Fluid dynamics

A vortex that blows up

In September 2026 a manuscript claimed a fluid flow that starts still, is driven by a smooth push, keeps finite energy, and still reaches infinite speed in finite time. This is its swirling core, running on your GPU in the coordinates where the collapse stands still.

Time left, τ = 1 − t: 0.1
Core radius: ×1
Swirl speed: ×1
Core energy: ×1

Camera

Speed (e-folds of τ per second)0.55

Exponent h (paper: h < 0.01)0.005

Swirl strength Ω6.0

Following the collapse: the camera zooms in as fast as the core shrinks, radius by √τ and height by τ^(½−h), so the vortex holds still while the readouts run.

What you are looking at

Particles riding the collapsing core

Each streak is a fluid particle carried by the leading-order vortex of the construction. Near the middle, fluid is drawn in, spun up and thrown out along the axis. The camera zooms in exactly as fast as the core shrinks, so the picture holds still while the clock runs towards the moment of blow-up. Switch the camera to Fixed to watch the core collapse to a point instead.

What is real, and what is illustrative

From the paper: the form of the leading field, the similarity coordinates, every scaling exponent, and the incompressibility formula that sets the radial flow.
Ours, and checked: the equations of motion in log time that the GPU integrates (proof below).
Illustrative: the profile shapes. The paper builds its profiles by an existence argument; ours have the same structure and decay. The pulses and higher-order corrections that make the flow an exact solution are not simulated.

The explainer

How can less energy go infinitely fast?

Pick a level on the diagram: Curious has no symbols, Builder has the code-level picture, and Mathematician has the statements. The pictures are the same at every level.

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The question
This whirlpool spins faster and faster, heading for infinite speed at one exact moment, while the energy in its core drains away towards nothing. How can less and less energy go infinitely fast?
Still water, a smooth push
It starts with perfectly still fluid and a push that is smooth and acts only inside a limited region. Nothing is infinite at the start, and the push never becomes infinite either. If anything blows up, the fluid did it to itself.
Spin-up
Near the middle, fluid spirals inwards. Like a skater pulling in their arms, the closer it gets to the axis, the faster it spins.
In at the middle, out along the axis
The spun-up fluid can’t pile up in the middle, because water can’t be squeezed. So it is thrown out up and down the axis, and the whole whirl gets thinner and shorter as it goes.
Pulses at the edge
A whirl like this can’t keep itself going. Where it meets the calmer fluid around it, it needs a push that would grow without limit. The trick is ripples: rings of fluid at the core’s edge oscillate, grow by feeding on the swirl, then die away, and on average their churning delivers the push. The outside force only has to start them.
The paradox in numbers
As the end approaches, the speed rises without limit and the core’s energy falls to zero, both at once.
Smaller beats faster
Here’s the click. Energy is roughly speed squared times how much fluid is moving. The core gets faster, but it also gets tiny, and it shrinks faster than its speed grows. So its energy drains away even as it races towards infinity.
Zoom with the collapse
Now zoom in at exactly the rate the whirl shrinks. The collapse freezes into the same shape while the clock creeps towards the end. That is how the live simulation at the top of this page works: the camera dives forever and the vortex holds still.
What is claimed
That’s the claim at the heart of a September 2026 paper: a smooth push on still water that makes the speed infinite in a finite time. Mathematicians are still checking it. It is also not something you could make in a lab: as a later study points out, real water would tear open into vapour bubbles, and air would form shock waves, long before.

The maths we checked

The collapse stands still in log time

Fix $0 < h < \tfrac12$ , $A = \tfrac12 + h$ , $D = \tfrac12 - h$ . For $\tau = 1 - t > 0$ let $q > |z|^{1/D}$ solve $q - z^2 q^{2h} = \tau$ , and put $\eta = z q^{-D}$ , $X = r^2/(2q)$ , $d = 1 - \eta^2$ , $L = 1 - 2h\eta^2$ . For smooth $U, V_0, F$ , consider the axisymmetric field

u_r = \frac{V_0(X,\eta)}{r}, \qquad u_\theta = q^{-A}\sqrt{2X}\,F(X,\eta), \qquad u_z = q^{-A}\,U(X,\eta).

Claim. Along every particle path, with $S = -\ln(1-t)$ ,

\frac{dX}{dS} = d\Big(V_0 + \frac{X(1-2\eta U)}{L}\Big), \qquad \frac{d\eta}{dS} = d\Big(U + \frac{D\eta(1-2\eta U)}{L}\Big), \qquad \frac{d\theta}{dS} = d^{1+h}e^{hS}F.

In particular the motion of (X, η) is an autonomous planar system.

Read the proof

Step 1: how q changes

Since $z = \eta q^{D}$ and $2D + 2h = 1$ , we have $z^2 q^{2h-1} = \eta^2$ . Differentiating $q - z^2 q^{2h} = 1 - t$ in $t$ and in $z$ :

q_t = -\frac{1}{L}, \qquad q_z = \frac{2 z q^{2h}}{L} = \frac{2\eta\, q^{1-D}}{L},

using $D + 2h = 1 - D$ . Here $L \ge 1 - 2h > 0$ , so the implicit function theorem applies.

Step 2: rates along a particle

The scale $q$ depends on $(z, t)$ only, so along a path $\dot q = q_t + u_z q_z = \big(-1 + 2\eta U q^{1-A-D}\big)/L = (2\eta U - 1)/L$ , because $A + D = 1$ . Then, with $r u_r = V_0$ :

\dot X = \frac{r u_r}{q} - \frac{X}{q}\dot q = \frac{1}{q}\Big(V_0 + \frac{X(1-2\eta U)}{L}\Big), \qquad \dot\eta = u_z q^{-D} - \frac{D\eta}{q}\dot q = \frac{1}{q}\Big(U + \frac{D\eta(1-2\eta U)}{L}\Big),

\dot\theta = \frac{u_\theta}{r} = \frac{q^{-A}\sqrt{2X}F}{\sqrt{2Xq}} = q^{-1-h}F.

Step 3: log time

$dS/dt = 1/\tau$ , and $\tau = q - z^2q^{2h} = q(1-\eta^2) = qd$ . Multiplying each rate by $\tau = qd$ gives the first two equations, and $d\theta/dS = d\,q^{-h}F$ . Finally $q = e^{-S}/d$ , so $q^{-h} = e^{hS}d^{h}$ . Nothing on the right-hand sides for $X$ and $\eta$ depends on $q$ or $S$ . ∎

Adversarially reviewed by four independent AI verifiers, each given only the statement and the proof with instructions to break it; all four found no hole. Also checked symbolically with SymPy: every residual vanishes to 40 significant digits at random points.

Sources

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