## A Gigasecond at Wooster

A second ago, I finished this blog entry. A kilosecond ago, I wrote it. A megasecond ago, I isolated myself against the 2020 pandemic. A gigasecond ago, I began my career at The College of Wooster, which I celebrate today, 31.7 years later.

Logarithmic timeline centered on my first gigasecond at Wooster

## Higgs Without Molasses

Although almost all ordinary mass effectively arises from the kinetic and binding energy of quarks and gluons bound to protons and neutrons in atomic nuclei, the Higgs mechanism does endow some particles like quarks and weakons with intrinsic masses. Here I gently introduce the Higgs mechanism without using loose analogies like vacuum field viscosity.

STATIONARY ACTION

For simple systems, the Lagrangian is the difference between the kinetic and potential energies,

$$L = T - V.$$

Demand that the action

$$S = \int dt \,L$$

be stationary and integrate by parts

$$0 = \delta S = \int dt \left(\frac{\partial L}{\partial x} \delta x + \frac{\partial L}{\partial \dot x} \delta \dot x \right) = \int dt \left(\frac{\partial L}{\partial x}- \frac{d}{d t}\frac{\partial L}{\partial \dot x} \right)\delta x,$$

for all variations $\delta x$, to find the Euler-Lagrange motion equation

$$\frac{\partial L}{\partial x} = \frac{d}{d t}\frac{\partial L}{\partial \dot x}.$$

CLASSIC OSCILLATOR

For the simple harmonic oscillator, the Lagrangian

$$L[x,\dot x] = \frac{1}{2} m \dot x^2 - \frac{1}{2} k x^2$$

implies the Euler-Lagrange equation

$$-kx = m \ddot x,$$

which recovers familiar laws of Hooke and Newton.

SPINLESS PARTICLE

In 1+1 dimensional spacetime, represent a spinless particle by the quantum of the field $\phi[t,x]$. In units where $c = 1$ and $\hbar = 1$, the Lagrangian

$$L = \int dx\, \mathcal L = \int dx \left( \mathcal{T} - \mathcal{V} \right)$$

and the Lagrangian density

$$\mathcal{L}[\phi, \partial_t \phi, \partial_x \phi] = \frac{1}{2} \left(\partial_t \phi \right)^2 -\frac{1}{2} \left( \partial_x \phi \right)^2 -\frac{1}{2} m^2 \phi^2,$$

where the signs of the kinetic energy terms reflect the signs of the spacetime interval $d\tau^2 = dt^2 - dx^2$. The corresponding Euler-Lagrange equation

$$\frac{\partial \mathcal{L}}{\partial \phi} = \frac{\partial}{\partial t}\frac{\partial \mathcal{L}}{\partial (\partial_t\phi)} + \frac{\partial}{\partial x}\frac{\partial \mathcal{L}}{\partial (\partial_x\phi)}$$

reduces to the Klein-Gordon equation

$$-m^2 \phi = \partial_t^2 \phi - \partial_x^2 \phi$$

or

$$\left(\partial_t^2 - \partial_x^2 + m^2 \right) \phi = 0.$$

Assuming wave-particle duality, seek plane wave solutions

$$\phi = e^{i(-Et+ px)} = e^{i(-Et+ px)/\hbar}= e^{i(-\omega t+ kx)}$$

to find

$$\left( (-iE)^2 - (i p)^2 + m^2 \right) \phi = 0$$

or

$$E^2 = p^2 + m^2,$$

where $m$ is the particle mass (and $E = m = m c^2$ for $p = 0$).

Nonzero mass determines the simple harmonic oscillator potential energy $\mathcal V = m^2 \phi^2 / 2 > 0$ and the curvature of the corresponding upright parabola. Zero mass collapses the Klein-Gordon equation to the electromagnetic wave equation.

SYMMETRY BREAKING

Consider the massless Lagrangian density

$$\mathcal{L}= \frac{1}{2} \left(\partial_t \phi \right)^\dagger \left(\partial_t \phi \right) -\frac{1}{2} \left( \partial_x \phi \right)^\dagger \left( \partial_x \phi \right) + \frac{a}{2} \phi^\dagger \phi - \frac{b}{4} \left( \phi^\dagger \phi \right)^2,$$

where $\phi = \phi_m e^{\phi_a} = \phi_r + i \phi_i$, the adjoint $\phi^\dagger = \phi^*$ reduces to complex conjugation, and the parameters $a,b > 0$. For all complex arguments $\phi_a$, the potential energy density

$$\mathcal{V} = - \frac{a}{2} \phi_m^2 + \frac{b}{4} \phi_m^4$$

depends only on the complex magnitude $\phi_m$. The vanishing derivative

$$0 = \frac{d \mathcal{V}}{d \phi_m} = - a \phi_m + b \phi_m^3= \phi_m(- a + b \phi_m^2)$$

implies local maximum and minimum at $\phi_m = 0$ and $\phi_m = \sqrt{a/b} = \phi_0$, as in the figure, where the circular symmetry reflects the invariance of the Lagrangian under global phase (or gauge) transformations $\phi \rightarrow \phi \, e^{i \Lambda} = \phi_m e^{i( \phi_a + \Lambda )}$.

Higgs potential with minimum at nonzero field (left) and circle of vacua (right).

If the potential is a sombrero, the circular brim are ground or vacua states of nonzero field

$$\phi_r^2 + \phi_i^2 = \phi_0^2.$$

Break the circular symmetry by fixing $\phi_r = \phi_0$ and $\phi_i = 0$, but allow the field to oscillate radially and circularly

$$\phi[t, x] = \phi_0 + \phi_1[t, x] + i \phi_2[t, x] .$$

The Lagrangian density becomes

$$\begin{array}{c}\displaystyle\mathcal{L}= \frac{1}{2} \left(\partial_t \phi_1 \right)^2 -\frac{1}{2} \left( \partial_x \phi_1 \right)^2 - a \phi_1^2 \\ \displaystyle+ \frac{1}{2} \left(\partial_t \phi_2 \right)^2 -\frac{1}{2} \left( \partial_x \phi_2 \right)^2 \\ \displaystyle- \sqrt{a b}\,\phi_1 \left(\phi_1 ^2+\phi_2 ^2\right) -\frac{b}{4} \left(\phi_1 ^2+\phi_2^2\right)^2\\ \displaystyle+ \frac{a^2}{4 b}\end{array}$$

or

$$\begin{array}{c}\displaystyle\mathcal{L}= \frac{1}{2} \left(\partial_t \phi_1 \right)^2 -\frac{1}{2} \left( \partial_x \phi_1 \right)^2 - m_1^2 \phi_1^2 \\ \displaystyle+ \frac{1}{2} \left(\partial_t \phi_2 \right)^2 -\frac{1}{2} \left( \partial_x \phi_2 \right)^2 - m_2^2 \phi_2^2 \\ \displaystyle+\text{interaction} + \text{constant},\end{array}$$

where $m_1 = \sqrt{a} > 0$ and $m_2 = 0$.

SUMMARY

Starting with a massless field $\phi$ with nonzero vacuum states, symmetry breaking creates a field $\phi_1$, corresponding to parabolic radial motion and a massive quantum $m_1 > 0$, and a field $\phi_2$, corresponding to constant circular motion and a massless quantum $m_2 = 0$ (a Goldstone boson). This is the Higgs mass endowment mechanism.

## Meeting 100+ years of experience in nonlinear dynamics

I met two scientists for my BZ-history project with a combined age of 177 years.

It was a great pleasure and honor to talk to them.

Meeting with Horst-Dieter Försterling

Meeting with Hermann Haken.

## The Tall Towers

In 1945, science fiction author Arthur C. Clarke published “Extra-Terrestrial Relays – Can Rocket Stations Give Worldwide Radio Coverage?” in Wireless World magazine. Clarke calculated a special orbit, about 36 000 km above the equator with a period of one sidereal day, in which artificial satellites would appear to hover motionless above Earth. The satellites would be like the tops of tall towers, able to relay wireless communications over the horizon. In 1964, NASA first realized Clarke’s concept with the Syncom 3 satellite, which telecast the 1964 Tokyo Olympiad from Japan to the United States.

Today, over 400 operational satellites occupy geosynchronous or Clarke orbit, peering at Earth from almost a tenth of the way to Luna. And this week for the first time, a servicing spacecraft, Northrop Grumman’s Mission Extension Vehicle-1 (MEV-1), rendezvoused and docked with one of them, Intelsat-901, which was nearly out of station-keeping fuel. The capture mechanism went through the throat of the Intelsat-901’s apogee engine, which was not designed for docking. MEV-1 will use its ion engines to orient the stack and extend Intelsat-901’s operation for another 5 years. MEV-1 will then move Intelsat-901 to a slightly higher graveyard orbit — before rendezvousing and docking with another satellite to extend its lifetime.

One of the “tall towers”, Intelsat-901 hovers above Earth a tenth of the way to Luna in a remarkable photo taken this week by MEV-1 shortly before its historic docking

## Hiking to conference

Last weekend, I attended a conference in Germany. I used the opportunity during my sabbatical to return to this conference series, which I attended the last time in 2002.

The conference takes place in a small village in the Harz, a Mittelgebirge (I didn’t know that this is an English word!) in Northern Germany.

Because I had time, I chose to walk/hike the 7 miles from the train station in the next larger town Goslar to the hotel. It was wonderful and my most relaxing travel to a conference site ever. Because of the rain in the last days, some parts where pretty wet and some streams got larger than usual. But other people ‘built’ already crossings.

One thing I remembered, after it was too late: A shortcut is not always the best path to take. If you save on distance traveled in a mountainous regions you could pay with an increase in slope! Not surprising but interesting to realize.

After my talk, listening to presentations, and talking to many friends and colleagues, I hiked back on Tuesday. I stoped at a lake and a bear cave until the sun started to get pretty low. This is when I took a picture of my gigantic shadow.

## Losing Betelgeuse

At my computer Tuesday evening, I receive a message from a university physics chat that is both thrilling and chilling: LIGO + Virgo report a “burst” gravitational wave event, possibly due to a core-collapse supernova (or a binary collision where one object is in the hypothetical “mass gap” between black holes and neutron stars). The burst event is in Orion — near Betelgeuse.

Betelgeuse is a red supergiant star evolving rapidly to an expected supernova. It has dimmed dramatically in recent months, and I’ve seen estimates of thousands to hundreds of thousands of years until it explodes catastrophically in a surge of neutrinos, but $10^{4\pm1}$ years is so soon astronomically. (In physics, the uncertainty is in the mantissa, but in astrophysics, the uncertainty is in the exponent.)

While a Betelgeuse supernova would irrevocably scar my favorite constellation, it would be the most dramatic astronomical event of my lifetime, outshining Earth’s moon for months before dimming to dark. For a minute or two I seriously contemplate losing Betelgeuse — and gaining a spectacular naked-eye supernova. I frantically search online for more information. Betelgeuse seems safe, but at least one astronomer walks outside to visually check. My breathing returns to normal. More time to prepare the next generation neutrino detectors.

Enjoy Orion while you can, because sometime soon, Betelgeuse is gonna blow.

A Betelgeuse supernova would irrevocably damage my favorite constellation, but it would be the most spectacular astronomical event of my lifetime

## Continental Bridge

I remember looking at a classroom map of Earth and thinking the continents seem like puzzle pieces, especially north and south America in the west and Europe and Africa in the east. I mentally fit them together. Later I learned about continental drift and plate tectonics, driven by gravity and mantle convection, with newborn crust at the Mid Atlantic Ridge recording reversals of Earth’s magnetic field in its cooling magma.

Last week I was in Iceland for the winter solstice in search of aurora. Came for astronomy; stayed for geology.

Iceland is a hot spot, probably the expression of an entrenched mantle plume, straddling the boundary between the North American and Eurasian tectonic plates. In southwest Iceland, on the lava-scarred Reykjanes peninsula, the Mid Atlantic Ridge is above ground. As the plates drift apart, several centimeters each year, a rift valley widens. Although the rift itself is several kilometers across, at Sandvik a small footbridge over a fissure in the rift symbolizes the connection between the continents.

Fissure in the rift between continents, Reykjanes peninsula, southwest Iceland, December 2019

Symbolic bridge between North America and Eurasia

## Table of Nuclides

As of 2019, we have identified or synthesized 118 distinct elements with Z protons, but about 2900 distinct nuclides with N neutrons (where atom is to element as nucleus is to nuclide).

The start of my version of the table of nuclides is below, where number of protons Z increases toward 2 o’clock, number of neutrons N increases toward 11 o’clock, and atomic mass A = Z + N increases toward 12 o’clock on average (because more neutrons than protons are needed to bind large nuclei). Rainbow colors code lifetimes t from short (violet) to long (red). For example, the heavy hydrogens are very short lived. The whole chart is a very tall 880 KB PDF table of nuclides. Enjoy!

Start of table of atoms. Rainbow colors code lifetimes, violet (short) to red (long). Number of protons increases toward 2 o’clock, number of neutrons increases toward 11 o’clock.

## Intrepid-Surveyor

Fifty years ago, Apollo 12 landed within sight of another spacecraft, a dramatic demonstration of pinpoint landing capability. While Dick Gordon orbited Luna in the command module Yankee Clipper, Pete Conrad and Al Bean left the lunar module Intrepid and walked over to the robotic Surveyor, which had landed over two years earlier. They retrieved parts of Surveyor and returned them to Earth for engineering analysis. Bean’s photograph of Conrad at Surveyor with Intrepid on the horizon is a spoce exploration icon. Recently, the Lunar Reconnaissance Orbiter photographed the landing site and revealed Surveyor and Intrepid’s descent stage connected by dark tracks in the lunar regolith left by the astronauts.

Al Bean photographed Pete Conrad at the Surveyor 3 spacecraft with the lunar module Intrepid on the horizon, November 20, 1969

2011 Lunar Reconnaissance Orbiter photograph of the Apollo 12 landing site including the astronauts’ tracks from two moonwalks

## Relaxing Fermat

In 1637, while reading a copy of Diophantus’s Arithmetica, Pierre de Fermat famously scribbled

“Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos & generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.”

which roughly translates to

“It is impossible to separate a cube into two cubes, or a quartic into two quartics, or in general, a power higher than the second into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.”

In modern notation, the equation

$$x^n + y^n = z^n$$

has no positive integer solutions for exponents $n > 2$. Although Fermat did leave a proof for the case $n = 4$, 358 years past before Andrew Wiles published his proof of the general case in 1995.

Relaxing Fermat’s constraints to allow non-integers greatly expands the number of solutions. The looping animation shows all solutions for $1 \le n < \infty$ and $1 \le \{x,y,z\} \le 11$. All points on the arcs

$$y_{z,n}[x] = (z^n - x^n)^{1/n}$$

are solutions, and red dots indicate integer solutions. Watch the the famous Pythagorean triple $\{3,4,5\}$ flash by for $n = 2$. Integer solutions are visibly harder for large finite $n$. Many more solutions exist for $n < 1$.

Points along arcs are solutions to the generalized Fermat equation; red points are integer solutions