Viscosity and Newton's Law of Viscosity

1. The thought experiment

Trap a fluid between two large plates a distance h apart. Hold the bottom plate still and drag the top one at speed U. By the no-slip condition the fluid touching each plate moves with it, so a linear velocity profile u(y) = Uy/h develops. Experiment shows the force per unit area needed to keep the top plate moving is proportional to U and inversely proportional to h:

This is Newton's law of viscosity: shear stress is proportional to the local velocity gradient. The constant μ is the dynamic viscosity, with units Pa·s. Water at 20 °C: μ ≈ 1.0×10⁻³ Pa·s; air: 1.8×10⁻⁵ Pa·s; honey: ~10 Pa·s.

2. Why the law holds — momentum diffusion

Molecules wander between layers. A molecule hopping from a fast layer into a slow one delivers excess streamwise momentum; one hopping the other way removes it. The net effect is a transfer of momentum down the velocity gradient — macroscopically, a shear stress. Viscosity is therefore momentum diffusivity, and dividing by density makes this explicit:

Kinematic viscosity ν has the same units as thermal diffusivity α and mass diffusivity D — the three diffusivities whose ratios give the Prandtl and Schmidt numbers.

3. Newtonian and non-Newtonian fluids

A fluid is Newtonian if μ is independent of the shear rate: water, air, glycerine. Many industrially and biologically important fluids are not. Shear-thinning fluids (paint, blood) show falling apparent viscosity; shear-thickening ones (cornstarch suspensions) the opposite; and yield-stress fluids flow only above a threshold stress. The Casson model used throughout the nanofluid literature is of this last type:

Key takeaway

τ = μ du/dy is the constitutive bridge between kinematics (velocity gradients) and dynamics (forces). Insert it into Newton's second law for a fluid element and the Navier–Stokes equations follow.

4. Extended worked examples

Example 1 — force on a sliding plate. Water (μ = 1.0×10⁻³ Pa·s) fills a 0.5 mm gap between plates; the top plate of area 0.2 m² moves at 1 m/s. Shear stress τ = μU/h = (10⁻³)(1)/(5×10⁻⁴) = 2 Pa; force F = τA = 0.4 N — a finger push. Replace the water with honey (μ ≈ 10 Pa·s): τ = 20 000 Pa and F = 4 000 N, the weight of a small car. Same geometry, four orders of magnitude in μ, four orders of magnitude in force — viscosity is not a small correction.

Example 2 — air is more viscous than water (kinematically). Dynamic viscosities: μ_air = 1.8×10⁻⁵, μ_water = 1.0×10⁻³ Pa·s — water wins by a factor 55. But divide by density: ν_air = 1.8×10⁻⁵/1.225 = 1.47×10⁻⁵ m²/s versus ν_water = 1.0×10⁻⁶ m²/s. Air's momentum diffusivity is 15 times larger. Since boundary-layer thickness grows like √(νt), velocity disturbances spread faster through air than water. Whenever a formula contains ν rather than μ — Reynolds number, boundary-layer growth — air behaves as the "more viscous" medium.

Example 3 — reading a viscometer. A rotational viscometer shears a sample at γ̇ = 50 s⁻¹ and measures τ = 60 mPa. Apparent viscosity μ = τ/γ̇ = 1.2×10⁻³ Pa·s. Repeating at γ̇ = 500 s⁻¹ gives τ = 480 mPa → μ = 0.96×10⁻³. The 20% drop with shear rate flags the sample as mildly shear-thinning — a single-point measurement would have missed the non-Newtonian character entirely, which is why rheometers sweep the shear rate.

Example 4 — temperature is not a footnote. Water: μ = 1.79×10⁻³ Pa·s at 0 °C, 1.00×10⁻³ at 20 °C, 0.28×10⁻³ at 100 °C — a factor of 6 across the liquid range, following an Arrhenius-type law μ ≈ A e^{B/T} (molecules must hop out of cages, and heat helps). Gases run the other way: Sutherland's law gives roughly μ ∝ T^{3/2}/(T + S), increasing with temperature, because faster molecules carry momentum between layers more effectively. An engine oil calculation done at the wrong temperature can be wrong by an order of magnitude.

5. Common misconceptions

"Viscous means dense." Mercury: ρ = 13 546 kg/m³, μ = 1.5×10⁻³ Pa·s — denser than steel, barely more viscous than water. Glycerine: ρ = 1 260, μ = 1.4 Pa·s — slightly denser than water, a thousand times more viscous. The two properties come from different physics (packing versus intermolecular friction) and are uncorrelated.

"τ = μ du/dy is a law of nature." It is a constitutive model — a definition of the class of Newtonian fluids, verified for them by experiment. Blood, paint, polymer melts and drilling mud simply refuse to obey it. Conservation of mass and momentum are laws; Newton's viscosity relation is a material choice.

"In an inviscid model there is no shear anywhere." The model sets μ = 0, so τ = 0 whatever the velocity gradient. The gradients themselves can be enormous — which is exactly the pathology at a wall that Prandtl's boundary layer repairs.

"Kinematic and dynamic viscosity are interchangeable." They answer different questions: μ sets the force for a given gradient; ν sets how fast momentum diffuses. Mixing them up in a Reynolds number (which needs μ with ρ, or ν alone) is one of the most common numerical errors in student work.

6. Where this shows up

Hydrodynamic lubrication keeps engine bearings alive: a film microns thick generates the pressure that separates metal from metal, and the load capacity is proportional to μ — the whole SAE oil-grade system is viscosity management. Blood rheology matters clinically: blood is shear-thinning (viscosity falls from ~60 mPa·s at low shear to ~3.5 mPa·s in arteries), so microcirculation models that assume a Newtonian fluid overestimate resistance. Polymer processing (extrusion, injection moulding) is a battle with viscosities of 10²–10⁵ Pa·s that also depend on shear rate and temperature simultaneously. And volcanology reads eruption style straight off magma viscosity: basalt at ~10²–10³ Pa·s flows in rivers; rhyolite at 10⁸ Pa·s plugs the vent and explodes.

7. Practice problems

P1. SAE-30 oil (μ = 0.29 Pa·s) fills a 1 mm gap; the upper plate moves at 3 m/s. Find τ. Solution: τ = 0.29 × 3/0.001 = 870 Pa.

P2. What force does P1's stress exert on a 0.5 m² plate, and what power is needed to keep it moving? Solution: F = 870 × 0.5 = 435 N; P = Fv = 1 305 W — dissipated entirely as heat in the film.

P3. Compute ν for mercury (data above) and compare with water. Solution: ν = 1.5×10⁻³/13 546 = 1.1×10⁻⁷ m²/s — one-ninth of water's. Kinematically, mercury is one of the least viscous liquids known.

P4. A fluid gives τ = 5 Pa at γ̇ = 10 s⁻¹ and τ = 9 Pa at γ̇ = 40 s⁻¹. Newtonian? Solution: apparent μ falls from 0.5 to 0.225 Pa·s — shear-thinning, not Newtonian; a power-law fit gives n ≈ 0.42.

8. Going deeper

In gases, kinetic theory derives μ ≈ ⅓ρc̄λ (mean speed c̄, mean free path λ) — and since λ ∝ 1/ρ, the density cancels: gas viscosity is independent of pressure over a wide range, a startling prediction Maxwell confirmed by experiment. In liquids no such simple theory exists; Eyring's rate theory pictures molecules hopping between cages, giving the exponential temperature law. The tensorial generalisation of Newton's relation, τ_ij = μ(∂u_i/∂x_j + ∂u_j/∂x_i), is the exact form that enters the Navier–Stokes equations — the 1-D law of these notes is its shadow in simple shear.

9. Historical context

The hypothesis is genuinely Newton's: Book II of the Principia (1687) proposes that the "resistance arising from the want of lubricity" between fluid layers is proportional to the velocity with which they separate — the linear law in seventeenth-century dress. It stayed a hypothesis for 150 years. Navier (1822) and Stokes (1845) built it into the full equations of motion; Poiseuille's meticulous capillary experiments (1840s), motivated by blood flow, gave the law its sharpest quantitative test — his measured fourth-power law is exactly what the linear stress relation predicts; and Maxwell's kinetic theory (1860s) derived μ for gases from first principles, including the shocking pressure-independence he then verified himself. The word "Newtonian fluid" thus honours a guess that took two centuries to become a theorem-for-gases and a superbly tested model for simple liquids.

10. Another way to see it: momentum diffusion

Rewrite the law per unit density: the flux of x-momentum in the y-direction is −ν ∂(ρu)/∂y. Compare with Fourier's heat conduction q = −k ∂T/∂y and Fick's diffusion j = −D ∂c/∂y: identical structure. Viscosity is diffusion of momentum, with ν the diffusivity in m²/s — the same units as thermal diffusivity α and mass diffusivity D. This single reframing pays constantly: diffusion spreads over distance √(νt), which instantly yields boundary-layer growth δ ~ √(νx/U) (replace t by the transit time x/U); the Prandtl number becomes a ratio of two diffusivities racing each other; and the "air is kinematically more viscous than water" surprise of Example 2 stops being a curiosity and becomes the statement that momentum diffuses faster through air. When an equation contains ν, think "spreading"; when it contains μ, think "force."

11. Frequently asked questions

Is glass a very viscous liquid — do cathedral windows flow? No. Glass below its transition temperature is an amorphous solid; measured relaxation times exceed the age of the universe. Thick-bottomed medieval panes reflect manufacturing, not millennia of creep. (Pitch, by contrast, genuinely is a ~10⁸ Pa·s liquid: the Queensland pitch-drop experiment has released nine drops since 1930.)

Why does the same oil feel thicker on a cold morning? The liquid Arrhenius law: cooling deepens the cage-hopping barrier exponentially. Multigrade oils (10W-40) blend polymers that uncoil when hot, deliberately flattening μ(T) so the engine sees adequate viscosity at both temperature extremes.

Can viscosity be zero? In superfluids, yes: helium-4 below 2.17 K flows through capillaries with strictly zero viscous resistance — a macroscopic quantum effect, and the exception that shows ordinary viscosity is molecular friction, absent when molecules condense into one quantum state.

What are the main non-Newtonian behaviours in one line each? Shear-thinning (paint, blood): μ falls with shear rate. Shear-thickening (cornstarch-water): μ rises — walk on it, don't stand. Bingham plastic (toothpaste, drilling mud): no flow until a yield stress. Viscoelastic (polymer melts): stress depends on deformation history, giving rod-climbing and die swell.

Which viscosity goes in the Reynolds number? Either, consistently: Re = ρvL/μ = vL/ν. Errors come from mixing — using ν where μ is expected mis-scales by a factor of ρ, three orders of magnitude for water.

12. Further practice

P5. Estimate how long momentum takes to diffuse across a 1 cm layer of (a) water, (b) air at rest. Solution: t ~ h²/ν: water 10⁻⁴/10⁻⁶ = 100 s; air 10⁻⁴/1.47×10⁻⁵ ≈ 7 s — the diffusion picture producing real numbers.

P6. A design mistakenly uses ν = 1.0×10⁻⁶ m²/s for air in Re = vL/ν with v = 10, L = 0.1. What Re does it report, and what is correct? Solution: reported 10⁶; correct 10/1.47×10⁻⁵×0.1 = 6.8×10⁴ — a fifteen-fold error, potentially flipping the predicted regime and every downstream correlation.

P7. From μ ≈ ⅓ρc̄λ and c̄ ∝ √T, λ ∝ T/p, show gas viscosity is pressure-independent and grows like √T. Solution: ρ ∝ p/T; product ρλ ∝ (p/T)(T/p) = const, leaving μ ∝ c̄ ∝ √T ∎ — Maxwell's two famous predictions in three lines.

13. Worked exam problem

Problem. A journal bearing: shaft radius 25 mm, length 50 mm, radial oil gap 50 μm, oil μ = 0.10 Pa·s, speed 3 000 rpm. Find (a) the shear stress in the film, (b) the friction force and torque, (c) the power dissipated as heat.

Solution. (a) Surface speed U = ωr = (3 000×2π/60)×0.025 = 7.854 m/s; treating the thin annulus as a plane film, τ = μU/h = 0.10×7.854/5×10⁻⁵ = 15.7 kPa. (b) Wetted area A = 2πrL = 2π×0.025×0.05 = 7.85×10⁻³ m²; F = τA = 123 N; torque T = Fr = 3.08 N·m. (c) P = Tω = 3.08×314.2 = 968 W — nearly a kilowatt of heat in a film thinner than a hair, which is why bearings have oil coolers. Every number came from τ = μ du/dy and geometry: the chapter's one law, earning its keep.

14. Key takeaways

Newton's law of viscosity is a constitutive model: shear stress proportional to velocity gradient, with μ the price of sliding layers past each other. Distinguish μ (force per gradient) from ν = μ/ρ (momentum diffusivity) — and remember air beats water kinematically. Liquids thin with heat, gases thicken; non-Newtonian fluids simply decline the model, and rheology begins where this chapter's straight line bends.

15. Where to go next

The tensor version of this law powers the Navier–Stokes equations; the ratio it forms with inertia becomes the Reynolds number; and its wall-hugging consequences fill the boundary-layer chapter. Compute stresses and diffusion times with the shear stress and kinematic viscosity calculators.