Question bank — Navier-Stokes equations — derivation from Newton's second law for fluid

False — a particle can still accelerate through the convective term $(\vec{u}\cdot \nabla )\vec{u}$, e.g. a leaf speeding up as it drifts into a narrowing river even though the flow at each fixed point never changes. (See the convective-acceleration figure.)

False — a uniform pressure pushes equally on all faces and cancels; only the gradient $-\nabla P$ (a difference across the blob) gives a net force density, so the value of $P$ is irrelevant to motion.

False — it vanishes when the velocity field is linear or uniform (zero curvature ${\nabla }^2\vec{u}=0$), regardless of speed; fast uniform flow has no viscous force, while slow curved flow can have a large one.

False — $\mu =0$ gives Euler's equation ($\rho D\vec{u}/Dt=-\nabla P+\rho \vec{g}$); you get hydrostatics only by additionally setting $\vec{u}=0$.

False — it is nonlinear (velocity multiplies its own gradient); this single term is why turbulence and most of the difficulty in fluid dynamics exist.

False — both are generally nonzero and balance: $\nabla P=\rho \vec{g}$ (with $\hat{z}$ up, this reads $\partial P/\partial z=-\rho g$). Zero net force does not mean the individual force densities vanish.

False — viscosity diffuses momentum; it can equally speed a slow layer up by dragging it along a faster one. It removes velocity differences, not speed per se.

False — it is momentum conservation, i.e. $\vec{F}=m\vec{a}$ per unit volume. Energy is a separate (derivable) balance.

False — $\nabla \cdot \vec{u}=0$ is incompressibility. A gas being compressed or expanded has $\nabla \cdot \vec{u}\ne 0$; density $\rho$ then varies in space and time, which is exactly what lets sound waves exist.

Error — for a field the particle moves to a new location where $\vec{u}$ differs, so you must add the convective term: true acceleration is $D\vec{u}/Dt=\partial \vec{u}/\partial t+(\vec{u}\cdot \nabla )\vec{u}$.

Error — dimensionally and physically wrong; force per volume needs a difference of pressure across the blob, which is the gradient $-\nabla P$ (units $\text{N}/{\text{m}}^3$). A single value $P$ (units $\text{N}/{\text{m}}^2$) cannot be a force density.

Error — $\tau =\mu \partial u_x/\partial y$ is only one component of the full viscous stress tensor ${\tau }_{ij}=\mu (\partial u_i/\partial x_j+\partial u_j/\partial x_i)$, valid here because we picked a simple shear. Also $\tau$ is a stress (force per area); the net force density is the difference of stress across the element, adding one more derivative, giving $\mu {\nabla }^2\vec{u}$ for incompressible flow.

Error — we drop $dV$ because it is a common factor on both sides after writing $\vec{F}=m\vec{a}$; the smallness justifies treating fields as differentiable, not the cancellation itself.

Error — gravity is a body force sitting on the force side of $\vec{F}=m\vec{a}$; the material derivative belongs only to the acceleration ($\vec{a}$) side.

Error — $\nabla \cdot \vec{u}=0$ is the incompressibility constraint used to simplify the viscous term; for compressible gases it fails, $\rho$ varies, and an extra term $(\mu /3+\zeta )\nabla (\nabla \cdot \vec{u})$ involving the bulk viscosity $\zeta$ appears. It is not optional bookkeeping.

Error — the first is (velocity dotted into the gradient) acting on $\vec{u}$ (a vector, the convective acceleration); the second is (divergence of $\vec{u}$, a scalar) times $\vec{u}$. Different operations entirely.

Because $\vec{F}=m\vec{a}$ is a law about matter (the same blob of mass), and that blob moves; watching a fixed point mixes different particles passing through it. Hence the material derivative $D/Dt$.

The stress on a face is a first derivative $\mu \partial u_x/\partial y$; the net force density is the difference of that stress across the blob, which is the derivative of a derivative — a second derivative $\mu {\nabla }^2\vec{u}$. (See the viscous-diffusion figure: force comes from the curvature of the profile.)

Fluid is pushed from high pressure toward low pressure; $\nabla P$ points toward increasing $P$, so the force density is $-\nabla P$, pointing toward decreasing pressure. (See the pressure-gradient figure.)

Steady, unidirectional flow with pressure gradient $\partial P/\partial x=-G$ reduces the $x$-equation to the boundary-value ODE $\mu \frac{d^2{u}_x}{d{y}^2}=-G$ on $0\le y\le h$ with no-slip $u_x(0)=u_x(h)=0$. Integrate once: $u_x^{\prime }=-\frac{G}{\mu }y+C_1$. Integrate again: $u_x=-\frac{G}{2\mu }{y}^2+C_{1}y+C_2$. Apply $u_x(0)=0\Rightarrow C_2=0$; apply $u_x(h)=0\Rightarrow C_1=\frac{Gh}{2\mu }$. This gives the downward parabola $u_x(y)=\frac{G}{2\mu }y(h-y)$, zero at both walls and maximal at the centre.

Estimate each term's size using a typical speed $U$ and length $L$. The inertial term scales as $\rho (\vec{u}\cdot \nabla )\vec{u}\sim \rho U^2/L$; the viscous term scales as $\mu {\nabla }^2\vec{u}\sim \mu U/L^2$. Their ratio is $\frac{\rho U^2/L}{\mu U/L^2}=\frac{\rho UL}{\mu }=\frac{UL}{\nu }\equiv \text{Re}$. Large $\text{Re}$ → inertia (and turbulence) wins; small $\text{Re}$ → viscosity smooths everything out. Note density $\rho$ is essential; dropping it gives wrong dimensions.

In the continuum picture the blob's mass is $\rho dV$, so its weight is $\rho \vec{g}dV$; dividing by volume gives the body-force density $\rho \vec{g}$ (units $\text{N}/{\text{m}}^3$) uniformly.

Removing viscosity deletes only the friction term; the fluid can still move and accelerate, so the inertia and convective terms survive — that is precisely the inviscid flow equation, not the rest equation.

${\nabla }^2\vec{u}=0$, so there is no viscous force — no velocity difference means no layer-on-layer drag, even though the fluid is moving.

The shear stress (first derivative) is zero there because velocity is maximal, but the viscous force density $\mu d^2{u}_x/d{y}^2$ is nonzero and constant everywhere — it is the curvature, not the slope, that supplies the force.

Velocity is zero at the wall, so there is no acceleration and no inertia there; the pressure gradient is balanced entirely by the viscous term, which is why wall friction is largest where the profile is steepest.

Only $0=-\nabla P+\rho \vec{g}$ remains — every velocity-dependent term (unsteady, convective, viscous) vanishes, recovering hydrostatics $dP/dz=-\rho g$ (with $\hat{z}$ up).

The viscous term $\mu {\nabla }^2\vec{u}$ becomes small compared to inertia, so the flow behaves approximately like Euler flow — except in thin boundary layers near walls where viscosity always matters.

No — $\partial \vec{u}/\partial t=0$ but the direction of $\vec{u}$ changes along the path, so $(\vec{u}\cdot \nabla )\vec{u}$ supplies a centripetal-style acceleration; steady never means static.

Both the inertia side and the gravity term vanish, leaving $0=-\nabla P+\mu {\nabla }^2\vec{u}$ — a degenerate massless balance; physically unrealistic, it signals that the density carries the "matter" that Newton's law needs. (It also makes $\nu =\mu /\rho$ blow up.)

Yes — sound is a small oscillation of density $\rho$ and pressure $P$. It requires $\nabla \cdot \vec{u}\ne 0$ (fluid locally compresses/expands), so it lives outside the incompressible model; the coupling of the momentum equation to a varying-$\rho$ continuity equation is what carries the wave.

No — the bulk-viscosity term is proportional to $\nabla (\nabla \cdot \vec{u})$, and incompressibility forces $\nabla \cdot \vec{u}=0$, so it drops out entirely. It only matters for compressible motions like shock waves and sound absorption.