Thinking in Waves Episode 40 / Part XIII — The Frontier, as Waves

Answering at half the driving period

Time Crystals as Waves —
What matters is not the half, but that it comes back Producing a subharmonic takes no more than the Mathieu equation (growth rate $h\omega_0/4$, matched to five digits).
★★ A time crystal is called a new phase because — you can mis-set the kick angle by 15 % and the response still sits at exactly $1/2$.
★ And switch the interactions off and that stubbornness disappears (the peak slides obediently to $(1-\varepsilon)/2$).

Tools needed: Episodes 1, 17, 18, 23, 32, 36Computation: kenshou/jikan2.js

Episode 36 looked at the Floquet multipliers of a periodically driven wave — $\det M=1$.
★ This episode is about what happens when that $\mu$ passes through $-1$: the sign flips in one period, so it takes two periods to come back — the period doubles.

01First, confirm where it is not

Feed $f_0$ into a linear time-invariant (LTI) system and measure whether $f_0/2$ comes out.

Component	Amplitude	dB (rel. $f_0$)
$f_0$ (input)	8.4064e−2	0.0
$f_0/2$ (subharmonic)	4.9078e−17	−304.7
$2f_0$ (harmonic)	3.5688e−17	−307.4

The point of §01

With the window an integer number of periods long, both fall to the rounding-error floor ($10^{-17}$).
★ Episode 1's "a linear time-invariant system cannot create a frequency" — a time crystal does not live here.
★ This is not a triviality but the starting point: the moment a subharmonic appears, the system has given up either linearity or time-invariance.

02Wobble the coefficient and the half appears

The Mathieu equation — the coefficient itself is a function of time $$\ddot x+\omega_0^2\bigl(1+h\cos\omega t\bigr)x=0,\qquad \omega=2\omega_0\ \text{is the strongest tongue}$$

$h$	growth rate (measured)	$h\omega_0/4$	ratio
0.02	0.00499994	0.00500000	0.99999
0.05	0.01249908	0.01250000	0.99993
0.10	0.02499268	0.02500000	0.99971
0.20	0.04994157	0.05000000	0.99883
0.40	0.09953644	0.10000000	0.99536

The point of §02

The drive is at $\omega=2$, the answer at $\omega_0=1$ — exactly half. The growth rate matches $h\omega_0/4$ to five digits.
★ It is the same mechanism as pumping a swing: raise and lower your weight twice per swing and the motion grows. Input at twice the frequency, answer at half.
★★ Note that linearity was not given up here — the equation is linear in $x$. What was given up is time-invariance.

03In Floquet multipliers — period doubling passes through $\mu=-1$

$h$	$\omega$	$\det M$	$\mathrm{tr}\,M$	$\|\mu\|_{\max}$	Verdict
0.1	1.6	1.000000000	−1.42385	1.00000	stable
0.1	1.9	1.000000000	−1.97970	1.00000	stable
0.1	2.0	1.000000000	−2.00617	1.08168	★ period doubling
0.1	2.1	1.000000000	−1.98310	1.00000	stable
0.3	2.0	1.000000000	−2.05548	1.26491	★ period doubling

The point of §03

$\det M=1.000000000$ — as in Episode 36 (a Hamiltonian system preserves phase-space area).
★ Where $\mathrm{tr}\,M$ drops below $-2$, $\mu$ passes through $-1$. $\mu=-1$ means "the sign flips in one period" — hence two periods to return.
★★ Episode 36 showed that $\det M=1$ forces amplification and decay to appear in pairs; period doubling is that pair being born on the negative real axis.

04A broken symmetry shows up as two states

Add dissipation (real systems are here) $$\ddot x+\gamma\dot x+\omega_0^2(1+h\cos2\omega_0t)\,x+\beta x^3=0,\qquad h_{\rm th}=\frac{2\gamma}{\omega_0}$$

$h$	final state from $x_0=+0.01$	from $x_0=-0.01$	phase difference
0.05	x=−0.00000	x=+0.00000	never starts
0.15	x=−0.27149	x=+0.27149	★ 3.141593 rad
0.30	x=−0.42198	x=+0.42198	3.141593 rad

The point of §04

Above $h_{\rm th}=2\gamma/\omega_0=0.1$ the oscillation starts, and the phase falls into one of two states $\pi$ apart (measured 3.141593 rad).
★ "There are two" is the symmetry breaking — shift time by one drive period and one turns into the other. The system picked one $\mathbb Z_2$ branch by itself.
★★ And to start, it must overcome the dissipation — Episode 32's "symmetry breaking needs dissipation," replayed in the time direction.

05★★ Rigidity — mis-set it, and does it come back?

Here is the real question. If you only want a half, §02 already did it. A time crystal earns the word "phase" because the half survives a mis-set drive.

We test it with a classical spin chain. Each period: (a) precession about $z$ by disordered Ising couplings, (b) a kick about $x$ by the angle $\pi(1-\varepsilon)$. With no interactions, the response should slide obediently to $(1-\varepsilon)/2$.

$\varepsilon$	$J=0$ (no interaction)	$J=0.6$ (with)	prediction $(1-\varepsilon)/2$
0.00	0.500000	0.500000	0.500
0.02	0.490000	★ 0.500000	0.490
0.05	0.475000	★ 0.500000	0.475
0.10	0.450000	★ 0.500000	0.450
0.15	0.425000	★ 0.500000	0.425
0.25	0.375000	0.437714	0.375

The point of §05 — ★★★ the core of this episode

Without interactions the peak slides obediently to $(1-\varepsilon)/2$.
With interactions it sits at exactly 0.500000 up to $\varepsilon=0.15$.

★ That is rigidity. Mis-set the kick by 15 % and the system cancels its own error and returns to the half.
★★ Switch the interactions off and it vanishes — so rigidity is not a property of one oscillator but of many. That is exactly where it differs from Episode 23's subharmonic in a single nonlinear oscillator.

06Measuring the width — where it breaks

$\varepsilon$	peak	$\|f-1/2\|$	amplitude at 1/2	Verdict
0.00	0.500000	0	0.992634	locked
0.05	0.500000	0	0.969319	locked
0.12	0.500000	0	0.845628	locked
0.18	0.498811	1.2e−3	★ 0.021244	★ breaking up
0.25	0.433234	6.7e−2	0.000711	unlocked
0.50	0.250000	0.25	0.000004	unlocked

The point of §06 — ★ honestly

By frequency alone, $\varepsilon=0.18$ still reads 0.4988 — but the amplitude has fallen from 0.846 to 0.021, a factor of 40.
★ So "when does it break" depends on what you look at — the frequency holds on to the last, and the amplitude dies first.
★★ In this calculation it is clearly locked for $\varepsilon\lesssim0.12$ and clearly gone for $\varepsilon\gtrsim0.25$. In between it is breaking up — there is no single threshold.
★ This width is what a nonlinear oscillator would call the width of the Arnold tongue (Episode 23).

07Why it cannot exist in equilibrium — in the language of passivity

Translated into Episodes 17 and 18 $$\text{passive}\ \Longleftrightarrow\ \mathrm{Re}\,Z\ge0\ \Longrightarrow\ \text{cannot deliver energy on its own}$$

For an oscillation to start spontaneously, something must have $\mathrm{Re}\,Z<0$ — negative resistance. And negative resistance cannot be made without feeding energy in from outside.

damping $\gamma$	threshold $h_{\rm th}=2\gamma/\omega_0$	Meaning
0.01	0.02	below this, nothing starts
0.05	0.10	below this, nothing starts
0.30	0.60	★ more dissipation needs a harder drive

The point of §07

★★★ You need both a drive (external) and dissipation — which is why an equilibrium time crystal is impossible.
★ Equilibrium means nothing comes in from outside and there is no net dissipation. There is no $h$ and no $\gamma$, so there is no threshold to cross.
★★ Episode 32's "symmetry breaking needs dissipation" is repeated here verbatim, in the time direction.

08Who can create a frequency — four tiers

System	Frequencies produced	Episode
Linear, time-invariant (LTI)	only the input frequency	Ep. 1
Nonlinear, time-invariant	$n\omega$ and intermodulation	Ep. 23
Linear, time-periodic (parametric)	$\omega/2$, rational multiples of $\omega$	Ep. 36, here
★ Nonlinear + periodic + dissipative + many-body	★ $\omega/n$, and rigidly	★ here

The point of §08

★ A time crystal is called a new phase because only the bottom tier "comes back when you push it."
★★ The three tiers above merely produce a half; detune the drive and they follow it obediently. The $J=0$ column of §05 was exactly that.

09What wave language can and cannot say

Item	Verdict	Basis
No subharmonic from LTI	◎ numerics	−305 dB (rounding floor)
Parametric drives produce it	◎ numerics	growth rate matches $h\omega_0/4$
$\det M=1$, doubling at $\mu=-1$	◎ numerics	continues Episode 36
Broken symmetry = two states	◎ numerics	phase difference exactly $\pi$
★ Rigidity (still 1/2 when detuned)	◎ numerics	★ only with interactions
The width of rigidity	△ numerics	★ not a single threshold
Impossible in equilibrium	◎ analytic	$\mathrm{Re}\,Z\ge0$ (Ep. 17, 18)
★ Quantum time crystals as such	× no	★ this article is classical only
★ Whether it lasts forever	× no	★ only finite time was simulated

The point of §09 — ★ honestly

This article computed only a classical spin chain. The actual discussion of time crystals centres on quantum many-body localization (the quantum version of Episode 37's localization), and that is outside these tools.
★ We also simulated only 4096 periods — "does it last forever" cannot in principle be settled numerically (the same limit as Episode 5's "anything longer than the age of the universe is indistinguishable from a constant").
★★ What can be said is one thing: the phenomenon of rigidity appears only when interactions are present.

Exercises

§02's Mathieu equation is linear in $x$, yet a subharmonic appeared. Does that contradict Episode 1's "linear systems cannot create a frequency"?
Show answer
No. Episode 1's claim was about "linear and time-invariant."
★ The Mathieu equation is linear, but its coefficient is a function of time — it gave up time-invariance.
★★ To create a new frequency it is enough to give up one of linearity and time-invariance. Nonlinearity (Ep. 23) drops the former, parametric driving the latter. And a time crystal drops both, and must additionally be many-body.
In §05 the interactions cancelled the error. By what mechanism?
Show answer
The kick error exactly cancels over two periods. With a kick of $\pi(1-\varepsilon)$, each period falls short by an angle $\pi\varepsilon$.
★ But the neighbours' $z$ components also flip sign every period, so the precession reverses every period too — the shortfall comes back with the opposite sign next period and averages out over two.
★★ This is a self-correcting feedback loop (Episode 18's negative feedback). Hence "switch off the interactions and it disappears" — the feedback path is gone. ★ If $\varepsilon$ is too large the feedback cannot keep up, and it collapses as in §06.
§07 said equilibrium is impossible. Is §05's calculation then not in equilibrium?
Show answer
It is not. §05's spin chain is kicked every single period — work is done on it periodically from outside.
★ In fact a continuously kicked system usually heats to infinite temperature, where no order survives. That is why localization enters the time-crystal story: to stop the heating.
★★ In this classical calculation the disordered couplings play part of that role, but we did not verify that the heating actually stops (only 4096 periods were simulated). ★ So it is held as pending — per Episodes 29–30, what has not been checked does not become a claim.

Summary — what matters is not the half, but that it comes back

First we confirmed where it is not — feed $f_0$ into a linear time-invariant system and $f_0/2$ comes out at −305 dB (the rounding floor). Exactly as Episode 1 said, an LTI system cannot create a frequency.

Wobble the coefficient and it appears — the Mathieu equation $\ddot x+\omega_0^2(1+h\cos2\omega_0t)x=0$ has a growth rate matching $h\omega_0/4$ to five digits. In Floquet terms $\det M=1.000000000$, and where $\mathrm{tr}\,M$ drops below $-2$, $\mu$ passes through $-1$ — sign flips in one period, returns in two (continuing Episode 36).

Add dissipation and a threshold $h_{\rm th}=2\gamma/\omega_0$ appears; above it the phase falls into one of two states $\pi$ apart (measured 3.141593 rad) — "there are two" is the broken symmetry; Episode 32 replayed in time.

★★★ The core — producing a half takes only the Mathieu equation.
A time crystal earns the word "phase" because you can mis-set the kick by 15 %
and the response still sits at exactly 0.500000.

And switch the interactions off and that stubbornness disappears — at $J=0$ the peak slides obediently to $(1-\varepsilon)/2$ (0.490, 0.475, 0.450, 0.425, as predicted). ★ Rigidity is not a property of one oscillator but of many. The mechanism is negative feedback that cancels the error over two periods (Episode 18).

★ The width is not a single threshold — at $\varepsilon=0.18$ the frequency still reads 0.4988, but the amplitude has already died first, 0.846 → 0.021. Clearly locked below $\varepsilon\approx0.12$; clearly gone above $\approx0.25$.

★ And it cannot exist in equilibrium — a passive system has $\mathrm{Re}\,Z\ge0$ and cannot deliver energy of its own (Episodes 17, 18); starting requires both drive and dissipation.

★ Honestly: this is classical only, with no quantum many-body localization. We simulated 4096 periods; "does it last forever" cannot be settled numerically (the same limit as Episode 5).

Next time Episode 35 showed that light's orbital angular momentum carries a winding number that no continuous deformation can change. Episode 37 showed that disorder stops a wave.
★ Next time those two collide — a wave that disorder cannot stop: topological edge states.
★★ If there is no backward road, there is nothing to backscatter into. We look, in wave language, for the conditions under which Episode 37's localization stops working.

This document is Episode 40 of the "Thinking in Waves" series. Computations are in kenshou/jikan2.js. On the numbers here: the LTI subharmonic measurement, the Mathieu growth rates, $\det M$ and $\mathrm{tr}\,M$ of the monodromy matrix, the two final states of the damped parametric oscillator, and the kicked spin chain's peak positions and rigidity width were all computed here. The Mathieu instability tongues and the growth rate $h\omega_0/4$, Floquet theory, the parametric threshold $2\gamma/\omega_0$, the role of rigidity in time crystals, the impossibility of equilibrium time crystals (the Watanabe–Oshikawa theorem and related results), and the requirement of interactions and localization for a discrete time crystal are all well-known results and none are discoveries of this article — what this article did was line them up along a single axis, "which of linearity and time-invariance was given up," and compare rigidity with and without interactions inside one and the same model. The classical spin chain of §05–06 is not a model of any actual quantum time crystal; it is the smallest model in which rigidity can be seen classically. Whether the heating stops was not verified. The "rigidity width" of §06 is specific to this model, these parameters and this observation time (4096 periods); it is not a universal number. The threshold in §04 is the small-damping expression. — To print, use your browser's Print and "Save as PDF."