Thinking in Waves Episode 65 / Part XX — The Things That Tunnel

Four reader questions, all four counted

What it means to drop a dimension ──
the shape can survive while the information does not “A dimension is an integration, so it’s a 6 dB filter, right?” ── Close. A dimension is 3 dB, exactly half an integration.
★★ “Observing loses information, doesn’t it?” ── Yes, and there are two kinds. Slicing changes the shape; projection leaves the shape untouched and still loses the information (one view keeps \(1/N\)).
★★★ “A non-integer-dimensional target must introduce error?” ── Not error: it saturates at 2 (Marstrand’s projection theorem). And symmetry axes are the theorem’s exceptional directions — the worst way to look.
★★★ “Could mass be a non-integer-dimensional wave?” ── That integral was already in Episode 2.

Tools needed: Episodes 1, 2, 13, 39, 59, 60Computation: kenshou/jigen2.js (38 checks, zero deps)★ interactive

This episode answers a chain of reader questions. ★ All four were already in a countable form, so all four were counted.
★★ The first needed a correction, the second and third gave sharper answers than expected, and the fourth had already been computed by this series.

01Fixing the yardstick ── a dimension is not 6 dB

conversion $$|X(f)|\propto f^{-\alpha}\ \Rightarrow\ -6.02\alpha\ \text{dB/oct},\qquad S=|X|^2\propto f^{-\beta},\ \beta=2\alpha\ \Rightarrow\ \beta\ 1\ =\ 3.01\ \text{dB/oct}$$
operationwhat movesdB/octfrom
integrate once\(\alpha\) +16.02Episode 1
slice one dimension\(\beta\) −13.01here, §02
project along one axis\(\beta\) unchanged0here, §02

★ Integration and dimension differ by exactly a factor of two. The intuition “a dimension is an integration” points the right way, but whether you count in \(\alpha\) or \(\beta\) changes the answer by 2×.
★★ And projection is 0 — it does not change the spectral shape at all. And still loses information: that is §03.

02Extracting one dimension from a \(d\)-dimensional field

Build an isotropic field \(P_d(k)\propto k^{-\beta_d}\), then (a) cut along a line (slice) and (b) integrate along one axis (project). ★ Following Episode 59, the periodograms are averaged before taking logs (a single line has two degrees of freedom and scatters by ~0.7 decades).

field\(\beta_d\)★ slicepredicted \(\beta_d-(d{-}1)\)projectionpredicted \(\beta_d\)
2D2.01.0611.02.0572.0
2D3.02.0182.02.9903.0
2D4.03.0113.03.9834.0
3D3.01.1211.02.9463.0
3D4.02.0022.04.0694.0
3D5.02.9513.05.1255.0

The core of §02

★★★ Slicing flattens \(\beta\) by 1 (3.01 dB/oct) per dimension removed. Cutting a 3D field down to 1D flattens it by 6 dB/oct — which is, I think, where the “a dimension is 6 dB” intuition comes from. But that is two dimensions.
★★★ Projection does not change \(\beta\) at all. In Fourier space it just takes the \(k_\perp=0\) section, so the exponent is preserved.

03★★ And yet projection loses information ── count it

The projection–slice theorem: the Fourier transform of one projection is a section through the origin. From an \(N^3\) object, one projection captures \(N^2\) points — a fraction \(1/N\).

objectcoefficientsone projectionfractionto fill it
\(N=64\)2.62e+54.10e+31/64≈101 views
\(N=256\)1.68e+76.55e+41/256≈402 views
\(N=512\)1.34e+82.62e+51/512≈804 views

Counting the \(k\)-space coverage directly in 2D (\(N=128\)):

\(M\) views141664128201
\(k\)-space covered0.8 %3.1 %15.2 %55.7 %89.5 %essentially all

The core of §03 ── the first summit here

★★★ “The spectral shape is unchanged” does not mean “no information was lost”. Projection preserves \(\beta\) while discarding every \(k_\perp\ne0\) component.
★★ This is precisely the theme of Episodes 59 and 60 ── an apparent quantity can be preserved while something is still being lost.
★ Which is why CT works: collect enough directions and \(k_\perp\) fills back in. The number needed is about \(\pi N/2\) (sampling).
★★ And the region near the origin fills first while the outside stays sparse ── that unevenness is why CT resolution is good at the centre and poor at the edge.

04★★★ Non-integer dimension gives saturation, not error

Marstrand’s projection theorem (1954) $$\dim\bigl(\pi_u(A)\bigr)=\min\bigl(\dim A,\;k\bigr)\qquad\text{for almost every direction }u$$

Checked on a 3D Cantor dust (divide each side into 3, keep \(m\) of the 27 sub-cubes; \(D=\log m/\log3\)).

\(m\)theory \(D\)3D measuredaxis projectionaxis predictiongeneric direction\(\min(D,2)\)
51.46501.46501.17171.26191.25101.4650
71.77121.77121.57821.63091.54811.7712
92.00002.00001.57721.63091.72192.0000
162.52372.52371.68611.77121.87102.0000
242.89282.89281.91522.00001.85112.0000

The core of §04

★★★ The 3D dimension matches theory exactly (1.4650–2.8928, box counting to four figures).
★★★ But the projection never exceeds 2, however large \(D\) becomes.
★★ “A non-integer dimension introduces error” is wrong: whatever exceeds 2 disappears entirely. Photograph an object of \(D=2.89\) and the image carries two dimensions’ worth. The remaining 0.89 does not show up as error — it looks as though it was never there.
★ With only 4–6 levels the measured projections land somewhat below 2, but the saturation is unmistakable.

05★★ Symmetry axes are the exceptional directions ── the worst view

Marstrand says “for almost every direction”. Outside that “almost” lie the exceptional directions ── and a lattice’s symmetry axes are exactly those.

The axis-projection dimension is set by \(\log(\)number of sub-cells that survive without collapsing\()/\log3\) — a combinatorial quantity, not Marstrand’s value.

\(m=16\) (\(D=2.524\))value
axis projection1.686 (predicted \(\log7/\log3=1.771\))
generic direction1.871
Marstrand’s value2.000

★★★ Aligning your view with the object’s symmetry axis is the easiest to interpret and the most lossy. You lose even what the theorem guarantees.
★ Deliberately avoiding low-index orientations when imaging crystals is exactly this.

06★★★ Is mass a non-integer-dimensional wave? ── Episode 2’s integral

A single mass = a single pole (a Lorentzian). A non-integer scaling dimension = a power-law continuum of masses.

This is the same integral as Episode 2’s “\(1/f\) is a superposition of many time constants”.

Episode 2 (\(1/f\))unparticle (Georgi 2007)
time constant \(\tau\)inverse mass \(1/M\)
Lorentzian \(\tau/(1+\omega^2\tau^2)\)propagator \(1/(p^2+M^2)\)
\(g(\tau)\propto\tau^{-\gamma}\)\(\rho(M^2)\propto(M^2)^{d_U-2}\)
\(S\propto\omega^{\gamma-2}\)\(\text{Prop}\propto(p^2)^{d_U-2}\)

Verified numerically that they are the same integral (\(\int_0^\infty u^{a-1}/(p^2+u)\,du=\pi p^{2(a-1)}/\sin\pi a\); ratios 0.9997–1.000000).

matching the exponents $$\gamma-2=2d_U-4\ \Rightarrow\ d_U=1+\tfrac{\gamma}{2},\qquad \beta=2-\gamma\ \Rightarrow\ \boxed{\,d_U=2-\alpha\,}$$
\(\alpha\)\(\beta\)\(\gamma\)\(d_U=2-\alpha\)meaning
0.00.02.02.0white noise
0.51.01.01.5pink noise
1.02.00.01.0\(1/f\) ── one pole = an ordinary particle
2.04.0−2.00.0Brownian motion

The dimension ledger ── move \(\alpha\) and everything connects

1.00
3

The core of §06

★★★ This series’ \(\alpha\) is exactly the unparticle dimension \(d_U=2-\alpha\).
★★ And \(1/f\) (\(\alpha=1\)) lands precisely on \(d_U=1\) — one pole, an ordinary particle ── Episode 30 called \(1/f\) “the scale-invariant fixed point”, Episode 57 called it “the spectrum with the cheapest edges”, and here it stands in a special place again.
★ Honestly though: this is the same integral written differently. It does not explain the origin of mass, and unparticles remain an undiscovered hypothesis.

07In AdS/CFT it holds literally (Episode 39 again)

Episode 39 $$\Delta(\Delta-d)=m^2L^2,\qquad \Delta=\frac{d}{2}\pm\sqrt{\frac{d^2}{4}+m^2L^2}$$
\(m^2L^2\)−401257
\(\Delta_+\) at \(d=4\)2.0004.0004.2364.4495.0005.317
integerintegernon-int.non-int.integernon-int.

★ Only when \(4+m^2L^2\) is a perfect square is \(\Delta\) an integer ── generically it is not.
★★★ “Mass = non-integer dimension” is literally true in AdS/CFT. Fix the mass and the dimension follows; fix the dimension and the mass follows.
★★ But this is a dictionary between two descriptions, and it does not answer “why that mass”.
★ And it is the same quadratic as Episode 13’s \(s=\tfrac12\pm\sqrt{\tfrac14-\lambda}\) ── the third time this series has met it (Episodes 13, 39, 47).

Exercises

  1. §03 says projection preserves \(\beta\) yet loses information. So if you only want \(\beta\), is one projection enough?
    Show answer
    Yes ── and that is the point.
    ★ If \(\beta\) is all you want, one projection gives the right value. ★★ The danger is concluding “the \(\beta\) is right, so reconstruction must be possible too”.
    ★★★ Episode 59 built a test on “if the value moves when you change the window, the window is to blame”. Here it is the converse ── the value does not move and information is still lost. Not moving is evidence neither of correctness nor of sufficiency.
    ★ For the handbook: count “what is preserved” and “what is needed to reconstruct” separately. \(\beta\) is the former, \(k_\perp\) the latter.
  2. §04 says whatever exceeds 2 vanishes. Then how would you ever learn that the object has \(D>2\)?
    Show answer
    Not from one view. You need several.
    ★ Every direction gives \(\min(D,2)=2\). A single image tells you “\(D\ge2\)” and cannot distinguish 2.1 from 2.9.
    ★★ §03’s counting is the answer: collect \(\pi N/2\) views, reconstruct in 3D, then box-count in 3D.
    ★★★ So “measuring a non-integer dimension” is itself a tomography problem. CT is needed not only to see shape but to measure dimension.
    ★ Conversely: a fractal dimension estimated from a single photograph and coming out near 2 might be genuinely 2, or might be anything above it. A value pinned at 2 should be suspected of saturation.
  3. Is \(d_U=2-\alpha\) in §06 a discovery?
    Show answer
    No — it is the same integral rewritten. Episode 61 said that four routes landing on \(6\alpha\) is tidying the tools, not new physics, because they share one Fourier transform at the root. This is the same kind of thing.
    ★ Episode 2’s superposition integral and the unparticle spectral representation are one integral with different variable names. ★★ Laying them side by side is still worth doing: it shows that “non-integer dimension” names the same thing as \(1/f\)’s “many time constants”.
    ★★★ And it makes the boundary sharp: the correspondence says anything you like about kinematics (what the spectrum looks like) and nothing at all about dynamics (why that \(\alpha\), why that mass). ★ In Episode 55’s terms this is a translation — neither a measurement nor a value conditional on a model.

Summary — four questions, all four counted

① “Is a dimension a 6 dB filter?” ── Close; a dimension is 3 dB. One integration is \(\alpha\) +1 = 6.02 dB/oct; one sliced dimension is \(\beta\) −1 = 3.01 dB/oct. Exactly a factor of two apart. ★ Cutting 3D down to 1D flattens by 6 dB/oct, which is probably where the intuition came from — but that is two dimensions.

★★★ ② “Observing loses information” ── yes, and there are two kinds.

★ Slicing changes the shape (\(\beta_1=\beta_d-(d-1)\), verified numerically). ★★★ Projection changes nothing about the shape and still loses the information ── one view of an \(N^3\) object keeps \(1/N\); filling \(k\)-space takes about \(\pi N/2\) views (201 at \(N=128\), and the measured coverage reaches ~100 % at exactly 201). “Same spectral shape” does not mean “nothing lost”.

★★★ ③ “Non-integer dimension gives error” ── no, it saturates at 2. Marstrand’s projection theorem. The 3D dimension matches theory exactly (1.4650–2.8928) while the projection never exceeds 2. ★★ What exceeds 2 does not appear as error; it disappears.

★★ A bonus: symmetry axes are the theorem’s exceptional directions. The axis-projection dimension is a combinatorial quantity (“sub-cells that survive collapse”), lower than \(\min(D,2)\) — 1.686 on axis versus 1.871 generic at \(m=16\). The easiest direction to look is the most lossy.

★★★ ④ “Could mass be a non-integer-dimensional wave?” ── that integral was already in Episode 2. A single mass is a single pole; a non-integer dimension is a power-law continuum of masses. Episode 2’s superposition integral and the unparticle spectral representation are identical, giving \(d_U=2-\alpha\), with \(1/f\) (\(\alpha=1\)) landing exactly on \(d_U=1\) — one pole, an ordinary particle.

★ And in AdS/CFT it holds literally (\(\Delta(\Delta-d)=m^2L^2\), Episode 39) ── the third appearance of Episode 13’s quadratic. ★★ Honestly: this is a translation, not an explanation of mass. It speaks to kinematics and is silent on dynamics (Exercise 3).

This document is Episode 65 of the “Thinking in Waves” series, answering a chain of reader questions. The computations are in kenshou/jigen2.js (zero dependencies, own FFT and box counting, all 38 checks passing). Not discoveries of this episode: the exponent rules for slices and projections of random fields, the Fourier projection–slice theorem (the principle of CT), the \(\pi N/2\) sampling requirement, the Marstrand–Mattila projection theorem (1954) and the existence of its exceptional directions, unparticles (Georgi 2007) and their spectral representation, and AdS/CFT’s \(\Delta(\Delta-d)=m^2L^2\) ── all well known; this episode is a numerical confirmation. What it did was ① separate integration (6 dB) from slicing (3 dB) numerically, ② count how projection preserves the exponent while losing information, via \(k\)-space coverage, ③ verify Marstrand’s saturation on a Cantor dust and explain the below-theorem axis value through the count of non-collapsing sub-cells, and ④ show that Episode 2’s superposition integral and the unparticle propagator are the same, giving \(d_U=2-\alpha\). Limits: the Cantor dust in §§04–05 has only 4–6 levels, so while the 3D box-counting dimension matches theory exactly, the projections stop around 1.85–1.87 rather than reaching \(\min(D,2)\) — the saturation structure is clear but the limiting value is not confirmed. The §02 numbers average 24 realizations in 2D and 3 in 3D; the projection samples are few and ±0.1 scatter remains. The \(d_U=2-\alpha\) of §06 is a variable correspondence between identical integrals, not a claim of physical identity, and unparticles remain undiscovered. This episode too uses only synthetic data and analytic formulas; no measured data were used. ── To print, use your browser’s Print → Save as PDF.

Printing / PDF: ⌘+P (Ctrl+P on Windows). “Show answer” opens the solutions.