Navier-Stokes Equation

Equivalent forms

\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v}\cdot\nabla\mathbf{v} = -\frac{1}{\rho}\nabla p + \nu\nabla^2\mathbf{v} + \mathbf{g}

\rho\frac{D\mathbf{v}}{Dt} = -\nabla p + \mu\nabla^2\mathbf{v}

p_{char} = \rho v^2

The whole of fluid motion compressed into a single line — inertia, pressure, friction, gravity.

Symbol	Name	SI unit	Dimension	Range
$\rho$	Density Fluid mass density	kg/m³	M·L⁻³	1 – 2000
$v$	Characteristic Velocity Representative flow speed used to non-dimensionalize the momentum equation	m/s	L·T⁻¹	0 – 50
$\mu$	Dynamic Viscosity Fluid dynamic viscosity	Pa·s	M·L⁻¹·T⁻¹	0.0001 – 10
$p_{char}$	Characteristic Dynamic Pressureoutput Inertial pressure scale ρv² that the pressure-gradient term must balance	Pa	M·L⁻¹·T⁻²	0 – 100000

Unit systems

Symbol	SI	CGS	Imperial
$\rho$	kg/m³	g/cm³	slug/ft³
$v$	m/s	cm/s	ft/s
$\mu$	Pa·s	P	lb/(ft·s)
$p_{char}$	Pa	dyn/cm²	psi

Where it holds

The incompressible form holds for Newtonian fluids at Mach number below

\sim 0.3

with constant density and viscosity. Compressible, non-Newtonian, or rarefied (high Knudsen number) flows require modified forms. Analytic solutions exist only for special geometries; general 3D smoothness is an open mathematical question.

Dimensional analysis

$[\rho ][v^{2}] = (M\cdot L^{-3})(L^{2}\cdot T^{-2}) = M\cdot L^{-1}\cdot T^{-2} \checkmark$ (pressure), matching every term in the equation

Discovery

Claude-Louis Navier & George Gabriel Stokes · 1845

Navier wrote the viscous momentum equations in 1822 from a molecular-force argument; Stokes re-derived them rigorously from continuum stress in 1845, putting viscous fluid dynamics on a firm footing. Existence and smoothness of solutions in 3D remains a Clay Millennium Prize problem worth $1,000,000.

Try this

One equation governs every river, hurricane, and bloodstream — so why can't we solve it?

Water (density 1000 kg/m³, viscosity 0.001 Pa·s) flows at 2 m/s past a 0.1 m obstacle. Estimate the characteristic dynamic pressure ρv² that the momentum equation must balance.

Research status: stable

Real-world applications

Weather and climate models (atmospheric and ocean circulation)
Aircraft and automobile aerodynamic design via CFD
Blood-flow modeling in arteries and aneurysms
Pipeline, pump, and turbine engineering

Common misconceptions

The equation is not unsolvable — it is solved numerically every day; only a general 3D existence-and-smoothness proof is missing
$v\cdot \nabla v$ is a nonlinear convective acceleration, not a viscous term — it is the source of turbulence and chaos
Pressure here is a Lagrange multiplier enforcing incompressibility, not a thermodynamic state variable

Experimental verification

Direct numerical simulations reproduce laboratory turbulence statistics; Poiseuille and Couette exact solutions match measured velocity profiles to better than 1%. Particle-image velocimetry of boundary layers confirms the predicted viscous sublayer and wake structures.

Derivation

Apply Newton's second law to an infinitesimal fluid element.

The material acceleration

Dv/Dt = \partial v/\partial t + v\cdot \nabla v

captures both unsteadiness and convection.

Surface forces are the divergence of the Cauchy stress tensor

\sigma

; for a Newtonian fluid

\sigma = -pI + \mu (\nabla v + \nabla v

ᵀ

) + \lambda (\nabla \cdot v)I

.

Taking the divergence and using incompressibility

(\nabla \cdot v = 0)

collapses the viscous term

to \mu \nabla ^{2}v

, giving

\rho Dv/Dt = -\nabla p + \mu \nabla ^{2}v + f

.

Limiting cases

Re \to 0 (\mu \nabla ^{2}v

dominates)⟶ Stokes / creeping-flow equation:

\nabla p = \mu \nabla ^{2}v

Inertial terms vanish; flow is reversible and linear, as around microorganisms and settling dust.

Re \to \infty (\mu \to 0)

⟶ Euler equation:

\rho (\partial v/\partial t + v\cdot \nabla v) = -\nabla p + f

Viscosity is negligible in the bulk; ideal inviscid flow, valid outside thin boundary layers.

\partial v/\partial t = 0

,

v\cdot \nabla v = 0

⟶ Stokes equation / hydrostaticsWith no acceleration the balance reduces to pressure versus viscous and body forces.

What if…

What if viscosity were exactly zero?

You recover Euler's equation. The no-slip boundary condition can no longer be satisfied, producing d'Alembert's paradox: zero drag on a body — which is why a thin viscous boundary layer always matters.

What if the flow is very slow and tiny (a swimming bacterium)?

Re ≪ 1 gives the linear Stokes equation: motion is reversible, so a reciprocal stroke produces no net displacement (the 'scallop theorem').

What if density varies with pressure?

You must use the compressible Navier-Stokes equations coupled to an energy equation and equation of state, essential above Mach 0.3.

1

Dynamic pressure scale past an obstacle

Given ·

\rho:: 1000
v:: 2

Find · p_{char}

Steps

Non-dimensionalize pressure in the momentum equation $by \rho v^{2}$
p_char $= 1000 \times (2)^{2}$
p_char $= 4000\,\mathrm{Pa}$ — the pressure swing the flow can produce

Result · p_char

= \rho v^{2} = 1000 \times 2^{2} = 4000\,\mathrm{Pa}

2

Estimating the viscous term magnitude

Given ·

\mu:: 0.001
v:: 2

Find · viscous stress scale

\mu v/L

for

L = 0.1\,\mathrm{m}

Steps

Compare inertial scale $\rho v^{2}$ with viscous scale $\mu v/L$
Ratio $= \rho vL/\mu$ = Reynolds number $= 1000\times 2\times 0.1/0.001 = 2\times 10^{5}$
High Re means the Euler limit governs the bulk flow

Result ·

\mu v/L^{2} \times L = \mu v/L = 0.001 \times 2 / 0.1 = 0.02\,\mathrm{Pa}

— far below the 4000 Pa inertial scale, so

Re \approx 2\times 10^{5} and

inertia dominates.

Reynolds Number→Bernoulli's Equation→Stokes' Law→Euler Equation (Inviscid Flow)→