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For a smooth planar incompressible flow, use $u={\psi }_y$, $v=-{\psi }_x$, so its scalar vorticity is $\omega =-\Delta \psi$. If $\omega =\lambda \psi$ for a constant $\lambda$, the flow solves the steady constant-density Euler equations with a suitable pressure.

Indeed the vector identity for the convective acceleration givesTherefore the pressure can be chosen so that the Bernoulli function satisfies $p/\rho +|\mathbf{u}{|}^2/2=C-\lambda {\psi }^2/2$. Its gradient is $-\omega \nabla \psi$, exactly the steady momentum equation. It is constant along each streamline, since the stream function is constant there.

For example $\psi =\sin x\cos y$ satisfies $\omega =2\psi$. Its pressure is $p_0-\rho {\sin }^{2}x/2+\rho {\sin }^{2}y/2$. Thus a rotational flow can have an explicit pressure and a streamline-dependent Bernoulli constant.

In steady flow, the velocity and other relevant flow fields are independent of time in the chosen reference frame. Streamlines then coincide with the paths of moving particles, while convective acceleration can still be nonzero.

In a steady inviscid planar incompressible flow, vorticity is constant along streamlines. A regular stream function supplies a local transverse coordinate, so locally $\Delta \psi$ depends only on $\psi$. A global single-valued function needs the same value on every component of each level set. This can fail: $\psi =x^3-x$ gives the steady Euler shear $(u,v)=(0,1-3x^2)$ with constant pressure, but $\psi (0)=\psi (1)=0$ while $\Delta \psi$ is respectively $0$ and $6$.

For steady flow of constant-density fluid in a uniformly rotating frame with conservative body forces, the displayed quantity is constant along each relative streamline. The centrifugal potential contributes the negative quadratic term, and the Coriolis force contributes no work because it is perpendicular to the relative velocity. The result follows by dotting the rotating Euler equations for an inviscid fluid with that velocity.

Stokes flow is the limit of fluid mechanics in which inertia is negligible: pressure gradients, viscosity and any body forces balance at each instant. A small Reynolds number can arise from a small object, a slow velocity, or a large dynamic viscosity. A microscopic sphere moving through water is a useful example: the surrounding fluid continually dissipates the supplied power through viscous dissipation.

The picture is a meridional slice of the three-dimensional axisymmetric flow around a sphere, with the sphere held fixed and the distant fluid moving rightward at speed $U$. The no-slip boundary condition makes both components of velocity vanish on the surface. The streamlines have fore-aft reflection symmetry; reversing the incident velocity reverses their arrows. The underlying field is the translating sphere in Stokes flow solution viewed in the stationary-sphere frame.

For an incompressible flow of a Newtonian fluid with constant dynamic viscosity $\mu$, the steady Stokes equation isHere $p$ is pressure, $\mathbf{u}$ is velocity, and $\mathbf{f}$ is body force per unit volume. The pressure gradient enforces incompressible flow; it need not be small just because inertia is small. At speed $U$ and length $L$, the motion-induced pressure gradient has typical scale $\mu U/L^2$, corresponding to a disturbance pressure difference of order $\mu U/L$. See Shreyas Mandre's fluid dynamics notes, sections 9.1.1–9.1.2.

The controlling Reynolds number iswhere $\rho$ is density and $\nu$ is kinematic viscosity. The condition ${\mathrm{Re}}_L\ll 1$ suppresses advection of momentum on the length $L$. Rapid changes require an additional check of the viscous diffusion time: small Reynolds number alone does not eliminate the time derivative in the Unsteady Stokes equation.

Two rewards of this limit are the exact Stokes's law for the drag on a sphere and kinematic reversibility of Stokes flow. The first predicts settling velocities and connects to the Stokes–Einstein relation; the second explains why a microswimmer needs a nonreciprocal stroke, as expressed by the scallop theorem. Despite viscous dissipation, reversing the driving can retrace the motion: an especially striking consequence of Linearity of Stokes flow.

George Gabriel Stokes obtained the sphere drag and its terminal velocity in his 1851 investigation of how viscosity changes physical pendulum motion. He also used the slow fall of small water droplets to discuss clouds. The paper was read on 9 December 1850; 1851 is its publication date. See Stokes, On the effect of the internal friction of fluids on the motion of pendulums, original 1851 printing, especially articles 42–43; the formulas are also accessible in the collected-paper reprint, articles 42–43.

A streakline at observation time $T$ is the set of present positions of all fluid elements that passed through one fixed release point $\mathbf{a}$ during a specified history of injection. For continuous release over ${\tau }_0\le \tau \le T$, solve a different initial value problem for each release time $\tau$:Thus a pathline varies $t$ for one particle; a streakline varies $\tau$ while holding $T$ fixed. Interrupted release gives only the corresponding subsets of the streakline. A moving nozzle requires a time-dependent release position instead, so the standard fixed-point definition should not silently be applied to it. Continuous release makes streaklines easy to observe and useful for showing where material from one source has reached. ^[ref]

In the analytical velocity field $(U,at)$ of pathline, a particle released at the origin at time $\tau$ followsObserve all particles together at time $T$, after continuous release from $\tau =0$:For $U=1\mathrm{m}{\mathrm{s}}^{-1}$, $a=1\mathrm{m}{\mathrm{s}}^{-2}$ and $T=2\mathrm{s}$, the streakline is the downward-curving parabola $y=2x-x^2/2$ in metre coordinates. Releases at $\tau =0,1,2\mathrm{s}$ occupy $(2,2)$, $(1,1.5)$ and $(0,0)\mathrm{m}$, respectively. At $x=1\mathrm{m}$, the streakline lies above the pathline $y=x^2/2$ and below the origin's streamline $y=2x$. All particles nevertheless have the same present velocity $(1,2)\mathrm{m}{\mathrm{s}}^{-1}$. The streakline slope $dy/dx=a\tau /U$ generally differs from the present velocity slope $aT/U$; they agree at the nozzle. The oldest particle is the downstream endpoint, which explains why the pathline and streakline meet there without being the same curve.

Each panel fixes one observation time. Coloured dots are the present positions of particles with the indicated release times; dashed pathlines show those particles' earlier histories. The black streakline connects present positions of the continuously released family, including release times between the sampled dots. After $T=1\mathrm{s}$, particles already present keep moving and new ones enter at the origin, so the later streakline is not a frozen record of the earlier one. Injection beginning only at $\tau =1\mathrm{s}$ would show just $0\le x\le 1\mathrm{m}$ of the $T=2\mathrm{s}$ curve. Both panels are diffusion-free analytical models, not experimental photographs.

Kline's contracting-channel demonstration switches from marking an element to continuous bubble production at a marking wire: the resulting trace is a streakline. The apparatus therefore changes the curve being displayed by changing the marking history, even with the same channel. A wire produces a family of traces; conceptually each fixed emitting location supplies its own streakline. Individual bubbles are markers rather than perfect fluid elements: Jean Hertzberg's teaching notes describe hydrogen production by electrolysis at a thin cathode wire and warn that larger bubbles have more buoyancy. ^{[ref] [ref]}

A dye version uses a small fixed injection port upstream in a transparent water channel, a dilute premix of dye with the channel water, illumination and a fixed optical camera. Start continuous injection at a recorded time ${\tau }_0$ and take a short exposure at $T$: the visible filament approximates a streakline, not the accumulated pathline of one particle. Repeat with the same port and imaging geometry after changing the flow with time; compare the filament against local directions estimated from short displacements of sparse markers at $T$. Agreement in a steady flow and possible disagreement when the flow changes are the predictions being tested; no measured curves are claimed here. The protocol follows Hertzberg's advice to use small ports, little injected volume and a dilute premix with matched density and viscosity. Check for nozzle disturbance by reducing injection rate, for buoyancy by watching the premix in still water, and for diffusion by comparing filament widths at different ages. Keep exposure short enough to suppress motion blur. Dye concentration approximates a passive scalar only when it has negligible feedback on the velocity field; finite port width, diffusion and out-of-plane motion limit how literally a visible filament can be read as a geometrical streakline. ^[ref]

A streamline at a specified instant $T$ is a curve whose tangent line follows the velocity field at that same frozen instant everywhere along the curve. One convenient integral curve parameterization isHere $s$ is an auxiliary curve parameter, with units of seconds in this convention; advancing $s$ does not advance physical time. Any regular reparameterization preserves the tangent geometry. Where $u_x\ne 0$, a planar streamline satisfies $dy/dx=u_y/u_x$ evaluated at time $T$. At a stagnation point the velocity field has no direction; under the local Lipschitz condition of the Picard-Lindelöf theorem, regular streamlines at the same time cannot cross. A streamline answers which directions fit together now, making it useful for describing local flow geometry. ^[ref]

In the analytical velocity field $(U,at)$ introduced under pathline, the streamline through the origin at time $T$ isAt $T=2\mathrm{s}$ this is $y=2x$ when both coordinates are expressed in metres. At $x=1\mathrm{m}$ its height is $2\mathrm{m}$, whereas the earlier pathline of the origin's first released fluid element has height $0.5\mathrm{m}$. Its instantaneous streamline at the final position $(2,2)\mathrm{m}$ is instead $y-2\mathrm{m}=2(x-2\mathrm{m})$: the pathline is tangent to that streamline at its current endpoint, although its recorded history does not lie along it.

The green streamline uses the velocity field at $T=2\mathrm{s}$ throughout; the orange pathline uses times $0\le t\le 2\mathrm{s}$; the purple streakline uses different release times at the single observation time $T=2\mathrm{s}$. The grey arrows all show the instantaneous velocity direction at $T$. These are exact analytical curves with chosen units, not measured dye shapes.

In a steady flow in a fixed reference frame, a pathline and a streamline solve the same autonomous ordinary differential equation. Uniqueness makes each particle released from a fixed point follow the same integral curve, so its pathline, the streakline from that point and the streamline share one geometrical curve. Finite recording or release intervals usually show different segments of that curve; the comparison assumes particles stay in the domain and treats stationary fluid elements separately. Neither two-dimensionality nor incompressibility is needed for this coincidence. Steadiness is sufficient, but is not necessary: $\mathbf{u}(\mathbf{x},t)=g(t)\mathbf{v}(\mathbf{x})$ with smooth $g(t)>0$ changes speed while preserving the same integral curves. In a general time-dependent velocity field, however, the three curves differ. Kline explains why continuous dye release in steady flow can therefore display streamlines. ^[ref]

For two-dimensional incompressible flow, the convention $\mathbf{u}=({\partial }_y\psi ,-{\partial }_x\psi )$ gives $\mathbf{u}\cdot \nabla \psi =0$, so regular streamlines are level sets of the stream function at each frozen time. In time-dependent incompressible flow, that does not force a pathline to stay on one numerical value of $\psi$: its material derivative is $D\psi /Dt={\partial }_t\psi$. The distinction is the difference between a geometric snapshot and a parcel's history.

A stream tube is a narrow tube whose side boundary consists of streamlines. In steady flow no fluid crosses its sides, so continuity makes its mass flux $A\rho u$ constant. If flow follows a magnetic field with constant magnetohydrodynamic mass loading, its normal area obeys $A|\mathbf{B}|=\mathrm{constant}$.

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.

Use vorticity to distinguish the local spin of a fluid element from the motion of its center along a pathline. Circular streamlines can occur both with and without vorticity: the two velocity fields below even have the same circulation around the unit circle.

For a continuously differentiable velocity field $\mathbf{u}=(u_x,u_y,u_z)$, vorticity is the curl vectorEach partial derivative divides velocity by distance, so vorticity has units ${\mathrm{s}}^{-1}$. In a planar velocity field independent of $z$, only ${\omega }_z={\partial }_x{u}_y-{\partial }_y{u}_x$ survives. Positive ${\omega }_z$ points along $+{\mathbf{e}}_z$: the right-hand rule associates it with counterclockwise spin viewed from positive $z$ toward the plane. ^[ref]

The local rotational angular velocity is half the vorticity, ${\mathbf{\Omega }}_{\mathrm{spin}}=\mathbf{\omega }/2$. To see the factor, split the velocity gradient $L_{ij}={\partial }_j{u}_i$ into its symmetric rate-of-strain tensor $E=(L+L^T)/2$ and antisymmetric spin tensor $W=(L-L^T)/2$. A nearby displacement $\delta \mathbf{x}$ has relative velocity $L\delta \mathbf{x}$; its rotational part satisfies $W\delta \mathbf{x}=(\mathbf{\omega }/2)\times \delta \mathbf{x}$. This is a proof outline of the local spin interpretation. The rate-of-strain tensor can also change the angle of a particular material line, so that line's angular velocity need not equal $\mathbf{\omega }/2$.

For solid-body rotation, $\mathbf{u}=(-\Omega y,\Omega x,0)$ gives $\mathbf{\omega }=(0,0,2\Omega )$ and ${\mathbf{\Omega }}_{\mathrm{spin}}=(0,0,\Omega )$. Taking $\Omega =1{\mathrm{s}}^{-1}$, the vorticity is $2{\mathbf{e}}_z{\mathrm{s}}^{-1}$, although the angular velocity is only $1{\mathbf{e}}_z{\mathrm{s}}^{-1}$. For the potential line vortex, $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$, the planar curl is $r^{-1}{\partial }_r(r{u}_{\theta })=r^{-1}{\partial }_{r}K=0$ wherever $r>0$. Its circular pathlines therefore coexist with zero local spin.

This distinction makes vorticity useful for recognizing shear flow as well as swirling motion; taking the curl of the Euler equations also removes a pressure gradient and leads to the vorticity equation. A historical landmark is Hermann von Helmholtz's 1858 paper on fluid vortex motion, an early study devoted to vortex dynamics. ^[ref]

A transported planar potential dipole follows the surrounding regular velocity, so its position satisfies the displayed equation after its own singular term is removed. The conjugation comes from the convention $w^{\prime }=v_x-i{v}_y$. Reflected image parameters are held fixed when taking the spatial derivative, then evaluated at the instantaneous dipole position.

For a unit circular wall, the Milne-Thomson circle theorem gives image potential $\bar{m}z/(1-\bar{a}z)$ and therefore $\dot{\bar{a}}=\bar{m}/(1-|a{|}^2{)}^2$. For a horizontal straight wall the image is $\bar{m}/(z-\bar{a})$, giving $\dot{\bar{a}}=-\bar{m}/(a-\bar{a}{)}^2$. For the line $y=-x$, reflected conjugation gives image $-i\bar{m}/(z+i\bar{a})$ and hence $\dot{\bar{a}}=i\bar{m}/(a+i\bar{a}{)}^2$. In each case the image lies outside the fluid region and makes the wall a streamline; differentiation of the regular image proves the motion formula. Several compatible straight-wall reflections can be summed before the same operation is applied.