Thinking in Waves Episode 6 ／ Part III · Physics as waves

Look at it as a wave and mass shows its true face

Mass is a cutoff frequency ──
the vacuum was a waveguide The Klein–Gordon dispersion relation is $\omega^2=c^2k^2+(mc^2/\hbar)^2$.
That is word for word the waveguide formula ── and $mc^2/\hbar$ is the cutoff frequency.
The range of a force, $E=mc^2$, and the relativistic energy relation all follow from it.

Prerequisites: Episode 1Computation: kenshou/shadan.js

In Episode 1 we said "differentiation and integration are filters". Then a wave equation is a filter's specification sheet.
Look at mass with that eye and a completely different face appears.

01A wave equation is a filter spec

Put a plane wave $e^{i(kx-\omega t)}$ into the equation and a relation between $\omega$ and $k$ falls out. That is the dispersion relation ── the transfer function deciding which $(\omega,k)$ get through.

Equation	Dispersion relation	Character
Wave equation	$\omega=ck$	every $\omega$ passes
Klein–Gordon	$\omega^2=c^2k^2+(mc^2/\hbar)^2$	★ there is a cutoff from below
Schrödinger	$\omega=\hbar k^2/2m$	non-relativistic; quadratic
Diffusion equation	$\omega=-iDk^2$	imaginary ── decays instead of propagating

02Read the second row with an acoustician's eye

The Klein–Gordon dispersion relation $$\omega^2=c^2k^2+\omega_c^2,\qquad \omega_c\equiv\frac{mc^2}{\hbar}$$

System	Dispersion relation	Character
Free wave (light in vacuum)	$\omega=ck$	no cutoff; any low $\omega$ passes
Waveguide	$\omega^2=c^2k^2+\omega_c^2$	nothing below $\omega_c$ passes
A field of mass $m$	$\omega^2=c^2k^2+(mc^2/\hbar)^2$	★ same form, with $\omega_c=mc^2/\hbar$

The heart of §02

Mass is the cutoff frequency of the vacuum, regarded as a waveguide.

A particle at rest ($k=0$) oscillates in place at $\omega=\omega_c$. Below that frequency, no wave propagates.

Particle	Mass [MeV]	Cutoff $\omega_c$ [rad/s]
electron	0.511	$7.763\times10^{20}$
muon	105.658	$1.605\times10^{23}$
pion	139.570	$2.120\times10^{23}$
proton	938.272	$1.425\times10^{24}$
W boson	80379	$1.221\times10^{26}$

The electron's cutoff is $7.8\times10^{20}$ rad/s. The top of human hearing is $2\times10^{5}$ rad/s — a factor of $10^{16}$.

03What happens below the cutoff ── this is the decisive part

For $\omega<\omega_c$ there is no real $k$ $$k^2=\frac{\omega^2-\omega_c^2}{c^2}<0\quad\Longrightarrow\quad k=i\kappa\ (\text{purely imaginary})$$

$e^{ikx}=e^{-\kappa x}$ ── exponential decay (an evanescent wave)

And for a static field ($\omega=0$):

Decay length $$\kappa=\frac{\omega_c}{c}=\frac{mc}{\hbar}\quad\Longrightarrow\quad \text{decay length}=\frac{\hbar}{mc}$$ $$V(r)\propto\frac{e^{-r/\lambda}}{r},\qquad \lambda=\frac{\hbar}{mc}\qquad\textbf{★ the Yukawa potential}$$

Particle	Mass [MeV]	Range $\hbar/mc$ [fm]	Note
electron	0.511	386.2	electron Compton wavelength
pion	139.570	1.414	★ the range of the nuclear force — Yukawa's prediction
proton	938.272	0.2103
W boson	80379	$2.455\times10^{-3}$	★ why the weak force is "weak"

The heart of §03

The range of a force is the evanescent decay length below the cutoff.
The pion's 1.41 fm is exactly the scale of a nucleus (compare the nuclear radius $1.2A^{1/3}$ fm).

And the answer to "why does electromagnetism reach infinity while the weak force does not" is ──
the photon has no cutoff; the W boson has one at $1.2\times10^{26}$ rad/s. Mass, or the lack of it. That is all.

04Velocity too comes from differentiating the spectrum

Two velocities from one dispersion relation $$v_p=\frac{\omega}{k}\quad(\text{a ratio}),\qquad v_g=\frac{d\omega}{dk}\quad(\textbf{★ a derivative})$$ $$\text{differentiate }\omega^2=c^2k^2+\omega_c^2:\quad 2\omega\frac{d\omega}{dk}=2c^2k\quad\Longrightarrow\quad \boxed{v_p\cdot v_g=c^2}$$

$v/c$	$v_g/c$	$v_p/c$	$v_p v_g/c^2$
0.100	0.1000	10.0000	1.000000
0.500	0.5000	2.0000	1.000000
0.999	0.9990	1.0010	1.000000

The phase velocity exceeds the speed of light

$v_p=c^2/v>c$ ── but it carries no information. What carries information is the group velocity, that is, "the derivative of the spectrum".
(This theme is treated in detail in the sister series "Understanding Refraction".)
★ Differentiation is at work here too ── "speed" is the derivative of a spectrum called the dispersion relation.

05Reading $E=mc^2$ in wave language

Just multiply the cutoff frequency by $\hbar$ $$\hbar\omega_c=\hbar\cdot\frac{mc^2}{\hbar}=mc^2$$

★ $E=mc^2$ is "the energy of the cutoff frequency"

And when $k\ne0$ $$(\hbar\omega)^2=(\hbar ck)^2+(mc^2)^2\qquad\Longleftrightarrow\qquad E^2=(pc)^2+(mc^2)^2$$

$p$ [MeV/c]	$E=\sqrt{p^2+m^2}$	$\hbar\omega$ [MeV]	Match
0.000	0.510999	0.510999	○
0.511	0.722662	0.722662	○
10.000	10.013047	10.013047	○
100.000	100.001306	100.001306	○

The heart of §05

The relativistic energy–momentum relation is the dispersion relation itself.
It looks like Pythagoras' theorem because it was a waveguide relation all along.

06All the faces of mass, together

Quantity	In wave language	Meaning
Mass $m$	cutoff frequency $\omega_c=mc^2/\hbar$	the spec of the vacuum seen as a waveguide
Rest energy	$\hbar\omega_c$	this is $E=mc^2$
$E^2=(pc)^2+(mc^2)^2$	$\omega^2=c^2k^2+\omega_c^2$	relativity ＝ the waveguide formula
Range of a force	$\hbar/(mc)$	evanescent decay length below cutoff
Particle speed	$d\omega/dk$	★ the derivative of the spectrum
Phase velocity	$\omega/k=c^2/v$	exceeds $c$ but carries no information

Conclusion of this episode

"Seeing mass as a wave" succeeds completely.
But the answer is not "an integral" ── it is "a cutoff frequency".

As we shall see next time, the route of treating mass as the integral of gravity is ruled out by experiment.
The wave route, by contrast, turns straight into relativity and quantum theory.

⇒ ★ Mass is not an integral but a filter characteristic.

Exercises

Why is a "particle at rest" oscillating?
Show answer
Because even at $k=0$, $\omega=\omega_c\ne0$.
For an electron that is $7.8\times10^{20}$ rad/s ── the Compton frequency, and the starting point of de Broglie's matter waves.
"At rest" is a spatial statement; in time it is oscillating as fast as anything can. That is the content of $E=mc^2$.
In this episode's language, how did Yukawa predict the meson mass?
Show answer
The range of the nuclear force was known to be about 1.4 fm. Range = evanescent decay length $\hbar/mc$, so invert it:
$m=\hbar/(c\times1.4\ \text{fm})=197.3/1.4\approx\mathbf{140}$ MeV.
The actual pion is 139.57 MeV. Yukawa did this in 1935 ── measure how far a force reaches and out comes the mass of an unseen particle.
What would happen if the photon had mass, in this framework?
Show answer
Electromagnetism would acquire a cutoff frequency and the Coulomb force would decay exponentially (the Proca equation).
Experimentally, the shapes of the Earth's and Jupiter's magnetic fields give bounds of order $m_\gamma<10^{-18}$ eV.
Converted to a range, $\hbar/m_\gamma c>2\times10^8$ m ── that inequality is the experimental content of "reaches infinity".

Summary　The vacuum was a waveguide

The Klein–Gordon dispersion relation $\omega^2=c^2k^2+(mc^2/\hbar)^2$ is word for word the waveguide formula, with $\omega_c=mc^2/\hbar$ the cutoff ── mass is the cutoff frequency of the vacuum.

Below cutoff the wavenumber turns imaginary and the field decays exponentially (evanescent). For a static field the decay length is $\hbar/mc$ ── the Yukawa potential, and the true identity of a force's range. 1.41 fm for the pion, $2.5\times10^{-3}$ fm for the W. "Why does the weak force not reach?" — because it has a cutoff.

Velocities follow too ── $v_g=d\omega/dk$ (the derivative of the spectrum), $v_p=\omega/k$, and $v_pv_g=c^2$ exactly.

$E=mc^2$ is "the energy of the cutoff, $\hbar\omega_c$", and $E^2=(pc)^2+(mc^2)^2$ is the dispersion relation itself. It looked Pythagorean because it was a waveguide formula.

Mass is not an integral but a filter characteristic.

Next time The wave route succeeded. What about the other two conjectures ──
"mass is the integral of the universe's gravity" and "the speed of light is the integral of information"?
Next time we compute both and score them. One of them gives exactly 1/2 ── and that turns out to be an identity.

This document is Episode 6 of the series "Thinking in Waves". The computation is in kenshou/shadan.js. On this episode: the phrasings "mass = cutoff frequency" and "vacuum = waveguide" are our own framing, but the dispersion relation itself is a standard result, and the formal match with a waveguide is widely known (the same thing field theory calls "a massive mode is gapped"). It is not a physical claim that the vacuum really is a waveguide ── nothing beyond "the formulas have the same shape" is asserted. The Yukawa derivation is the static limit ($\omega=0$); the dynamic case needs separate treatment. The relation $v_pv_g=c^2$ in §04 holds for a free particle and fails in a medium. The photon-mass bound is a published value, ranging from $10^{-18}$ to $10^{-27}$ eV depending on method. The relation between phase velocity and information transfer is treated in detail in the sister series "Understanding Refraction". ── To print, use your browser's Print → Save as PDF.

System	Dispersion relation	Character
Free wave (light in vacuum)	\(\omega=ck\)	no cutoff; any low \(\omega\) passes
Waveguide	\(\omega^2=c^2k^2+\omega_c^2\)	nothing below \(\omega_c\) passes
A field of mass \(m\)	\(\omega^2=c^2k^2+(mc^2/\hbar)^2\)	★ same form, with \(\omega_c=mc^2/\hbar\)