Thinking in Waves Episode 27 ／ Part XI · Settling the hypothesis

The last mystery left by the previous episode gets an answer

Where does log-uniform come from? ──
what nature had ready was the exponential To make 1/f you do not need to arrange a power-law distribution. An ordinary quantity sitting inside an exponent is enough.
★ And because $\beta-1$ is a ratio of two decade-counts, the exponent pins itself near 1 on its own.
And one prediction failed ── discrete scale invariance leaves a 3 dB trace.

Prerequisites: Episodes 2, 13, 19, 26Computation: kenshou/taisuu.js

Last time we concluded: "a fractional order appears when the coarse-grained degrees of freedom are distributed scale-invariantly."
And only 1/f noise stayed open, in a changed form ── ★ why are log-uniform distributions of time constants so common in nature?
This episode answers it.

01First a check ── does log-uniform really give 1/f?

Superposing first-order (Lorentzian) relaxations $$S(\omega)=\int g(\tau)\,\frac{2\tau}{1+\omega^2\tau^2}\,d\tau,\qquad \text{log-uniform}\ \Leftrightarrow\ g(\tau)=\frac{A}{\tau}$$

In this case the integral can be done by hand

The result $$\int \frac{A}{\tau}\cdot\frac{2\tau}{1+\omega^2\tau^2}\,d\tau=\frac{2A}{\omega}\Big[\arctan\omega\tau\Big]\ \xrightarrow[\ \omega\tau_1\ll1\ll\omega\tau_2\ ]{}\ \frac{\pi A}{\omega}$$

Checked numerically ($\tau\in[10^{-6},10^{6}]$ s, 12 decades, sliced logarithmically).

$\omega$ [rad/s]	$S(\omega)$	Slope [/dec]	Prediction
$10^{-4}$	3.1216e+4	──	outside the band
$10^{0}$	3.141589	−1.0000	★ $\pi A/\omega=3.141593$ (error 1.3e−6)
$10^{3}$	3.1396e−3	−1.0002	−1
$10^{4}$	3.1216e−4	−1.0025	outside the band

The heart of §01

Log-uniform ⇒ 1/f, confirmed (slope exactly −1 inside the band, amplitude matching $\pi A/\omega$ to 6 digits).
★ And outside the band it flattens ── 1/f always has an upper and a lower corner. It does not continue forever.

02★ The real question ── where does log-uniform come from?

Looked at directly in $\tau$, $g(\tau)\propto1/\tau$ seems like a special shape. But ── change one variable and it becomes the most ordinary shape there is.

Thermally activated processes (Arrhenius) $$\tau=\tau_0\,e^{E/kT}\qquad\Longrightarrow\qquad \ln\tau=\ln\tau_0+\frac{E}{kT}$$

$\ln\tau$ is a linear function of the barrier $E$

Write the Jacobian $$g(\tau)\,d\tau=D(E)\,dE,\qquad dE=kT\frac{d\tau}{\tau}\qquad\Longrightarrow\qquad \boxed{\ g(\tau)=\frac{kT\,D(E)}{\tau}\ }$$

If $D$ is flat, you get $1/\tau$ directly

The heart of §02

★★ $E$ uniform ⟺ $\ln\tau$ uniform ⟺ $g(\tau)\propto1/\tau$

No power-law distribution needs arranging. All that is needed is "whatever sits in the exponent is flat" ── an entirely ordinary situation.

How many decades does that buy ($\tau_0=10^{-13}$ s, room temperature 300 K)? The barrier per decade is $kT\ln10=0.05953$ eV.

Spread in barrier $E$ [eV]	Spread in $\tau$ [decades]	$\tau_{\max}$
0.10	1.68	5.2e−7 s
0.36	★ 6.0	──
0.70	11.76	6.3e+3 s = 1.75 hours
1.00	16.80	6.9e+8 s

★ Six decades of 1/f cost only 0.36 eV of scatter in the barrier. That is far too cheap.
And with $E_{\max}=1.0$ eV, $\tau_{\max}=1.75$ hours ── 1/f should turn over below $2.5\times10^{-5}$ Hz. If an experiment sees no turnover down to $10^{-6}$ Hz, the barriers reach beyond 1.3 eV.

03Why "1"? ── the exponent is a ratio of two decade-counts

What if $D(E)$ is not flat? Take $D(E)\propto e^{cE}$ and let $D$ vary by 10× and 1000× across the band.

Variation of $D(E)$	Measured $\beta$	Predicted $1+kT\,d\ln D/dE$
flat	0.99998	1.00000
10×	1.08499	1.08504
1000×	1.25485	1.25511

★ Why the "1" in 1/f does not move $$\beta-1=kT\,\frac{d\ln D}{dE}=\frac{\text{decades over which }D\text{ varies}}{\text{decades over which }\tau\text{ spreads}}$$

The heart of §03

Here $\tau$ spreads over 11.8 decades, so a 10× variation in $D$ gives $\beta-1=1/11.8=0.085$, and even 1000× gives only 0.255.
★★ The denominator is large, so $\beta$ pins itself near 1.
That measured 1/f exponents scatter over $\beta=0.8$–$1.4$ is exactly this ratio. ── ★ 1/f is not the result of fine-tuning. It is harder to avoid.

04Example ① $\alpha$ decay ── the exponential map at its most extreme

In Episode 19 we saw that a decay constant is an imaginary frequency. That decay constant is the textbook case of an exponential map ── $\lambda\propto e^{-G}$, with the Gamow factor $G\propto Z/\sqrt Q$.

Nuclide	$Q$ [MeV]	Half-life	$\log_{10}T_{1/2}$	Line prediction
$^{232}$Th	4.0816	4.43e+17 s	17.647	17.784
$^{238}$U	4.2700	1.41e+17 s	17.149	17.778
$^{226}$Ra	4.8710	5.05e+10 s	10.703	10.867
$^{222}$Rn	5.5900	3.30e+5 s	5.519	5.688
$^{214}$Po	7.8340	164.3 µs	−3.784	−3.700
$^{212}$Po	8.9540	0.299 µs	−6.524	−6.552

Least squares over 9 nuclides ── the Geiger–Nuttall law $$\log_{10}T_{1/2}=1.5065\,\frac{Z}{\sqrt Q}-47.835,\qquad R^2=0.9955$$

Residual standard deviation 0.575 decades, against a 24-decade range

The heart of §04

★★ $Q$ varies by only 2.19×, and the half-life by 24.2 decades.

★ The link to Episode 19: there we said a decay constant is an imaginary frequency, so the observation-window wall does not apply. Here we go one level deeper ── decay constants fill in continuously from the age of the universe down to microseconds because $Q$, an utterly ordinary quantity, sits inside an exponent.

05Example ② Discrete scale invariance ── ★ where a prediction failed

There is a second way to be log-uniform ── a geometric ladder $\tau_n=\tau_0\lambda^n$. Episode 13's Efimov states ($\lambda=e^{\pi/s_0}=22.69$) are exactly this. Does 1/f come out, and if so how does it differ?

Ratio $\lambda$	Rungs per decade	Ripple [dB p-p]	Predicted $2/\cosh(\pi^2/\ln\lambda)$
2	3.32	0.00004	0.00005
3	2.10	0.00871	0.00872
10	1.00	0.95608	0.95662
22.69 (Efimov)	0.74	★ 2.951	2.967
100	0.50	8.301	8.700

Every $\lambda$ gives 1/f. The difference is a log-periodic ripple, larger for larger $\lambda$ ── and it agrees with the predicted formula to 3 digits for $\lambda\le22.7$.

The heart of §05 ── ★ a failed prediction

The Efimov ratio leaves a ripple of 2.95 dB. That is the opposite of Episode 13's "log-periodicity is hard to see".
★ What vanished in Episode 13 was the impulse response of an imaginary-order filter (crushed by $e^{-\beta\pi}$).
★ What survives here is the ripple a discrete ladder of time constants carves into the spectrum.
── "Discrete scale invariance" and "imaginary order" show up differently.

That turns into a testable prediction ── if measured 1/f spectra show no log-periodic ripple, the ladder ratio must be small.

Ripple detection limit	Largest allowed $\lambda$
below 1.00 dB	10.24
below 0.10 dB	★ 4.52
below 0.03 dB	3.57

★★ If 1/f is measured flat to 0.1 dB, the ladder ratio of time constants is at most 4.5. ── So what lies behind 1/f is a continuous (dense) distribution, not an Efimov-style coarse ladder. That is consistent with the activated processes of §02.

06Control experiment ── the condition was looser still

Take $\ln\tau$ Gaussian with width $\sigma$ decades. What we measure is not the slope ── at $\omega\tau=1$ even a single Lorentzian has slope exactly −1 (the corner point). What to measure is the width: over how many decades of $\omega$ does the slope stay within $-1\pm0.1$?

$\sigma$ [decades]	Width that looks like 1/f [decades]	Verdict
≈0 (single $\tau$)	0.10	★ just the corner. Not 1/f
0.5	0.26	intermediate
1.0	0.62	intermediate
2.0	2.02	★ indistinguishable from 1/f
4.0	7.58	★ indistinguishable from 1/f

The heart of §06

Widen $\sigma$ and the 1/f-looking width grows roughly in proportion ($\sigma=4$ decades gives 7.6).
── ★ It need not even be log-uniform. Wide on a log axis is enough to look like 1/f. What nature has to supply is looser still.

07Grading the mechanisms side by side

Mechanism	Form	What it needs	Log-uniform?
Activation (Arrhenius)	$\tau=\tau_0e^{E/kT}$	barriers $E$ scattered flatly	◎ near-automatic
Tunnelling	$\tau\propto e^{2\kappa d}$	distances $d$ scattered flatly	◎ near-automatic
$\alpha$ decay (Ep. 19)	$\lambda\propto e^{-G}$	just line up $Q$	◎ measured (24 decades)
Multiplicative random process	lognormal	$\sigma$ above 2 decades suffices	○ conditional
Discrete scale invariance	$\tau_n=\tau_0\lambda^n$	Episode 13's ladder	◎ but a 3 dB ripple
Criticality	$\tau\propto\xi^z$	correlation length diverging	○ ★ needs tuning
Uniform spring chain (Ep. 26)	$\tau$ constant	──	× stays integer order

The heart of §07

The top three are all just "an ordinary quantity sitting inside an exponent". No power-law distribution to arrange, no critical point to tune to.
★ Criticality, by contrast, requires the temperature to sit exactly at $T_c$ ── self-organised criticality drew attention because it is a way for a system to do that tuning by itself.
But as an explanation of 1/f, activation is far cheaper.

Exercises

In some systems the 1/f exponent $\beta$ approaches 1 as the temperature rises, in others it moves away. How does the formula here read that?
Show the answer
Because both sides of $\beta-1=kT\,d\ln D/dE$ move with temperature. The numerator $kT$ grows with $T$, so if $D(E)$ is sloped, $\beta$ moves away from 1.
On the other hand, raising $T$ shifts the barrier window that matters at the observation frequency, $E^\ast=kT\ln(1/\omega\tau_0)$, to higher barriers ── and if $D$ happens to be flat there, $\beta$ moves toward 1.
★ So measuring $\beta(T)$ amounts to scanning $D(E)$ ── this is the idea behind the Dutta–Horn analysis. This article only confirmed the shape numerically; it analysed no measured data.
What follows from a report that 1/f continues without a turnover down to $10^{-6}$ Hz?
Show the answer
That the barriers extend beyond 1.3 eV. With $\tau_{\max}=\tau_0e^{E_{\max}/kT}$, $\tau_0=10^{-13}$ s at 300 K: $E=1.0$ eV gives 1.75 hours ($2.5\times10^{-5}$ Hz), $E=1.3$ eV gives $6.9\times10^{8}$ s.
★ And "1/f continuing forever" does not exist ── the total power $\int(1/f)df$ diverges logarithmically, so there must be a corner somewhere. Not seeing it just means the measurement was not long enough.
Episode 13 concluded that log-periodicity is hard to see, yet here a 3 dB ripple appears. Is that a contradiction?
Show the answer
Not a contradiction ── they measure different things. What was crushed in Episode 13 was the impulse response of an imaginary-order filter, whose amplitude falls as $e^{-\beta\pi}$ ($10^{-6}$ at $\beta=3$).
What survives here is the periodic undulation a discrete ladder of time constants carves into the spectrum, of size $2/\cosh(\pi^2/\ln\lambda)$.
★ The first is a log-oscillation made by a continuous imaginary order; the second is one made by discrete real orders ── both live on the $\ln\omega$ axis, but they come from different places. And the second is much easier to find.

Summary　What nature had ready was not a power law but an exponential

We answered the question Episode 26 left ── "why are log-uniform distributions of time constants so common in nature?"

First, log-uniform ⇒ 1/f confirmed (slope exactly −1 inside the band, amplitude matching $\pi A/\omega$ to 6 digits). Outside the band it flattens ── 1/f always has an upper and a lower corner.

★ And the answer was the exponential map. With $\tau=\tau_0e^{E/kT}$, $\ln\tau$ is linear in $E$, so $g(\tau)=kT\,D(E)/\tau$ ── a flat $D$ alone gives $1/\tau$. No power-law distribution needs arranging. The barrier per decade is 0.0595 eV at room temperature, so 0.36 eV of scatter buys 6 decades of 1/f.

★★ Why the exponent pins near 1 came out too ── $\beta-1$ = (decades over which $D$ varies) ÷ (decades over which $\tau$ spreads). Even a 1000× variation in $D$ gives $\beta-1=0.25$. The 0.8–1.4 scatter of measured exponents is exactly this ratio.

$\alpha$ decay is this map at its most extreme: $Q$ varies by 2.19× while the half-life varies by 24.2 decades (Geiger–Nuttall, $R^2=0.9955$) ── which is why Episode 19's decay constants fill in continuously.

★ And one prediction failed. Discrete scale invariance ($\tau_n=\tau_0\lambda^n$) does make 1/f, but at the Efimov ratio 22.69 it leaves a 2.95 dB log-periodic ripple (matching $2/\cosh(\pi^2/\ln\lambda)$ to 3 digits). What was crushed in Episode 13 was the imaginary-order impulse response ── a different thing. Turned around, this is a testable prediction ── 1/f flat to 0.1 dB means a ladder ratio of at most 4.5, i.e. a continuous distribution behind it.

★ 1/f is not the result of fine-tuning but a phenomenon that is hard to avoid.
Any system whose rates are exponentials of ordinary quantities becomes log-uniform on its own.

★★ And Episode 26's door closes ── Episode 26: "fractional order ⇐ scale-invariant degrees of freedom"; Episode 27: "scale-invariant degrees of freedom ⇐ the exponential map". The origin of fractional order was the exponential function.

Next episode This episode showed that a flat quantity inside an exponent is all it takes.
★ Next time we chase where that exponent comes from ── tunnelling, thermal activation, the Boltzmann factor: all of the form $e^{-(\text{something})}$. Why is nature so full of exponentials?
★ Seen as waves the answer is one thing ── because every decaying wave is an exponential. Episode 6's evanescent waves come back here.

This document is Episode 27 of the "Thinking in Waves" series. The computations are in kenshou/taisuu.js. On the numbers here: the spectral superpositions, the $\beta$ from barrier distributions, the ladder ripple and the lognormal widths were all computed for this article. The $Q$ values and half-lives of the 9 $\alpha$ emitters are literature values (representative PDG/NNDC numbers); the least-squares slope 1.5065 and $R^2=0.9955$ are this article's calculation. The Geiger–Nuttall law, the derivation of $g(\tau)=kT\,D(E)/\tau$ from activation (the McWhorter model) and the Dutta–Horn $\beta(T)$ analysis are all well-known results and not discoveries of this article ── what this article did was line them up as an answer to the question Episode 26 left, and check numerically that $\beta-1$ is a ratio of decade-counts and how large the ripple is. The ripple formula $2/\cosh(\pi^2/\ln\lambda)$ is the $k=1$ term of a Mellin resummation; the 5 % discrepancy at $\lambda=100$ comes from the higher terms ($k\ge2$). The origin of 1/f noise is still debated system by system; this article only claims that many systems are explained by activated processes. ── To print, use your browser's Print → Save as PDF.