Visual walkthrough — Navier-Stokes equations — derivation from Newton's second law for fluid

Step 1 — WHAT is our object, and what is a "velocity field"?

WHAT. Picture the river as a grid of arrows. At every point in space there is a little arrow telling you which way the water there is moving and how fast (arrow length). That collection of arrows is called a velocity field, written $\vec{u}(\vec{r},t)$.

Let's decode that notation, symbol by symbol:

• $\vec{u}$ — the little arrow: a velocity (has direction + size). The tiny arrow on top means "this is a vector, not just a number."
• $\vec{r}$ — where you are standing (a position). The field's value depends on where you look.
• $t$ — when you look (time). The field can change moment to moment.

WHY. A fluid is not one particle — it is a continuum. We cannot write $\vec{F}=m\vec{a}$ for every molecule. So we cut out one tiny cube of fluid (volume $dV$, mass $dm=\rho dV$, where $\rho$ = mass packed per unit volume, the density) and we will do Newton's law on that one cube.

PICTURE.

Step 2 — WHAT is the cube's acceleration? (the material derivative)

WHAT. Acceleration means "how fast the velocity is changing for the cube itself." Here is the trap: the cube moves. As it drifts, it lands in a spot where the water has a different arrow. So its velocity changes for two separate reasons.

WHY two reasons. Watch a leaf enter a narrowing channel:

1. The whole river might be speeding up over time → the arrow at your fixed point grows. Call this the local (unsteady) part.
2. Even if the river never changes in time, the leaf is carried from wide slow water into narrow fast water → its own speed rises. Call this the convective (carried) part.

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HOW (chain rule, from scratch). The cube sits at position $\vec{r}(t)=(x(t),y(t),z(t))$ that changes with time. Its velocity is $\vec{u}(x(t),y(t),z(t),t)$. Differentiating this "function-inside-a-function" with the chain rule:

$\frac{d\vec{u}}{dt}=\underset{\text{time changes}}{\underbrace{\frac{\partial \vec{u}}{\partial t}}}+\underset{\text{position changes}}{\underbrace{\frac{\partial \vec{u}}{\partial x}\frac{dx}{dt}+\frac{\partial \vec{u}}{\partial y}\frac{dy}{dt}+\frac{\partial \vec{u}}{\partial z}\frac{dz}{dt}}}$

Now the key substitution: the rate the cube's $x$-position changes is its $x$-velocity, so $\frac{dx}{dt}=u_x$, and likewise $\frac{dy}{dt}=u_y$, $\frac{dz}{dt}=u_z$. Plug those in and the three position terms bundle into one compact object:

$\boxed{\frac{D\vec{u}}{Dt}=\frac{\partial \vec{u}}{\partial t}+(\vec{u}\cdot \nabla )\vec{u}}$

Step 3 — WHAT pushes the cube? Force #1: gravity

WHAT. Every scrap of mass is pulled down by gravity. The pull on our cube is its mass times $\vec{g}$ (the downward gravitational acceleration, about $9.8{\text{m/s}}^2$).

$d{\vec{F}}_{\text{body}}=\underset{\text{density}}{\underbrace{\rho }}\underset{\text{down pull per kg}}{\underbrace{\vec{g}}}\underset{\text{cube volume}}{\underbrace{dV}}$

WHY it's called a body force. It acts on the whole body of the cube at once, not on its surface. Contrast with pressure and viscosity, which act only on the cube's faces.

PICTURE.

Step 4 — Force #2: pressure needs a difference, not a value

WHAT. Pressure $P$ is water squeezing inward on each face of the cube. Left face gets pushed right; right face gets pushed left.

WHY only differences matter. If the squeeze is the same on both sides, the two pushes cancel — zero net force. To move the cube, the pressure on one face must exceed the pressure on the opposite face. So the force depends on how pressure changes across the cube.

HOW (derivation). Push on the left $x$-face (area $dydz$) is $P(x)dydz$ pointing $+x$. Push on the right face is $P(x+dx)dydz$ pointing $-x$. Net $x$-force:

$d{F}_x=[P(x)-P(x+dx)]dydz=-\underset{\text{slope of }P}{\underbrace{\frac{\partial P}{\partial x}}}\underset{dV}{\underbrace{dxdydz}}$

The bracket is minus the change in $P$ over the tiny step $dx$ — that is the derivative $\partial P/\partial x$. Doing all three directions:

$\boxed{d{\vec{F}}_{\text{pressure}}=-\nabla PdV}$

The minus sign says: force points from high pressure toward low pressure (downhill on the pressure landscape). See Hydrostatic pressure.

PICTURE.

Step 5 — Force #3: viscosity, and WHY it's a second derivative

WHAT. Neighbouring fluid layers moving at different speeds rub each other — molecules from the fast layer drag the slow layer forward and vice-versa. This sideways drag is viscous friction. See Viscosity & Newton's law of viscosity.

Newton's law of viscosity gives the stress (force per area) one layer exerts on the next: $\tau =\mu \frac{\partial u_x}{\partial y}$ where $\mu$ ("mu") is the fluid's stickiness (viscosity) and $\partial u_x/\partial y$ is how fast the flow speed changes as you climb in $y$.

WHY second derivative. Here is the subtle jump. $\tau$ is only the pull on one face. The cube feels the pull on its top face minus the pull on its bottom face. A force needs the difference in stress across the cube — that is a derivative of the stress, which was already a derivative. Derivative of a derivative = second derivative.

HOW (derivation). Net $x$-force from the two $y$-faces: $\left[\mu \frac{\partial u_x}{\partial y}{|}_{y+dy}-\mu \frac{\partial u_x}{\partial y}{|}_y\right]dxdz=\mu \underset{\text{difference of slopes}}{\underbrace{\frac{{\partial }^2{u}_x}{\partial y^2}}}dV$

Add the contributions from all faces (and use incompressibility to drop extra pieces):

$\boxed{d{\vec{F}}_{\text{visc}}=\mu {\nabla }^2\vec{u}dV}$

PICTURE.

Step 6 — Assemble everything: $\vec{F}=m\vec{a}$ for the cube

WHAT. Stack the three forces on the left, put mass × acceleration on the right.

$\underset{\text{gravity}}{\underbrace{\rho \vec{g}dV}}\underset{\text{pressure}}{\underbrace{-\nabla PdV}}+\underset{\text{viscous}}{\underbrace{\mu {\nabla }^2\vec{u}dV}}=\underset{m}{\underbrace{\rho dV}}\underset{\vec{a}}{\underbrace{\frac{D\vec{u}}{Dt}}}$

WHY divide by $dV$. Every term carries the cube's volume $dV$. Cancel it, and we get a statement per unit volume — true no matter how tiny the cube. Then expand $\frac{D\vec{u}}{Dt}$ from Step 2:

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Step 7 — Edge case A: switch off viscosity ($\mu =0$)

WHAT. An idealised fluid with no stickiness. Delete the $\mu {\nabla }^2\vec{u}$ term.

$\rho \frac{D\vec{u}}{Dt}=-\nabla P+\rho \vec{g}$

WHY it matters. This is the Euler equation — a result we already trust. A correct derivation must collapse to it in this limit. It does. ✓

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Step 8 — Edge case B: switch off motion ($\vec{u}=0$)

WHAT. Fluid at rest. Every velocity term vanishes ($\partial \vec{u}/\partial t=0$, convective $=0$, viscous $=0$).

$0=-\nabla P+\rho \vec{g}\Rightarrow \nabla P=\rho \vec{g}\Rightarrow \frac{dP}{dz}=-\rho g$

WHY. With nothing moving, pressure must exactly hold up the weight. Integrating gives $P=P_0+\rho gh$ ($h$ measured downward) — plain Hydrostatic pressure. Our most basic fluid fact falls straight out. ✓

PICTURE.

Step 9 — Edge case C: steady flow between plates (Poiseuille)

WHAT. Water pushed through the gap between two fixed plates at $y=0$ and $y=h$, driven by a pressure that drops steadily along $x$: $\partial P/\partial x=-G$ (with $G>0$). Steady, straight flow: only $u_x(y)$ survives.

WHY these cancellations. Steady ⟹ $\partial \vec{u}/\partial t=0$. Straight parallel flow ⟹ convective term $=0$. Only the $x$-equation is left:

$0=G+\mu \frac{d^2{u}_x}{d{y}^2}\Rightarrow \frac{d^2{u}_x}{d{y}^2}=-\frac{G}{\mu }$

Integrate twice, then apply no-slip ($u_x=0$ right at each wall, because fluid clings to solid):

$u_x(y)=\frac{G}{2\mu }y(h-y)$

A parabola: zero at both walls, maximum $\frac{G{h}^2}{8\mu }$ in the middle. This parabola is the fingerprint of viscous pressure-driven flow. See Poiseuille flow.

PICTURE.

The one-picture summary

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