Thinking in Waves Episode 29 ／ Part XI · Settling the hypothesis

Grading one of our own episodes again

Why is the wave equation second order? ──
second order meant "both directions" First order cannot oscillate, third order runs away, and fourth order houses negative energy.
And the band of width 2 is set by ★ passivity and Ostrogradsky taking alternate turns ── odd orders fall to the first, even orders to the second.
★ This episode corrects our own Episode 15 in two places.

Prerequisites: Episodes 6, 11, 12, 15, 28Computation: kenshou/nikai.js

What was left at the bottom of the chain in Episode 28 was this ── ★ why is the wave equation second order?
Episode 15 said "physics lives in the narrow rectangle $\alpha\in[-1/2,\,2]$", with causality at the lower edge and Ostrogradsky at the upper. This time we measure that again.

01What happens at first order ── diffusion is not a wave

The archetype of first order in time ── the diffusion equation $$\frac{\partial u}{\partial t}=D\frac{\partial^2u}{\partial x^2}\quad\xrightarrow{\ e^{i(kx-\omega t)}\ }\quad -i\omega=-Dk^2\quad\Longrightarrow\quad \omega=-iDk^2$$

$\omega$ is purely imaginary ── no real part

The heart of §01

With no real part in $\omega$, nothing can oscillate. Only decay.
At second order ($\partial_t^2u=c^2\partial_x^2u$), $\omega^2=c^2k^2$ gives $\omega=\pm ck$ ── real, so it can oscillate.

The other symptom is infinite propagation speed. The solution $u=e^{-x^2/4Dt}/\sqrt{4\pi Dt}$ is nonzero at every $x$ for any $t>0$.

$t$ [s]	$x$ [m]	$x^2/4Dt$	Order of magnitude
$10^{-9}$	1.0	2.5e+12	$10^{-1.1\times10^{12}}$
$10^{-3}$	1.0	2.5e+6	$10^{-1.1\times10^{6}}$
$10^{0}$	1.0	2.5e+3	$10^{-1086}$

★ Mathematically "infinitely fast". But 1 m away after 1 ns it is $10^{-1.1\times10^{12}}$ ── not an observable quantity (the same point made for heat conduction in Episode 11).

In fairness, though ── first order in both time and space does give waves. $\partial_tu+c\partial_xu=0$ gives $\omega=ck$, real, and it propagates. ★ But in one direction only.

02★ What second order really is ── "both directions"

The wave operator factorizes into two first-order operators $$\frac{\partial^2}{\partial t^2}-c^2\frac{\partial^2}{\partial x^2}=\left(\frac{\partial}{\partial t}-c\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\right)$$

rightward advection × leftward advection

$(x,t)$	$\Box g$ directly	from the factorization	Rel. difference
(0.2, 0.5)	2.62360253	2.62359274	3.7e−6
(1.0, 0.3)	−3.25306111	−3.25305532	1.8e−6
(−0.7, 1.2)	−0.39724006	−0.39723333	1.7e−5

They agree to within the finite-difference truncation error. The consequence is that one initial bump always splits into two halves (d'Alembert). Solving the wave equation numerically:

$c=1$, after 5 seconds $$\text{right bump: height }0.49996,\ \text{at }+5.000\qquad \text{left bump: height }0.49996,\ \text{at }-5.000$$

Height 0.5 each, positions $\pm ct=\pm5.0$, exactly as predicted

The heart of §02

★★ Second order meant "both directions".

At first order (advection) the bump does not split; it keeps height 1 and travels one way. ── Second order is what it costs to be able to go both ways.

03The one way to stay first order ── take a square root, and spin appears

So is "first order and both directions" impossible? There is exactly one route ── take the square root of the wave operator.

Dirac's move $$(i\gamma^\mu\partial_\mu)^2=\Box\qquad\Longleftrightarrow\qquad \{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}I$$

No ordinary numbers satisfy this. Matrices are required ── the question is how large

Try $2\times2$ first. The Pauli matrices $\sigma_1,\sigma_2,\sigma_3$ do mutually anticommute (exactly 0 numerically). Can a fourth one, $M=aI+b\sigma_1+c\sigma_2+d\sigma_3$, be built?

$\{\cdot,\cdot\}$	$\sigma_1$	$\sigma_2$	$\sigma_3$
$I$	$2\sigma_1$	$2\sigma_2$	$2\sigma_3$
$\sigma_1$	$2I$	0	0
$\sigma_2$	0	$2I$	0
$\sigma_3$	0	0	$2I$

Imposing $\{M,\sigma_1\}=0$ kills everything in the $\sigma_1$ column, so $a=b=0$; repeating with $\sigma_2,\sigma_3$ gives $c=d=0$. ★ The fourth matrix can only be zero ── $2\times2$ tops out at three.

Checking $4\times4$ (Dirac representation) ── all 10 pairs $$\max_{\mu\nu}\big|\{\gamma^\mu,\gamma^\nu\}-2\eta^{\mu\nu}I\big|=0.00$$

Even the rounding error is zero (every entry is 0 or ±1)

The heart of §03

★★ Four components ── that is spin.

The price of "stay first order and still go both ways" was four internal components. Antiparticles (two of those components) come along in the same package.

04What breaks at third order ── Abraham–Lorentz

Write radiation reaction classically and a third derivative appears $$m(\ddot x-\tau_0\dddot x)=F,\qquad \tau_0=\frac{2e^2}{3\cdot4\pi\varepsilon_0mc^3}=\frac{2r_e}{3c}=6.2664\times10^{-24}\ \text{s}$$

Even with $F=0$ a solution $\ddot x=Ae^{t/\tau_0}$ survives ── the runaway.

Elapsed	$t/\tau_0$	Amplification
$\tau_0\ln2=4.34\times10^{-24}$ s	0.693	2.000
1 as	159.6	$10^{69.3}$
1 ps	$1.6\times10^{11}$	$10^{6.9\times10^{10}}$

Numerical integration (RK4) puts the doubling at $4.3458\times10^{-24}$ s against the predicted $\tau_0\ln2=4.3436\times10^{-24}$, agreeing to 5e−4. Fix the boundary condition in the future to avoid the runaway and you get pre-acceleration instead (Episode 12).

The heart of §04 ── in fairness

Against $\tau_0=6.27\times10^{-24}$ s, the electron's Compton time is $\hbar/mc^2=1.29\times10^{-21}$ s. A ratio of 206 ── $\tau_0$ lies deep inside the region where classical electromagnetism does not apply.
★ So it is not that "causality broke"; it is that the theory was used where it must not be.

05★ Read on the ladder, the third-order failure is one line

The work fed into the system per cycle, measured in Episode 12 $$W=\pi k\,\omega^\alpha|x|^2\sin\frac{\alpha\pi}{2},\qquad F=k\,D^\alpha x$$

The impedance order $\beta$ of $Z=F/v$ is $\beta=\alpha-1$ (since $v=i\omega x$)

$\alpha$	$\beta$	Element	$W$ (numeric)	Meaning
−0.5	−1.5	(forbidden region)	−2.22144	★★ supplies energy (active)
0	−1	spring	0.00000	lossless
0.5	−0.5	★ Warburg	2.22144	dissipative
1	0	damper	3.14159	dissipative
2	1	mass	0.00000	lossless
3	2	★ radiation reaction	−3.14159	★★ supplies energy (active)
4	3	Pais–Uhlenbeck	0.00000	lossless

The heart of §05

★★ At $\alpha=3$ the work is negative ── the system keeps receiving energy. That is what the $e^{t/\tau_0}$ of §04 really was.
Episode 12's $\sin(\alpha\pi/2)$ and the Abraham–Lorentz runaway are one and the same thing.
★ And $\alpha=4$ gives $W=0$ ── the sign of the dissipation does not catch it. That is §07.

06★★ Measuring Episode 15's rectangle again

Passivity ── it may give energy out, but never conjure it $$\operatorname{Re}Z(\omega)\ge0\ \ (\forall\omega),\qquad Z=(i\omega)^\beta\ \Longrightarrow\ \operatorname{Re}Z=\omega^\beta\cos\frac{\beta\pi}{2}$$

The sign is decided by the cosine alone

Here is the point of this episode ── passivity does not allow a single band.

★ Bands of width 2, every 4 $$\cos\frac{\beta\pi}{2}\ge0\quad\Longleftrightarrow\quad \beta\in[-1,1]\cup[3,5]\cup[7,9]\cup\cdots$$

$\beta=3$ (i.e. $\alpha=4$) is lossless, so passivity cannot eliminate it. Ostrogradsky does. ★ Two theorems take alternate turns.

Order $\alpha$	Example	Passivity	Ostrogradsky	Verdict
0	spring	◯ passes	◯ passes	allowed
1	damper	◯ passes	◯ passes	allowed
2	mass / wave equation	◯ passes (edge)	◯ passes	★ allowed
3	radiation reaction	★ kills it	──	★ forbidden (runaway)
4	Pais–Uhlenbeck	◯ lets it through	★ kills it	★ forbidden (ghost)
5	──	★ kills it	──	forbidden

The heart of §06

★★ Odd orders fall to passivity, even orders to Ostrogradsky.

Neither alone is enough. Episode 15 attributed the upper edge to Ostrogradsky alone, but it is passivity that kills $\alpha=3$.

With that, we grade Episode 15.

Episode 15's claim	Verdict	Reason
Physics lives in a band of width 2	◎ correct	in one convention, $\beta\in[-1,+1]$
Upper edge is 2	△ wrong units	$\alpha=2$ is the equation order; in $\beta$ it is $+1$
Lower edge is $-1/2$	× wrong	★ $-1/2$ is an impedance order, and it is not an edge at all
The lower edge is set by causality	× wrong	★ passivity, not causality. $(i\omega)^{-2}$ is causal but active
The upper edge is set by Ostrogradsky	△ half	$\alpha=3$ is killed by passivity

What had gone wrong

★ The two edges of the rectangle were measured in different units.
Lower edge $-1/2$: Warburg ── an impedance order $\beta$
Upper edge $+2$　: the wave equation ── an equation order $\alpha$ (which is $+1$ in $\beta$)
★ Warburg's $\beta=-1/2$ is not an edge; it sits around the middle of the band. It looked like an edge only because it was the most negative thing on hand ── an observation, not a theorem.

★★ Rewritten band: $\beta\in[-1,+1]$ (i.e. $\alpha\in[0,2]$), imaginary part ≈ 0.
Both edges lossless, the interior dissipative, outside is not physics. Width exactly 2.

07Who kills fourth order ── Ostrogradsky, numerically

The Pais–Uhlenbeck oscillator (the archetype of fourth order) $$L=\tfrac12\big[\ddot x^2-(\omega_1^2+\omega_2^2)\dot x^2+\omega_1^2\omega_2^2x^2\big]\ \Longrightarrow\ x''''+(\omega_1^2+\omega_2^2)\ddot x+\omega_1^2\omega_2^2x=0$$ $$H=\tfrac12\ddot x^2-\dot x\,\dddot x-\tfrac12(\omega_1^2+\omega_2^2)\dot x^2-\tfrac12\omega_1^2\omega_2^2x^2\ =\ \tfrac12(\omega_1^2-\omega_2^2)(A^2\omega_1^2-B^2\omega_2^2)$$

State ($\omega_1=2,\omega_2=1$)	$H$ (numeric)	$H$ (analytic)	Sign
mode 1 only, $A=1$	6.000000	6.000000	positive
mode 1 only, $A=2$	24.000000	24.000000	positive
mode 2 only, $B=1$	−1.500000	−1.500000	★★ negative
mode 2 only, $B=2$	−6.000000	−6.000000	★★ negative

★ Mode 2 carries negative energy, so raising $B$ drives $H$ arbitrarily low ── unbounded below. That is the Ostrogradsky instability.

Does it actually run away? We added a $\lambda x^3$ coupling and looked (from $A=B=1$, 400 s, "runaway" = $|x|>10^6$. At $\lambda=0$, $H$ is conserved to $10^{-12}$, so the integration is trustworthy).

$\omega_1/\omega_2$	$\lambda=0.01$	$\lambda=0.1$	$\lambda=1$
2.00	— never	— never	★ 6.8 s
1.50	— never	— never	★ 5.0 s
1.10	★ 28.6 s	★ 10.2 s	★ 4.7 s
1.01	★ 24.9 s	★ 9.9 s	★ 4.6 s

The heart of §07 ── ★ a prediction that failed

At $\omega_1/\omega_2=2$ with weak coupling, nothing ran away in 400 s. But bring $\omega_1$ close to $\omega_2$ and even weak coupling blows up. ★ Near-degenerate is the dangerous case ── when the positive and negative sectors are close in frequency, the exchange bites.

★ Stated honestly: "fourth order always runs away immediately" is false. What is certain is that $H$ is unbounded below.
── And unbounded below is enough. There is no reason to keep sitting on a chair that merely happens not to have collapsed yet.

Exercises

If "second order is for both directions", how is that different from an electromagnetic wave having two polarizations?
Show the answer
They are different counts. "Second order = both directions" is about the time axis: the two factors of $\Box=(\partial_t-c\partial_x)(\partial_t+c\partial_x)$ are the right- and left-moving waves.
The 2 of polarization is a transverse spatial freedom (3 dimensions minus the direction of travel) and has nothing to do with the order.
★ Same numeral, different origins ── one counts time derivatives, the other counts spatial dimensions. As in Episode 14's "three reasons an exponent leaves the integers", when you see a number, check where it came from.
If Warburg ($\beta=-1/2$) is mid-band, what real element sits at the edge $\beta=-1$?
Show the answer
A plain spring (or capacitor). $Z=k/(i\omega)$ gives $\beta=-1$ and $\operatorname{Re}Z=0$ ── the lossless edge. The opposite edge $\beta=+1$ is a mass (or inductor), also lossless.
★ Both edges of the band are elements that lose nothing; everything inside loses something.
★ And Episode 2's Warburg sits exactly between spring and damper ($\beta=-1/2$) ── the element that loses half.
Was Episode 15's error a miscalculation? If not, what was it?
Show the answer
Not arithmetic but units ── what gets called $\alpha$. Every number was right: Warburg is $-1/2$ and the wave equation is 2. What was wrong was putting those two on the same axis.
★ This is the most common kind of error in physics ── Episode 7's "only dimensionless quantities mean anything" and Episode 14's $|\det M|$ were both about the same bookkeeping: what did you divide by?
★ And it was caught only because the calculation was pushed out to $\alpha=3$ and $\alpha=4$. Going past the edge is what revealed that the edge had never been defined.

Summary　Second order means both directions; the width-2 band is set by two alternating theorems

We answered what Episode 28 left: why second order?

First order cannot oscillate (diffusion gives $\omega=-iDk^2$, purely imaginary). It also propagates infinitely fast, but 1 m away after 1 ns is $10^{-1.1\times10^{12}}$ ── not an observable. First order in both time and space does give waves, but one way only.

★ Second order turned out to mean "both directions". It factorizes as $\Box=(\partial_t-c\partial_x)(\partial_t+c\partial_x)$, and numerically a bump splits into two halves of height 0.5 travelling to $\pm ct$. To stay first order and still go both ways you must take a square root, and $2\times2$ admits only three anticommuting matrices (shown in the table), so $4\times4$ is required ── ★ and that price is spin.

Third order runs away. $\tau_0=6.27\times10^{-24}$ s, $e^{t/\tau_0}$, doubling in $4.34\times10^{-24}$ s. But that is 1/206 of the Compton time, so it is only classical electromagnetism used where it must not be. ★ And on the ladder it is one line ── $W\propto\sin(\alpha\pi/2)$ turns negative at $\alpha=3$. Episode 12's formula and the Abraham–Lorentz runaway were the same thing.

★★ The width-2 band is set by two theorems taking turns. Passivity, $\cos(\beta\pi/2)\ge0$, allows $\beta\in[-1,1]\cup[3,5]\cup\cdots$ ── bands every 4 ── so $\alpha=4$ ($\beta=3$) survives passivity. Odd orders fall to passivity, even orders to Ostrogradsky. The fourth-order $H=\frac12(\omega_1^2-\omega_2^2)(A^2\omega_1^2-B^2\omega_2^2)$ is unbounded below, though whether it actually runs away depends on the frequency ratio ── at $\omega_1/\omega_2=2$ with weak coupling, nothing happened in 400 s.

★ And this episode corrected our own Episode 15 in two places.
① The two edges of the rectangle were measured in different units ($\beta$ and $\alpha$).
② The lower edge is set by passivity, not causality ── and $-1/2$ was never an edge.

Having declared the series falsifiable, its own episodes are fair game too. The correct band is $\beta\in[-1,+1]$, width exactly 2 ── one order on each side.

Next episode So the band has width 2. How does that relate to the dimension of space?
★ Next time we move the dimension ── solve the wave equation in $d$ dimensions and in even dimensions waves leave a tail (Huygens' principle fails). Why do ripples on a 2D pond linger while 3D sound cuts off sharply?
★ And we count, in the language of orders, why $d=3$ looks special.

This document is Episode 29 of the "Thinking in Waves" series. The computations are in kenshou/nikai.js. On the numbers here: the numerical wave-equation solution, the operator factorization check, the Pauli/Dirac anticommutators, the Abraham–Lorentz runaway, the per-cycle work $W$, and the Pais–Uhlenbeck $H$ and runaway scan were all computed for this article. D'Alembert's solution, the Dirac equation and its $\gamma$ matrices, the Abraham–Lorentz $\tau_0$, the relation between passivity and positive-real functions, Ostrogradsky's theorem and the Pais–Uhlenbeck oscillator are all well-known results and not discoveries of this article ── what this article did was line them up on a single axis, the order, and re-grade the claims of Episode 15. The runaway scan in §07 used one specific coupling ($\lambda x^3$) and one initial condition ($A=B=1$); it does not exhaust the parameter space. The existence of apparently stable islands is reported in the literature, but this article does not classify them. The §03 statement "$2\times2$ tops out at three" shows only that the four matrices needed in 3+1 dimensions cannot be built there; the general Clifford-algebra count ($2\log_2N+1$ for $N\times N$) is not derived. The pre-acceleration in §04 is stated from the equation, not computed here. ── To print, use your browser's Print → Save as PDF.

\(t\) [s]	\(x\) [m]	\(x^2/4Dt\)	Order of magnitude
\(10^{-9}\)	1.0	2.5e+12	\(10^{-1.1\times10^{12}}\)
\(10^{-3}\)	1.0	2.5e+6	\(10^{-1.1\times10^{6}}\)
\(10^{0}\)	1.0	2.5e+3	\(10^{-1086}\)

\(\{\cdot,\cdot\}\)	\(\sigma_1\)	\(\sigma_2\)	\(\sigma_3\)
\(I\)	\(2\sigma_1\)	\(2\sigma_2\)	\(2\sigma_3\)
\(\sigma_1\)	\(2I\)	0	0
\(\sigma_2\)	0	\(2I\)	0
\(\sigma_3\)	0	0	\(2I\)