Thinking in Waves Episode 26 ／ Part XI · Settling the hypothesis
Grading Episode 2's hypothesis with 25 episodes' worth of tools
Is a fractional order always the result of coarse-graining? ──
we found a counterexample
Integer-order parts alone can build a half order. A fractional order always decomposes into first-order relaxations.
And yet ── there is a real system that coarse-grains down to nothing but an integer order.
★ What was missing is self-similarity. A fractional order is the trace of scale-invariant degrees of freedom that were removed.
In Episode 2 we wrote this ── [Hypothesis] A system that shows a fractional order is always the result of coarse-graining (integrating out) hidden degrees of freedom.
And also: "one counterexample would refute it ── this is falsifiable." Twenty-five episodes have supplied the tools, so let us actually check.
01Splitting it in three ── which way does "always" point?
The sentence quietly mixes two directions. Without separating them we cannot grade it.
| Test | Question | Direction |
|---|---|---|
| ① | Can a system built only from integer-order parts produce a fractional order? | existence |
| ② | Does a fractional order always decompose into integer-order pieces? | fractional → coarse-graining |
| ③ | ★ Does coarse-graining always give a fractional order? | coarse-graining → fractional |
The heart of §01
What the hypothesis meant was ② (see a fractional order, and there are removed degrees of freedom behind it).
★ But read another way it also claims ③ ── and the real contest is in ③.
02Test ① Building a half order from integer-order parts only
Make a ladder out of resistors and capacitors. Both parts are integer order ── a resistor is \(Z=R\) (order 0), a capacitor \(Z=1/(i\omega C)\) (order \(-1\)). There is no half-order component anywhere.
\(R=1,\ C=1,\ \omega=10^{-4}\)
| Stages \(N\) | \(|Z|\) | Phase [deg] | Offset from \(-45^\circ\) |
|---|---|---|---|
| 1 | 10000.000 | −89.9943 | −44.9943 |
| 10 | 1000.010 | −89.7794 | −44.7794 |
| 50 | 201.012 | −85.1018 | −40.1018 |
| 200 | 89.697 | −46.8584 | −1.8584 |
| 1000 | 100.3542 | −44.7975 | 0.2025 |
The semi-infinite ladder has an exact solution ── the fixed point of the continued fraction (substitute \(Z_\infty\) into itself and solve):
At \(\omega=10^{-4}\): \(|Z|=100.354181\), phase \(-44.7974^\circ\) ── matching the \(N=1000\) numbers
The remaining \(0.2^\circ\) is the \(R/2\) term (it shrinks at lower frequency). Now the amplitude slope.
| \(\omega\) | \(|Z|\) | Slope [dB/dec] | Half-order prediction |
|---|---|---|---|
| \(10^{-4}\) | 100.354 | −9.977 | −10 |
| \(10^{-3}\) | 31.978 | −9.934 | −10 |
| \(10^{-2}\) | 10.360 | −9.790 | −10 |
The heart of §02
★★ Built from integer-order parts only ── and out comes a half order.
\(-10\) dB/dec \(=-3.01\) dB/oct \(=6\alpha\) gives \(\alpha=-1/2\), and the phase \(-45^\circ=90\alpha\) degrees ── exactly the half-rung of Episode 3's ladder.
This is what the Warburg impedance of Episode 2 really is. Coarse-graining (folding infinitely many stages into one impedance) genuinely can create a fractional order.
03Test ② The other direction ── does a fractional order always decompose?
Now reverse it. Can a fractional-order response be written as a superposition of ordinary first-order relaxations?
The \(1/(1+i\omega\tau)\) on the right is a first-order relaxation of time constant \(\tau\) ── plain integer order
The answer is known (for \(0<\alpha<1\)). We check it numerically, slicing \(\tau\) logarithmically and summing.
| \(\alpha\) | \(\omega\) | Superposition | \((i\omega)^{-\alpha}\) | Rel. error |
|---|---|---|---|---|
| 0.25 | \(10^{-1}\) | 1.641178 − 0.680518\(i\) | 1.642916 − 0.680518\(i\) | 9.8e−4 |
| 0.50 | \(10^{0}\) | 0.707104 − 0.707104\(i\) | 0.707107 − 0.707107\(i\) | 3.4e−6 |
| 0.75 | \(10^{1}\) | 0.068052 − 0.164118\(i\) | 0.068052 − 0.164292\(i\) | 9.8e−4 |
The heart of §03
★★ Any fractional order with \(0<\alpha<1\) can be written as a continuum of first-order relaxations.
This is effectively a proof of claim ② ── behind any dissipative fractional-order response, a distribution of hidden degrees of freedom can always be assigned.
★ And that distribution is \(g(\tau)\propto\tau^{\alpha-1}\) ── a power law, i.e. scale invariant. This one line is what matters later.
04Test ③ Does coarse-graining always give a fractional order?
★ This is the real question. We go looking for a counterexample. Take the most innocent coarse-graining there is ── a semi-infinite chain of masses and springs, of which we keep only the end mass.
Band edge \(\omega_{\max}=2\sqrt{k/m}\); with \(m=k=1\), \(\omega_{\max}=2\)
| \(\omega/\omega_{\max}\) | \(|Z|\) | Phase (rel. to force) | What it is |
|---|---|---|---|
| 0.001 | 1.000000 | 0° | pure resistance |
| 0.100 | 1.000000 | ≈ 0° | pure resistance |
| Low-frequency limit | \(\sqrt{km}=1\) | 0° | ★ ordinary friction |
The heart of §04 ── the counterexample
★★ Phase 0° ── integer order. Not fractional.
Infinitely many degrees of freedom were removed and what came out is a plain friction coefficient \(\sqrt{km}\).
★ "Remove degrees of freedom and you always get a fractional order" is false. Episode 2's hypothesis does not stand as written.
05So what sets the order? ── it was the distribution
① and ③ are both "fold infinitely many degrees of freedom into one impedance". The difference was how they were laid out.
| System | Layout of the degrees of freedom | Density | Result |
|---|---|---|---|
| ① RC ladder | ★ self-similar (every stage the same shape) | power law | ★ \(\alpha=-1/2\) |
| ② superposed relaxations | \(g(\tau)\propto\tau^{\alpha-1}\) | power law | ★ fractional \(\alpha\) |
| ③ spring chain | uniform (every stage identical) | constant | integer order |
Generalise. If the removed degrees of freedom (the bath) have spectral density \(J(\omega)\propto\omega^s\), what order does the self-energy \(\Sigma(\omega)\) of the surviving degree of freedom have?
| \(s\) | \(\omega\) | Slope [/dec] | Phase [deg] | \((i\omega)^s\) predicts |
|---|---|---|---|---|
| 0.5 | \(10^{-3}\) | 0.5090 | 45.56 | 45.0 |
| 1.0 | \(10^{-3}\) | 1.0000 | 89.96 | 90.0 |
| 1.5 | \(10^{-3}\) | 1.4941 | 133.82 | 135.0 |
The heart of §05
★★ The exponent of the removed degrees of freedom becomes the order itself (\(\alpha=s\)).
\(s=1\) (Ohmic) → ordinary friction. \(s=1/2\) → half order. \(s=2\) → integer again.
★ The spring chain was integer order because at low frequency \(J(\omega)\propto\omega\) ── a uniform chain just happens to be Ohmic.
06Grading the hypothesis, and rewriting it
| Claim | Verdict | Evidence |
|---|---|---|
| fractional → explainable by coarse-graining | ◎ holds | Test ②. For \(0<\alpha<1\) a spectral representation always exists |
| coarse-graining → always fractional | × fails | ★ Test ③. The spring chain is integer order |
| coarse-graining can create a fractional order | ◎ holds | Test ①. Half order from an RC ladder |
A fractional order appears when the coarse-grained degrees of freedom are
★ distributed scale-invariantly (as a power law).
Coarse-graining alone is not enough. Self-similarity is required. ── And here the whole series joins up.
| Episode | What it found | Shared key |
|---|---|---|
| 13 | imaginary order = generator of scale transformations | scale invariance |
| 22 | wavelets = uniform resolution on \(\ln\omega\) | scale invariance |
| This one | condition for a fractional order = the hidden degrees of freedom are scale invariant | scale invariance |
The heart of §06
★★ A fractional order is the trace of scale-invariant degrees of freedom that were removed.
07Re-sorting real systems by this criterion
| System | Order | Degrees of freedom removed | Self-similar? |
|---|---|---|---|
| Warburg (Ep. 2) | \(-1/2\) | semi-infinite diffusion | ★ diffusion is self-similar (\(\sqrt t\)) |
| Viscoelasticity (Ep. 10) | 0–1 | polymer relaxation times | ★ chain lengths are power-law distributed |
| Heat conduction (Ep. 11) | \(-1/2\) | semi-infinite medium | ★ diffusive, hence self-similar |
| Cole–Cole | 0–1 | dielectric relaxation spread | ★ power-law distributed |
| Ordinary friction | 1 | Ohmic bath | not self-similar → integer order |
| 1/f noise (Ep. 2) | \(-1/2\) | log-uniform time constants | ★ why log-uniform is unsolved |
The heart of §07 ── Episode 15's open doors
★ "Is a fractional order always the result of coarse-graining?" ── closed, with a condition.
★ "The origin of 1/f" ── still open. But the question has changed shape:
"Why are log-uniform distributions of time constants so common in nature?"
Exercises
- If each stage of the RC ladder were scaled stage by stage as \(R\to2R,\ C\to C/2\), what happens to the order (qualitatively)?
Show the answer
It moves off the half order. The half order here comes from every stage being identical in shape ── the product \(RC\) is the same at every stage (the simplest form of self-similarity).
Make the time constants \(\tau_n=R_nC_n\) a geometric sequence and the removed degrees of freedom become evenly spaced on a log axis, moving the order to a different value set by the ratio.
★ The §05 conclusion ── "the order is set by the distribution of the hidden time constants" ── applies directly. - How would you rebuild the spring chain (Test ③) so that it does give a fractional order?
Show the answer
Let the mass or the spring constant vary as a power of position (e.g. \(m_n\propto n^{p}\)). The low-frequency density of states then becomes \(J(\omega)\propto\omega^s\) with \(s\ne1\), and the §05 table gives order \(s\).
Or make it branch (a fractal chain) ── branching gives a density of states whose power is set by the fractal dimension.
★ The uniform chain is integer order because it is uniform, not because it was coarse-grained. - Was Episode 2's hypothesis "wrong"? What is the scientifically accurate thing to say?
Show the answer
One direction was right, the other wrong. "Fractional order → a distribution of hidden degrees of freedom can be written" is effectively proved by Test ②. "Coarse-graining → always fractional" is refuted by the counterexample in Test ③.
★ And being refuted let us rewrite the hypothesis into a stronger form ── "when the removed degrees of freedom are distributed scale-invariantly."
★ On Episode 15's scoreboard this would read "◎ holds, conditionally". A hypothesis declared falsifiable was actually falsified, and got sharper for it ── that is this episode's result.
Summary The hypothesis was half right, and got rewritten
We split Episode 2's hypothesis ── "a system that shows a fractional order is always the result of coarse-graining hidden degrees of freedom" ── into three and graded each.
Test ①: stack 1000 stages of a ladder made only of resistors and capacitors (both integer order) and the phase is \(-44.80^\circ\) with a slope of \(-9.98\) dB/dec ── a half order. It matches the exact semi-infinite solution \(Z_\infty=R/2+\sqrt{R^2/4+R/(i\omega C)}\). Integer-order parts alone can make a fractional order.
Test ②: any fractional order with \(0<\alpha<1\) can be written as a continuum of first-order relaxations (\(g(\tau)=\sin(\alpha\pi)/\pi\cdot\tau^{\alpha-1}\), relative error \(10^{-3}\)–\(10^{-6}\)) ── see a fractional order and a distribution of hidden degrees of freedom can always be assigned.
Test ③ ── ★ the counterexample: keep only the end mass of a uniform semi-infinite mass–spring chain and the impedance is a pure resistance \(\sqrt{km}\) at phase 0°. Infinitely many degrees of freedom removed, and still integer order. "Coarse-graining → always fractional" is false.
The difference was how the removed degrees of freedom were distributed. With a bath of spectral density \(J(\omega)\propto\omega^s\) the order is exactly \(s\) (slope \(s\), phase \(90s\) degrees, confirmed numerically). A uniform chain is Ohmic (\(s=1\)) ── which is why it was integer order.
★ Rewritten hypothesis: a fractional order appears when the coarse-grained
degrees of freedom are distributed scale-invariantly (as a power law).
Together with Episode 13 (imaginary order = scale transformation) and Episode 22 (wavelets = log axis) ── all three turned on scale invariance. A fractional order is the trace of scale-invariant degrees of freedom that were removed. One of Episode 15's open doors closes, conditionally; only the origin of 1/f stays open, and in a changed form ── "why are log-uniform distributions of time constants so common in nature?"
★ Next time we go after where that distribution comes from ── looking at the mechanisms that make distributions (criticality, self-organisation, diffusion-limited growth) as waves, and testing numerically which of them emits log-uniform.
Episode 13's discrete scale invariance (the \(e^{\pi/s_0}=22.7\) staircase) comes back here.
This document is Episode 26 of the "Thinking in Waves" series. The computations are in kenshou/sokushi.js. On the numbers here: the RC ladder values, the spectral-representation comparison, the spring-chain impedance and the bath self-energy were all computed for this article (in dimensionless units \(R=C=m=k=1\)). The exact semi-infinite ladder solution and the spectral-representation \(g(\tau)\) are known results; this article only confirms them numerically. The correspondence between bath spectral density and order (\(\alpha=s\)) is a well-known result for Caldeira–Leggett-type models and is not a discovery of this article ── what this article did was use it to grade Episode 2's hypothesis. The 1–4° phase offset from the prediction at \(s=1.5\) comes from keeping the cutoff \(\omega_c\) finite (higher-order terms in \(\omega/\omega_c\)). The "self-similar?" column of §07 is a qualitative summary, not something computed here for each system. The origin of 1/f noise is an open problem; this article only changed the shape of the question. ── To print, use your browser's Print → Save as PDF.