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In a steady adiabatic inviscid fluid with a conservative body force, the Bernoulli invariant is the displayed specific energy, constant along each streamline. Dotting the Euler equations with velocity and using the specific enthalpy identity gives $\mathbf{u}\cdot \nabla b=0$. In ideal magnetohydrodynamics, magnetic work need not vanish; the planar ideal-MHD Bernoulli invariant includes the corresponding electromagnetic correction.

The Bernoulli principle relates pressure, kinetic energy per unit volume and Newtonian gravitational potential along a streamline of steady flow with inviscid flow. For constant mass density, the Bernoulli equation gives $p/\rho +u^2/2+\Phi$ constant along that streamline. Constancy across different streamlines requires further hypotheses, such as irrotational flow.

Precisely, suppose a fluid of constant positive mass density obeys the Euler equations for an inviscid fluid, its velocity field and pressure are steady, and the body force per unit mass is $-\nabla \Phi$. ThenEvery term has units of energy per unit mass. For a uniform gravitational field, Newtonian gravitational potential is $\Phi =gz$, up to an additive constant. The pressure term represents work by surrounding fluid, so acceleration can exchange pressure work with kinetic energy. Dotting the steady Euler equations with $\mathbf{u}$ gives the displayed directional derivative: this is why constancy follows a streamline. ^[ref]

Continue the horizontal water constriction example from incompressible flow. Assume negligible viscous losses, steady flow and approximately uniform flow at both measurement sections, so the streamline speeds there equal the mean speeds $0.20$ and $0.80\mathrm{m}{\mathrm{s}}^{-1}$. Equal height cancels the Newtonian gravitational potential terms. The ideal pressure drop isThis is the Venturi effect in its ideal limit: the gain in kinetic energy density is supplied by pressure work. Cambridge's treatment uses this relation to connect a measured pressure difference to volumetric flow rate. ^[ref]

The $300\mathrm{Pa}$ result is conditional, rather than a prediction of the actual viscous pipe. Mass conservation fixes $Q$ and the ratio of mean speeds even with viscosity, but nonuniform velocity profiles require an integrated energy balance, and wall friction dissipates mechanical energy. A pump would supply additional work. In particular, constant mean speed in a straight pipe does not imply constant pressure: Poiseuille flow needs a continuing pressure gradient to overcome viscosity. ^[ref]

For the circular topographic anticyclone in a positive uniform current $U$, the opposing disturbance speed is at most $Ca=fϵa/2$. Below the displayed threshold $u$ remains positive everywhere, so no streamline can close. Above it an interior stagnation point at $(0,-U/C)$ is a center, while the exterior point $(0,-C{a}^2/U)$ is a saddle. At equality the two meet at the rim. This is the onset criterion; once closed contours form, the potential vorticity of trapped fluid is not determined solely by its present upstream conditions.

For an incompressible irrotational two-dimensional potential flow, an analytic complex potential of a two-dimensional potential flow combines the velocity potential $\Phi$ and streamfunction $\Psi$. The Cauchy-Riemann equations give $u={\Phi }_x={\Psi }_y$ and $v={\Phi }_y=-{\Psi }_x$, hence $W^{\prime }(z)=u-iv$. Curves of constant $\Psi$ are streamlines. An analytic coordinate change $z=f(\zeta )$ transforms the complex velocity by $dW/dz=(dW/d\zeta )/f^{\prime }(\zeta )$ wherever $f^{\prime }\ne 0$.

For a fixed-composition inviscid fluid, specific enthalpy satisfies $dw=Tds+dp/\rho$. Combining this identity with the vorticity cross-product identity gives the displayed equation, with $\mathbf{\omega }=\nabla \times \mathbf{u}$ and $\epsilon =u^2/2+\Phi +w$. It exposes entropy gradients as the thermodynamic source of a mismatch between Bernoulli function gradients and the vorticity term. In a steady adiabatic process, its projection along a streamline makes $\epsilon$ constant there.

Dynamic pressure is the bulk kinetic energy per unit volume associated with fluid motion at speed $U$ in a specified reference frame:Here $\rho =m/V$ is mass density. The word "pressure" identifies its units and its use as a pressure scale; $q$ is not an extra isotropic stress to add to the local static pressure in every surface traction. A stationary fluid can have large static pressure with $q=0$. Conversely, knowing $U$ and $\rho$ fixes $q$ without fixing static pressure. The definition remains useful in viscous fluid flow and compressible flow, even when the simplest Bernoulli equation is inapplicable. ^[ref]

For stipulated mass density ${\rho }_{\infty }=1.20\mathrm{kg}{\mathrm{m}}^{-3}$ and reference speed $U_{\infty }=20\mathrm{m}{\mathrm{s}}^{-1}$, $q_{\infty }=240\mathrm{Pa}$. At $40\mathrm{m}{\mathrm{s}}^{-1}$ with the same mass density, $q_{\infty }=960\mathrm{Pa}$. This quadratic scaling makes dynamic pressure a useful normalization for the lift coefficient, drag coefficient and pressure coefficient; it predicts four times a particular force only if its corresponding coefficient remains unchanged.

The connection with stagnation pressure requires hypotheses. For steady flow, constant mass density, inviscid flow, and gravity as the only body force, the Bernoulli equation gives, along one streamline,Here $p$ is static pressure, $g$ is gravitational acceleration, and $z$ is height. Bring that flow to rest without losses at the same height and the limiting static pressure is the stagnation pressure $p_0$:There must be no intervening pump supplying work or other energy input. Constancy along one streamline does not assert equal $B$ on all streamlines; irrotational flow in a connected region supplies that additional conclusion. Changing height requires retaining $\rho gz$. In compressible flow the stopping pressure difference generally differs from $\rho U^2/2$, and viscous dissipation can reduce recovered stagnation pressure. ^[ref]

A Pitot-static tube turns this relation into a measurement: its upstream-facing opening senses stagnation pressure, while separate side openings sample static pressure. Connect these to the positive and negative ports of a differential pressure transducer or manometer. Under the hypotheses above, the measured difference $\Delta p$ givesThus the stipulated $240\mathrm{Pa}$ and $1.20\mathrm{kg}{\mathrm{m}}^{-3}$ imply $20\mathrm{m}{\mathrm{s}}^{-1}$, not an independently measured speed. Align the probe with the incoming flow and keep its passages clear: blockage and pinched tubing invalidate the comparison. At very low speed, the difference can be smaller than instrument error; at high Mach number, compressibility and possibly a shock wave require a different reduction. ^[ref]

A useful control is differential-pressure calibration. For a digital manometer, zeroing establishes the reading at equal port pressures; it does not establish the gain away from zero. The manufacturer's Series 475 Handheld Digital Manometer: Specifications - Installation and Operating Instructions by Dwyer Instruments, Inc. directs the user to vent both ports to the atmosphere before zeroing. Its stated accuracy is a fraction of full scale, not of the reading, and depends on instrument temperature. Select a suitable range and record the instrument variant and units; the manual covers several ranges.^[ref]

An independent physical check can use a liquid-column manometer: apply a stable pressure difference to both instruments and compare the digital reading with the hydrostatic pressure inferred from the vertical level difference. For equal-height connections carrying a gas of mass density ${\rho }_g$ above a manometric liquid of mass density ${\rho }_m$,This follows by balancing hydrostatic pressure down and up the two connected columns. Use vertical height, even if a tube is inclined. Known mass densities, measured height and gravitational acceleration set the comparison scale. Check several differences across the working range and reverse the port connections to check the sign. This is a proposed independent check, not a claim that the cited teaching experiment performed a full instrument calibration. Recheck zero after reconnecting the sealed tubing; a zero adjustment alone cannot detect a blocked passage or an incorrect scale.

Electric flux measures how much electric field crosses a chosen oriented surface, with a sign distinguishing the two crossing directions. A strong electric field running along a surface contributes nothing to its electric flux: only the normal component counts. This lets Gauss's law turn a potentially complicated electric field into one number recording the net enclosed electric charge.

For a surface $S$ with chosen unit normal $\mathbf{n}$, the electric flux is the surface integralHere $dA$ is positive surface area, while $d\mathbf{S}$ is the vector area element. The dot product makes electric flux a signed scalar, rather than a vector like the electric field. Its SI unit is $\mathrm{N}{\mathrm{m}}^2/C$: electric field measured using the newton per coulomb, multiplied by surface area in square metres. Reversing the orientation of a surface replaces $\mathbf{n}$ by $-\mathbf{n}$ and reverses the electric flux. Calling it electric flux does not imply that electric charge is moving through the surface. ^[ref].

For an open patch in a plane, with surface area $A$ and a constant electric field, the unit normal is constant and the surface integral simplifies toThe angle $\theta$ is between the electric field and the chosen unit normal, not between the electric field and the plane. The cosine describes the signed projected surface area seen by the electric field. Thus electric flux is positive for $0\le \theta <90^{\circ }$, zero at $90^{\circ }$, and negative for $90^{\circ }<\theta \le 180^{\circ }$. ^[ref].

For example, take $E=100N/C$ and $A=0.02{\mathrm{m}}^2$, so $EA=2\mathrm{N}{\mathrm{m}}^2/C$. Three choices of angle giveKeeping the $60^{\circ }$ patch fixed and reversing its unit normal gives $120^{\circ }$ and changes $+1$ to $-1\mathrm{N}{\mathrm{m}}^2/C$. The electric field and physical patch have not changed. If the electric field is instead tangent to the patch, its dot product with the unit normal vanishes: electric flux is zero even though $E=100N/C$.

For a closed surface enclosing a volume $V$, choose the outward unit normal. Gauss's law givesHere ${\epsilon }_0$ is the vacuum permittivity, and $Q_{\mathrm{enclosed}}$ is the signed total electric charge inside. The electric field in the surface integral includes all sources, including external electric charges. Assume no ideal point charge lies on the integration surface. Zero enclosed electric charge implies zero total electric flux; it does not imply zero electric field everywhere on the closed surface or inside it. Positive and negative contributions to the surface integral can cancel. ^[ref].

To see the cancellation with sources inside, place point charges $+q$ at $(-a,0,0)$ and $-q$ at $(a,0,0)$, where $q>0$, $a=0.30\mathrm{m}$, and both lie inside the sphere of radius $R=1\mathrm{m}$ centered at the origin. This separated pair has the electric dipole geometry. Its enclosed electric charge is $q-q=0$, so its total outward electric flux is zero. At the point $P=(0,0,R)$ on the sphere, Coulomb's law and the superposition principle give the individual electric fieldsThe $z$ components cancel and the positive $x$ components add:For the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $q=1\mathrm{nC}$, this is about $4.740N/C$ along positive $x$. At $P$, the outward unit normal is $\hat{\mathbf{z}}$, so this nonzero electric field is tangent to the sphere. At other points of the sphere, its normal component can be positive or negative. The source-field formula applies away from the two point charges, where the electric field has zero divergence; it is singular at the sources. ^[ref].

The diagram shows streamlines of the three-dimensional electric field in the section $y=0$. The circle is a cross-section of the sphere, not the integration surface: Gauss's law uses the entire three-dimensional surface integral. A two-dimensional line integral of outward field around that circle would be a different quantity, with units $\mathrm{N}m/C$ rather than the electric flux units $\mathrm{N}{\mathrm{m}}^2/C$.

An equipotential curve is a planar level curve of a scalar potential. Where the gradient is nonzero, its tangent is perpendicular to that gradient. In potential flow, it is therefore perpendicular to the velocity field; this distinguishes equipotential curves from streamlines.

A free streamline is a streamline that forms an unknown boundary of a fluid region, typically against a region of prescribed pressure. In steady inviscid potential flow with gravity and surface tension neglected, the Bernoulli equation makes its speed constant when its pressure is constant. It is simultaneously a level set of the stream function. These two boundary conditions take particularly simple forms in the complex potential of a two-dimensional potential flow and the hodograph plane.

The method of images enforces impermeability at a plane wall by reflecting a line vortex across it and reversing its circulation. The two logarithmic stream functions cancel on the wall, which is consequently a streamline. For a vortex of circulation $\Gamma$ at $(0,b)$ above $y=0$, the image of circulation $-\Gamma$ at $(0,-b)$ induces velocity $(\Gamma /(4\pi b),0)$ at the real vortex. The signed drift parallel to the wall is therefore $\Gamma /(4\pi b)$.

Laminar flow is a regime of fluid flow in which the motion remains organized, with neighboring fluid parcels following smooth, coherent paths rather than the irregular interacting fluctuations characteristic of turbulence. In a straight pipe, the simplest example consists of coaxial layers sliding past one another. This describes the character of the motion, not a requirement that every parcel have the same velocity. Nor does it require steady flow: a smooth profile can evolve in time as viscosity transmits motion from a newly moving wall into a resting fluid. The unsteady wall-driven solution provides an explicit example. ^[ref]

The Reynolds pipe-and-dye experiment makes this distinction visible. A thin injected dye band can remain coherent in laminar flow, whereas irregular motion can fold it and carry dye across the pipe. However, molecular diffusion can spread dye even in laminar flow, and an inlet disturbance can move a coherent band without producing turbulence. A dye band is a streakline, not a measurement of the entire velocity field; in steady flow an ideal passive tracer's streakline coincides with a streamline. Reynolds explicitly distinguished a wandering band from one broken into eddies. ^[ref]

For a circular pipe of internal diameter $D$, define the mean axial velocity $U$ using the volumetric flow rate $Q$ and cross-sectional area $A$:Here $\rho$ is mass density, $\mu$ is dynamic viscosity and $\nu =\mu /\rho$ is kinematic viscosity. This particular Reynolds number uses the diameter and mean velocity, not the radius or centreline velocity. Small Reynolds number generally favors laminar flow, but a single universal critical Reynolds number does not distinguish it from turbulence in all geometries and experiments. Even in a given pipe, inlet conditions and the magnitude of disturbances affect the observed onset. Reynolds demonstrated this dependence by deliberately inserting a wire coil. ^[ref]

The quantitative example above assumes steady flow, incompressible flow, a Newtonian fluid with constant dynamic viscosity, a straight horizontal circular pipe of radius $R=D/2$, stationary walls satisfying the no-slip boundary condition, and fully developed flow. The last assumption means that the axial velocity profile no longer changes downstream. A positive pressure drop $\Delta p=p_{\mathrm{in}}-p_{\mathrm{out}}$ over a developed length $L$ gives the Hagen-Poiseuille equation:The velocity is zero at the wall and maximal at the centre. Averaging over the circular area, whose annuli have area $2\pi rdr$, gives $U=u(0)/2$; averaging along one diameter would give a different answer. These assumptions produce a parabola, rather than every laminar flow having a parabolic profile. ^[ref]

As a stipulated calculation, take $D=0.010\mathrm{m}$, $U=0.10\mathrm{m}{\mathrm{s}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =0.0010\mathrm{Pa}\mathrm{s}$ and $L=1.0\mathrm{m}$, and assume the developed laminar flow above. Then $\nu =10^{-6}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, ${\mathrm{Re}}_D=1000$, $u(0)=0.20\mathrm{m}{\mathrm{s}}^{-1}$, $Q=7.85\times 10^{-6}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ (about $0.471\mathrm{L}\mathrm{mi}{\mathrm{n}}^{-1}$), and $\Delta p=32\mu LU/D^2=32\mathrm{Pa}$. These are predictions from chosen parameters, not measurements or a claim that ${\mathrm{Re}}_D=1000$ guarantees laminar flow in every apparatus.

The beauty of this example is that a spatially varying velocity field still gives an exact relation between two accessible measurements, pressure drop and volumetric flow rate. It predicts the hydraulic resistance $\Delta p/Q=8\mu L/(\pi R^4)$. Doubling the radius therefore multiplies $Q$ by sixteen at fixed pressure drop, length and dynamic viscosity, provided the model remains applicable. This makes the laminar flow model useful for sizing narrow tubes or determining dynamic viscosity. It also explains why a persistent narrow dye band can be informative without implying that the fluid is moving as a uniform plug.

The incompressible planar field $(\alpha x-\beta y,\beta x-\alpha y)$ has the displayed stream function under $(u,v)=({\psi }_y,-{\psi }_x)$ and vorticity $2\beta$. In rotated coordinates $s=(x+y)/\sqrt{2}$, $t=(x-y)/\sqrt{2}$, $\psi =((\alpha -\beta )s^2-(\alpha +\beta )t^2)/2$. For positive parameters its nonstationary streamlines are closed ellipses if $\alpha <\beta$, hyperbolas and straight separatrices if $\alpha >\beta$, and parallel straight lines with a stationary diagonal if $\alpha =\beta$.

In the impermeable strip $x>0$, $0<y<a$, a line source of strength $Q$ at an interior point $c$ has the displayed complex potential of a two-dimensional potential flow, with locally consistent complex logarithm and $f(z)=\cosh (\pi z/a)$. The hyperbolic-cosine map of a half-strip sends the walls to the real axis. An equal image line source at $\overline{f(c)}$ makes that axis a streamline. Since $f^{\prime }(c)\ne 0$, the local singular part remains $Q\log (z-c)/(2\pi )$, proving preservation of strength. With no extra imposed flow, the complex velocity tends to $Q/a$ as $x\to \infty$, so the outgoing volume flux is $Q$. The source position $c$ has positive imaginary part; assuming it real would place it on a wall.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in a two-dimensional potential flow. Suppose an incident complex potential of a two-dimensional potential flow $f$ is analytic on and inside $|z|=a$, with its physical singularities outside that disk. Then the displayed $W$ has constant stream function on $|z|=a$. The proof is the identity $a^2/\bar{z}=z$ on the circle: there $W=f(z)+\overline{f(z)}$ is real. Hence the circle is a streamline and satisfies the no-penetration boundary condition. The reflected term has its image singularities inside the obstacle. Additional imposed circulation must be specified separately. The theorem is named after Louis Melville Milne-Thomson.

For a steady inviscid barotropic fluid in a conservative gravitational potential, define $h(\rho )=\int dp/\rho$. Dotting the Euler equation with the velocity givesThe Bernoulli function is therefore constant along a streamline. It balances kinetic energy, pressure work or enthalpy, and gravitational potential energy per unit mass. Different streamlines can have different constants; for an irrotational flow, the identity $\nabla H=\mathbf{u}\times (\nabla \times \mathbf{u})$ makes the constant common throughout a connected region.

For an incompressible fluid, $h=p/\rho$ up to a constant, giving $u^2/2+p/\rho +\varphi =C$. For a constant-entropy perfect-gas streamline, $h=\gamma p/[(\gamma -1)\rho ]$. Together with mass conservation, this explains conversion of enthalpy into kinetic energy in a nozzle and the need to include gravity in an escaping wind. Viscous losses, heating and entropy production require their corresponding energy balance. Across an adiabatic shock the total stagnation enthalpy can still be conserved, but the upstream isentropic pressure-density law cannot simply be continued through the entropy jump.

For a barotropic fluid, define $h(\rho )=\int dp/\rho$. Dotting the steady Euler equation with $\mathbf{u}$ givesThus the Bernoulli function $H=u^2/2+h+\psi$ is constant along each streamline. Different streamlines can have different constants; steady flow alone does not imply one spatially uniform $H$.

In a quasi-one-dimensional flow through area $A(z)$ with negligible gravity, the axial momentum equation and $dp=c_s^{2}d\rho$ giveMeanwhile the mass flow rate $\dot{M}=\rho uA$ is constant, so $d\ln \rho =-d\ln u-d\ln A$. Substitution proves the nozzle area-velocity relationA subsonic flow accelerates where $A$ decreases, whereas a supersonic flow accelerates where $A$ increases. A smooth accelerating transonic branch in a de Laval nozzle therefore crosses $u=c_s$ at a minimum-area throat. Suitable reservoir and downstream pressures are also required: a converging-diverging shape by itself does not exclude a wholly subsonic solution or a downstream shock.

For the stipulated constant-$K$ polytropic flow, $c_s^2=\gamma K{\rho }^{\gamma -1}$. At the throat $u_m=c_{s,m}$, soThe specific enthalpy is $h=\gamma K{\rho }^{\gamma -1}/(\gamma -1)=c_s^2/(\gamma -1)$. Evaluating $H$ at the throat and substituting $\rho =\dot{M}/(Au)$ elsewhere givesOn the outward supersonic branch, $A\sim C{r}^2$ lets the gas expand and its specific enthalpy tend to zero for $\gamma >1$. ConsequentlyFor $\gamma =5/3$, $u_{\infty }=2c_{s,m}$. Also $T\propto {\rho }^{\gamma -1}\propto r^{-2(\gamma -1)}$. These are the adiabatic nozzle terminal speed and expansion laws for a lossless escaping branch; a subsonic branch could instead approach finite pressure with $u\propto A^{-1}$. A pressure-shaped astrophysical jet is being approximated by a nozzle here: the assumed geometry does not by itself explain the confinement or account for magnetic forces or relativistic motion in an actual active galactic nucleus.

Assume an inviscid barotropic fluid acted on by a conservative body force $-\nabla \psi$. Define its barotropic enthalpy function byThe Euler equations for an inviscid fluid are then ${\partial }_t\mathbf{v}+(\mathbf{v}\cdot \nabla )\mathbf{v}=-\nabla (H+\psi )$. Using the vorticity $\mathbf{\omega }=\nabla \times \mathbf{v}$ and the vector identitygives ${\partial }_t\mathbf{v}+\nabla B=\mathbf{v}\times \mathbf{\omega }$, where $B=v^2/2+H+\psi$.

For steady flow, dotting this equation with $\mathbf{v}$ yields $\mathbf{v}\cdot \nabla B=0$. Thus Bernoulli's equation isThe constant need not agree between different streamlines. If the flow is also irrotational, the vector equation gives $\nabla B=0$, so one common constant applies throughout a connected flow region.

For an isentropic process with $p=K{\rho }^{\gamma }$, $H=\gamma p/[(\gamma -1)\rho ]$ up to an additive constant, equal to the thermodynamic specific enthalpy. For an externally maintained isothermal gas, $H=a^2\ln (\rho /{\rho }_{\mathrm{ref}})$; this pressure potential is not the constant thermodynamic enthalpy of a fixed-temperature ideal gas.

If steady flow is not assumed, a useful additional case is unsteady potential flow $\mathbf{v}=\nabla \varphi$. The same vector equation integrates spatially toGeneral unsteady barotropic flow does not possess the steady streamline constant without further conditions.

Set ${\partial }_t\mathbf{u}=0$ and $\nu =0$ in the Navier-Stokes equation. Dotting the resulting Euler equation with a displacement $d\mathbf{r}$ tangent to a streamline givesHere mass density $\rho$ is constant, and $B$ is constant along each streamline. The terms are kinetic energy per unit mass, the pressure work per unit mass available from pushing a parcel through the flow, and gravitational potential energy per unit mass. Their sum is conserved mechanical energy per unit mass; $p/\rho$ is not an additional independently conserved parcel energy. Different streamlines can have different $B$ unless, for example, the flow is also irrotational in a connected region.

Put the stagnation point at the origin and the local tangent plane at $z=0$. A regular Taylor expansion has no linear term because $\mathbf{u}(0)=0$. The impermeable tangent plane imposes ${\partial }_z\varphi =0$ at $z=0$ to leading order, excluding mixed quadratic terms $xz,yz$. For the illustrated axisymmetric approach, rotational symmetry in the tangent plane makes the $x^2$ and $y^2$ coefficients equal and removes $xy$. Finally the Laplace equation fixes the normal coefficient to minus twice the tangential one. Up to an irrelevant additive constant,For incoming flow from $z>0$, $W>0$. This is a local leading form under the stated symmetry, not the exact universal potential near every possible wall stagnation point; a generic tangential strain can have unequal principal rates.

A streamline solves $\dot{x}=2Wx$, $\dot{y}=2Wy$, $\dot{z}=-4Wz$, hence $x=A{e}^{2Wt}$, $y=B{e}^{2Wt}$, $z=C{e}^{-4Wt}$. Eliminating the parameter givesEach streamline lies in a meridional plane. With cylindrical radius $s=\sqrt{x^2+y^2}$, its shape is $z=C_1/s^2$. The symmetry axis and wall are limiting invariant streamlines. A particle exactly on the incoming axis approaches the stagnation point asymptotically.

On the surface, $u_r=0$ and $u_{\theta }=(3v/2)\sin \theta$. The only surface stagnation points are thereforeThere are no other stagnation points in the exterior: away from the axis the polar component is nonzero, and on the axis the radial component vanishes only at $r=R$.

Near the front point put $(X,Y,Z)=(x,y,z-R)$. Expanding the potential from solution givesso the linear velocity field is $(3vX/(2R),3vY/(2R),-3vZ/R)$. This agrees with solution for $W=3v/(4R)$.

At the rear point, using $Z=z+R$ instead, the quadratic term has the opposite sign: $W=-3v/(4R)$. Incoming axial flow and tangential outflow at the front become tangential inflow and axial outflow at the back. The local streamline geometry is unchanged while its direction reverses.