Written and reviewed September 16, 2026 · An introductory explanation with original worked calculations
A complicated equation with a familiar job
The Navier–Stokes momentum equation applies Newton’s second law to a moving fluid. Its acceleration terms describe how a fluid parcel changes velocity; its pressure, viscosity, and body-force terms describe what causes that change. A separate mass-conservation equation completes the incompressible model.
Imagine following a tiny marker in water. It can accelerate because the current changes around it, or because it moves into a region with a different current. That distinction is the key to reading the left side of the equation.
1. Choose the model before reading the symbols
We use a Newtonian, incompressible fluid with constant density and constant viscosity, in an inertial reference frame. Under these assumptions:
\[ \frac{\partial\mathbf u}{\partial t} +(\mathbf u\cdot\nabla)\mathbf u =-\frac{1}{\rho}\nabla p +\nu\nabla^2\mathbf u+\mathbf b, \qquad \nabla\cdot\mathbf u=0. \]
Here \(\mathbf u=(u_x,u_y,u_z)\) is the velocity field in metres per second, \(t\) is time, \(p\) is pressure in pascals, \(\rho\) is density in kilograms per cubic metre, and \(\mathbf b\) is body force per unit mass in metres per second squared. Kinematic viscosity is \(\nu=\mu/\rho\), where \(\mu\) is dynamic viscosity in pascal-seconds; \(\nu\) has units of square metres per second.
The vector equation represents three momentum equations, one for each spatial direction. The Clay Mathematics Institute’s mathematical formulation uses pressure scaled by density, so its pressure term looks slightly different. Always check a source’s definitions before comparing notation.
What do the triangle-shaped symbols do?
\(\nabla\), pronounced “del,” collects spatial derivatives. The pressure gradient \(\nabla p\) points toward increasing pressure. Divergence \(\nabla\cdot\mathbf u\) measures local net outflow per unit volume. The Laplacian \(\nabla^2\mathbf u\) sums second spatial derivatives of each velocity component. These operators have different jobs despite sharing the same symbol.
2. Local acceleration: the current changes at a fixed point
\[ \text{Local acceleration}=\frac{\partial\mathbf u}{\partial t}. \]
Hold a velocity sensor at one position. If its measured velocity changes with time, this term is nonzero. For example, a spatially uniform field \(\mathbf u=(kt,0,0)\), with constant \(k\), has local acceleration \((k,0,0)\). Because this field has no spatial variation, its convective acceleration is zero. A force balance must still supply the acceleration.
3. Convective acceleration: the parcel moves through a changing field
\[ (\mathbf u\cdot\nabla)\mathbf u =u_x\frac{\partial\mathbf u}{\partial x} +u_y\frac{\partial\mathbf u}{\partial y} +u_z\frac{\partial\mathbf u}{\partial z}. \]
A flow can be steady at every fixed position while individual parcels accelerate. As an original local example, take \(\mathbf u=(ax,-ay,0)\), where \(a\) is constant. Its local acceleration is zero, but its convective acceleration is \((a^2x,a^2y,0)\). At \(a=2\,\mathrm{s^{-1}}\) and \((x,y)=(0.10,0)\,\mathrm m\), the acceleration is \((0.40,0,0)\,\mathrm{m/s^2}\).
This example is incompressible because \(a-a=0\). It can satisfy the momentum equation locally with zero body force and \(p=p_0-\rho a^2(x^2+y^2)/2\). It is not a finite-energy flow over all of infinite space: the velocity grows with distance. That distinction keeps a useful classroom example separate from a global existence problem.
Put the two acceleration terms together
If a parcel follows \(\mathbf X(t)\), then \(d\mathbf X/dt=\mathbf u(\mathbf X(t),t)\). Applying the chain rule to its velocity gives:
\[ \frac{d}{dt}\mathbf u(\mathbf X(t),t) =\frac{\partial\mathbf u}{\partial t} +(\mathbf u\cdot\nabla)\mathbf u =\frac{D\mathbf u}{Dt}. \]
\(D/Dt\) is the material derivative: the rate of change while following the fluid. This is why the complete left side, rather than the time derivative alone, is the parcel’s acceleration.
4. Pressure gradient: unequal pushes create a net force
\[ \text{Pressure acceleration}=-\frac{1}{\rho}\nabla p. \]
Uniform pressure pushes on all sides of a tiny parcel without a net pressure force. A pressure difference produces an imbalance. The minus sign makes the pressure contribution point toward lower pressure; it does not guarantee that the actual velocity points that way, because other forces and existing motion also matter.
For example, a pressure decrease of 200 pascals over 2 metres gives \(dp/dx=-100\,\mathrm{Pa/m}\). At density \(1000\,\mathrm{kg/m^3}\), the pressure contribution is \(+0.10\,\mathrm{m/s^2}\) in the \(x\) direction. This is one contribution, not necessarily the net acceleration.
5. Viscosity: momentum spreads between neighboring layers
\[ \text{Viscous acceleration}=\nu\nabla^2\mathbf u. \]
Viscosity couples neighboring layers through shear stress. NASA’s fluid-equation overview describes the distinction between transport by the flow and viscous diffusion. The simple Laplacian form used here relies on our constant-viscosity, incompressible assumptions.
Viscosity is not always a local brake
A slow layer can gain momentum from a faster neighbor. The viscous term is therefore not simply a force opposite to the local velocity. In a linear shear flow \(u_x=cy\), the second derivative is zero, so the interior viscous acceleration is zero even though the shear stress \(\mu c\) is nonzero. Stresses can act on a parcel and still balance.
6. Body force: forces acting throughout the fluid
\(\mathbf b\) is an acceleration-like quantity because we wrote the equation per unit mass. For gravity with upward coordinate \(z\), \(\mathbf b=(0,0,-g)\). Multiplying the momentum equation by \(\rho\) turns every term into force per unit volume; the body-force term becomes \(\rho\mathbf b\).
Check the simplest case: still water. With velocity zero, pressure and gravity balance, giving \(dp/dz=-\rho g\). Pressure increases downward. A fluid at rest can therefore have large forces whose sum is zero.
7. Incompressibility: volume is conserved, not velocity
\[ \nabla\cdot\mathbf u =\frac{\partial u_x}{\partial x} +\frac{\partial u_y}{\partial y} +\frac{\partial u_z}{\partial z}=0. \]
For the constant-density model, inflow and outflow balance for an infinitesimal volume. A parcel can stretch in one direction while contracting in another. It may accelerate, turn, or rotate without changing volume. Do not replace incompressibility with “the velocity is the same everywhere.”
8. Worked solution: pressure-driven flow between two plates
Now use every term to solve a specific problem. Consider a steady, laminar, fully developed flow between stationary parallel plates at \(y=-H\) and \(y=H\), idealized as very wide. Let \(\mathbf u=(U(y),0,0)\), with no body-force component along \(x\), and constant pressure gradient \(dp/dx=-G\), where \(G>0\). Impose no slip: \(U(-H)=U(H)=0\).
Step 1: remove terms for stated reasons
Steady flow removes the local time derivative. Convective acceleration vanishes because velocity varies only across the gap, while parcels move only along it. Incompressibility holds because \(\partial U/\partial x=0\). The remaining \(x\)-momentum balance is:
\[ 0=\frac{G}{\rho}+\nu\frac{d^2U}{dy^2}, \qquad \mu\frac{d^2U}{dy^2}=-G. \]
Step 2: integrate and use both wall conditions
\[ U(y)=-\frac{G}{2\mu}y^2+C_1y+C_2, \qquad \boxed{U(y)=\frac{G}{2\mu}(H^2-y^2)}. \]
The two wall conditions give \(C_1=0\) and \(C_2=GH^2/(2\mu)\). The result is a parabola: zero velocity at either wall and maximum speed at the center. This is plane Poiseuille flow under the assumptions just stated.
Step 3: check a numerical example
Choose \(H=0.001\,\mathrm m\), \(\mu=0.001\,\mathrm{Pa\,s}\), \(\rho=1000\,\mathrm{kg/m^3}\), and \(G=4\,\mathrm{Pa/m}\). These are illustrative model inputs. The center speed is \(GH^2/(2\mu)=0.002\,\mathrm{m/s}\), or 2 millimetres per second.
The pressure contribution is \(G/\rho=0.004\,\mathrm{m/s^2}\). Since \(U''=-G/\mu=-4000\,\mathrm{m^{-1}s^{-1}}\) and \(\nu=10^{-6}\,\mathrm{m^2/s}\), the viscous contribution is \(-0.004\,\mathrm{m/s^2}\). They cancel exactly. Fluid keeps moving without accelerating because the forces balance.
9. What does it mean to solve Navier–Stokes?
A flow problem needs a domain, material properties, forcing, boundary conditions, and—when time dependent—initial conditions. Solving means finding velocity and pressure fields that satisfy all of them. Recognizing a formula without checking its assumptions is not enough.
The plate example is solvable because its symmetry and assumptions simplify the equations. It does not settle what happens to every three-dimensional flow. For the separate research story, see our explanation of OpenAI’s Navier–Stokes proof claim and its stated scope.
Common questions and mistakes
Does steady flow mean zero acceleration?
No. It means the velocity at a fixed point does not change with time. Convective acceleration can remain, as the stretching-flow example demonstrates.
Are dynamic and kinematic viscosity interchangeable?
No. Use \(\mu\) in the force-per-volume form and \(\nu=\mu/\rho\) in the acceleration form. Checking units catches a missing density factor.
Can I solve the momentum equation without the continuity condition?
Not for this incompressible model. The velocity must also preserve local volume. Pressure and velocity have to be found consistently with that constraint and the boundary conditions.
Key takeaways
Read the left side as the acceleration of a moving parcel. Read the right side as pressure, viscous, and body-force contributions per unit mass. Then enforce incompressibility and the problem’s conditions. The most useful question is not “Which term can I cross out?” but “Which physical assumption makes this term zero?”
For the introductory physics behind the terms, review pressure and fluids and Newton’s laws. These AP lessons provide prerequisites; they are not a full course on viscous-flow partial differential equations.