Euler's Equation

Equation

\[\boxed{\;\rho\frac{D\mathbf{v}}{Dt} = \rho\left[\frac{\partial \mathbf{v}}{\partial t} + (\mathbf{v}\cdot\nabla)\mathbf{v}\right] = -\nabla p + \rho\mathbf{g}\;}\]

Physical meaning

Newton's second law for an inviscid fluid element: mass × acceleration (following the particle — hence the material derivative) = pressure-gradient force + body force. Fluid accelerates from high toward low pressure, and falls under gravity, and that is all — no friction.

Variables

Symbol	Meaning	SI unit
\(\rho\)	density	kg m⁻³
\(\mathbf{v}\)	velocity field	m s⁻¹
\(p\)	pressure	Pa
\(\mathbf{g}\)	body force per unit mass	m s⁻²

Assumptions

Inviscid: \(\mu = 0\) (no shear stresses) — valid far from walls at high Reynolds number
Continuum; pair with continuity (and an energy/state equation if compressible) for a closed system

Derivation

For a fluid element, the surface force is the net pressure push. Component \(z\) across a box \(dx\,dy\,dz\): \(\left(p - \frac{\partial p}{\partial z}\frac{dz}{2}\right)dxdy - \left(p + \frac{\partial p}{\partial z}\frac{dz}{2}\right)dxdy = -\frac{\partial p}{\partial z}dV\). Collecting components: \(\delta\mathbf{F}_S = -\nabla p\, dV\). Adding weight \(\rho\mathbf{g}\,dV\) and equating to \((\rho\,dV)\,D\mathbf{v}/Dt\) gives the equation. Equivalently: momentum via the Reynolds transport theorem with stress tensor \(\mathbf{T} = -p\mathbf{I}\).

Special cases

Hydrostatics (\(\mathbf{v} = 0\)): \(\nabla p = \rho\mathbf{g}\) → hydrostatic equilibrium, \(p = p_0 - \rho g z\).
Rigid-body motion: \(\nabla p + \rho g\hat k = -\rho\mathbf{a}\) → tilted and parabolic free surfaces.
Steady flow along a streamline: integrates to Bernoulli's equation.
Irrotational flow: \(\mathbf{v} = \nabla\phi\) → potential flow with a global Bernoulli constant.

Worked example: converging nozzle

Steady horizontal nozzle, sections 1 → 2. Euler along the streamline reduces to Bernoulli: \(p_1 - p_2 = \frac{\rho}{2}(v_2^2 - v_1^2)\); with continuity \(v_2 = v_1 A_1/A_2 > v_1\), so \(p_2 < p_1\) — pressure falls in the direction of acceleration, exactly as the equation's \(-\nabla p\) demands.

Limitations

No boundary layers, no drag (d'Alembert's paradox), no dissipation — add \(\mu\nabla^2\mathbf{v}\) to get Navier–Stokes. Across shocks the differential form fails; use jump conditions.

Navier–Stokes equation — Euler + viscosity
Bernoulli's equation — first integral
Continuity equation — always solved alongside

Quiz

Q1 (conceptual). Euler's equation contains no viscosity, yet predicts pressure perfectly well in a static fluid. Why is the inviscid assumption harmless there?

Answer

Viscous stress is proportional to velocity gradients; at rest they vanish identically, so the viscous term would contribute nothing anyway.

Q2 (multiple choice). In steady flow, a fluid particle in a horizontal plane accelerates only if:

(a) \(\partial\mathbf{v}/\partial t \neq 0\) (b) a pressure gradient (or body force) acts
(c) the flow is compressible (d) vorticity is nonzero