Viscosity — dynamic μ, kinematic ν = μ - ρ; Newtonian vs non-Newtonian: Question bank

Before we start, one reminder so every symbol is earned:

$τ$ (Greek "tau") = the shear stress, the sideways force per unit area one fluid layer feels from its neighbour. Units pascals (Pa).
$μ$ ("mu") = dynamic viscosity, the fluid's internal-friction constant. Units Pa·s.
$ν$ ("nu") = kinematic viscosity $= μ / ρ$ , where $ρ$ ("rho") is density. Units m²/s.
$\overset{γ}{˙} = d u / d y$ = shear rate, how fast the flow speed $u$ changes as you move across the flow (direction $y$ ). Units s⁻¹.

True or false — justify

True or false: A river moving as a solid block at 10 m/s carries a large viscous stress inside it.

False. Viscous stress depends on the velocity gradient

d u / d y

, not the speed

u

. If every layer moves at the same 10 m/s,

d u / d y = 0

, so

τ = μ \cdot 0 = 0

everywhere inside.

True or false: A fluid at rest has zero viscosity.

False. Viscosity

μ

is a fixed material property that exists whether or not the fluid moves. What is zero at rest is the viscous stress, because the shear rate is zero — but

μ

itself is unchanged.

True or false: If

μ_{A} > μ_{B}

then fluid A always spreads a push more slowly than B.

False. How fast a push (momentum) spreads is set by

ν = μ / ρ

, not

μ

. Air has smaller

μ

than water yet larger

ν

(because air's

ρ

is ~800× smaller), so air spreads momentum faster per unit mass.

True or false: Non-Newtonian means "extremely viscous."

False. Non-Newtonian means

μ

is not constant as the shear rate changes. Water (very low

μ

) and honey (very high

μ

) are both Newtonian because their

μ

stays fixed; ketchup is non-Newtonian even though it isn't especially thick.

True or false: Doubling the plate speed in Couette flow doubles the viscous stress.

True for a Newtonian fluid. The profile stays linear so

\overset{γ}{˙} = U / h

doubles, and

τ = μ \overset{γ}{˙}

is linear, so

τ

doubles. For a non-Newtonian power-law fluid (

τ = K \overset{γ}{˙}^{n}

) this fails — stress scales as

2^{n}

, not

2

True or false: Kinematic viscosity

ν

and a diffusion coefficient share the same units.

True. Both are m²/s. This is why

ν

is called the diffusivity of momentum: in time

t

, momentum spreads a distance of order

ν t

, exactly like heat or a dye spreading.

True or false: Oobleck (cornstarch + water) has a well-defined single viscosity number.

False. It is shear-thickening: its apparent viscosity

μ_{a pp} = τ / \overset{γ}{˙}

rises as you shear it faster (that's why you can run across it but sink if you stand still). No single

μ

describes it — you must state the shear rate.

Drag comes from the gradient near the surface, not the speed itself. The air far away moves fast too, but the stress is set by how sharply speed drops to zero at the car's skin — the boundary-layer gradient. See Boundary Layer.

Find the flaw: "Ketchup is non-Newtonian, so its apparent viscosity increases when I squeeze the bottle harder."

Ketchup is shear-thinning: harder squeezing means higher shear rate, so

μ_{a pp}

drops and it flows more easily. The student reversed the direction — that reversal describes shear-thickening.

Find the flaw: "For a Bingham plastic like toothpaste,

τ = τ_{0} + μ_{p} \overset{γ}{˙}

, so even at zero applied stress a tiny flow occurs."

No — below the yield stress

τ_{0}

the material does not flow at all (

\overset{γ}{˙} = 0

). The formula only applies after

τ

exceeds

τ_{0}

; there is a genuine threshold, which is why toothpaste holds its shape.

Find the flaw: "Units of

μ

must be Pa because it is a kind of stress."

μ

is not a stress; it is the ratio of stress to shear rate. Deriving from

μ = τ / \overset{γ}{˙}

gives

Pa / s^{- 1} = Pa\cdots

, i.e. kg/(m·s), not Pa.

Find the flaw: "The power-law

τ = K \overset{γ}{˙}^{n}

with

n = 1

describes a shear-thickening fluid."

n = 1

is exactly Newtonian (constant

μ = K

). Shear-thickening needs

n > 1

; shear-thinning needs

n < 1

Why questions

Why does the shear stress depend on the difference in layer speeds rather than the speeds themselves?

Viscous stress is molecular momentum-swapping between layers. If both layers already move at the same speed, an exchanged molecule carries the same momentum both ways, so there is no net transfer — hence stress needs a speed difference (a gradient).

Why do we bother defining

ν

when we already have

μ

Newton's second law says acceleration = force/mass, so predicting motion needs viscous force compared to inertia.

ν = μ / ρ

folds in the density (inertia), telling us how fast viscosity actually accelerates and spreads the flow — see Momentum Diffusion.

Why is

ν

the natural quantity inside the Reynolds number rather than

μ

The Reynolds number compares inertial to viscous effects, and that comparison is dimensionless only when viscosity appears as

ν = μ / ρ

(m²/s) alongside a length and a speed. See Reynolds Number.

Why does honey feel far more viscous than air yet air's

ν

is larger?

"Feel" tracks

μ

(force to shear), where honey wins hugely. But

ν

divides by density; air's density is hundreds of times smaller, so

μ / ρ

ends up larger for air even though its

μ

is tiny.

Why must Newton's law use the local gradient

d u / d y

rather than the overall ratio

U / h

U / h

only equals the gradient when the velocity profile is a straight line. In curved profiles (like Poiseuille Flow in a pipe) the gradient varies with position, so stress must be computed point-by-point from the true local

d u / d y

Why does a shear-thinning paint help both on the brush and on the wall?

On the brush the shear rate is high, so

μ_{a pp}

is low and the paint spreads thinly. On the wall the shear rate is near zero, so

μ_{a pp}

is high and the paint clings without dripping. One material serves two opposite needs.

Edge cases

Edge case: What is the viscous stress when

\overset{γ}{˙} = 0

for any fluid — Newtonian or not?

Zero. For Newtonian

τ = μ \cdot 0 = 0

; for a power-law fluid

τ = K \cdot 0^{n} = 0

(for

n > 0

). The only exception is a Bingham plastic, where a nonzero yield stress

τ_{0}

can be present without flow.

Edge case: As

\overset{γ}{˙} \to 0

for a shear-thinning power-law fluid (

n < 1

), what happens to the apparent viscosity?

μ_{a pp} = K \overset{γ}{˙}^{n - 1}

with

n - 1 < 0

blows up to infinity as

\overset{γ}{˙} \to 0

. Physically the paint becomes effectively "solid" at rest — which is why it doesn't drip. Real fluids cap this with a finite zero-shear plateau.

Edge case: As

\overset{γ}{˙} \to \infty

for a shear-thickening fluid (

n > 1

), what happens to

μ_{a pp}

μ_{a pp} = K \overset{γ}{˙}^{n - 1}

with

n - 1 > 0

grows without bound, so the fluid behaves ever more like a solid under fast impact — the "run on oobleck, sink standing still" effect.

Edge case: For an ideal (inviscid) fluid,

μ = 0

. What is

ν

, and what breaks?

ν = 0/ ρ = 0

: momentum does not diffuse at all, so no boundary layer forms and layers slide with zero drag. This is a useful idealisation but cannot satisfy the no-slip condition at a real wall — see Stokes Drag and Terminal Velocity for where finite

μ

is essential.

Edge case: A sphere falls in oil and reaches steady speed. Which viscosity governs the final velocity, and why is it constant?

Dynamic viscosity

μ

governs the drag force (Stokes drag

\propto μ

). At terminal velocity the upward drag exactly balances gravity minus buoyancy, so net force is zero and speed stops changing. See Terminal Velocity.

Edge case: If two fluids have identical

μ

but different

ρ

, do they generate the same wall shear stress in the same Couette setup?

Yes — wall shear stress is

τ = μ d u / d y

and does not contain

ρ

. Density only enters when you ask about acceleration or diffusion of momentum (via

ν

), not the instantaneous stress.

Recall One-line survival kit

Stress needs a gradient; $μ$ is force-strength; $ν = μ / ρ$ is spreading-speed; "Newtonian" means constant $μ$ , nothing about magnitude.

Question bank — Viscosity — dynamic μ, kinematic ν = μ - ρ; Newtonian vs non-Newtonian

True or false — justify

Spot the error

Why questions

Edge cases