Jeromie Beasley · Compound Eye · Filament Operator Mixer

A Stable Filament That Still Amplifies

Five coordinates, one invariant plane, and a Jordan block inside it. Every eigenvalue says decay, yet a well-aimed disturbance grows first. Turn the dials and watch the states actually move: the stretch, the kick, the pseudospectral disk, and the two ways eigenvectors collapse.

Built from research revision 4 of the paper (25 September 2026) and its proof package. Every curve and number here is computed live from the paper's formulas or from the full five-by-five matrix.
5coordinates, one invariant plane V
κ > 2growth for the best initial state
κ > 2.48527growth after an interface kick
0unstable eigenvalues anywhere
√(ε|μ|+ε²)exact pseudospectral radius
27 / 27independent checks pass
Theorem 4.1 · the operator norm of the Jordan semigroup

Every eigenvalue decays. The states grow anyway.

Inside the invariant plane V = span{x₂, r} the operator is J = [[b, μ], [0, b]], and states evolve as u(t) = e−Jtu(0) = e−bt(x₂ − μt·r, r). The interface coordinate r only decays, but it keeps pushing the shape coordinate x₂. With κ = |μ|/b, the best initial state grows above its starting size exactly when κ > 2. Time runs in units of bt.

state plane V · the unit circle moves
size of the state over time
λ-plane of −J
κ = |μ|/b
best state: peak gain Mmax
peak time bt*
t* = √(κ²−4)/|μ|
interface kick: peak
eigenvalues of −J
−b, −b
stable, and they never move
initial growth rate of M
every enabled eye along the same clock
peak gain against κ · the two thresholds

Gold: the best initial state, Mmax from (4.2). Teal: a kick confined to the interface coordinate, the peak of (4.3). Between κ = 2 and κr = 2.48527 a carefully aimed state grows while an interface kick does not.

What the λ-plane shows: the double eigenvalue sits at −b for every κ, so the spectrum predicts pure decay. The violet disk is the numerical range of −J, centred at −b with radius |μ|/2 (a standard fact, not stated in the paper). Its right edge, b(κ/2 − 1), is the initial growth rate of M(t) read off the proof of Theorem 4.1. The disk crosses into the right half-plane exactly when κ > 2.
Equation (4.3) · the interface impulse threshold · the trap snapshot (6.1)

Kick the interface and watch the shape answer

An impact is not free to pick the most amplified direction. If it lands in the interface coordinate, the state starts at er and follows e−bt(−μt, 1): the interface glow fades while the shape bends, then everything decays. This input rises above its starting size only for κ > κr = 2.48526751266, where 1 − x = e−2x at the tangency. This is the model test that applies if an impact projects onto the interface coordinate [H].

cartoon · bend = shape coordinate x₂ · glow = interface coordinate r
energy after the kick · E(t)/E₀ = Ar(t)²
why 2.48527 · the tangency
peak E/E₀ after the kick
peak time bt
x₊ = (1 + √(1 − 4/κ²))/2
spectral condition
Re σ(−K) = −b
stable for every κ
trap snapshot in window
energy now
The filament picture is a cartoon of two model coordinates, not a plasma simulation. A physical fracture would also need a calibrated map from the model energy to magnetic stress, emission or shape [H].
Equations (3.1) to (3.3) · the exact pseudospectral disk

Shake the matrix and watch the eigenvalues scatter

For the reduced block the resolvent norm depends only on the distance from the eigenvalue, so every ε-pseudospectrum is exactly a disk: rε = √(ε|μ| + ε²). Small perturbations move the eigenvalue by about √(ε|μ|), far more than ε. The growth test belongs to −J, whose disk is centred at −b; it reaches the right half-plane once ε > εcrit = σmin(J). Here b = 1.

complex plane · resolvent norm of −J
radius against ε · square root, then linear
ε
disk radius rε
εcrit = σmin(J)
(√(μ²+4b²) − |μ|)/2
farthest dot / rε
dots drawn
eigenvalues of −J + E
A pseudospectral incursion and actual transient growth are related but distinct tests, and neither establishes an astronomical rupture. The sign matters: the test that bears on growth of u̇ = −Ku uses the pseudospectrum of −K.
Proposition 5.1 · detuning versus reverse coupling

Two roads into the same defect, at different speeds

Approach the Jordan block two ways and watch the eigenvectors close. Detune the diagonal by δ and the angle shrinks linearly, sin θ = |δ|/√(μ² + δ²). Unfold with a small reverse coupling γ and it shrinks like a square root, sin θ = √(γ/(μ + γ)). At the defect itself the eigenvector frame loses a direction while the invariant plane V stays two-dimensional.

diagonal detuning Jδ = [[b+δ, μ], [0, b]]
reverse coupling Jγ = [[b, μ], [γ, b]]
eigenvector angle against control · slopes 1 and ½
control
δ = γ
detuning: sin θ
linear, (5.1)
reverse coupling: sin θ
square root, (5.2)
ratio of the two angles
grows without bound as the control → 0
eigenvalue split
δ versus 2√(μγ)
These are statements about an eigenbasis. Neither rate implies that plasma streamlines turn into sheets, that a boundary becomes inviscid, or that a precursor is suppressed; those would need a field model and a measurement map.
Theorem 2.1 · the exact interface embedding · Section 7

All five coordinates, and the plane that holds the block

The full operator K₅(γ) couples a shape pair (x₁, x₂), a shear pair (s₁, s₂) and an interface coordinate r. The columns for x₂ and r have no entries outside rows x₂ and r, so V = span{x₂, r} is exactly invariant and carries Jγ. The whole five-coordinate norm is therefore never below the reduced one. Kick a coordinate and watch the signal flow through the couplings.

couplings · signal flowing through K₅
the matrix K₅(γ) · V highlighted
full norm ‖e−K₅t‖ against the reduced ‖e−Jt‖
spectrum of −K₅ from the factorization (2.3)
spectral condition
peak full gain
numerical, 5×5 matrix exponential
peak reduced gain
the block Jγ alone
min over t of (full − reduced)
Riesz range Ran Π
span{e₂, e₅}, fixed
so g = Ω = 0 in these coordinates
Five is sufficient, not necessary (Remark 2.2): a four-coordinate family with one shear mode carries the same Jordan block. And because V never moves in these coordinates, the projector geometry of Section 7 is flat here; a geometric signal needs an embedding that actually varies.
Section 1 · Figure 1 · what the observations say

The Snake: a kink blamed on a candidate impact

The Galactic Centre Snake is a radio filament about 30′ (roughly 70 pc) long and under 1 pc wide. At its major kink sits G359.13142−0.20005, a compact radio source with an X-ray counterpart. Its discoverers propose, and a 2025 NASA Chandra release restates, that this likely pulsar struck the filament at roughly 500–1000 km/s and distorted its magnetic field. No radio pulses have been detected from it. The confirmed millisecond pulsar PSR J1744−2946 lies in the Sunfish, 49″ due west, and its localization rules out an association with the Snake.

around the major kink · Galactic offsets, l increasing to the left
candidate at the kink
G359.13142−0.20005
radio + X-ray · no pulses detected
PSR J1744−2946
49″ due west
ΔRA·cos δ = −49.1″ · ΔDec = +1.4″
projected separation
1.9 pc
at 8 kpc, 1″ ≈ 0.0388 pc
proposed impact speed
500–1000 km/s
scenario estimate · direction not given

Source offsets are to scale, computed from the coordinates in the two source names. The filament traces are schematic, drawn by hand rather than traced from an image. North and east arrows come from the equatorial-to-Galactic rotation at the candidate.

On the current account the Snake is a forced disruption: its kink is attributed to an outside object whose nature and speed are unconfirmed. It is not an observed spontaneous rupture, and archival images cannot show whether a precursor existed. Other readings remain open: filaments lit by relativistic electrons from a nearby source, a stellar trail, or a filament that bends on its own through a kink instability.
Evidence grades · T proved here · C cited · V checked by script · H modelling or interpretation

What is proved, what is cited, and what is a hypothesis

Each row is one claim of the paper with its grade. Open a row to see it move in its lens.

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