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For a complex potential of a two-dimensional potential flow $W=\varphi +i\psi$ in the coordinate $z=x+iy$, the Cauchy-Riemann equations give $W^{\prime }={\varphi }_x-i{\varphi }_y=u-iv$. A material point with physical velocity $(u,v)$ therefore satisfies $\dot{z}=\overline{W^{\prime }(z)}$. Complex conjugation matters when deriving a pathline from a complex potential of a two-dimensional potential flow.

Use irrotational flow to identify where a velocity potential can describe the velocity field locally. An irrotational flow has $\mathbf{\omega }=\nabla \times \mathbf{u}=0$ throughout the fluid region. For a continuously differentiable velocity field, the Poincaré lemma supplies a velocity potential on sufficiently small neighborhoods. A global single-valued velocity potential requires zero circulation around every closed curve; a simply connected domain is a sufficient domain assumption.

The potential line vortex $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$ is an irrotational flow on the punctured plane, or on an annulus $a<r<b$ with $a>0$, even though its pathlines are circular. Every sufficiently small loop bounding a disk in that fluid has zero circulation, by Stokes theorem. A loop winding once counterclockwise around the excluded origin has circulation $2\pi K$. Thus local irrotational flow and nonzero circulation around a hole are compatible: the enclosing disk passes outside the domain of the velocity field.

An irrotational flow is a kinematic condition on a velocity field, independent of whether a particular model neglects viscosity or assumes incompressible flow. Adding incompressible flow makes the local velocity potential satisfy the Laplace equation. This reduces the number of unknowns and makes potential flow a powerful starting point for flow around bodies and for gravity waves. ^[ref]

For a bottom-forced gravity wave with signed surface coefficient $b$, evaluate the first-order velocity field at a parcel's mean position $(\bar{x},\bar{y})$ and integrate, with $\Theta =k(\bar{x}-ct)$. The horizontal and vertical displacement coefficients areThus the pathline is an ellipse with horizontal and vertical semiaxes $|A|,|B|$ whenever both are nonzero, and a line segment if one vanishes. The identity $A^2-B^2=b^2[(g/(k{c}^2){)}^2-1]$ determines which axis is longer. Also $B(0)=b$ and $B(-d)=a$, verifying the surface and bottom displacements. Closed first-order orbits do not rule out a higher-order mean drift.

Following a parcel. Let $\mathbf{X}(t)$ be a fluid pathline, with $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The multivariable chain rule gives the material derivative,The first term measures change at fixed position; the advection term accounts for the parcel moving through spatial gradients. The derivative is also called the Lagrangian derivative.

The forces on a parcel. A material volume $dV$ has mass $\rho dV$. The pressure forces on its boundary sum to $-\nabla pdV$, while the gravitational acceleration is $-\nabla \psi$. Newton's second law therefore gives the Euler equations for an inviscid fluid,This neglects viscosity and other body forces. The left side is the parcel's acceleration, not just the time derivative at a fixed spatial location.

Bernoulli conservation. The claimed streamline result needs steady flow and a barotropic fluid, $p=p(\rho )$, or an equivalent enthalpy relation along each isentropic streamline. These hypotheses are necessary; the unsteady Euler equation alone does not imply the stated conservation law. Define the barotropic enthalpy functionDot the steady Euler equation with $\mathbf{u}$, using $\mathbf{u}\cdot [(\mathbf{u}\cdot \nabla )\mathbf{u}]=\mathbf{u}\cdot \nabla (u^2/2)$. This givesThus the Bernoulli function isEvery term has units of energy per mass, ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Different streamlines can have different constants.

The spherical accretion equation. Write the inward speed as $u(r)>0$, so the velocity vector is $-u(r)\hat{\mathbf{r}}$ and the positive mass accretion rate isAssume a prescribed central potential $\psi =-GM/r$ and negligible gas self-gravity. The steady radial Euler equation isFor a barotropic equation of state, $dp/dr=c_s^{2}d\rho /dr$, with $c_s^2=dp/d\rho$. Eliminate the density derivative to obtainA smooth sonic point requires the right side to vanish where $u=c_s$, henceFor a nonisothermal gas this is an implicit relation involving the local sound speed. In the isothermal case $c_s$ is constant and this is the Bondi sonic point.

The isothermal integral. For $p=c_s^2\rho$, choose the enthalpy reference so that $h_b=c_s^2\ln (\rho /{\rho }_{\infty })$. The reservoir is at rest at infinity and $\psi (\infty )=0$, making its Bernoulli constant zero. ThusorThe logarithm has a dimensionless argument. Constant temperature during compression generally requires heat removal, so this is Isothermal Bondi accretion, rather than adiabatic accretion with $\gamma >1$.

The critical accretion rate. Select the smooth reservoir-connected transonic branch. Evaluate the last relation at $u_s=c_s$ and $GM/r_s=2c_s^2$:Mass conservation then gives the isothermal Bondi accretion rate,The units are $\mathrm{kg}{\mathrm{s}}^{-1}$. Equivalently $\dot{M}=4\pi \lambda G^2{M}^2{\rho }_{\infty }/c_s^3$, with $\lambda =e^{3/2}/4\simeq 1.12$.

The inner-boundary selection of isothermal Bondi flow is important: a standard absorbing accretor inside the sonic radius selects this critical solution. The steady equations and ${\rho }_{\infty }$ alone also admit subcritical subsonic branches supported by a different inner boundary; they do not by themselves fix a unique rate. The usual answer presumes transonic capture.

Following a fluid element. Let $\mathbf{X}(t)$ be a pathline, satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The material derivative, also called the Lagrangian time derivative, is the rate observed while following that element. The multivariable chain rule givesThe partial derivative holds position fixed; the advection term accounts for motion through a spatially varying field. For a vector field, the same formula applies component by component in fixed Cartesian axes.

Momentum balance. For an inviscid flow, pressure is the only surface stress. A small material volume $dV$ has net pressure force $-\nabla pdV$, obtained by the divergence theorem. Its mass is $\rho dV$ and its material acceleration is $D\mathbf{u}/Dt$. Applying Newton's second law gives the Euler equations for an inviscid fluid,This is the local momentum conservation equation when viscosity and physical body forces are absent. A prescribed body force per unit mass would add $\rho \mathbf{f}$ on the right.

Changing to rotating coordinates. In a rotating frame with constant angular velocity $\mathbf{\Omega }$, the lab acceleration is the rotating-frame material acceleration plus $2\mathbf{\Omega }\times \mathbf{u}+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. ThereforeFor uniform rotation,With constant ${\rho }_0$, this gives the Euler equation in a uniformly rotating frame,The term subtracted from $p$ is the equilibrium centrifugal contribution. Dividing the bracket by ${\rho }_0$ defines a reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. The Coriolis acceleration remains a velocity-dependent term.

Linear disturbances. Linearize about rest in the rotating coordinates, with the equilibrium part of $\Pi$ removed. The advective acceleration is quadratic in disturbance amplitude, so the linearized Euler equations and incompressible flow condition areTaking the curl removes the pressure gradient. The identitygives the inertial-wave vorticity equation,

Plane-wave frequencies. Let $\mathbf{k}\ne 0$ and use the given plane wave convention. Incompressible flow requires $\mathbf{k}\cdot {\mathbf{u}}_0=0$, and the vorticity equation givesFor a nonzero frequency, apply $\mathbf{k}\times$ again, using $\mathbf{k}\times (\mathbf{k}\times {\mathbf{u}}_0)=-k^2{\mathbf{u}}_0$. It follows thatThis is the inertial-wave dispersion relation. If $\theta$ is the angle of the wavevector to the rotation axis, the positive angular frequency is $2|\Omega \cos \theta |$. It is largest, $2|\Omega |$, for a parallel wavevector, decreases to zero for a perpendicular one, and is independent of the magnitude of the wavenumber at fixed direction. The two signs describe the two oscillatory polarization branches. When $k_z=0$, the zero-frequency result is obtained directly from the linear equations rather than by dividing by $\omega$; it describes the nonoscillating balanced limit.

The local conservation laws. The first equation is the radial Euler equation: the material acceleration $Du/Dt={\partial }_{t}u+u{\partial }_{r}u$ is caused by the pressure gradient force per unit mass, $-{\partial }_{r}p/\rho$. It neglects gravity and viscosity. The second equation is mass conservation in spherical coordinates,The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.

To obtain the third equation, let $e=p/[(\gamma -1)\rho ]$ be the specific internal energy of an ideal gas with constant heat capacity ratio $\gamma$. The first law of thermodynamics for a smooth adiabatically moving element isDifferentiating $e$ and collecting terms givesand henceThus the entropy parameter $p/{\rho }^{\gamma }$ is conserved along each smooth pathline; for fixed composition, specific entropy differs from $c_v\ln (p/{\rho }^{\gamma })$ only by a constant. This is local isentropic flow behind the shock. A physical shock wave itself generates entropy and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.

The front-radius scaling. In the self-similar blast wave approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a{\rho }_0^b{t}^d$. Using $[E]=\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$ and $[{\rho }_0]=\mathsf{M}{\mathsf{L}}^{-3}$, dimensional analysis givesThus $a=1/5$, $b=-1/5$, $d=2/5$ andThe dimensionless coefficient ${\xi }_0$ depends on $\gamma$ and the resolved similarity profiles; dimensions alone do not fix it. Define the length scale $\ell (t)=(E{t}^2/{\rho }_0{)}^{1/5}$, so $\xi =r/\ell$ and the front is at the fixed coordinate $\xi ={\xi }_0$.

Kinetic energy. Substitute $\rho ={\rho }_0{f}_{\rho }(\xi )$ and $u=(r/t)f_u(\xi )$ into the volume integral. Since $r=\ell \xi$ and $dr=\ell d\xi$,But ${\rho }_0{\ell }^5/t^2=E$, so the first of the spherical blast-wave similarity energy integrals isThe conclusion requires a fixed, integrable similarity profile, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.

Pressure and internal energy. The derivatives of the similarity variable areWriting a prime for differentiation with respect to $\xi$, the material acceleration becomesThe radial momentum equation therefore requires a pressure gradient on the scale ${\rho }_0\ell /t^2$. Integrating over radius gives pressure similarity in a spherical blast wave:whereThe prefactor has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, as required for pressure. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.

For constant $\gamma >1$, the internal energy is consequentlyIf $E$ is the total excess explosion energy, its normalization isThe two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.

The streamlines are the level curves of the streamfunction, so $\theta$ is constant: radial rays directed outwards from the origin. Their arrows point outwards because $Q>0$. A complete straight line through the origin comprises two distinct rays separated by the excluded line source. Equivalently, $r\dot{\theta }=u_{\theta }=0$ and $\dot{r}=Q/(2\pi r)>0$ along each pathline; the steady flow makes these the same curves as the streamlines.

Consider the nonresonant traveling forced solution, with $k>0$ and $c\ne 0$. Put $x=\bar{x}+\xi$, $y=\bar{y}+\upsilon$, where $(\bar{x},\bar{y})$ is the mean position of a fluid element. At linear order, evaluate the velocity field at this mean position: substituting $\xi ,\upsilon$ into its spatial arguments would change the already first-order velocity only at second order. With $\Theta =k(\bar{x}-ct)$, differentiation of the velocity potential givesSince $\dot{\Theta }=-kc$, integration, with zero mean displacements, yieldsIf both coefficients are nonzero, eliminating the wave phase proves that the pathline is an ellipse:Its principal axes of an ellipse are horizontal and vertical, with respective semi-axesThus the full axis lengths are $2|A|$ and $2|B|$. The signed coefficients also determine the direction of traversal. The useful checks $B(0)=b$ and $B(-d)=a$ reproduce the vertical displacements of the free surface and bottom.

To identify the major axis without comparing the two expressions separately, use ${\cosh }^{2}s-{\sinh }^{2}s=1$:Consequently the major axis is horizontal if $c^2<g/k$ and vertical if $c^2>g/k$; equality gives circles. If $A$ or $B$ vanishes at a particular depth, the first-order orbit degenerates to a line segment. These are linearized particle orbits of a bottom-forced gravity wave, not exact closed trajectories at all orders: a higher-order calculation may introduce a mean drift.

Adopt the streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Integrating the first component of the velocity field in $y$ gives $\psi =-xy+\frac{1}{2}\Omega e^{-2t}{y}^2+f(x,t)$. The second component then requires $f_x=\Omega e^{2t}x$. Thus, omitting an arbitrary function of time,At a fixed instant, the streamlines are the connected curves $\psi =K$. To classify this quadratic form, introduce the invertible scaled and rotated coordinatesThenFor $\Omega >1$, both coefficients are positive. Every $K>0$ therefore gives an ellipse, and $K=0$ is the stagnation point at the origin. For $0<\Omega <1$, the quadratic form is an indefinite quadratic form: each $K\ne 0$ gives a hyperbola. The level $K=0$ consists of the straight separatriceswhose rays approach or leave the origin; the origin itself is a stagnation point. At the transition $\Omega =1$, $\psi =d^2$, so the nonstationary streamlines are parallel straight lines and $d=0$ is a line of stagnation points. These are instantaneous streamlines; the time dependence means that they need not coincide with pathlines.

Follow a fluid particle released at time $s$, and use $\tau$ for its running time. Its pathline satisfies $dx/d\tau =1$, $dy/d\tau =\tau x$ and $dz/d\tau =0$, with zero position at $\tau =s$. Hence $x(\tau )=\tau -s$, $z(\tau )=0$, andAt observation time $t$, set $x=t-s$. Substitution gives the streaklineThe curve is parameterized by release times $s\le t$. If release starts at time zero, only $0\le x\le t$ is occupied. This is a streakline, the positions of different particles released at one fixed point, rather than an instantaneous streamline of this unsteady flow.

A particle pathline follows the time-dependent velocity field, so its coordinates satisfy $\dot{x}-x=\sin t$ and $\dot{y}=-y$. The second equation and the initial data give $y=e^{-t}$. Using the integrating factor $e^{-t}$ in the first,Therefore the complete pathline isIf $x_0+\frac{1}{2}\ne 0$, its first coordinate grows without bound in magnitude. If $x_0=-\frac{1}{2}$, the first coordinate remains bounded and periodic while $y\to 0$. The unique release coordinate giving a bounded particle path is $x_0=-1/2$.

For a two-dimensional incompressible flow $\mathbf{u}=(u,v)$, the continuity equation is $u_x+v_y=0$. Locally, or globally on a simply connected fluid region, this permits a streamfunction withIndeed, the one-form $-vdx+udy$ is closed by incompressibility and hence is locally the differential of $\psi$. Conversely these expressions automatically give zero divergence by equality of mixed derivatives.

A streamline at a fixed time is tangent to the instantaneous velocity field. Along it,Thus streamlines are connected curves on which the streamfunction is constant; wherever $\nabla \psi \ne 0$, the level curves have exactly the required tangent direction. At a stagnation point, the tangent is degenerate and distinct branches of a level set need not form a single streamline. For unsteady flow this argument fixes time: it does not identify streamlines with particle pathlines.

The divergence is ${\partial }_x{u}_x+{\partial }_y{u}_y=1-1=0$, while the scalar vorticity is ${\partial }_x{u}_y-{\partial }_y{u}_x=0$. The flow is therefore incompressible flow and irrotational flow. Integrating the streamfunction equations givesThe time dependence must be retained: no time-independent function $\psi (x,y)$ represents this unsteady field for every $t$. At $t=0$ the streamlines are $xy=\mathrm{constant}$, with the coordinate axes as separatrices. Their arrows point towards increasing $|x|$ and decreasing $|y|$, as shown below.

A pathline solves the material ODEs, givingFor $x_0\ne 0$, elimination of time yieldsIf $x_0=0$, the pathline stays on the $y$ axis and the parametric expression remains valid; it cannot in general be written as a single-valued $y=f(x)$ along that vertical trajectory. For $x_0=y_0=1$, the path is $y=\log x-1+2/x$: it initially descends, reaches its minimum $(2,\log 2)$, then rises. Unlike a frozen streamline, it records the time-varying flow.

A fluid is described on a continuum scale by density, pressure and velocity fields. An Eulerian field is measured at fixed position; a material element changes at rate $D/Dt={\partial }_t+\mathbf{v}\cdot \nabla$. mass conservation requiresFor incompressible flow this becomes $\nabla \cdot \mathbf{v}=0$. A streamline is instantaneously tangent to velocity, whereas a pathline follows a particle through time; they coincide in a steady flow but need not in an unsteady one.

For an ideal inviscid fluid, pressure and body forces give the Euler equation. In the constant-density gravitational case,A stationary impermeable boundary requires zero normal velocity, but inviscid theory does not impose no slip. Taking its steady projection along a streamline gives $p/\rho +v^2/2+gz=\mathrm{constant}$. If the flow is also irrotational in the connected region, that constant is shared between streamlines. Inviscid does not mean irrotational: rotating ideal-fluid flows are possible. A velocity potential exists locally for an irrotational field, and in incompressible flow it obeys Laplace's equation. The Bernoulli assumptions must be checked before converting a pressure difference into a speed.^[ref]

Real fluids transmit shear stress. An incompressible Newtonian fluid with constant dynamic viscosity $\eta$ obeysThe constitutive shear stress is proportional to strain rate, and stationary solid boundaries usually impose no slip. viscosity diffuses vorticity and converts organized mechanical energy into heat; its local dissipation $2\eta e_{ij}{e}_{ij}$ is nonnegative. It cannot generally be included merely by changing the constant in a lossless Bernoulli equation.

In a long straight circular tube, steady fully developed laminar flow hasThis Hagen-Poiseuille equation follows by balancing the axial pressure gradient with viscous shear and setting $v_z(a)=0$. The parabolic profile, the fourth-power radius sensitivity and the pumping power $\Delta pQ$ are experimentally distinct consequences. Entrance regions, rough walls, non-Newtonian response or a changing temperature require additional modelling.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscosity. Small values favor creeping-flow balances; high values permit thin boundary layers, separation and turbulence, although transition depends on geometry and perturbations, not one universal critical number. At high Reynolds number an outer flow can be approximately inviscid while near-wall viscosity still determines drag. This resolves why a zero-drag ideal potential-flow model cannot predict the resistance of an ordinary moving body. In turbulent flow, fluctuating Reynolds stresses redistribute momentum and increase the required pressure drop; averaging the equations introduces closure requirements.

A controlled tube experiment measures pressure difference and collected volume per time while changing forcing. Use tracers or a velocity probe to test profile shape, allow sufficient entrance length, and monitor radius and temperature independently. Repeat with reversed flow and at several rates to test offsets and linearity. Dye mixing alone does not measure viscosity or establish a fully developed velocity profile. The original figure below separates a specified laminar velocity profile from an independent pressure/discharge measurement.

Flow visualization and the idealization. A streamline is everywhere tangent to the instantaneous velocity field at one fixed time. A streakline is the locus, at one observation time, of particles that previously passed through a specified injection point. A marked particle's own trajectory is a pathline. These coincide geometrically in steady flow, but need not do so in an unsteady draining flow.

The intended ideal liquid is approximately incompressible and inviscid: viscosity and associated shear stresses are negligible on the scales and times being considered, the Mach number is small, and the density is effectively constant. This does not require zero vorticity; the initially rotating state is already a counterexample. It also does not justify ignoring viscosity in an arbitrarily thin boundary layer or at a real drain.

Spiralling flow and angular momentum. In an axisymmetric inviscid flow, pressure and gravity exert no torque about the vertical axis. A parcel therefore conserves axial specific angular momentum,The late, shallow remnant in the intended draining picture is supplied by parcels initially near the outer rim, so $r_0\simeq R$. The slow-drain angular-momentum amplification then givesThe assumption about the parcels' origin is necessary: torque conservation by itself does not assign $r_0=R$ to every parcel at every time.

A continuous dye feed near the outer free surface produces inward spirals that wind increasingly rapidly near the drain. The figure uses a late quasisteady model, $u_r=-D/r$ and $u_{\varphi }=C/r$ with positive constants $D,C$, giving $\varphi ={\varphi }_0-(C/D)\ln (r/r_0)$. Its streaklines and streamlines approximately coincide over an observation interval short compared with the overall draining time. Strictly, individually marked particles at the initial surface trace pathlines, not a continuously injected streakline.

The free-surface depression. Let $z=H(r)$ be the surface. Slow radial draining makes radial inertial corrections small compared with the azimuthal centripetal acceleration. The radial Euler equation and vertical hydrostatic equilibrium giveThus $g{H}^{\prime }(r)=u_{\varphi }^2/r=C^2/r^3$, with $C={\omega }_0{R}^2$. Integrating from $r$ to the rim givesFor $r\ll R$ this becomesThe surface rises outward because the inward pressure gradient force must sustain the circular motion. The dimensions are metres, and the $1/r^2$ profile is not valid inside the finite drain or vortex core.

The terrestrial estimate. Use the vertical component of Earth's rotation as the ideal initial angular speed, ${\omega }_0\simeq {\Omega }_E\sin \lambda$. At Cambridge, taking $\lambda \simeq 52.2^{\circ }$ and ${\Omega }_E=7.292\times 10^{-5}{\mathrm{s}}^{-1}$, this gives ${\omega }_0\simeq 5.76\times 10^{-5}{\mathrm{s}}^{-1}$. Hence, for $R=1\mathrm{m}$, $r=0.01\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Taking the whole terrestrial angular speed rather than its local vertical projection would give about $2.7\mu \mathrm{m}$. The Cambridge estimate uses the projection appropriate to the vertical vortex. This very small head assumes terrestrial rotation supplies the only initial swirl; imposed residual rotation would change it quadratically.

Curl of the azimuthal velocity. For axisymmetric $u_{\varphi }=r\omega (r)$ with no vertical dependence, the rotational part of the vorticity is vertical:The ideal quasisteady radial sink has no curl outside its core. The source's symbol $\Omega$ denotes this vertical vorticity component, not the fluid's angular velocity. In general its magnitude is the absolute value of the bracket.

The two limiting states. For initial uniform rotation, $\omega ={\omega }_0$, whereas the late approximation has $\omega =C/r^2$. ThereforeThe late flow can have large angular velocity and nonzero circulation while being locally irrotational. The drain core is excluded from the domain of the $1/r$ velocity field.

What is actually conserved. Inviscid does not mean that each parcel's vorticity magnitude is fixed in three dimensions. For an incompressible barotropic liquid with conservative body forces, the vorticity equation isVertical compression of a material column changes its vorticity through vortex stretching, even while its volume is conserved by horizontal expansion. In the columnar shallow-layer limit, vorticity dilution in a shallow draining layer givesThus a remnant much shallower than its original column can have much smaller vertical vorticity. At finite depth there is generally a residual value; the exactly irrotational late profile is a leading limiting approximation, not an exact parcelwise conversion of nonzero vorticity to zero in finite time.

Kelvin's circulation theorem conserves circulation around a material loop, not around a fixed spatial circle. A loop whose original radius is near $R$ has circulation about $2\pi {\omega }_0{R}^2$, consistent with the late $u_{\varphi }=C/r$ field. Fluid also leaves through the drain, so the remaining tank contents are not a fixed material system. If the calculation is expressed explicitly in the Earth-fixed rotating frame, the corresponding shallow-water potential vorticity is $(\zeta +f)/H$, with $f=2{\Omega }_E\sin \lambda$; the conserved quantity involves absolute vorticity.

A pathline records the motion of one identified fluid element: keep its starting position $\mathbf{a}$ and starting time $t_0$ fixed, and let physical time $t$ advance. In a velocity field $\mathbf{u}(\mathbf{x},t)$ its Lagrangian trajectory $\mathbf{X}(t;t_0,\mathbf{a})$ solves the initial value problemA streamline instead freezes the time argument of the velocity field; a streakline fixes the observation time and varies the release time of many fluid elements. A pathline is useful for answering where a particular parcel goes and how long it takes, rather than merely which way the fluid is moving now. ^[ref]

For a concrete analytical model, take spatial coordinates $x,y$ in metres, time $t\ge 0$ in seconds, and the spatially uniform but time-dependent velocity fieldIts numerical components in these units are $(1,t)$. A fluid element starting at the origin at $t_0=0$ has the pathlineAt $t=1\mathrm{s}$ it is at $(1,0.5)\mathrm{m}$; at $t=2\mathrm{s}$ it is at $(2,2)\mathrm{m}$. The parabola records earlier, shallower velocity directions as well as the final direction. Differentiating gives $(\dot{x},\dot{y})=(U,at)$ and acceleration $(0,a)$, so this curved pathline needs no spatial variation of the velocity field. More generally the chain rule along a pathline gives $df(\mathbf{X}(t),t)/dt={\partial }_{t}f+\mathbf{u}\cdot \nabla f$, the material derivative of a transported scalar field.

A motivating tracer demonstration appears in S. J. Kline's film notes: water passes through a contracting channel, and hydrogen bubbles mark a small element at a wire at time $t_0$. The illustrated element is shown again $4.7\mathrm{s}$ later, with its accumulated pathline superposed. Following that same marked element across successive film frames records a history; the entire bubble pattern in one frame has a different meaning. ^[ref]

For a modern version of the readout, use sparse small markers in a transparent water channel, illuminate the viewed plane, and keep an optical camera fixed with a ruler in that plane. Record timestamps and track the same isolated marker between frames; successive positions estimate its pathline and $[\mathbf{X}(t+\Delta t)-\mathbf{X}(t)]/\Delta t$ estimates its velocity. A long exposure records a pathline segment but loses the timing within it, so its total length divided by exposure time gives only mean speed. These are proposed measurement steps based on Jean Hertzberg's teaching guidance, rather than additional measurements from Kline's film. Markers leaving the viewed plane, overlapping tracks and perspective can break particle identification; repeat with a thinner viewed slice and a lower marker concentration. ^[ref]

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.