The Wine-Bottle Potential — Spontaneous Breaking of a Continuous Symmetry

Concept: A complex scalar field with V(φ) = μ²|φ|² + λ|φ|⁴ (λ = 1 here), evolved as a unit-mass point in the plane (Re φ, Im φ) — i.e. with the normalisation L = ½(∂φ₁)² + ½(∂φ₂)² − V, so that m² is exactly the curvature of V and exactly ω² of the small oscillation you watch. For μ² > 0 there is one symmetric vacuum at φ = 0. Driving μ² negative turns the origin into a hilltop and produces a whole ring of equivalent vacua at |φ|min = √(−μ²/2λ); the field's arbitrary choice of a point on that ring spontaneously breaks the continuous U(1) symmetry. Curvature across the valley, V″(|φ|min) = −4μ², is the mass² of the radial (Higgs-like) mode; motion along the valley feels no force at all — the massless Goldstone mode. Both the 3-D vertical scale and the bottom cross-section (V along Re φ at Im φ = 0, showing the double well) use FIXED axes (V from −1.05 to +1.55 in the 3-D view, −1.3 to +1.2 on the cross-section), so sweeping μ² visibly changes the depth; the rim simply leaves the frame. The plot boundary at |φ| = 1.47 is a hard elastic wall, not physics.
Serves: Introductory particle physics / QFT: the spontaneous-symmetry-breaking unit, immediately before gauging the symmetry (Higgs mechanism). Also useful in condensed matter for superfluid phase modes.
Inquiry: A five-card guided inquiry drives the scene: symmetric bowl → gated hilltop prediction (the stage pauses, the answer releases it) → the massive radial kick → gated ring-push prediction (the answer fires the push) → gated μ²-sweep question that unmasks the mass table. The cross-section strip is the operational check: during a Goldstone coast the slice dot glides at constant height (V never changes) while the 3-D ball circulates; during a radial kick the dot climbs the well walls. Nothing is disabled — the reveals are withheld, not the controls.
Misconception: "Anything displaced oscillates back." A restoring force requires curvature of V; along a symmetry direction the potential is exactly flat, so the mode has zero frequency and zero mass. The misconception is encoded in the "springs back" choice of the ring-push prediction card; choosing it draws a dashed ghost marker that reverses at the push point while the real ball coasts straight past it.

Principle	Where it lives here
Predict before reveal (order + pause, never disabling)	The gated inquiry cards pause the stage for each commitment and the answer itself triggers the reveal; Play and every control stay live from the first frame.
Misconception targeted structurally	The "springs back" chip spawns a dashed ghost that oscillates back while the real field coasts on — the rebuttal is drawn, not asserted.
Invariant made visible	Fixed V axes in both views: sweeping μ² visibly moves the depth, \|φ\|min and the radial m², while the ring m² row stays pinned at 0.
Operational, not vibes	\|φ\|min, V_min, V″ and the period 2π/m reproduce the engine values exactly; the fixed-scale slice shows V constant during the Goldstone coast.
Formal layer on demand	U(1) transformation, the kinetic normalisation, vacuum formulas and mode masses live behind the ⚛︎ The Physics button, which opens a chaptered revision sheet; the curvature fit is drawn on the slice once the mass table is unmasked.

The Wine-Bottle Potential

Teach me Hide Text

Speed

Wine-bottle potentialV(φ) = μ²|φ|² + λ|φ|⁴, drag to orbit, click the surface to place φ

Guided Inquiry

1 · A quartic potential

Let’s get to know the landscape. The potential V(φ) is a quartic, and μ² is the coefficient of the quadratic term — it can be positive or negative. Have a play: what happens when you change μ²?

2 · At the centre

Now let μ² = −2. The potential looks like the bottom of a wine bottle, and the field sits right at the centre, φ = 0. Do you think it will stay there?

3 · The massive mode

Here’s the first thing the valley gives us. Small oscillations about the degenerate global minima, the equally deep lowest points all around the valley, are responsible for a massive mode in the spectrum of the theory.

4 · A push along the valley

Now let’s give it a nudge. The field rests at some point on the circle of degenerate global minima, where every point is equally deep. If you push the field along the green arrow, what do you think happens?

5 · What remains?

One last experiment. When μ² is varied but kept less than zero, which of the following modes do you think remains?

Listen

Potential

μ² −2.00

Ring of vacua λ = 1

Push the field

damping γ 0.00

Readouts

\|φ\|min	—
Vmin	—
radial m²	—
ring m²	—
radial period	—

V(φ) = μ²|φ|² + λ|φ|⁴, φ → e^(iα)φ (global U(1))

L = ½(∂φ₁)² + ½(∂φ₂)² − V, φ = φ₁ + iφ₂ (normalisation used here)

μ² < 0: |φ|min = √(−μ²/2λ), V_min = −μ⁴/4λ

m²_radial = V″(|φ|min) = −4μ² = 8λ|φ|²min, m²_ring = 0

With the normalisation above (φ₁, φ₂ canonically normalised) the curvature of V is exactly ω², so m² = ω² and the radial period is 2π/m. In the textbook convention L = |∂φ|² − V one rescales φ → φ/√2, giving m²_radial = −2μ² and v_std = √(−μ²/λ); the physics — one massive radial mode, one massless Goldstone per broken continuous symmetry — is identical. λ = 1 here.

The Physics

Spontaneous symmetry breaking · what this sim shows, the physics behind it, and every control

At a glance

A complex field φ moves on the potential V(φ) = μ²|φ|² + λ|φ|⁴, and the sign of μ² decides what that landscape looks like. This sim lets you reshape it, place the field on it and push it, so you can see how the field behaves and how stiff each direction of motion is.

What you'll learn

How the shape of V depends on the sign of μ², and what the phase of φ has to do with it.
What happens to a field placed exactly at φ = 0.
How the curvature of V in a direction sets the mass of that mode.
How to read |φ|min, V_min and the mode masses from the Readouts panel.

The field φ lives in a plane. A push can point away from the centre (orange) or sideways around it (green).

What's on screen

Red ball and trail: the field φ = φ₁ + iφ₂, drawn as a point in the plane.
Surface: V(φ) as a height over the plane, coloured on a fixed scale. The V axis (0 and ±1) is at the left.
Orange arrow: the radial direction, away from the centre. Its pill shows the radial m² once revealed.
Green arrow: the direction around the centre. Its pill shows the ring m² once revealed.
Strip below the stage: a cross-section of V along Re φ, with a dot for where φ sits.
Readouts panel: the minimum's radius and depth, and the mode masses.

See itSet μ² = +1 and release the field from rest at φ = 0.85. Watch for: the ball swinging through the centre, and the dot on the strip rocking up and down the same curve.

The controls

Press ⌖ on a row to see where that control sits in the sim.

Control	What it changes	Range
μ²	The coefficient of the quadratic term in V. Its sign sets the shape of the landscape.	−2 to +1 opens at +1.00
Phase tag	Names the case: Ring of vacua for μ² < 0, One vacuum otherwise.	label
Ring push	Kicks the field sideways, around the centre, at speed 0.55. At the centre the direction is random.	button
Radial kick	Kicks the field away from the centre, along the line to it, at speed 0.9. At the centre the direction is random.	button
damping γ	Slows the field as e^−γt. It is added by hand and is not part of V.	0–0.6 opens at 0.00
Readouts	\|φ\|min, V_min, radial m², ring m² and the radial period. They read — until they are revealed in the Guided Inquiry; Free exploration shows them at once.	panel
Stage	Drag to orbit the view. Click the surface to place φ there, at rest.	canvas
▶ Play / ⏸ Pause	Runs the motion, or freezes it.	top bar
Speed	Plays the motion faster or slower on screen. Model time, and so every period, stays the same.	0.25×–4× opens at 1×
↻ Reset	Returns to the opening setting: μ² = +1, γ = 0.	top bar

Fixed: λ = 1 and unit mass. An elastic wall at |φ| = 1.47 keeps the field on screen.

The sign of μ² reshapes the potential

The sim's potential. Turning φ by any phase α leaves V unchanged: a global U(1) symmetry.

V depends only on the size |φ|. The two terms compete:

μ² > 0: both terms raise V, so there is one minimum, at φ = 0.
μ² < 0: the quadratic term pulls V down near the centre and the quartic term pulls it up far out. V dips below zero and rises again, so φ = 0 becomes a hilltop. The lowest points sit at the same distance from the centre in every direction, a ring of minima.

V along Re φ at the two ends of the μ² slider.

The ring's radius and depth, both read from the Readouts panel. At μ² = −2 and λ = 1: |φ|min = 1 and V_min = −1.

See itSweep μ² from +1 down to −2 with the field at rest at the centre. Watch for: the single bowl on the strip turning into a double-welled curve, and the tag changing to Ring of vacua.

A field at the centre does not stay when μ² < 0

φ = 0 is an equilibrium for every μ², because V is flat there. Whether it is a stable equilibrium depends on the curvature. Near the centre the equation of motion becomes φ̈ ≈ −2μ²φ.

μ² > 0: a displacement oscillates about the centre.
μ² < 0: a displacement grows as e^κt with κ = √(−2μ²), which is 2 per model second at μ² = −2.

The equation of motion the sim integrates, with unit mass and λ = 1.

The sim starts the field at exactly φ = 0, where nothing would move. So when μ² < 0 it adds one small nudge (speed 0.05, random direction) as a stand-in for a real fluctuation. V prefers no direction, so the direction is an accident: run it again and the ball leaves another way.

See itPut the field at rest at φ = 0 with μ² = −2. Watch for: the note “vacuum fluctuation — random direction”, then the ball rolling off the hilltop and down into the valley.

Two ways to move away from a minimum

In polar form the phase θ never appears in V.

From a point on the ring of minima there are two directions to go:

Across the valley (ρ changes): V rises on both sides, so there is a restoring force and the field oscillates.
Along the valley (θ changes): V is exactly constant, so nothing pulls the field back. A push keeps it going round.

Two directions out of the same minimum, with very different costs.

In the sim a Ring push at μ² = −2 lets the radius creep out from 1.00 to about 1.06 and back, raising V by under 2% of the well depth. That is centrifugal, not a restoring force.

See itAt μ² = −2, kick the field across the valley. Watch for: the dot on the strip rocking back and forth across the well.

See itAt μ² = −2, push the field along the valley. Watch for: the dot on the strip staying level while the ball circles the ring.

Curvature sets the mass of each mode

With this sim's kinetic term, ½(∂φ₁)² + ½(∂φ₂)², the curvature of V in a direction is the ω² of small oscillations that way. So m² = V″ and the period is 2π/m.

Across the valley V″ = −4μ². Along it V does not change, so m² = 0.

μ² γ

These sliders move the sim too.

Halving |μ²| halves the radial m². The ring m² stays at zero. 2π/m holds for small oscillations; the sim's radial kick is large, so at μ² = −2 the real swing takes about 2.4 instead of 2.2.

See itCompare μ² = −2 with μ² = −1, field on the ring. Watch for: radial m² falling from 8.00 to 4.00 in the Readouts panel while ring m² stays at 0.00.

Go deeper: m² depends on the convention

Some textbooks write L = |∂φ|² − V, with no ½. The same V then gives m²_radial = −2μ², half the sim's value, and the minimum is written v/√2 with v = √(−μ²/λ). Only the kinetic normalisation differs: the ring mode is massless in both.

The law is symmetric, the vacuum is not

Turning the phase of φ changes neither V nor the equation of motion. Yet the field rests at a single point of the ring, at one particular phase θ₀. Every other point is just as good, and any of them could have been chosen.

Both statements are true at once.

This is spontaneous symmetry breaking: the law keeps its symmetry and the state does not. The rotation carries one minimum into another, and the ring is the set of all of them. In the sim the random nudge from section 04 decides θ₀.

See itDrop the field on the hilltop twice, from the same start. Watch for: the two roll-off directions reported on the card; they are different, and neither is built into V.

Where it matters

Superfluids and magnets. In superfluid helium-4 the phase of the condensate picks a point on its own circle, and the massless mode is sound (the phonon). In a ferromagnet the spins pick a direction, and the massless mode is the magnon.
Pions. The strong interaction has an approximate symmetry that the vacuum breaks, and quark masses break it slightly too. The three pions are therefore light (about 135 to 140 MeV, against the proton's 938 MeV) but not massless.
The Higgs field. The Standard Model's Higgs potential has this wine-bottle shape, with more components and the symmetry gauged. The three would-be massless modes become the extra polarisation of the W⁺, W⁻ and Z, and the radial mode is the 125 GeV Higgs boson. This sim's U(1) is global, so nothing is absorbed.

Go deeper: counting the massless modes

Goldstone's counting, for relativistic theories.

Three vacuum manifolds. The superfluid's circle is this sim's ring; the magnet breaks two generators, but they pair up into one magnon.

Summary

In one line. When μ² < 0 the field rests on a ring of equal minima; across the ring V curves up, so that mode is massive (m² = −4μ²), and along it V is flat, so that mode is massless.

Key points

For μ² < 0, V has a ring of minima at |φ| = √(−μ²/2λ), and φ = 0 is an unstable hilltop.
The law is symmetric under phase rotation; the state picks one phase, and nothing in V decides which.
m² is the curvature of V in that direction: −4μ² across the valley, 0 along it.

Common mix-ups

"A field released near the centre falls back, like a ball in a bowl." Only when μ² > 0. For μ² < 0 the centre is a hilltop.
"μ² < 0 means a negative or imaginary mass." μ² is a coefficient in V. The mass comes from the curvature at the minimum, which is positive (−4μ²).
"My textbook says m² = −2μ², so the sim is wrong." Both are right: the kinetic term is normalised differently.

Things to try on your own

Set μ² = −1, then −2. Read |φ|min and the radial m² from the Readouts panel and check them against √(−μ²/2) and −4μ².
At μ² = −2 give a Ring push with γ = 0.3. Note what happens to the circulation, and why V is not the cause.
Set μ² = 0 and press Radial kick. Compare what the Readouts panel shows with μ² = −1.