Continuous-time flow matching

Hook

A diffusion model's noise schedule — how much noise to add at each of dozens or hundreds of steps — is its own small research problem, tuned separately from the model itself. What if the path from noise to data were just... a straight line?

Intuition

x_t = 1.500 at t = 0.50 — velocity = 7 (same at every t)

t0.50

Drag $t$ along the path. The velocity readout never changes — a straight line has exactly one slope, everywhere along it, which is the entire simplification this chapter is built on.

Formalize

Flow matching defines the path between a noise sample $x_0$ and a data sample $x_1$ as a straight line, and trains a model to predict that line's velocity:

x_t = (1-t)x_0 + tx_1, \qquad \frac{dx_t}{dt} = x_1 - x_0

$x_t$ — the point on the path between noise and data at time $t$ .
$x_0$ — the noise sample, the path's starting point (at $t=0$ ).
$x_1$ — the data sample, the path's endpoint (at $t=1$ ).
$t$ — the time parameter, ranging from $0$ (pure noise) to $1$ (data).
$dx_t/dt$ — the velocity along the path: the quantity a neural network is trained to predict.

Velocity is constant along the path
It doesn't depend on $t$ at all — it's the same number at $t=0.01$ as at $t=0.99$ .
Trained as a plain regression target
A neural network $v_\theta(x_t, t)$ is trained to regress directly onto $x_1 - x_0$ for many sampled pairs — a plain regression target, not a multi-step noise-removal problem.
No schedule, no reverse-process approximation
There's no schedule to design and no approximation of a reverse diffusion process — just "given where you are and what time it is, which way is data?"

Play

1 giant step

5.0000

50 tiny steps

5.0000

both land on x1 = 5 exactly — no discretization error to trade off, because the true path has no curvature to miss

A single Euler step and fifty small ones land in the exact same place. That's not true for a curved path (diffusion's reverse process genuinely curves, which is why it needs many small steps to trace accurately) — it's only true because this path is perfectly straight, with zero curvature to approximate away.

Worked example

Three independent (noise, data) pairs, evaluated at $t=0.5$ :

Pair (-2, 5)
$x_{0.5} = 0.5(-2) + 0.5(5) = 1.5$ . Velocity $= 5-(-2)=7$ .
Pair (3, -7)
$x_{0.5} = 0.5(3) + 0.5(-7) = -2$ . Velocity $= -7-3=-10$ — a different pair, a completely different constant velocity.
Pair (0, 4)
$x_{0.5} = 0.5(0) + 0.5(4) = 2$ . Velocity $= 4-0=4$ .

Checkpoint

Find the (x0, x1) pair, among the three candidates, whose position at t = 0.5 is negative.

Pick a pair to try it

Summary

x_t = (1-t)x_0 + tx_1, \qquad v_\theta(x_t, t) \approx x_1 - x_0

Each individual pair's path is a straight line with one velocity — but a real flow-matching model sees many noise-data pairs during training, and at a given $(x,t)$ different pairs' paths can pass nearby with different velocities. What the model actually learns is the average velocity across every pair consistent with being at $x$ at time $t$ , which is no longer perfectly straight in general — the simplification this chapter isolates is the per-pair target, not the learned field as a whole. The next chapter asks what happens if, instead of one endpoint reached by integrating many small steps, a model is trained to jump straight from anywhere on a path to its endpoint in one shot.