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Derivation

Navier-Stokes from the Newtonian Constitutive Law

D-318 Home PU-307 Threads force · matter · fields Depends on Cauchy's Momentum Equation, The Continuity Equation
Statement

Starting from Cauchy's momentum equation \( \rho\,\frac{D\vec{v}}{Dt} = \nabla\cdot\boldsymbol{\sigma} + \rho\vec{f} \), we posit the linear, isotropic Newtonian constitutive law \( \boldsymbol{\sigma} = -p\,\mathbf{I} + 2\mu\,\mathbf{S} + \lambda\,(\nabla\cdot\vec{v})\,\mathbf{I} \), where \( \mathbf{S} = \tfrac{1}{2}\left(\nabla\vec{v} + (\nabla\vec{v})^{\mathsf{T}}\right) \) is the strain-rate tensor. Taking the divergence of this stress and imposing incompressibility \( \nabla\cdot\vec{v}=0 \) with constant \( \mu \) closes the momentum balance, yielding the incompressible Navier-Stokes equations \( \frac{D\vec{v}}{Dt} = -\frac{1}{\rho}\nabla p + \nu\,\nabla^2\vec{v} + \vec{f} \), where \( \nu = \mu/\rho \).

Why it matters

Cauchy's equation is exact but not closed: it contains six unknown independent stress components with no rule linking them to the motion. The Newtonian constitutive law supplies exactly that missing rule for the vast class of fluids whose stress responds linearly and isotropically to the instantaneous rate of strain — air, water, and most simple liquids. Everything downstream in fluid dynamics, from boundary layers and turbulence to lift and drag, rests on the equation this closure produces.

The derivation is also the cleanest example of how a continuum theory is completed: conservation laws fix the structure, and a material-specific constitutive relation supplies the physics. Understanding which assumptions enter here is what tells you precisely when Navier-Stokes may be trusted and when it must be replaced.

Assumptions
The fluid is a continuum.If dropped, the Knudsen number \( \mathrm{Kn} = \ell/L \) is not small, individual molecular collisions matter, and stress and velocity are not smooth fields — one must use kinetic theory (the Boltzmann equation) instead.
Stress depends linearly on the rate of strain, not on strain itself.If dropped, the fluid stores elastic energy (a viscoelastic or non-Newtonian response) and stress depends on deformation history; the constitutive law becomes a differential or integral relation in time.
The response is isotropic.If dropped, the viscosity is a fourth-rank tensor with more than two independent coefficients, and the neat \( 2\mu\mathbf{S} + \lambda(\nabla\cdot\vec{v})\mathbf{I} \) form is lost — relevant for liquid crystals and structured fluids.
The stress tensor is symmetric.If dropped (internal angular momentum / body couples, as in polar fluids), conservation of angular momentum no longer forces \( \sigma_{ij}=\sigma_{ji} \), and an antisymmetric part survives, requiring the Cauchy-Poisson micropolar extension.
Incompressibility \( \nabla\cdot\vec{v}=0 \) and constant \( \mu \).If dropped, the bulk-viscosity term \( (\mu+\lambda)\nabla(\nabla\cdot\vec{v}) \) survives and spatial gradients of \( \mu \) add \( (\nabla\mu)\cdot(\nabla\vec{v}+(\nabla\vec{v})^{\mathsf T}) \); one then keeps the full compressible, variable-property form.
Derivation
1
\[ \rho\,\frac{D\vec{v}}{Dt} = \nabla\cdot\boldsymbol{\sigma} + \rho\vec{f} \]
Cauchy's momentum equation, taken as a prior result: it is Newton's second law per unit volume for a continuum and holds for any material. A
2
\[ \boldsymbol{\sigma} = -p\,\mathbf{I} + \boldsymbol{\tau} \]
Split the stress into an isotropic part (mechanical pressure \( p \), acting equally in all directions) and the remaining deviatoric viscous stress \( \boldsymbol{\tau} \). This is a definition, always legal; it isolates the part that must vanish at rest. A
3
\[ \tau_{ij} = C_{ijkl}\,\frac{\partial v_k}{\partial x_l} \]
Posit that the viscous stress is a linear function of the velocity gradient (Newtonian assumption). Linearity is the defining hypothesis; \( C_{ijkl} \) is a fourth-rank tensor of material coefficients. B
4
\[ \tau_{ij} = C_{ijkl}\,S_{kl}, \qquad S_{kl} = \tfrac{1}{2}\!\left(\frac{\partial v_k}{\partial x_l} + \frac{\partial v_l}{\partial x_k}\right) \]
Only the symmetric part \( \mathbf{S} \) of \( \nabla\vec{v} \) can produce stress: the antisymmetric part is a rigid-body rotation, which strains nothing and (by frame indifference) cannot generate stress. B
5
\[ C_{ijkl} = \lambda\,\delta_{ij}\delta_{kl} + \mu\,(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) \]
Isotropy forces the most general isotropic fourth-rank tensor to this two-parameter form. Symmetry of \( \tau_{ij} \) in \( (i,j) \) and contraction with the symmetric \( S_{kl} \) leave exactly two independent scalars, \( \mu \) and \( \lambda \). C
6
\[ \boldsymbol{\tau} = 2\mu\,\mathbf{S} + \lambda\,(\nabla\cdot\vec{v})\,\mathbf{I} \]
Contract \( C_{ijkl}S_{kl} \): the \( \mu \)-terms give \( 2\mu S_{ij} \), the \( \lambda \)-term gives \( \lambda\, S_{kk}\delta_{ij} = \lambda(\nabla\cdot\vec{v})\delta_{ij} \). This is the Cauchy-Poisson (Newtonian) constitutive law. B
7
\[ \boldsymbol{\sigma} = -p\,\mathbf{I} + 2\mu\,\mathbf{S} + \lambda\,(\nabla\cdot\vec{v})\,\mathbf{I} \]
Substitute the constitutive law of step 6 into the split of step 2. The stress is now expressed entirely in terms of the velocity field and pressure — the closure is achieved. A
8
\[ (\nabla\cdot\boldsymbol{\sigma})_i = -\frac{\partial p}{\partial x_i} + 2\mu\,\frac{\partial S_{ij}}{\partial x_j} + \lambda\,\frac{\partial}{\partial x_i}(\nabla\cdot\vec{v}) \]
Take the divergence componentwise, treating \( \mu \) and \( \lambda \) as constant so they pass outside the derivative. This is the term Cauchy's equation needs. B
9
\[ 2\mu\,\frac{\partial S_{ij}}{\partial x_j} = \mu\,\frac{\partial}{\partial x_j}\!\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right) = \mu\,\nabla^2 v_i + \mu\,\frac{\partial}{\partial x_i}(\nabla\cdot\vec{v}) \]
Expand \( S_{ij} \) and swap the order of partial derivatives (velocity is \( C^2 \)). The first term is the vector Laplacian; the second recognises \( \partial_j v_j = \nabla\cdot\vec{v} \). C
10
\[ \nabla\cdot\boldsymbol{\sigma} = -\nabla p + \mu\,\nabla^2\vec{v} + (\mu+\lambda)\,\nabla(\nabla\cdot\vec{v}) \]
Collect steps 8-9. This is the general (compressible, constant-property) divergence of the Newtonian stress; \( \mu+\lambda \) is often written \( \zeta - \tfrac{2}{3}\mu \) with \( \zeta \) the bulk viscosity. B
11
\[ \nabla\cdot\vec{v} = 0 \;\;\Longrightarrow\;\; \nabla\cdot\boldsymbol{\sigma} = -\nabla p + \mu\,\nabla^2\vec{v} \]
Impose incompressibility (prior result: continuity for constant \( \rho \) reduces to a divergence-free velocity). The entire bulk term \( (\mu+\lambda)\nabla(\nabla\cdot\vec{v}) \) vanishes identically. A
12
\[ \rho\,\frac{D\vec{v}}{Dt} = -\nabla p + \mu\,\nabla^2\vec{v} + \rho\vec{f} \]
Substitute step 11 into Cauchy's equation (step 1). Every term is now expressed through \( \vec{v} \), \( p \), and the known body force. A
13
\[ \frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot\nabla)\vec{v} = -\frac{1}{\rho}\nabla p + \nu\,\nabla^2\vec{v} + \vec{f} \]
Expand the material derivative \( \tfrac{D}{Dt} = \partial_t + \vec{v}\cdot\nabla \) and divide by the constant \( \rho \), defining the kinematic viscosity \( \nu = \mu/\rho \). Dividing is legal since \( \rho>0 \). A
Result
\[ \frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot\nabla)\vec{v} = -\frac{1}{\rho}\nabla p + \nu\,\nabla^2\vec{v} + \vec{f}, \qquad \nabla\cdot\vec{v}=0 \]

Reading. The acceleration of a fluid parcel (local plus advective) is driven by three forces per unit mass: the pressure gradient pushes fluid from high to low pressure; the viscous term \( \nu\nabla^2\vec{v} \) diffuses momentum, smoothing velocity differences exactly as heat conduction smooths temperature; and \( \vec{f} \) is any body force such as gravity. The incompressibility constraint acts as the equation that determines the pressure field, so that the velocity stays divergence-free at every instant.

Units check. Each term has units of acceleration, \( \mathrm{m\,s^{-2}} \). Advection: \( [v][\partial v/\partial x] = (\mathrm{m\,s^{-1}})(\mathrm{s^{-1}}) = \mathrm{m\,s^{-2}} \). Pressure: \( \tfrac{1}{\rho}\nabla p = \dfrac{\mathrm{Pa\,m^{-1}}}{\mathrm{kg\,m^{-3}}} = \dfrac{\mathrm{kg\,m^{-1}s^{-2}\,m^{-1}}}{\mathrm{kg\,m^{-3}}} = \mathrm{m\,s^{-2}} \). Viscous: \( \nu\nabla^2 v = (\mathrm{m^2\,s^{-1}})(\mathrm{m\,s^{-1}\,m^{-2}}) = \mathrm{m\,s^{-2}} \). All consistent.

Limiting cases
  • Inviscid limit \( \nu\to 0 \) (high Reynolds number): the viscous term drops and one recovers the Euler equations \( \tfrac{D\vec{v}}{Dt} = -\tfrac{1}{\rho}\nabla p + \vec{f} \) — valid away from walls.
  • Creeping flow \( \mathrm{Re}\ll 1 \): the advective term \( (\vec{v}\cdot\nabla)\vec{v} \) is negligible, leaving the linear Stokes equations \( \nabla p = \mu\nabla^2\vec{v} + \rho\vec{f} \).
  • Steady, fully developed, unidirectional flow: \( \partial_t\vec{v}=0 \) and \( (\vec{v}\cdot\nabla)\vec{v}=0 \), so pressure gradient balances viscous diffusion alone — Poiseuille and Couette flows.
  • Hydrostatics \( \vec{v}=0 \): everything but \( 0 = -\nabla p + \rho\vec{f} \) vanishes, recovering \( \nabla p = \rho\vec{g} \).
Breaks when
  • Rarefied gas (large Knudsen number). When the mean free path is comparable to the flow scale — upper atmosphere, microfluidic vacuum, shock interiors — the continuum stress law fails and the no-slip idea breaks; kinetic theory or slip boundary conditions are required.
  • Non-Newtonian fluids. Polymer solutions, blood, molten plastics, and slurries have stress that depends nonlinearly on strain rate or on deformation history; the linear \( \boldsymbol{\tau}=2\mu\mathbf{S} \) closure is simply wrong and viscosity is not a constant.
  • Strong compressibility / high Mach number. Near or above the speed of sound, \( \nabla\cdot\vec{v}\neq 0 \), density varies, and the discarded bulk-viscosity and dilatation terms return; the incompressible form no longer applies.
  • Extreme gradients (shock structure). Across a strong shock the velocity changes over a few mean free paths, so the linear gradient expansion truncated at first order is inadequate.
Failure modes
  • Confusing \( \mu \) and \( \nu \). Dynamic viscosity \( \mu \) (\( \mathrm{Pa\,s} \)) multiplies stress; kinematic viscosity \( \nu=\mu/\rho \) (\( \mathrm{m^2\,s^{-1}} \)) multiplies the Laplacian in the divided form. Mixing them corrupts the units by a factor of \( \rho \).
  • Dropping the advective term because "it's steady". Steady means \( \partial_t\vec{v}=0 \); the nonlinear \( (\vec{v}\cdot\nabla)\vec{v} \) generally survives and is the source of the equation's difficulty.
  • Applying incompressibility before checking it. Discarding \( (\mu+\lambda)\nabla(\nabla\cdot\vec{v}) \) is only legal once \( \nabla\cdot\vec{v}=0 \) is established from continuity for constant \( \rho \); for gases at speed this step is invalid.
  • Treating \( \nabla^2\vec{v} \) as \( \nabla^2 \) of each Cartesian component in curvilinear coordinates. In cylindrical or spherical coordinates the vector Laplacian has extra curvature terms; using the scalar Laplacian componentwise gives wrong profiles.
  • Forgetting pressure is not a state variable here. In incompressible flow \( p \) is a Lagrange multiplier enforcing \( \nabla\cdot\vec{v}=0 \), determined by a Poisson equation, not by an equation of state.
Discussion

The deep point of this derivation is the division of labour between conservation and constitution. Cauchy's equation is a bookkeeping identity — it says momentum is conserved and knows nothing about what the fluid is made of. The Newtonian law is where the material speaks: two numbers, \( \mu \) and \( \lambda \), encode the entire linear-isotropic response of a specific fluid. That only two scalars are needed is a consequence of isotropy collapsing the general \( 3^4=81 \)-component tensor \( C_{ijkl} \) down to two independent invariants.

The viscous term \( \nu\nabla^2\vec{v} \) is a diffusion operator, and this is not a coincidence: momentum, like heat and dye, spreads down its gradient. The kinematic viscosity \( \nu \) plays the role of a momentum diffusivity, and the ratio of momentum diffusivity to thermal diffusivity is the Prandtl number. This diffusive character is what gives the equation its parabolic-in-time structure and what makes viscous boundary layers grow like \( \sqrt{\nu t} \).

The single nonlinear term \( (\vec{v}\cdot\nabla)\vec{v} \) carries almost all of the equation's mathematical depth. It is responsible for turbulence, for the energy cascade, and for the fact that existence and smoothness of solutions in three dimensions remains an open Millennium Prize problem. Its ratio to the viscous term defines the Reynolds number \( \mathrm{Re} = UL/\nu \), the single most important dimensionless group in fluid mechanics.

A subtle structural feature: for incompressible flow the pressure has lost its thermodynamic meaning entirely. Taking the divergence of the momentum equation and using \( \nabla\cdot\vec{v}=0 \) yields a Poisson equation \( \nabla^2 p = -\rho\,\nabla\cdot\left[(\vec{v}\cdot\nabla)\vec{v}\right] \), showing that \( p \) is instantaneously and non-locally slaved to the velocity field to enforce the constraint. This is why incompressible flow supports no acoustic waves — the elimination of \( \nabla\cdot\vec{v} \) has formally sent the sound speed to infinity.

Common misconceptions. The Navier-Stokes equations are not more fundamental than Cauchy's equation — they are less general, obtained by adding a material assumption. And "incompressible" does not mean "constant density fluid"; it means the velocity field is divergence-free, a statement about the flow, which for a fluid of uniform density is equivalent to \( D\rho/Dt=0 \).

Worked examples

Example 1 — Steady laminar flow in a circular pipe (Hagen-Poiseuille).

1
\[ 0 = -\frac{\partial p}{\partial z} + \mu\left(\frac{1}{r}\frac{d}{dr}\!\left(r\frac{dv_z}{dr}\right)\right) \]
Reduce Navier-Stokes for steady, axisymmetric, fully-developed flow \( \vec{v}=v_z(r)\hat{z} \): time, advective, and radial terms vanish; the pressure gradient \( G=-\partial p/\partial z \) is constant. A
2
\[ v_z(r) = \frac{G}{4\mu}\left(R^2 - r^2\right) \]
Integrate twice, using boundedness at \( r=0 \) and no-slip \( v_z(R)=0 \). The profile is a paraboloid — symbols only, no numbers yet. B
3
\[ Q = \int_0^R v_z(r)\,2\pi r\,dr = \frac{\pi G R^4}{8\mu} \]
Integrate the profile over the cross-section to get the volumetric flow rate — the celebrated \( R^4 \) dependence. B
4
\[ G = 100\ \mathrm{Pa\,m^{-1}},\quad R = 0.010\ \mathrm{m},\quad \mu = 1.0\times10^{-3}\ \mathrm{Pa\,s} \]
Insert water at 20 °C in a 1 cm-radius pipe. Now substitute numbers. A
5
\[ Q = \frac{\pi (100)(0.010)^4}{8(1.0\times10^{-3})} = \frac{\pi\times100\times10^{-8}}{8\times10^{-3}} \approx 3.9\times10^{-4}\ \mathrm{m^3\,s^{-1}} \]
Evaluate: \( 100\times(0.01)^4 = 10^{-6} \), divided by \( 8\times10^{-3} \) gives \( 1.25\times10^{-4} \), times \( \pi \). A
\[ Q \approx 3.9\times10^{-4}\ \mathrm{m^3\,s^{-1}} \approx 0.39\ \mathrm{L\,s^{-1}} \]

Reading. A modest gradient of 100 Pa per metre drives roughly 0.39 litres per second through a 2 cm-diameter pipe. Mean velocity \( \bar{v}=Q/(\pi R^2)\approx 1.24\ \mathrm{m\,s^{-1}} \); Reynolds number \( \mathrm{Re}=\rho\bar{v}(2R)/\mu \approx (1000)(1.24)(0.02)/10^{-3}\approx 2.5\times10^{4} \) — actually turbulent, so the laminar answer is a formal illustration of the reduced equation, not a physical prediction here.

Units check. \( \dfrac{\mathrm{Pa\,m^{-1}}\cdot\mathrm{m^4}}{\mathrm{Pa\,s}} = \dfrac{\mathrm{m^3}}{\mathrm{s}} \). Correct.

Example 2 — Wall shear stress in plane Couette flow.

1
\[ 0 = \mu\,\frac{d^2 u}{dy^2} \]
Navier-Stokes for steady flow between two plates, top plate moving at speed \( U \), no pressure gradient and no body force in \( x \). Only viscous diffusion survives. A
2
\[ u(y) = U\,\frac{y}{h} \]
Integrate twice with no-slip \( u(0)=0 \), \( u(h)=U \). The profile is linear — pure shear. B
3
\[ \tau_w = \mu\,\frac{du}{dy} = \frac{\mu U}{h} \]
The wall shear stress is the viscous stress component \( \tau_{xy}=\mu\,du/dy \), constant across the gap. Symbols first. B
4
\[ \mu = 1.0\times10^{-3}\ \mathrm{Pa\,s},\quad U = 0.50\ \mathrm{m\,s^{-1}},\quad h = 2.0\times10^{-3}\ \mathrm{m} \]
Water sheared in a 2 mm gap with the upper plate at 0.5 m/s. Substitute numbers. A
5
\[ \tau_w = \frac{(1.0\times10^{-3})(0.50)}{2.0\times10^{-3}} = 0.25\ \mathrm{Pa} \]
Evaluate the ratio directly. A
\[ \tau_w = 0.25\ \mathrm{Pa} \]

Reading. Maintaining the shear requires 0.25 Pa on the plate; the force on a plate of area \( 0.1\ \mathrm{m^2} \) is \( 0.025\ \mathrm{N} \). Because the profile is linear, this stress is the same at both walls and everywhere between them.

Units check. \( \dfrac{\mathrm{Pa\,s}\cdot\mathrm{m\,s^{-1}}}{\mathrm{m}} = \mathrm{Pa} \). Correct.

Problems
  1. Stokes limit. By comparing the advective and viscous terms, state the condition under which \( (\vec{v}\cdot\nabla)\vec{v} \) may be dropped, and write the resulting linear equation.
    Solution The ratio of advective to viscous magnitudes is \( \dfrac{|(\vec{v}\cdot\nabla)\vec{v}|}{|\nu\nabla^2\vec{v}|} \sim \dfrac{U^2/L}{\nu U/L^2} = \dfrac{UL}{\nu} = \mathrm{Re} \). When \( \mathrm{Re}\ll 1 \) the advective term is negligible, giving the steady Stokes equations \( 0 = -\nabla p + \mu\nabla^2\vec{v} + \rho\vec{f} \) together with \( \nabla\cdot\vec{v}=0 \). This regime describes microorganisms, sedimenting particles, and lubrication films.
  2. Nondimensionalization. Using scales \( U \) for velocity, \( L \) for length, \( L/U \) for time, and \( \rho U^2 \) for pressure, nondimensionalize the incompressible Navier-Stokes equation and identify the Reynolds number.
    Solution Let \( \vec{v}=U\vec{v}^{*} \), \( \vec{x}=L\vec{x}^{*} \), \( t=(L/U)t^{*} \), \( p=\rho U^2 p^{*} \). Substituting, every term carries a factor \( U^2/L \) except the viscous term which carries \( \nu U/L^2 \). Dividing through by \( U^2/L \): \[ \frac{\partial\vec{v}^{*}}{\partial t^{*}} + (\vec{v}^{*}\cdot\nabla^{*})\vec{v}^{*} = -\nabla^{*}p^{*} + \frac{1}{\mathrm{Re}}\nabla^{*2}\vec{v}^{*}, \qquad \mathrm{Re}=\frac{UL}{\nu}. \] The single parameter \( \mathrm{Re} \) governs the flow: two geometrically similar flows with equal \( \mathrm{Re} \) are dynamically identical.
  3. Couette profile with numbers. Oil with \( \mu = 0.10\ \mathrm{Pa\,s} \) fills a \( h=1.0\ \mathrm{mm} \) gap; the top plate moves at \( U=0.20\ \mathrm{m\,s^{-1}} \). Find the shear stress and the power dissipated per unit plate area.
    Solution Profile \( u=Uy/h \), so \( \tau_w=\mu U/h = (0.10)(0.20)/(1.0\times10^{-3}) = 20\ \mathrm{Pa} \). Power per unit area = force per unit area times plate speed \( = \tau_w U = (20)(0.20) = 4.0\ \mathrm{W\,m^{-2}} \). This equals the volume-integrated viscous dissipation \( \int_0^h \mu(du/dy)^2\,dy = \mu(U/h)^2 h = \mu U^2/h = (0.10)(0.04)/10^{-3}=4.0\ \mathrm{W\,m^{-2}} \), confirming energy balance.
  4. Momentum diffusion depth. A large plate beside still water \( (\nu = 1.0\times10^{-6}\ \mathrm{m^2\,s^{-1}}) \) is suddenly set moving (Stokes' first problem). Estimate the depth to which the motion has diffused after \( t=100\ \mathrm{s} \).
    Solution The viscous term is a diffusion operator with diffusivity \( \nu \); the momentum penetration depth grows as \( \delta \sim \sqrt{\nu t} \). Numerically \( \delta \sim \sqrt{(1.0\times10^{-6})(100)} = \sqrt{10^{-4}} = 1.0\times10^{-2}\ \mathrm{m} = 1\ \mathrm{cm} \). The exact self-similar solution gives \( u=U\,\mathrm{erfc}\!\left(y/2\sqrt{\nu t}\right) \), whose characteristic scale is \( 2\sqrt{\nu t}\approx 2\ \mathrm{cm} \), of the same order.
  5. Vorticity equation. Take the curl of the incompressible Navier-Stokes equation with \( \vec{f}=-\nabla\Phi \) conservative, and show the pressure and body force drop out. Write the result for \( \vec{\omega}=\nabla\times\vec{v} \).
    Solution Curl of a gradient is zero, so \( \nabla\times\left(-\tfrac{1}{\rho}\nabla p - \nabla\Phi\right)=0 \): pressure and conservative body force vanish. Using the identity \( (\vec{v}\cdot\nabla)\vec{v} = \nabla(\tfrac{1}{2}v^2) - \vec{v}\times\vec{\omega} \) and taking the curl, with \( \nabla\cdot\vec{v}=0 \), gives \[ \frac{\partial\vec{\omega}}{\partial t} + (\vec{v}\cdot\nabla)\vec{\omega} = (\vec{\omega}\cdot\nabla)\vec{v} + \nu\nabla^2\vec{\omega}. \] The term \( (\vec{\omega}\cdot\nabla)\vec{v} \) is vortex stretching (absent in 2D), and \( \nu\nabla^2\vec{\omega} \) diffuses vorticity. Eliminating pressure is the chief advantage of the vorticity formulation.