Fluid kinematics — Eulerian vs Lagrangian description: Visual walkthrough

We are chasing one number: how fast does a quantity change for a lump of fluid that is being carried along by the flow? Call the quantity $ϕ$ (say, temperature). Call the answer $\frac{D ϕ}{D t}$ . That is all $D / D t$ means — a rate of change felt while riding along.

Step 1 — Two honest pictures of the same river

WHAT. Before any maths, pin down the two ways of watching a fluid, because the whole derivation is a conversion between them.

WHY. Our instruments (thermometers, pressure gauges) sit still and watch fluid go past — that is the Eulerian picture, a field $ϕ (x, y, t)$ pinned to locations. But Newton's laws talk about a specific particle — that is the Lagrangian picture, following one tagged lump. We need a bridge between them, and to build a bridge you first draw both banks.

PICTURE. On the left, a person on a bridge reads whatever water passes below (Eulerian, fixed dot). On the right, a tagged red parcel drifts downstream, carrying its own history (Lagrangian, moving dot).

Step 2 — A field is a landscape of numbers

WHAT. Draw the Eulerian field $ϕ (x, y, t)$ at one frozen instant as a coloured landscape — a value at every point.

WHY. You cannot talk about "moving into a warmer region" until you can see that regions differ. A field is just "a number stapled to every point of space." Picturing it as terrain (or shading) makes the coming ideas — gradient, moving uphill — literal.

PICTURE. Shading goes from cool (light) to warm (dark). The single red dot is our parcel sitting on this landscape. Nothing moves yet; we are only learning to read the map.

Step 3 — Reason ①: the landscape itself changes in time

WHAT. Freeze the parcel at one point and let time run. The whole landscape can rise or fall — imagine a tap being opened, warming the entire room.

WHY. This is the change you would measure even if you never moved. It is the honest Eulerian rate at a fixed point. We give it the symbol $\frac{\partial ϕ}{\partial t}$ .

PICTURE. Same location (red dot fixed), two snapshots: the shading darkens everywhere between $t$ and $t + Δ t$ . The parcel's value climbs without it moving.

Step 4 — Reason ②: the parcel walks to a new value

WHAT. Now freeze time and let the parcel move. Even on a perfectly unchanging landscape, walking from a low spot to a high spot changes the value you sit on.

WHY. The parcel does not choose to stay put — the flow carries it. So a second contribution appears purely from displacement. This is the term that catches beginners off guard, so we build it from a right triangle of small steps.

PICTURE. The parcel moves a tiny arrow $Δ x$ to the right and $Δ y$ up. Along the way the shading changes. We decompose the walk into a horizontal leg and a vertical leg.

Why this tool? Because "how much does $ϕ$ change when I take a small step in some direction" is answered by combining the two steepnesses. A step of $(Δ x, Δ y)$ changes $ϕ$ by $Δ ϕ_{from moving} \approx \frac{\partial ϕ}{\partial x} Δ x + \frac{\partial ϕ}{\partial y} Δ y .$ Each product reads "steepness in that direction $\times$ how far I went that way."

Step 5 — Both reasons happen at once: add the tiny changes

WHAT. In one real tick $Δ t$ , the parcel experiences both: the landscape shifts (Step 3) and it walks (Step 4). Add the two tiny changes.

WHY. They are independent causes acting in the same interval, so their small effects simply sum. This is the heart of the chain rule, stated in plain arithmetic before any symbol soup.

PICTURE. Two arrows stack tip-to-tail: a vertical "landscape rose" arrow ( $\partial ϕ / \partial t Δ t$ ) plus the "walked uphill" arrow from Step 4. Their total is the change the red parcel truly feels.

Δ ϕ_{felt} \approx landscape changed \frac{\partial ϕ}{\partial t} Δ t + parcel walked \frac{\partial ϕ}{\partial x} Δ x + \frac{\partial ϕ}{\partial y} Δ y

Step 6 — Divide by $Δ t$ : turn displacements into velocities

WHAT. Divide every term by the elapsed time $Δ t$ and let $Δ t \to 0$ so the " $\approx$ " becomes " $=$ ".

WHY. We want a rate (per second), not a lump change. Dividing a displacement $Δ x$ by $Δ t$ is exactly the definition of the parcel's velocity — that is why the velocity components appear, not by decree but by division.

PICTURE. The stacked arrows of Step 5 get "per second" tags. $\frac{Δ x}{Δ t}$ shrinks to the horizontal speed $u$ ; $\frac{Δ y}{Δ t}$ shrinks to the vertical speed $v$ .

Putting it together:

\frac{D ϕ}{D t} = \frac{\partial ϕ}{\partial t} + \frac{\partial ϕ}{\partial x} u + \frac{\partial ϕ}{\partial y} v .

Step 7 — Name the result and read every symbol

WHAT. Collect the moving terms into one compact package.

WHY. The two "walk" terms both multiply a velocity by a steepness — that pattern is the dot product of the velocity arrow with the gradient arrow, written $(v \cdot \nabla) ϕ$ . The dot product is the right tool because it answers "how much of the uphill direction am I actually walking along?"

This is the same formula the parent note wrote — now every symbol was earned by a picture.

Step 8 — The edge cases (never leave a scenario unshown)

Four situations decide which terms survive. Each has its own panel below.

① Standing still in a steady field ( $v = 0$ and $\partial / \partial t = 0$ ): both terms zero, $\frac{D ϕ}{D t} = 0$ . The parcel feels nothing. (Trivial, but it anchors the others.)
② Unsteady, but flat field ( $\nabla ϕ = 0$ , tap turning): only the local term. The landscape is level everywhere, so walking does nothing; only its rising-over-time matters. $\frac{D ϕ}{D t} = \frac{\partial ϕ}{\partial t}$ .
③ Steady, but sloped field ( $\partial / \partial t = 0$ , $v \neq = 0$ ): only the convective term. This is the headline case — nothing changes at any fixed point, yet the parcel accelerates because it walks into a steeper/faster region. $\frac{D ϕ}{D t} = (v \cdot \nabla) ϕ \neq = 0$ .
④ Moving along a level of the landscape ( $v ⊥ \nabla ϕ$ ): the dot product is zero even though the parcel moves and the field slopes — because it walks across the hill, never up it. Convective term $= 0$ .

Step 9 — Set $ϕ = v$ : the acceleration of a fluid particle

WHAT. Feed the velocity itself into the machine.

WHY. Newton needs acceleration = rate of change of the parcel's velocity, which is exactly $D v / D t$ .

a = \frac{D v}{D t} = \frac{\partial v}{\partial t} + (v \cdot \nabla) v

The convective term now has $v$ multiplying derivatives of $v$ — it is nonlinear. That single fact is the seed of turbulence and is why the Navier-Stokes equations are so hard. This acceleration is exactly what plugs into Euler's equation of motion.

The one-picture summary

Everything on one canvas: the red parcel on a sloped, time-changing landscape; the two arrows (landscape rising = local; walking uphill = convective) adding to the total change it feels.

Recall Feynman retelling (say it out loud, no symbols)

Imagine you're a tiny cork floating down a river. You carry a thermometer. Two things can make your reading change. One: somebody upstream turns on a hot spring and the whole river slowly warms — you'd feel that even if you were glued in place. That's the local change. Two: the water near the bank is cold and the middle is warm, and the current sweeps you from bank to middle — you warm up just by being moved, even if the river's temperature map never changes. That's the convective change. Your total felt change is simply those two added up. Written down, that sum is the material derivative $D / D t$ . And the sneaky lesson: a river can be perfectly "steady" — every fixed point reading constant forever — and yet you, the cork, still speed up and warm up, because you are the thing that moves.

Recall Term-by-term self test

Big- $D$ derivative means whose rate? ::: The moving parcel's (Lagrangian). Curly- $\partial t$ derivative holds what fixed? ::: The position (Eulerian, a fixed spot). Which term dies in steady flow? ::: The local term $\partial / \partial t$ . Which case gives acceleration in steady flow? ::: Convective, parcel walking into a faster region. When is the convective term zero even though the parcel moves? ::: When it moves along a level of the field, i.e. $v ⊥ \nabla ϕ$ . Why is fluid acceleration mathematically hard? ::: $(v \cdot \nabla) v$ is nonlinear in $v$ .

Related: Read this in Hinglish → · Steady vs unsteady flow · Streamlines, pathlines and streaklines · Continuity equation · Reynolds transport theorem