might as well turn out to be simply a correction factor for gravity on large-scale objects, rather than anything really existing). Correction factors may also play a key role for understanding mysterious numbers like the fine-structure constant, or even the muon anomaly.

In this short article, we examine such an unexpected correction factor that is required when switching from a discrete model to its real-valued counterpart, using an annulus as a very simple example. In summary, the example in this article demonstates the pitfalls that you may encounter when using contiguous mathematics (or, more specifially, the calculus) in the domain of discrete structures such as in quantum physics.

Notational Preliminaries

For the remainder of this article, \(ℕ\) denotes the set of natural numbers, that is, positive integer numbers.

We use the letter \(A\) to denote the area of some shape. For example, we use the notation \(A_\circ(r) = \pi{}r^2\) to denote the area of a circle with radius \(r\). The floor function \(\lfloor{}x\rfloor\) is used to convert real numbers into integer numbers: it denotes the largest integer number that is not above the argument \(x\).

Next, let us take a look at the real-valued annulus and then at the discrete annulus.

Contiguous Annulus

In contiguous or real-valued mathematics, an annulus is the difference between the area of two circles of different size, located in the same plane and sharing the same center point.

Fig. 1: Annulus Radius and Width — Fig. 1: Annulus with
Width w and Radius r=3w

For simplicity of our model, rather than dealing with to separate radiuses \(r_1\) and \(r_2\) for the smaller and larger circle, respectively, we instead define our annulus by a single radius \(r\) that refers to the center line of our annulus, and the width \(w\), that represents the stroke width of our annulus. Given \(r\) and \(w\), we define \(A_c(r, w)\) to denote the area of the contiguous annulus with radius \(r\) and width \(w\). Since \(r\) refers to the center line of our annulus, we essentially have \(r_1 = r - \frac{1}{2}w\) for the smaller circle, and \(r_2 = r + \frac{1}{2}w\) for the larger circle.

Area \(A_c\) of our contiguous annulus resolves as the difference between the larger and the smaller circle as follows:

\( \begin{array}{rcl} A_c(r, w) & = & A_\circ(r_2) - A_\circ(r_1)\newline & = & \pi{}r_2^2 - \pi{}r_1^2\newline & = & \pi((r + \frac{1}{2}w)^2 - (r - \frac{1}{2}w)^2)\newline & = & \pi(r^2 + rw + \frac{1}{4}w^2 - r^2 + rw - \frac{1}{4}w^2)\newline & = & \pi(rw + rw)\newline & = & 2\pi{}rw. \end{array} \)

In our graphically depicted example, we chose \(r=3w\), such that for this specific example,

\(A_c(3w, w) = 2\pi{}3w^2 = 6\pi{}w^2\).

Discrete Annulus

Now imagine we want to draw the same annulus on a rasterized background, just like drawing on a bitmap image. That is, we construct an approximation of the previously contiguous annulus by putting square tiles on the raster. We assume that the square tiles have exactly the same width (and also height, since we assume that they are squares) as the stroke width that we used for the contiguous annulus. Given again \(r\) and \(w\), we define \(A_d(r, w)\) to denote the area of the discrete annulus with radius \(r\) and width \(w\) that best approximates its contiguous counterpart. The best approximation that we can find for our depicted specific example consists of exactly \(16\) tiles, as you can see if you count the tiles in the figure.

Let us now compute the area for this discrete annulus. Since we assumed that each tile has width and height \(w\), the area of a single tile is \(w^2\). Since the best approximation that we could find for our specific example has exactly \(16\) tiles, the overall area is

\(A_d(3w, w) = 16w^2\)

in our specific example.

Specific Ratio \(C_3 = A_d(3w, w) / A_c(3w, w)\)

Naively, we would expect, that the contigous annulus and the discrete annulus will have approximately the same area, at least on the large scale for large radiuses. So, let us compute the ratio between the two areas, expecting a values close to \(1.0\), that converges to exactly \(1.0\) for large radiuses.

Comparing the discrete annulus’ area to that of the contiguous for the same width and radius for our two specific depipted examples (remember, both with \(r=3w\)), we get a ratio or correction factor \(C_3\) of

\( \begin{array}{rcl} C_3 & = & A_d(3w, w) / A_c(3w, w)\newline & = & 16w^2 / (6\pi{}w^2)\newline & = & \frac{16}{6\pi}\newline & = & \frac{8}{3\pi}\newline & \approx & 0.8488. \end{array} \)

That is, the area of our approximated rastered annulus is more than \(15\%\) smaller compared to the area of the contiguous annulus, at least for our specific example with \(r=3w\).

Counting Tiles with Bresenham

For \(r=3w\) we get ratio \(C_3\approx0.8488\). What about larger values of \(r\)? Can we generalize ratio \(C_n\) for \(r=nw\) for any positive integer value of \(n\)? Does the ratio converge to the value \(1.0\), as we would naively expect? If not, does the value of \(C_n\) at least converge to a stable correction factor for \(n\to\infty\)?

Yes, we can generalize our result! All we have to know is how many tiles we need to rasterize an annulus with \(r=nw\) for any \(n\in{}ℕ\). For counting tiles, we first have to specify how a annulus is rasterized at all. We choose to apply Bresenham’s algorithm adapted for annuli.

Fig. 3: Raster Width of South East Octant — Fig. 3: Raster Width of
South East Octant

We devide the annulus into \(8\) octants of equal (as close as rasterization permits) length and have a look at the octant that starts at the bottom point of the annulus, following its contour counterclockwise to the right until we have covered \(1/8\) of the annulus. Note that everywhere in this octant, the slope is not larger than \(1.0\). That is, when we move along the octant from left to right, raster column by raster column, height may change at most by one raster row. That is, there are not gaps in the raster with respect to height, and it is sufficient (but also necessary) to plot a single tile for each raster column that we cross from left to right in this octant in order to draw a contour without gaps, but also without drawing the same tile multiple times. So, the number of tiles in the octant that we draw equals the horizontal extent of the octant, that is the number of raster columns that the octant covers. The width of an octant of an annulus with radius \(r\) is \(\frac{1}{2}\sqrt{2}r\). Accordingly, for an annulus’ radius of \(n\) raster columns, the octant horizontally covers

\(\lfloor{}n\sin(\pi/4)\rfloor = \lfloor{}n\cos(\pi/4)\rfloor = \lfloor{}n\frac{1}{2}\sqrt{2}\rfloor\)

raster columns. Actually, when rounding to the floor value, we risk a single tile gap between two adjacent octants, while, when rounding to the ceiling value, we risk to draw the same tile twice, once for each of the two adjacent octants. Therefore, in sum for all of the \(8\) octants, our calculation of tiles for the rasterized annulus may be off by at most 8 tiles. However, when looking at the asymptotic behavior for \(n\to\infty\), it does not make any difference, if we use the floor or ceiling function for rounding.

So, we have calculated the number of tiles for a single octant. Since we have to consider all of the \(8\) octants to get a full annulus, we have to overall draw \(8\lfloor{}n\frac{1}{2}\sqrt{2}\rfloor\) tiles for an annulus with radius \(r=nw\).

General Ratio \(C_n = A_d(nw, w) / A_c(nw, w)\)

Now we can implement a program that computes the ratio values for any \(n\). Starting with \(n=2\) and iterating until \(n=49\), we see that the ration wildly jumps in the range of roughly \(0.6366 \dots 0.8988\).

Java Code for Linear Scale Ratios for \(n\in\lbrace1 \dots 49\rbrace\)

public class AnnulusRatioLinear
{
  private static final double lp = 1.0;
  private static final double halfSquareRoot = Math.sqrt(0.5);

  public static void main(final String argv[])
  {
    for (int i = 2; i < 50; i++) {
      final long r = i;
      final double a1 = 2 * Math.PI * r * lp;
      final double a2 = 8 * ((long)(halfSquareRoot * r / lp)) * lp * lp;
      System.out.printf("%d %f%n", r, (a2 / a1));
    }
  }
}

Fig. 4: Ratio Convergence for Small-Scale Linear Radiuses — Fig. 4: Ratio Convergence for
Small-Scale Linear Radiuses

It even seems unclear if the ratio converges against some limit for large values of \(n\). Therefore, we change the loop to produce exponentially growing radiuses for the range \(n\in\lbrace2^1, \dots, 2^{59}\rbrace\). We do not go beyond values of around \(2^{60}\), since we will see rounding errors or even number overflows when we get too close to the numeric resolution limit of \(64\) bits of double values.

Java Code on Logarithmic Scale Ratios for Exponential \(n\in\lbrace2^1, \dots, 2^{59}\rbrace\)

public class AnnulusRatioLogarithmic
{
  private static final double lp = 1.0;
  private static final double halfSquareRoot = Math.sqrt(0.5);

  public static void main(final String argv[])
  {
    for (int i = 1; i < 60; i++) {
      final long r = 1l << i;
      final double a1 = 2 * Math.PI * r * lp;
      final double a2 = 8 * ((long)(halfSquareRoot * r / lp)) * lp * lp;
      System.out.printf("%d %f%n", r, (a2 / a1));
    }
  }
}

Fig. 5: Ratio Convergence for Large-Scale Logarithmic Radiuses — Fig. 5: Ratio Convergence for
Large-Scale Logarithmic Radiuses

As we can see now on large scale, the computation indeed seems to converge to some value of approximately \(0.9003163161571062\). Can we provide an analytic expression for this numeric value?

Yes we can! Without loss of generality, we again choose \(w=1.0\) and \(r=nw, n\in{}ℕ\). Then the ratio \(A_d(r, w) / A_c(r, w)\) evaluates as:

\( \begin{array}{rcl} C_\infty{} & = & \lim_{n\to\infty}A_d(r, w) / A_c(r, w)\newline & = & \lim_{n\to\infty}A_d(nw, w) / A_c(nw, w)\newline & = & \lim_{n\to\infty}A_d(n, 1) / A_c(n, 1)\newline & = & \lim_{n\to\infty}\frac{8\lfloor{}n\frac{1}{2}\sqrt{2}\rfloor}{2\pi{}n}\newline & = & \lim_{n\to\infty}\frac{8n\frac{1}{2}\sqrt{2}}{2\pi{}n}\newline & = & \lim_{n\to\infty}\frac{2\sqrt{2}}{\pi}\newline & = & \frac{2\sqrt{2}}{\pi}\newline & \approx & 0.900316316, \end{array} \)

Note that we can drop the floor function \(\lfloor{}\centerdot{}\rfloor\) for the real-value to integer conversion when considering \(n\to\infty\), since the off-by-less-than-one becomes neglectable for large values of \(n\). The result that we get is just about the value that our algorithm returned for extremely high values of \(n\). In other words, our initial problem of determining

\(C_\infty = \lim_{n\to\infty}A_d(r=nw, w) / A_c(r=nw, w)\)

resolves to the expression

\(C_\infty = \frac{2\sqrt{2}}{\pi}\).

Variation of \(w\)

Fig. 6: Discrete Annulus Radius and Width — Fig. 6: Alternately Rasterized Annulus
with Width w and Radius r=2w

So far, we assumed that our tiles have width and height \(w\), which is also the stroke width of the annulus. This assumption was reasonable in that tiles with width and height \(w\) are the best possible match at the topmost, bottommost, leftmost and rightmost place of the annulus.

Alternative, one may seek for a bestmost match on the diagonal positions on the annulus. This approach leads to tiles with diagonal diameter of \(w\). How does this change affect or modify our previous results?

For simplicity, we keep the tiling of our example with \(16\) tiles exactly the same, with still \(16\) tiles.
Our new \(w\) is now the diagonal diameter of the tiles. Consequently, the new width and height of our tiles is now \(\frac{1}{2}\sqrt2w\), and the new area of our tiles is the square of the width or height, that is \(\frac{1}{2}w^2\).
The total area of our varied discrete annulus with still \(16\) tiles is now \(A_d(r, w) = 16\frac{1}{2}w^2 = 8w^2\). Effectively, in terms of \(w\), the area of each tile, and thus the area of our complete discrete annulus has been halved.
If it does not make any sense for you that the area has been halved, since on figure, the area looks still the same, then rather think of changing the unit, just as if you had been measuring your garden so far in square meters, and now you switch to square inches. The area as drawn on paper is still the same, but its value changes due to changing the unit (the \(w\) is the unit that we changed).
For our contiguous annulus, the area is still \(2\pi{}rw\), since this is true for any contiguous annulus with radius \(r\) and stroke width \(w\). However, note that the radius (in terms of \(w\)) has been changed by our modification: The inner circle has been slightly moved towards the center, such that the tiles’ diagonal diameter completely fits into the annulus. Initially, our example was constructed with \(r=3w\). With our change, we now have \(r=2w\), as the light blue tiles indicate (even if there placement is off by exactly half a column and row of the raster). That is, the area of the contiguous annulus in our alternate scenario now is \(A_c(r, w) = 2\pi{}rw = 2\pi{}2w^2 = 4\pi{}w^2\).

In summary, we get:

\( \begin{array}{rcl} C_2 & = & A_d(r, w) / A_c(r, w)\newline & = & A_d(2w, w) / A_c(2w, w)\newline & = & 8w^2 / (4\pi{}w^2)\newline & = & \frac{8}{4\pi}\newline & = & \frac{2}{\pi}\newline & \approx & 0.6366. \end{array} \)

This result seems to be even lower than our previous \(C_3\), but is not directly comparable, as we have seen that convergence for low radiuses looks rather chaotic, and we now have the case \(r=2w\) instead of the original \(r=3w\). Hence, we need again to look at Bresenham.

So, once again, without loss of generality, we choose \(w=1.0\) and \(r=nw, n\in{}ℕ\). Since \(w\) now designates the diameter of our tiles, its width and height is each \(\frac{1}{2}\sqrt2w\), and its area \(\frac{1}{2}w^2\). Then the ratio \(A_d(r, w) / A_c(r, w)\) evaluates similar to above, but with the tiles having half the size, as:

\( \begin{array}{rcl} C_\infty{} & = & \lim_{n\to\infty}A_d(r, w) / A_c(r, w)\newline & = & \lim_{n\to\infty}A_d(nw, w) / A_c(nw, w)\newline & = & \lim_{n\to\infty}A_d(n, 1) / A_c(n, 1)\newline & = & \lim_{n\to\infty}\frac{\frac{1}{2}8\lfloor{}n\frac{1}{2}\sqrt{2}\rfloor}{2\pi{}n}\newline & = & \lim_{n\to\infty}\frac{4n\frac{1}{2}\sqrt{2}}{2\pi{}n}\newline & = & \lim_{n\to\infty}\frac{\sqrt{2}}{\pi}\newline & = & \frac{\sqrt{2}}{\pi}\newline & \approx & 0.450158158, \end{array} \)

which is, now that the tiles have half of their original size, not surprisingly also half of the above value.

Smooth Contour

Fig. 7: Adding Additional Tiles — Fig. 7: Adding Additional
Tiles

Following Bresenham’s algorithm, our rastered annulus so far was drawn with the minimal number of tiles needed such that there are no gaps between the tiles. However, this method has the effect that horizontally adjacent tiles on different rows will touch only in a single corner point, creating hard edges in the contour, rather than touching the next tile with always a full adjacent edge. To achieve the latter, one needs to add additional tiles that, in effect, make the stroke of the annulus somewhat thicker and result in a larger area. How many additional tiles will this approach produce, and what is the effect on our correction factor?

Remember that we are looking at an octant with slope not more than \(1.0\). The width of the octant is \(\sqrt\frac{1}{2}r\), and its height is \((1 - \sqrt\frac{1}{2})r\). Again following the contour of the octant counterclockwise, we draw a a sequence of sections of horizontally connected tiles, and every next section starts one row above. Therefore, additionally to the \(\lfloor\sqrt\frac{1}{2}r\rfloor\) tiles, one for each column that we pass, we additionally have to insert \(\lfloor(1 - \sqrt\frac{1}{2})r\rfloor\) tiles, one tile for each change of row. In summary, this makes \(\lfloor{}r\rfloor\) tiles for \(r\to\infty\).

This affects the ratio as follows:

\( \begin{array}{rcl} C_\infty{} & = & \lim_{n\to\infty}A_d(r, w) / A_c(r, w)\newline & = & \lim_{n\to\infty}A_d(nw, w) / A_c(nw, w)\newline & = & \lim_{n\to\infty}A_d(n, 1) / A_c(n, 1)\newline & = & \lim_{n\to\infty}\frac{8\lfloor{}n\rfloor}{2\pi{}n}\newline & = & \lim_{n\to\infty}\frac{8n}{2\pi{}n}\newline & = & \lim_{n\to\infty}\frac{4}{\pi}\newline & = & \frac{4}{\pi}\newline & \approx & 1.273239545, \end{array} \)

Smooth Contour and Variation of \(w\)

For smooth contour, we get \(C_\infty{} = \frac{4}{\pi}\). The variation of \(w\) as discussed above, in effect halves the correction factor. That is, when applying both, smooth contour and variation of \(w\), we get a correction factor of

\(C_\infty{} = \frac{2}{\pi} \approx 0.636619772\).

Summary

When approximating an annulus of width \(w\) on a raster (like a bitmap) with tiles, the area of the original annulus and the approximation differ by a factor (that we denote as correction factor) that converges for large radiuses to some value other than \(1.0\). The exact value depends on whether the raster tiles have either width and height \(w\) or diameter \(w\), but also on the exact method of tiling (either minimal number of tiles without producing any gaps, or minimal number of tiles that always face a complete edge to its neighbour, rather that just touching it just with one of its corners). For the different modus operandi, we get the following correction factors:

Square Tiles with	Width and Height \(w\)	Diameter \(w\)
Touching Corners Sufficient	\(C_\infty{} = \frac{2\sqrt{2}}{\pi} \approx 0.900316316\)	\(C_\infty{} = \frac{\sqrt{2}}{\pi} \approx 0.450158158\)
Touching Edges Required	\(C_\infty{} = \frac{4}{\pi} \approx 1.273239545\)	\(C_\infty{} = \frac{2}{\pi} \approx 0.636619772\)

Future Work

Of course, further variations are imaginable, for example tiles that are not square-shaped, or variations of the algorithm that selects the tiles on the raster for approximating the contiguous annulus.

Conclusion

We don’t solve the problem of dark matter, nor do we directly contribute to solve the mysteries of the fine structure constant or muon anomaly.

However, this article shows on a very specific example that, when using real-valued maths in the context of quantized (rasterized) structures, we may need to apply a correction factor to compute valid results – even and in particular on the large scale, where you naively would expect that you can neglect quantization. Generally, you can’t neglect quantization even on the large scale, as the example in this article shows!

Making Of

This is somewhat ridiculous: Drawing a sketch of what I am writing in this post (including authoring the Java code and compile all equations and derivations) took me, maybe, half an hour of work time or alike. In contrast, expanding my sparse notes and elaborating them into this full-blown blog post, including drawing all figures, took me almost a whole weekend of work, and I still see text passages that would deserve more elaboration and explanation and that show up major flaws with respect to consistency in notation especially for unexperienced readers. Maybe I need a secretary with sufficient mathematical background for blowing up my sparse notes to full-blown blog posts and for proof-reading? 🤷

Note that this article contains maths that you should be familiar with if you have visited an ordinary middle school, or in Germany something like gymnasiale Mittelstufe (actually, we do not even require any knowledge on infinitesimal calculus throughout this whole article). All of the maths in this article were also well known at least to the ancient Greeks, if not already to the ancient Egypts. There is not at all any new or rocket science in this article. Still, all of these trivial facts presented in this article seem to be unknown to many (hobby) physicists that put their heart and soul into discussion on topics like dark matter or alike. Actually, such discussions where motivation and inspiration for me to write this article, such that, in the future, I just can post a link to this article, rather than to have to explain over and over again the most trivial stuff to people that want to discuss physics but do not seem to be aware of the most basic facts. 🙂