Thinking in Waves Episode 12 ／ Part V · Physical quantities as waves

The reason the sky is blue lies in the order

Radiation comes from the second derivative ──
which is why the sky is blue Expand the field of a charge in distance and only the $1/r$ term is proportional to acceleration and carries energy away.
That is why radiation needs acceleration ── a question of order.
And squaring a second derivative gives $\omega^4$ ── blue is scattered 4.4 times more than red.

Prerequisites: Episodes 1, 9Computation: kenshou/housha.js

Last time heat was the thing that does not travel. This time, the thing that certainly does ── electromagnetic waves.
And the question is: why does radiation require acceleration?

01Only the far field picks up the second derivative

Term	Distance dependence	Proportional to	Order	Carries energy?
Electrostatic field	$1/r^2$	charge $q$	0	×
Induction field	$1/r^2$	current $\dot q$	1	×
Radiation field	$1/r$	acceleration $\ddot q$	2	★ ○

Why only $1/r$ survives ── just counting $$\text{energy flux}\propto E^2,\qquad \text{sphere area}\propto r^2\quad\Longrightarrow\quad \text{total radiated power}\propto E^2r^2$$ $$E\propto1/r^2\ \Rightarrow\ E^2r^2\propto r^{-2}\to0,\qquad E\propto1/r\ \Rightarrow\ E^2r^2=\text{const}\ \textbf{★ survives}$$

Conclusion of §01

A field falling faster than $1/r$ never reaches infinity.
And the $1/r$ term appears only in the second derivative (acceleration) ──
★ which is why "radiation requires acceleration". It was a question of order.

02The Larmor formula ── acceleration squared

Radiated power $$P=\frac{q^2a^2}{6\pi\varepsilon_0c^3}$$

★ Take the second derivative, then square it
In Episode 1's language: tilt the spectrum by $+12$ dB/oct, then turn it into power

Situation	Acceleration [m/s²]	Radiated power [W]
Electron falling under gravity	9.8	$5.5\times10^{-52}$
Electron in a hydrogen atom	$9.0\times10^{22}$	$4.7\times10^{-8}$
LHC proton (2.8 km radius)	$3.2\times10^{13}$	$5.9\times10^{-33}$

Ten orders of magnitude in acceleration means twenty in radiation ── because of the square.

03Why the classical atom collapses

The electron in a hydrogen atom moves in a circle. Circular motion is accelerated motion ── so classical electromagnetism says it must radiate and lose energy.

First, naively $$a=\frac{v^2}{a_0}=9.04\times10^{22}\ \text{m/s}^2\ \Rightarrow\ P=4.67\times10^{-8}\ \text{W}$$ $$\frac{|E|}{P}=\frac{2.18\times10^{-18}}{4.67\times10^{-8}}=4.7\times10^{-11}\ \text{s}$$

But as the radius shrinks the acceleration rises and the radiation accelerates. Integrating properly:

Fall time $$t=\frac{a_0^3}{4r_e^2c}=1.556\times10^{-11}\ \text{s}$$

$r_e=2.818\times10^{-15}$ m (classical electron radius)

Elapsed time [s]	Radius $r/a_0$	Acceleration ratio	Power ratio
0	1.0000	1	1
$7.8\times10^{-12}$	0.7937	1.59	2.52
$1.4\times10^{-11}$	0.4642	4.64	21.5
$1.555\times10^{-11}$	0.1000	100	$10^4$

Conclusion of §03

In about 16 picoseconds the electron falls into the nucleus.
In classical physics, atoms cannot exist. This was one of the reasons quantum mechanics became necessary.

And in the last 0.1 % of that time the radiated power rises $10^4$-fold ── the collapse accelerates.

04An oscillating charge radiates as $\omega^4$ ── this is what matters

Just square the second derivative $$P\propto\langle\ddot p^2\rangle=\frac{\omega^4p_0^2}{2}$$

★ The second derivative gives $\omega^2$; squaring gives $\omega^4$

This is Rayleigh scattering: the cross-section goes as $\omega^4\propto1/\lambda^4$.

Colour	Wavelength [nm]	Ratio of $1/\lambda^4$ (red = 1)
violet	400	6.97
blue	450	4.35
green	550	1.95
red	650	1.00

The heart of §04

The sky is blue because we squared a second derivative.

Air molecules are shaken by light and re-radiate in proportion to their acceleration. That strength goes as $\omega^4$, so shorter wavelengths scatter more ── blue 4.4 times more than red.

And a red sunset is the same formula turned inside out. With the sun low, light travels far through the atmosphere, the blue is scattered away first, and only red gets through.

05The radiation spectrum ＝ the acceleration spectrum

Episode 1's view, applied directly $$\text{radiated amplitude}\propto a(\omega)=(i\omega)^2x(\omega)=-\omega^2x(\omega)$$

★ The radiation spectrum is "the orbit's spectrum tilted by $+12$ dB/oct"

Form of the acceleration	Radiation spectrum	Reason
Charge in simple harmonic motion	line spectrum	that frequency only; atomic spectra
Bremsstrahlung	broad spectrum	★ a short pulse is broadband (Episode 9's theorem)
Synchrotron radiation	a comb of harmonics	relativistic circular motion
Thermal motion (blackbody)	continuous spectrum	superposition of random accelerations

Why X-rays appear is the bandwidth theorem again

If an electron stops in $10^{-18}$ s, the bandwidth is $\Delta\omega\gtrsim1/(2\Delta t)=5\times10^{17}$ rad/s.
In photon energy, $\hbar\Delta\omega=329$ keV ── the X-ray band falls out.
★ Bremsstrahlung is an X-ray source as a direct consequence of Episode 9's "short pulses have broad bandwidth".

06Rearranging the ladder of orders, electromagnetically

Order	Quantity	Field produced	Name
0	$q$ (charge)	electrostatic $\propto1/r^2$	Coulomb
1	$\dot q$ (current)	magnetic / induction $\propto1/r^2$	Biot–Savart, Faraday
2	$\ddot q$ (acceleration)	radiation $\propto1/r$	Larmor, Rayleigh, bremsstrahlung
3	$\dddot q$ (jerk)	radiation reaction	★ Abraham–Lorentz force

§06 ── go to third order and causality breaks

The radiation reaction is $F=\dfrac{q^2}{6\pi\varepsilon_0c^3}\dddot x$ ── third order. Therefore:

・it falls under Episode 2's Ostrogradsky instability
・and indeed there are solutions where acceleration begins before the force is applied (pre-acceleration)
・★ which breaks Episode 8's causality

The timescale of pre-acceleration is $\tau=2r_e/(3c)=6.3\times10^{-24}$ s ── about the light-crossing time of a classical electron radius.
★ "Raise the order too far and causality breaks" actually happens inside classical electromagnetism. This is a failure of the classical theory; quantum electrodynamics resolves it.

Exercises

In this episode's language, why does a charge in uniform motion not radiate?
Show answer
Because the acceleration is zero, so the $1/r$ term never arises. There are only the electrostatic ($1/r^2$) and induction ($1/r^2$) fields, and for both $E^2r^2\to0$, so no energy reaches infinity.
(The usual explanation — that uniform motion can be transformed to rest — says the same thing.)
Explain a red sunset from $\omega^4$.
Show answer
The scattering probability goes as $\omega^4$, so the longer the path through the atmosphere, the sooner the blue is lost.
With the sun low, the path is tens of times longer than at the zenith, attenuating blue by the equivalent of $e^{-4.35\times}$ and letting only red through.
★ Blue sky and red sunset are the two sides of one formula ── whether you look at the scattered light or at what is left.
Where does radiation reaction's "pre-acceleration" sit in this series?
Show answer
It is a case where Episode 2 (Ostrogradsky: order ≥ 3 is unstable) and Episode 8 (causality) break in the same place at the same time.
Classical electromagnetism carried a third-order term inside itself ── and, as predicted, produced both an instability and a non-causality.
★ It is the most concrete evidence for Episode 2's hypothesis that the order has a ceiling.

Summary　Radiation is a second derivative, squared

Expanding the field of a charge, only the $1/r$ term is proportional to acceleration and carries energy (it is the only one for which $E^2r^2$ stays finite). "Radiation requires acceleration" was a question of order.

From the Larmor formula $P=q^2a^2/(6\pi\varepsilon_0c^3)$, the classical hydrogen atom collapses in 16 picoseconds ($t=a_0^3/4r_e^2c=1.556\times10^{-11}$ s) ── one of the reasons quantum mechanics was needed.

And squaring the second derivative gives $\omega^4$ ── Rayleigh scattering. Blue scatters 4.4 times more than red. Blue sky and red sunset are two sides of one formula.

The radiation spectrum is "the orbit's spectrum tilted by $+12$ dB/oct", and bremsstrahlung is broadband by Episode 9's bandwidth theorem (stop in $10^{-18}$ s and you get 329 keV of bandwidth ── X-rays).

And at third order, pre-acceleration appears and causality breaks.
── The most concrete evidence for Episode 2's "the order has a ceiling".

Next time In Episode 2 we extended the order $\alpha$ to the reals. What if we make it imaginary?
Compute $(i\omega)^{i\beta}$ and the amplitude becomes a constant independent of $\omega$, with only the phase rotating as $\ln\omega$ ── an all-pass filter.
★ If real orders are the tool of time translation, imaginary orders are the tool of scale transformation. And that has been measured, as the Efimov effect.

This document is Episode 12 of the series "Thinking in Waves". The computation is in kenshou/housha.js. On the numbers: the Larmor formula is non-relativistic and must be replaced by the Liénard form as $v\to c$ ── the LHC proton entry uses the non-relativistic formula as a reference value (real synchrotron radiation is $\gamma^4$ larger, and for protons $\gamma$ is small enough that it is far weaker than for electrons). The classical fall time $t=a_0^3/(4r_e^2c)$ is the standard result under the quasi-static circular-orbit approximation. Rayleigh's $1/\lambda^4$ applies when the scatterer is much smaller than the wavelength; aerosols and droplets give Mie scattering with much weaker wavelength dependence (which is why an overcast sky is white). Sky colour also involves atmospheric absorption, ozone and multiple scattering, so it is not determined by $\omega^4$ alone. The pre-acceleration of the Abraham–Lorentz force is a known pathology of the classical theory, not something found here. ── To print, use your browser's Print → Save as PDF.