Navier–Stokes stress and viscosity
Build the Newtonian constitutive law, keep shear and bulk viscosity distinct, and connect stress gradients with momentum transport.
On this page
The question for today
What you’ll understand
- Distinguish rigid rotation from deformation when constructing viscous stress.
- Derive the constant-coefficient momentum equation and identify its incompressible limit.
- Calculate local shear stress and separate it from the net viscous force density.
A fluid can have a large velocity without experiencing viscous resistance: moving the entire laboratory at one constant speed cannot change its internal physics. The useful question is how neighboring fluid elements move relative to one another, and whether that relative motion changes their shape. We will add a constitutive law to momentum conservation, rather than introduce viscosity as a universal force opposite velocity. This distinction explains both why rigid rotation need not generate Newtonian shear stress and why a nonzero shear stress can coexist with zero net viscous force in a region.
Construct a frame-independent stress law
Write the Cauchy stress as pressure plus a viscous contribution. The traction on a surface with outward unit normal n is the stress tensor multiplied by n; pressure supplies a compressive normal traction, whereas viscosity can supply tangential traction. For an ordinary isotropic Newtonian fluid without couple stresses, the viscous contribution is linear in the symmetric velocity gradient. The antisymmetric part describes local rigid rotation, which changes orientation without changing the shape of an infinitesimal material element.
The displayed form uses dynamic shear viscosity mu and a second viscosity coefficient lambda. The bulk viscosity is zeta = lambda + 2 mu/3, so lambda itself need not be positive. Nonnegative shear and bulk viscosities give nonnegative viscous dissipation in this model. Neither incompressibility nor isotropy proves the extra assumption often called the Stokes hypothesis, zeta = 0. A material whose stress depends nonlinearly on deformation rate or on its deformation history requires a different constitutive description.
Take the divergence before simplifying
Apply momentum balance to a small material volume: inertia equals the divergence of stress plus the body force per unit volume. For spatially constant viscosity coefficients, differentiating the constitutive law gives the momentum equation below. The acceleration is the material derivative, including advection. Here b denotes force per unit mass, such as gravitational acceleration. If viscosity changes substantially with temperature or position, keep the divergence of the full viscous stress; pulling coefficients through derivatives silently removes additional terms.
For an incompressible flow, theta vanishes and the constant-mu viscous contribution reduces to mu times the vector Laplacian of velocity. Dividing by density introduces the kinematic viscosity nu = mu/rho, with units of square meters per second. It controls the diffusion time across a length scale. Pressure remains essential even when density is fixed: it adjusts to enforce the velocity constraint and boundary conditions. The continuity equation still accompanies momentum conservation; one vector equation cannot replace the complete fluid model.
Distinguish transport, dissipation, and boundary data
A shear stress transports momentum across an internal surface. Only its spatial variation supplies a net force to a small volume, so uniform shear stress does not imply local acceleration. With constant mu, a linear shear profile has nonzero stress but zero velocity Laplacian. Mechanical work can still enter through the moving boundaries and become internal energy. The dissipation density for an incompressible Newtonian fluid is twice mu times the sum of squared deformation-tensor components; a complete thermal calculation accounts for that heating.
At an ordinary impermeable no-slip wall, the fluid velocity equals the wall velocity. These conditions are modeling assumptions supported in many continuum-scale experiments, with exceptions such as rarefied gases or measurable interfacial slip. Initial velocity and suitable pressure or boundary data complete the problem. Before solving, check the continuum scale, constitutive range, and thermal assumptions. A compact equation is not permission to omit the mechanisms that determine its coefficients or boundary behavior.
Work it through
A curved channel profile and its local forces
Consider steady incompressible flow between stationary plates at y = 0 and y = 0.02 m, with velocity u = 4y − 200y² in meters per second and dynamic viscosity 0.1 Pa s. There are no streamwise body forces. Find the shear stress at y = 0.005 m and the pressure gradient needed for this profile.
Differentiate the prescribed velocity: du/dy = 4 − 400y. At y = 0.005 m the gradient is 2 inverse seconds. Multiplying by 0.1 Pa s gives the stress component tau_xy = 0.2 Pa.
Differentiate again to obtain d²u/dy² = −400 per meter per second. The viscous force density is therefore −40 N per cubic meter. This is a derivative of stress, rather than the local value 0.2 Pa.
The fully developed profile has zero material acceleration. Its streamwise balance is 0 = −dp/dx − 40 N per cubic meter, so dp/dx = −40 Pa per meter. Pressure falls downstream and supplies the positive force that balances viscosity.
The local shear stress is 0.2 Pa, while the pressure gradient is −40 Pa/m. Both walls satisfy zero velocity, and the stress changes sign across the channel center. Nonzero transport through an internal surface and net force on a volume are different quantities.
The living notebook
Explore the model
Interactive lab · fields
Rotation does not automatically imply viscous strain
Set Source strength q to zero and vary Rotation Ω. Predict that arrows get longer as the magnitude of Ω increases, while their circular arrangement persists. The displayed field is then u = Ω(−y,x), a rigid rotation whose symmetric velocity gradient is zero. The lab draws this prescribed field; it does not solve Navier–Stokes or calculate pressure, stress, heating, or wall interactions.
- Fx at (1, 0)
- 0.833
- Fy at (1, 0)
- 0.5
- Field magnitude there
- 0.972
The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forVisible rotation alone does not prove Newtonian viscous deformation. For this q = 0 field, the antisymmetric velocity gradient changes orientation while the symmetric part vanishes; stress requires the constitutive calculation, not an impression of swirling motion.
Pause & reason
Check your understanding
Your turn
Try it yourself
A Newtonian incompressible fluid rotates as u_x = −3y and u_y = 3x, with coordinates in meters and velocities in meters per second. Calculate D_xx, D_yy, and D_xy, and decide whether this ideal rigid rotation has a viscous shear stress.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Compute both cross derivatives before symmetrizing the gradient. A nonzero curl is not the same object as a nonzero symmetric rate of deformation.
Compare with the solution
The diagonal derivatives are both zero. The cross derivatives are du_x/dy = −3 and du_y/dx = 3 per second, so D_xy = (−3 + 3)/2 = 0. All components of D vanish, as does the divergence. Consequently the Newtonian viscous stress vanishes, although the vorticity is nonzero. A pressure gradient is still needed to provide centripetal acceleration if this field is a physical steady flow.
Keep the idea
Take it with you
- Newtonian viscous stress depends on deformation rate, not on absolute velocity or rotation alone.
- For constant coefficients, compressible momentum balance contains a gradient-of-divergence term as well as a velocity Laplacian.
- Local stress, net force density, and conversion of mechanical energy into heat answer different questions.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.