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For a uniform-density sphere with incompressible flow of radius $R$ with one surface spherical harmonic displacement ${\xi }_r(R)=lA{R}^{l-1}{Y}_l^m$, the bulk density perturbation vanishes but the surface contributes $\delta \rho =\rho {\xi }_r(R)\delta (r-R)$. Continuity of the Newtonian gravitational potential and the Poisson equation give $[{\partial }_r\delta \Phi {]}_{\mathrm{out}-\mathrm{in}}=4\pi G\rho {\xi }_r(R)$. The regular interior and decaying exterior solutions areThe coefficient follows because the derivative jump is $-(2l+1)C{R}^{l-1}{Y}_l^m$. This surface source is why neglecting the bulk density perturbation does not justify neglecting the perturbed gravity.

For a bounded conducting sphere with uniform magnetic diffusivity and an insulating exterior, an incompressible flow tangent to concentric spheres cannot sustain dynamo action. Neither the velocity nor the magnetic field needs to be axisymmetric. The radial magnetic scalar has a homogeneous advection-diffusion equation, so the poloidal field decays. Once it is absent, the toroidal magnetic potential also has a homogeneous scalar advection-diffusion equation. This is a theorem about spherical flow geometry, not about a general cylindrical flow with an axial component.

An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.

The turbulent energy spectrum $E(k)$ distributes kinetic energy over wavenumber shells, normalized by ${\int }_0^{\infty }E(k)dk=⟨|\mathbf{u}{|}^2⟩/2$. For isotropic incompressible flow its tensor form is ${\Phi }_{ij}=E(k)({\delta }_{ij}-k_i{k}_j/k^2)/(4\pi k^2)$. Its low-wavenumber behavior records large-eddy momentum statistics.

For inviscid planar incompressible flow, write velocity as $(-{\psi }_y,{\psi }_x)$ and vorticity as $\zeta ={\nabla }^2\psi$. Taking curl of the Euler equations for an inviscid fluid removes pressure and gives the displayed transport equation, with $J(\psi ,\zeta )={\psi }_x{\zeta }_y-{\psi }_y{\zeta }_x$. The Poisson equation then recovers the streamfunction at each instant. A velocity tangent to the plane cannot stretch its perpendicular vorticity vector.

For an ideal incompressible flow rotating about the axis of the field $\mathbf{B}=(J/2)s\hat{\mathbf{\varphi }}$, use cylindrical Fourier modes $e^{\sigma t+im\varphi }$. The azimuthal derivative of a cylindrical vector Fourier mode reduces induction to $\sigma \mathbf{b}=imJ\mathbf{u}/2$. If the boundary eigenproblem gives $\lambda =i\delta \nu$, with real $|\delta |<1$, eliminating the perturbation gives the displayed dispersion relation. Here $J=\nabla \times \mathbf{B}\cdot \hat{\mathbf{z}}={\mu }_0{J}_{\mathrm{phys}}$ for SI current density; the formula is conditional on that boundary eigenrelation.

The velocity divergence is the trace of the velocity gradient. It is the local fractional rate of volume expansion of a moving material element.

The velocity gradient tensor is $A_{ij}={\partial }_j{u}_i$. Its symmetric part is the rate-of-strain tensor $S$ and its antisymmetric part describes local rotation. For incompressible flow, $\mathrm{tr}A=\mathrm{tr}S=0$. The sign convention for a cubic invariant must be specified explicitly.

Use a velocity potential to reconstruct an irrotational flow from one scalar field. A velocity potential $\varphi (\mathbf{x},t)$ satisfiesIt has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is unchanged as a description of the velocity field when any function of time alone is added. For twice continuously differentiable $\varphi$, the identity curl of a gradient proves zero vorticity. In incompressible flow, taking the divergence also gives ${\nabla }^2\varphi =0$, the Laplace equation. This turns a three-component velocity field into a single harmonic function with appropriate boundary conditions. ^[ref]

For uniform flow $\mathbf{u}=(U,0,0)$, take $\varphi =Ux$. With $U=2\mathrm{m}{\mathrm{s}}^{-1}$, the change from $x=0$ to $x=3\mathrm{m}$ is $\Delta \varphi =6{\mathrm{m}}^2{\mathrm{s}}^{-1}$. By the fundamental theorem for line integrals, the line integral of velocity between these points equals that change along any path in the domain; a closed curve gives zero circulation.

For the potential line vortex, choose a local angle branch in polar coordinates and set $\varphi =K\theta$. Then $u_r={\partial }_r\varphi =0$ and $u_{\theta }=r^{-1}{\partial }_{\theta }\varphi =K/r$. One counterclockwise turn changes the continued velocity potential by $2\pi K$, although the velocity field returns to its original value. With $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$, this change is $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$, exactly the circulation. A full annulus therefore admits local angle branches and a multivalued velocity potential, but no global single-valued velocity potential when $K\ne 0$. Cutting the annulus along a radial line allows a single-valued angle branch on the remaining domain. The obstruction is path independence, not a failure of the local gradient formula.

Vertical shear instability destabilizes differential rotation that varies with height, when buoyancy restoration is absent or sufficiently weakened. In the homogeneous incompressible flow model with background $U_y=-3\Omega x/2+q\Omega z$, an axisymmetric perturbation with wavevector $(k_x,0,k_z)\ne 0$ hasThe pressure-free radial/vertical momentum equation and the azimuthal equation give this result after using incompressibility. Growth requires $2q{k}_x{k}_z>k_z^2$, so a small vertical shear favors nearly radial wavevectors. Here $q$ is the vertical shear amplitude, distinct from the orbital shear parameter $3/2$. Stratification, cooling and boundaries can change this simplified criterion.

The vertical velocity is the component of a fluid velocity field along the selected upward vertical direction. In incompressible flow, impermeable horizontal boundaries impose $w=0$. Eliminating pressure often yields a closed hydrodynamic stability equation for this component.

A convex scalar function $\Omega (D)$ on symmetric rate-of-strain tensors generates stress by $S=\partial \Omega /\partial D$. For incompressible flow, its argument is trace free and a pressure reaction must be added to the stress. At nondifferentiable points the law is $S\in \partial \Omega (D)$, where the subdifferential is defined by $\Omega (E)-\Omega (D)\ge S:(E-D)$ for every admissible $E$. This is a strain-rate potential, in contrast with potentials parameterized by thermodynamic forces.

For constant mass density $\rho$, kinematic viscosity $\nu >0$, ideal induction, zero boundary velocity and tangent magnetic field, put $E={\int }_D[\rho |\mathbf{v}{|}^2/2+|\mathbf{B}{|}^2/(2{\mu }_0)]dV$. Multiplying the momentum equation by $\mathbf{v}$ and integrating cancels pressure work and advective flux. The magnetic-energy derivative is minus the integral of Lorentz force density dotted with velocity, canceling the same work in the kinetic-energy equation. Integration by parts then gives the displayed identity. For incompressible flow with zero boundary velocity, $\int |\nabla \mathbf{v}{|}^2=\int |\nabla \times \mathbf{v}{|}^2$.

The viscous stress tensor is the part of the Cauchy stress tensor associated with deformation rates. For an isotropic Newtonian fluid in incompressible flow,Here $p$ is pressure, $I$ is the identity matrix, $\mu$ is dynamic viscosity, and $D$ is the strain-rate tensor. The convention is $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$; $T$ denotes transpose and $\mathrm{tr}$ denotes trace. Across a plane with unit normal vector $\mathbf{n}$, the viscous traction is ${\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}$. The tensor makes this rule work for any plane orientation. ^[ref]

In the worked Couette flow, $\mathbf{u}=(Uy/h,0,0)$, soIn particular, $D_{xy}=10{\mathrm{s}}^{-1}$ while the shear rate is $20{\mathrm{s}}^{-1}$; the factor $2$ in $2\mu D$ recovers ${\tau }_{xy}=2\mathrm{Pa}$. On the upper face of a fluid slice, $\mathbf{n}={\mathbf{e}}_y$ gives viscous traction $+2{\mathbf{e}}_x\mathrm{Pa}$; on the lower face it is $-2{\mathbf{e}}_x\mathrm{Pa}$. Constant stress can therefore transmit a force between the plates even though its divergence, and the net viscous force on each slice, vanish.

Compare planar extension $\mathbf{u}=(ax,-ay,0)$ with solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$:Extension changes lengths and produces viscous normal stresses $\pm 2\mu a$. Rotation preserves distances and gives zero viscous stress, despite its nonzero velocity gradient and vorticity $2\Omega {\mathbf{e}}_z$. For $a=\Omega =1{\mathrm{s}}^{-1}$ and $\mu =0.10\mathrm{Pa}\mathrm{s}$, extension gives ${\sigma }_{\mathrm{visc},xx}=0.20\mathrm{Pa}$ and ${\sigma }_{\mathrm{visc},yy}=-0.20\mathrm{Pa}$; rotation gives zero. A rotating fluid may still have a pressure gradient supplying its centripetal acceleration.

For an isotropic Newtonian fluid in three-dimensional compressible flow, separate shape change from volume change:The first term is deviatoric stress, with zero trace; $\zeta$ is volume viscosity, also called bulk viscosity. For uniform expansion $\mathbf{u}=b\mathbf{x}$, $D=bI$ and $\nabla \cdot \mathbf{u}=3b$, so the shear contribution vanishes and the viscous stress is $3\zeta bI$. Thus dynamic viscosity and volume viscosity describe distinct responses, and incompressible flow removes only the volume-change term. ^[ref]

The volumetric flow rate is the volume of fluid crossing a section per unit time, $Q={\int }_A\mathbf{u}\cdot \mathbf{n}dA$.