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For steady, inviscid, incompressible flow, Euler's equation integrated along a streamline givesThis is Bernoulli's equation. The constant is common to different streamlines if the flow is also irrotational; incompressibility alone does not imply that. Viscous loss or work by a pump requires additional terms.

A Pitot-static tube compares stagnation pressure $p_0$ at a forward-facing opening with nearby static pressure $p$. For equal height, the stagnation point has $v=0$, soThis measures flow speed when compressibility corrections are negligible.

In a flowmeter using the Venturi effect, continuity gives $A_1{v}_1=A_2{v}_2$. For a horizontal constriction with $A_2<A_1$, Bernoulli gives $p_1-p_2=\rho (v_2^2-v_1^2)/2$, henceThe fluid accelerates and static pressure falls in the throat. This is a second practical application of the conversion between pressure and kinetic energy; real instruments need calibration for flow losses.

Assume all rings have the same circulation $\Gamma$ and orientation, and that there is no imposed background flow. Vortex lines describe the direction of vorticity, not the direction of velocity. For an incompressible fluid, the velocity induced by a thin vortex filament has the same Biot-Savart integral form as the magnetic field of a current filament:A closely spaced coaxial array of vortex rings is therefore analogous to a solenoid. The velocity field passes along the common axis through the rings and returns outside a finite array. Its orientation reverses if every circulation is reversed. In a long array, far from its ends and vortex cores, the velocity field of a dense coaxial vortex-ring array is approximatelyThe exterior field is exactly zero only in the ideal continuous infinitely long limit, with the background velocity set to zero. In a finite array the streamlines spread out at one end, run back outside, and converge into the other. A large number of closely spaced rings alone does not guarantee that the total array length greatly exceeds $r$; if it does not, the end and return flows remain substantial. Different unspecified ring circulations would require superposing their actual signed contributions.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

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.

For the intended ideal Venturi effect, take steady flow, incompressible flow, negligible viscous losses, and the same height at the two stations shown. The Bernoulli equation along a streamline givesUsing water mass density $\rho \simeq 1000\mathrm{kg}{\mathrm{m}}^{-3}$,Thus the outlet pressure is lower by $16\mathrm{kPa}$. The increased kinetic energy per unit volume comes from the drop in pressure.

The source's assumption of laminar flow alone does not imply negligible viscosity or losses. The numerical result uses the usual lossless Bernoulli equation approximation, with the quoted speeds interpreted as representative streamline speeds. A real tube with significant viscous losses requires additional data; in a uniform-velocity energy balance its pressure drop would include an extra positive loss term.

The streamlines are contours of $\psi$. Since $A<0$, the exact streamfunction is negative inside the rectangle and zero on its perimeter. It has one interior minimum, with closed anticlockwise streamlines: a broad northward interior flow returns southward along the west wall; flow is westward near the north edge and eastward near the south edge. The western parts of the streamlines are closely packed because the western boundary current is much faster.

The figure uses the exact ordinary differential equation solution, so it also remains consistent with both wall conditions when the asymptotic scale separation is only moderate. It depicts the stipulated dynamical model, not observed winds or a reconstruction of the source illustration.

In steady flow, the supplied Euler equations for an inviscid fluid givesTaking its scalar product with $\mathbf{u}$ gives $\mathbf{u}\cdot \nabla H=0$. Along a streamline with tangent $\mathbf{u}$, this is precisely zero derivative of $H$. Hence $H$ is constant along each streamline, with possibly different constants on different streamlines.

If the flow is also irrotational, $\mathbf{\omega }=0$ and thus $\nabla H=0$. Therefore $H$ has one constant value throughout each connected fluid region. The stronger conclusion uses vanishing of the entire gradient, not just its component along the velocity.

Milne-Thomson's circle theorem states that if $f(z)$ is a complex potential of a two-dimensional potential flow whose singularities lie outside $|z|=a$, thenhas the circle as a streamline, with the original exterior singularities unchanged. Additive constants in the potential are immaterial.

Take $f(z)=\Gamma (2\pi i{)}^{-1}\log (z-z_1)$, using the convention that positive $\Gamma$ is positive circulation. Reflection in the circle gives, up to a constant,The reflected vortex at $a^2/{\bar{z}}_1$ has strength $-\Gamma$, while the central term has strength $+\Gamma$. Their combined circulation around the cylinder is zero; this is the circulation normalization implicit in the requested equation.

The vortex moves with the velocity induced by the other singularities, excluding its own singular term. Since complex velocity is $dw/dz=u-iv$, its value at the vortex is $d{\bar{z}}_1/dt$. ThusWriting $z_1=r{e}^{i\theta }$ and multiplying this equation by $z_1$ givesIts real part implies $\dot{r}=0$, and its imaginary part givesThe radius is constant and the angular velocity is constant on that orbit. For positive $\Gamma$ the orbit is clockwise with this convention. Retaining the conjugate on the time derivative is essential for the sign.

For incompressible planar flow define the streamfunction by $u={\psi }_y$, $v=-{\psi }_x$. A velocity potential satisfies $u={\varphi }_x$, $v={\varphi }_y$ when the flow is irrotational; the differential condition is $v_x-u_y=0$. The complex potential of a two-dimensional potential flow is $w=\varphi +i\psi$, with derivative $w^{\prime }=u-iv$.

For $w=m\log (z-z_0)/(2\pi )$, its derivative is $m/[2\pi (z-z_0)]$. This gives outward radial velocity $m/(2\pi r)$, whose flux across a circle is $2\pi r\cdot m/(2\pi r)=m$. It is therefore the stated line source.

Superposition givesFor positive $U,m$, the stagnation points areThey lie on the real axis, one upstream of the source and the other downstream of the sink. Streamlines are level sets of $\psi =Uy+(m/2\pi )[\arg z-\arg (z-1)]$. The dividing streamline passes through both stagnation points and forms a Rankine oval enclosing the source and sink; outside, streamlines tend to uniform horizontal flow.

Writing $z=r{e}^{i\theta }$ gives $\varphi =\Gamma \theta /(2\pi )$ and $\psi =-\Gamma \log r/(2\pi )$. Constant streamfunction means constant radius, so streamlines are circles. The tangential velocity is $v_{\theta }=\Gamma /(2\pi r)$, and its anticlockwise circulation is ${\int }_0^{2\pi }{v}_{\theta }rd\theta =\Gamma$.

Use an opposite vortex at $z=-1$ as the image of the vortex at one:On $z=iy$ the two distances are equal, so $\psi =0$ and the wall is a streamline. Its velocity follows from $w^{\prime }=-i\Gamma /[\pi (z^2-1)]$: at the wall $u=0$ and $v=-\Gamma /[\pi (1+y^2)]$.

The Bernoulli equation gives $p-p_{\infty }=-\rho {\Gamma }^2/[2{\pi }^2(1+y^2{)}^2]$. Relative to the ambient pressure force, the fluid's excess force on the wall is $-\int (p-p_{\infty })dy$ in the positive $x$ direction. Using the supplied integral,Thus the wall is attracted toward the fluid and vortex. Subtracting the uniform ambient pressure isolates the finite force due to the vortex.

The momentum of a material volume $V(t)$ is ${\int }_{V(t)}\rho \mathbf{u}dV$. Newton's second law equates its time derivative to the pressure force and gravity:Apply the given Reynolds transport theorem componentwise to the left side, and the divergence theorem to the pressure term. ThenSince this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive density gives the Euler equations for an inviscid fluidAt a fixed physical time $t_0$, a streamline is a curve tangent to $\mathbf{u}(\cdot ,t_0)$, which may be parametrized by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t_0)$. The chain rule gives $df(\mathbf{x}(s),t_0)/ds=\mathbf{u}\cdot \nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.

A streamline is a curve whose tangent is parallel to the instantaneous fluid velocity field. In steady flow one may parametrize it by $d\mathbf{r}/ds=\mathbf{u}(\mathbf{r})$. Dotting the steady Euler equations with $\mathbf{u}$ eliminates the cross-product term, since $\mathbf{u}\cdot ((\nabla \times \mathbf{u})\times \mathbf{u})=0$. Hencewhich is Bernoulli's principle along each streamline.

For the supplied field, $xu+yv=0$, so along a streamlineWriting $r^2=x^2+y^2$, the tangential speed is $v_{\theta }=-br+c/r$ and the radial speed is zero. Both $r$ and $|\mathbf{u}{|}^2=(-br+c/r{)}^2$ are constant around each nonstagnant circular streamline. The Bernoulli quantity is constant there, so the pressure is constant on every such circle.

If a circle is wholly stagnant, the same conclusion follows directly from the steady Euler equations: on it $\mathbf{u}=0$ and $\nabla |\mathbf{u}{|}^2=0$, hence $\nabla p=0$. Equivalently the radial momentum equation gives $\nabla p=\rho (-br+c/r{)}^2{\mathbf{e}}_r/r$ and no angular pressure variation, including at zero tangential speed. Thus $p$ is constant on $x^2+y^2=a^2$ for every nonzero $a$, with no exception for a stagnation circle.

Orient $C$ counterclockwise around the solid body and let $\mathbf{n}$ point from the solid into the fluid. The pressure force is $-{\int }_{C}p\mathbf{n}ds$. Since $\mathbf{n}ds=(dy,-dx)$,For steady irrotational flow, Bernoulli's equation gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant term integrates to zero. The boundary is a streamline, so $dw=w^{\prime }dz$ is real there, giving $w^{\prime }dz=\overline{w^{\prime }}d\overline{z}$. Thereforewhich is Blasius force theorem with the specified orientation.

The prescribed stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. The zero-circulation solution from the Milne-Thomson circle theorem isOn $|z|=a$ the two terms are conjugates, so the boundary stream function is constant. The added term has no singularities in the exterior and does not change the prescribed growing far-field part. Additive constants are immaterial.

Differentiating givesThe coefficient of $z^{-1}$ in $(w^{\prime }{)}^2$ is $36{\alpha }^{2}c{a}^2(a^2+c^2)$. The residue theorem and Blasius force theorem giveThe zero-circulation convention is relevant: if an independently imposed circulation $\Gamma$ is allowed, adding $\Gamma \log z/(2\pi i)$ changes $F_y$ to $-3\rho \alpha \Gamma c^2$. Thus the claimed zero transverse force belongs to the standard circle-theorem construction, or to a single-valued exterior complex potential of a two-dimensional potential flow, which excludes that extra circulation.

The circle theorem says that if a complex potential $f$ has no singularities on or inside $|z|=a$ (and has the required analytic continuation there), thenadds image singularities inside the circle and makes it a streamline. On the circle $a^2/\bar{z}=z$, so $w=f(z)+\overline{f(z)}$ is real and its imaginary part is constant. This gives zero normal velocity. The reflected term is holomorphic where defined, as follows from its power series in $1/z$, and its singularities are precisely the reflected ones; the external original singularities are retained. This proves the construction, with an additive constant and any independently prescribed circulation handled separately.

For the fluid vortex use $f(z)=-i\Gamma \log (z-b)/(2\pi )$. Reflecting gives an opposite image at $d=a^2/b$ and also a term $-i\Gamma \log z/(2\pi )$, up to constants. The zero-circulation circle-theorem construction has both these terms. To impose the specified circulation $-\Gamma$ about the obstacle, add $+i\Gamma \log z/(2\pi )$, canceling the central term. ThusIt has the stated circulation and the same impermeable boundary, since the added central-fluid vortex streamfunction is constant on the circle.

For the given Blasius integral take the obstacle contour counterclockwise. Inside it the sole singularity ofis at $d$. Its residue is ${\Gamma }^2/[2{\pi }^2(d-b)]$, from the cross term. The residue theorem then givesTherefore $\boxed{F_x=\rho {\Gamma }^{2}b/[2\pi (b^2-a^2)],F_y=0}$. The force on the obstacle points toward the external fluid vortex.

On the circle put $\zeta =a{e}^{i\theta }$. The Joukowski map gives $z=2a\cos \theta$, so its image is the slit $S=[-2a,2a]$. Away from that circle, the inverse equation is ${\zeta }^2-z\zeta +a^2=0$. For $z\notin S$, neither root has modulus $a$, and their product is $a^2$, so exactly one root has modulus greater than $a$. Selecting that root defines the exterior inverse.

More explicitly, define $s(z)=\sqrt{z^2-4a^2}$ to be the branch analytic on $\mathbb{C}\setminus S$ normalized by $s(z)/z\to 1$ at infinity. It is positive for real $z>2a$ and negative for real $z<-2a$. ThenExistence and analyticity of this branch also follow from the exterior inverse: $J^{\prime }(\zeta )=1-a^2/{\zeta }^2$ is nonzero for $|\zeta |>a$, so the selected simple root varies analytically; $s=2g-z$ has the stated normalization. The exterior map is injective because $J({\zeta }_1)=J({\zeta }_2)$ with distinct roots would imply ${\zeta }_1{\zeta }_2=a^2$, impossible for two exterior points. Thus $g$ is a conformal bijection throughout $\mathbb{C}\setminus S$.

The proposed potential is a uniform stream plus its cylinder doublet and the circulation term. On $\zeta =a{e}^{i\theta }$, the first two terms sum to the real quantity $2Ua\cos (\theta -\alpha )$, while the imaginary part of the logarithmic term is the constant $-\Gamma \log a/(2\pi )$. Thus the cylinder is a streamline. At infinity the complex velocity tends to $U{e}^{-i\alpha }$ and the circulation is $\Gamma$, verifying both far-field conditions. Logarithms may be chosen locally; their derivative gives a single-valued velocity.

Composition therefore gives the plate potentialAt the edge $z=2a$, corresponding to $\zeta =a$, the physical complex velocity is $W^{\prime }(\zeta )/J^{\prime }(\zeta )$. Since the denominator has a simple zero, finiteness requireswhich is exactlyWith this cancellation, the limiting ratio is $W^{\prime \prime }(a)/J^{\prime \prime }(a)$. The Kutta value of the circulation gives $W^{\prime \prime }(a)=2U\cos \alpha /a$ and $J^{\prime \prime }(a)=2/a$. Hence the finite edge velocity is

For the approximate similarity profile, $F^{\prime }=A\mathrm{erf}(\xi /\sqrt{2})$, and thereforeFor the prescribed undisturbed horizontal streams, a suitable choice is $D=0$, which gives zero transverse velocity at both far-field limits. Then $V=-\Delta U\sqrt{CS}B{e}^{-{\xi }^2/2}$ inside the layer. The nonzero internal crossflow is consistent with the streamwise widening of the velocity transition.

Another possible convention is to make the line $y=0$ a streamline, imposing $V(0)=0$. This requires $D=-B$, equivalently $F(0)=0$, and gives $V(\pm \infty )=\Delta U\sqrt{CS}B$. It therefore requires a uniform outer crossflow relative to the coordinates used for the centred error function profile. A single constant cannot impose both zero far-field crossflow and $V(0)=0$ unless $\Delta U=0$. The physical lateral boundary conditions, or the chosen material centreline, determine which convention is appropriate. Since $B,D=O(r)$, these transverse velocities are of order $\Delta U^2/U_c$ for fixed $C,S$.

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]

The region bounded by the dividing streamlines of a two-dimensional point source superposed on uniform flow. Its stagnation point is $-Q/(2\pi U)$ upstream and its far downstream width is $Q/U$. The boundary encloses fluid originating at the source rather than a rigid obstacle; the upstream stagnation point is reached only asymptotically by the upstream source trajectory.

A uniform planar stream combined with an equal source and sink produces a closed dividing streamline enclosing them. Outside it the flow approaches the uniform stream; inside it streamlines connect the source and sink. For positive $U,m$ and separation one, the complex velocity is $U-m/[2\pi z(z-1)]$ and the stagnation points are $[1\pm \sqrt{1+2m/(\pi U)}]/2$. Both lie on the dividing streamline, which has zero streamfunction with consistent logarithm branches.

Using $\mathbf{u}=(-{\psi }_y,{\psi }_x)$, the displayed stream function gives ${\mathbf{u}}_0=(\sin x\cos y,-\cos x\sin y)$. Its streamlines are level sets of ${\psi }_0$. At $(n\pi ,m\pi )$, the Jacobian matrix is $\mathrm{diag}((-1{)}^{n+m},-(-1{)}^{n+m})$, so the points are saddle equilibria. At $(\pi /2+n\pi ,\pi /2+m\pi )$ the eigenvalues are $\pm i$, giving center equilibria surrounded by closed streamlines. The grid lines $x=n\pi$ and $y=m\pi$ form heteroclinic orbits between adjacent saddles; circulation alternates between cells.