The Navier-Stokes equation is one of the most fundamental equations in fluid mechanics, describing how fluids (liquids and gases) move. Its complexity can be intimidating, but at its core, the equation balances forces in a fluid. Let’s break down how to derive the Navier-Stokes equation as if you were scribbling it on a napkin at a coffee shop.

Step 1: Understanding the Physical Context

The Navier-Stokes equation is derived from:

• Conservation of Mass: Fluids cannot magically appear or vanish.
• Conservation of Momentum: Newton’s second law applied to fluids (Force = mass × acceleration).

We’ll focus on the momentum conservation, as it forms the basis of the Navier-Stokes equation.

Step 2: Start with Newton’s Second Law

In fluid mechanics, we consider a small fluid element of mass mmm. Newton’s second law states:F=m⋅aF = m \cdot aF=m⋅a

We express acceleration aaa as the material derivative of velocity u\mathbf{u}u:a=DuDt=∂u∂t+u⋅∇ua = \frac{D \mathbf{u}}{Dt} = \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u}a=DtDu​=∂t∂u​+u⋅∇u

Thus, the equation becomes:mDuDt=Sum of Forcesm \frac{D \mathbf{u}}{Dt} = \text{Sum of Forces}mDtDu​=Sum of Forces

Step 3: Consider the Forces Acting on the Fluid

We account for three main forces:

1. Body Force (Gravity):fb=ρg\mathbf{f}_b = \rho \mathbf{g}fb​=ρg
2. Pressure Force:
   Pressure force acts on the fluid due to the surrounding pressure field:fp=−∇p\mathbf{f}_p = -\nabla pfp​=−∇p
3. Viscous Force (Friction):
   Viscous forces arise from internal friction, modeled using a viscosity constant μ\muμ:fv=μ∇2u\mathbf{f}_v = \mu \nabla^2 \mathbf{u}fv​=μ∇2u

Step 4: Combine the Forces

The sum of forces becomes:ρDuDt=−∇p+μ∇2u+ρg\rho \frac{D \mathbf{u}}{Dt} = -\nabla p + \mu \nabla^2 \mathbf{u} + \rho \mathbf{g}ρDtDu​=−∇p+μ∇2u+ρg

Expanding the material derivative:ρ(∂u∂t+u⋅∇u)=−∇p+μ∇2u+ρg\rho \left(\frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} \right) = -\nabla p + \mu \nabla^2 \mathbf{u} + \rho \mathbf{g}ρ(∂t∂u​+u⋅∇u)=−∇p+μ∇2u+ρg

Step 5: The Final Navier-Stokes Equation

Thus, the incompressible Navier-Stokes equation becomes:∂u∂t+u⋅∇u=−1ρ∇p+ν∇2u+g\frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} = -\frac{1}{\rho} \nabla p + \nu \nabla^2 \mathbf{u} + \mathbf{g}∂t∂u​+u⋅∇u=−ρ1​∇p+ν∇2u+g

Where:

• u\mathbf{u}u: Fluid velocity vector
• ppp: Pressure
• ρ\rhoρ: Fluid density
• ν=μρ\nu = \frac{\mu}{\rho}ν=ρμ​: Kinematic viscosity

Bonus: Continuity Equation (Conservation of Mass)

For an incompressible fluid, the mass conservation equation is:∇⋅u=0\nabla \cdot \mathbf{u} = 0∇⋅u=0

Final Thoughts

By combining Newton’s second law with physical forces acting on a fluid element, we derived the Navier-Stokes equation in its simplest form—right on a napkin. While real-world applications involve additional complexities (compressibility, turbulence, boundary conditions), this derivation shows that the core idea is straightforward: balance forces and track how fluids move.

Next time you’re at a coffee shop, grab a napkin and impress your friends with this essential equation from fluid dynamics!