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For uniform flow $\mathbf{U}$ plus a line vortex of counterclockwise circulation $\Gamma$, the outward advective momentum flux around a large circle is $\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The Bernoulli equation supplies an equal pressure contribution. Their sum is the force on the fluid; the force on an obstacle per unit span is its negative, $\rho \mathbf{U}\times \mathbf{\Gamma }$. Retaining only the advective term loses half the lift and obscures its sign.

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]

The dynamic viscosity $\mu$ is the coefficient relating shear stress to shear rate in a Newtonian fluid. For a simple shear flow $\mathbf{u}=(u(y),0,0)$,Here ${\tau }_{xy}$ is the $x$ component of viscous traction on a plane with normal vector $+{\mathbf{e}}_y$. The shear rate $\dot{\gamma }=du/dy$ has units ${\mathrm{s}}^{-1}$. Thus dynamic viscosity measures the stress required to sustain a given rate of deformation. It is the shear coefficient; the separate coefficient for volume change is volume viscosity. ^[ref]

For a worked Couette flow example, place a Newtonian fluid of constant dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$ between parallel plates separated by $h=0.0010\mathrm{m}$. Hold the lower plate fixed and move the upper plate at $U=0.020\mathrm{m}{\mathrm{s}}^{-1}$. Assume steady flow, no streamwise pressure gradient, and the no-slip boundary condition. ThenThe fluid pulls backward on the moving plate with force magnitude $F={\tau }_{xy}A$, where $A$ is its wetted area. For $A=0.010{\mathrm{m}}^2$ this is $0.020\mathrm{N}$. Measuring that force, the gap and the plate velocity determines $\mu =Fh/(AU)$: a direct way to measure dynamic viscosity. The linear profile and opposing wall tractions are the standard plane Couette flow solution. ^[ref]

The velocity gradient matters, rather than the velocity itself. Adding the same constant velocity to both plates and the fluid leaves $du/dy$, and hence shear stress, unchanged. In uniform flow, every layer translates together and $du/dy=0$. More generally, even a nonzero velocity gradient need not generate viscous stress: solid-body rotation has zero strain-rate tensor, as the viscous stress tensor example shows. Thin gaps amplify shear stress for a fixed plate-speed difference, making dynamic viscosity useful for predicting the force needed to move lubricated surfaces.

In a planar potential flow, superpose a positive-$x$ uniform flow of speed $U$ and a line source of strength $Q=2\pi m$ at the real point $b>1$. Introduce an impermeable unit circular cylinder with zero circulation. The Milne-Thomson circle theorem gives the complex potential of a two-dimensional potential flowThe displayed horizontal force per unit length acts on the cylinder, and $F_y=0$ by reflection symmetry. To prove it, let $q=w^{\prime }$ be the complex velocity. At infinity, $q=U+m/z+O(z^{-2})$, so the sum of all finite residues of $q^2$ is $2Um$. At the external pole $b$, write $q=m/(z-b)+H(z)$; thenThe residue theorem and the counterclockwise Blasius force theorem give $F_x-i{F}_y=2\pi \rho m[H(b)-U]$, proving the formula. With $m>0$ and $U>0$, its sign changes at $b=(m+\sqrt{m^2+4U^2})/(2U)$.

The Kutta condition selects the circulation of an idealized potential flow past a body with a sharp trailing edge by requiring finite limiting velocity at that edge. For the Joukowski map $z=\zeta +a^2/\zeta$, the trailing edge $z=2a$ corresponds to a simple zero of $dz/d\zeta$ at $\zeta =a$. If $W$ is the complex potential in the circle plane, then $dw/dz=W^{\prime }(\zeta )/(1-a^2/{\zeta }^2)$ is finite there only if $W^{\prime }(a)=0$. For a uniform flow of speed $U$ at angle $\alpha$ and a point vortex of circulation $\Gamma$, this gives $\Gamma =-4\pi Ua\sin \alpha$, and cancellation of the simple zeros gives the tangential trailing-edge velocity $U\cos \alpha$. The opposite endpoint can retain a leading-edge singularity.

For planar incompressible flow, $u_x+w_z=0$. Locally this permits a streamfunction with $u={\psi }_z$, $w=-{\psi }_x$, because equality of the mixed derivatives enforces incompressibility. The remaining component of the curl is $u_z-w_x={\psi }_{zz}+{\psi }_{xx}$, so irrotational flow requires Laplace's equation for $\psi$.

Put $K=\pi /b$. The sinusoidal contribution to the supplied streamfunction has ${\psi }_{xx}=U_{0}a{K}^2\cos (Kx)e^{-Kz}$ and ${\psi }_{zz}=-U_{0}a{K}^2\cos (Kx)e^{-Kz}$; the linear term is also harmonic. HenceAs $z\to \infty$, these approach the specified uniform flow, $(u,w)=(U_0,0)$.

The impermeability kinematic boundary condition is $w=u{z}_s^{\prime }(x)$ on $z=z_s(x)$, equivalently $\psi$ is constant on the wall. For $aK\ll 1$, evaluating the perturbed velocities at $z=0$ gives $w=-U_{0}aK\sin (Kx)=U_0{z}_s^{\prime }$ to first order. The residual is of order $U_0(aK{)}^2$. Equivalently $\psi (x,z_s)=U_{0}a\cos (Kx)[1-e^{-Ka\cos (Kx)}]=O(U_{0}K{a}^2)$. The surface velocity isThe first expression is the supplied approximate field evaluated exactly at the wall; its accuracy as a wall-flow solution is still only first order in $a/b$.

A uniform flow with physical velocity $(U\cos \alpha ,U\sin \alpha )$ has complex velocity $U{e}^{-i\alpha }$ and incident potential $f(\zeta )=U{e}^{-i\alpha }\zeta$. Apply the Milne-Thomson circle theorem with $R=1$:Indeed, on $\zeta =e^{i\theta }$ it is $W=2U\cos (\theta -\alpha )$, which is real, so the boundary is a streamline. At infinity, $W^{\prime }(\zeta )\to U{e}^{-i\alpha }$, verifying the specified incoming uniform flow.

The interface is material. Applying the kinematic boundary condition to $y-\eta (x,t)=0$ gives, exactly,The disturbance velocities and $\eta$ are small. Products of two disturbance quantities are second order, and a Taylor expansion moves their evaluation to $y=0$ at linear order. HenceThe extra upper-fluid term is advection of the displaced interface by the base uniform flow.

Use the Unsteady Bernoulli equation separately on each side, with constant mass density $\rho$. There are no restoring terms specified at this internal interface, and its pressure is continuous. Choose the time-dependent potential gauges so that the undisturbed pressure is $p_0$ on both sides. Subtracting the base values from Bernoulli's principle gives, at linear order,The quadratic disturbance-velocity terms have been discarded, while the cross term with the base velocity is first order and must remain. Equating the perturbation pressures yields the third condition:

The incident uniform flow $U{\mathbf{e}}_z$ has spherical components $u_r=U\cos \theta$, $u_{\theta }=-U\sin \theta$. Integrating the Stokes streamfunction formulas, with the additive constant chosen to vanish on the axis, givesIt is regular at the origin. Applying the sphere theorem for a Stokes streamfunction yieldsThe resulting velocity components areTo find the velocity potential, integrate ${\varphi }_r={\tilde{u}}_r$ and verify $r^{-1}{\varphi }_{\theta }={\tilde{u}}_{\theta }$. This givesThe first term is the incident uniform flow, and the second is a potential dipole located inside the excluded sphere. At $r=a$, the radial velocity vanishes, while the far-field velocity tends to $U{\mathbf{e}}_z$. The poles are stagnation points; the surface tangential speed reaches $3U/2$ at the equator. These formulas describe uniform potential flow around a fixed sphere.

Assume $q(z)=w^{\prime }(z)$ is single-valued and a holomorphic function between $C$ and a sufficiently large enclosing circle $|z|=R$. There are no fluid singularities in that region, so Cauchy's integral theorem permits contour deformation:Use the exterior Laurent series about zero, valid for $|z|$ larger than every finite singularity:It converges uniformly on the chosen circle, permitting termwise contour integration. For integer $n$, parameterization by $z=R{e}^{i\theta }$ givesOnly the $z^{-1}$ coefficient survives, provingThus if $b_1=\alpha +i\beta$, then $\Gamma =-2\pi \beta$ and $M=2\pi \alpha$. A uniform flow contributes a constant to $q$ and changes neither integral. The Laurent series here is an exterior expansion; it need not converge in a punctured neighborhood of zero if other singularities intervene.

For planar incompressible flow and irrotational flow, combine a velocity potential and a streamfunction into the complex potential of a two-dimensional potential flow:The chosen conventions giveThese are the Cauchy-Riemann equations, so $w$ is a holomorphic function wherever the flow is regular and a consistent local potential is chosen. Its complex derivative is the complex velocityA spatially uniform flow has $w=(U-iV)z+$ a constant. The following contour integral identities also apply to nonuniform potential flow, including the specified line vortex. A complex potential of a two-dimensional potential flow may be multivalued around a source or vortex even though its derivative is single-valued.

Use $w=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so its complex velocity is $q=w^{\prime }=u-iv$. The incident uniform flow and line source haveThe coefficient of the complex logarithm is one because a line source of strength $Q$ has complex potential of a two-dimensional potential flow $(Q/(2\pi ))\log (z-b)$, and here $Q=2\pi$. Applying the Milne-Thomson circle theorem to the unit circle givesSince $1/z-b=-b(z-1/b)/z$, locally consistent branches of the complex logarithm giveChanges of branch only alter the constant in a local expression; $w^{\prime }$ is single valued away from its singularities. The added terms represent a two-dimensional potential doublet at the origin, an image line source at $1/b$, and a compensating line sink at the origin. The image line source and line sink have equal and opposite strengths, so no net volume flux enters the cylinder.

On $|z|=1$, the Milne-Thomson circle theorem makes $w=f(z)+\overline{f(z)}$ up to a constant. Consequently the streamfunction is constant there, establishing the no-penetration boundary condition. All line source strengths are real, and the two-dimensional potential doublet has zero circulation, so the circulation around the cylinder is zero. There is no additional line vortex term.

With polar unit vectors ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, the second term is $\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$. It is tangent to circles, independent of $z$, and has zero divergence and zero vorticity away from the axis. Its singular axis carries a line vortex; adding the constant velocity field gives a uniform flow plus that line vortex.

The counterclockwise circulation around a positively oriented loop is $\oint \mathbf{u}\cdot d\mathbf{r}$. For a radius-$R$ circle, the uniform flow contributes zero and the line vortex contributesAny loop winding once around the axis has the same circulation, by Stokes theorem in the nonsingular intervening region.

On $C_R$, $\mathbf{n}={\mathbf{e}}_r$, $\mathbf{u}\cdot \mathbf{n}=U\cos \theta$, and $dl=Rd\theta$. ThusIn fact this equality holds for every $R>0$ for the stated field, since the remaining vector integral is $(0,\pi ,0)$.

This is the advective part of the momentum flux. To obtain the force, include pressure. For steady irrotational flow, the Bernoulli equation givesThe constant terms integrate to zero against $\mathbf{n}$; the remaining pressure contribution is $\oint p\mathbf{n}dl=\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The total outward momentum flux plus pressure term is therefore $\rho \mathbf{\Gamma }\times \mathbf{U}$, the force of the obstacle on the fluid. The opposite force, of fluid on the obstacle per unit span, isThis is the Kutta–Joukowski theorem, with positive $\Gamma$ defined counterclockwise: positive $U$ and $\Gamma$ give force in the negative $y$ direction. For an actual two-dimensional aerofoil, faster-decaying far-field terms do not change this limit. The displayed advective integral supplies half the lift; pressure supplies the other half.

Use the convention that positive vortex strength is counterclockwise and complex velocity is $u-iv$. A complex potential of a two-dimensional potential flow for the superposed uniform flow and two point vortices isLocal choices of logarithm branch change no velocity. Differentiation givesFor nonzero vortex strength and $U>0$, its two zeros areThe number under the square root is nonzero, so these are distinct stagnation points, neither at a vortex.

A free point vortex moves with the velocity induced by the other vortex and the background, omitting its own singular field. At either vortex, the other induces vertical velocity $-\Gamma /(4\pi a)$, while the background contributes $(U,0)$. Their velocities are therefore equal. Their separation stays $(2a,0)$, making this induced velocity constant for all time. Both travel along parallel straight lines with

On $\zeta =a{e}^{i\theta }$, the Joukowski map gives $z=2a\cos \theta$, so the image is the segment $S=[-2a,2a]$. Solving the resulting quadratic gives the inverse candidatesOn $D=\mathbb{C}\setminus [-2a,2a]$, choose the analytic square root $s(z)$ of $z^2-4a^2$ characterized by $s(z)/z\to 1$ at infinity. One explicit definition iswhere the square root on the right is the principal square root. Its argument never lies on the nonpositive real axis for $z\in D$, so this defines the required holomorphic function. Then $g(z)=(z+s(z))/2\sim z$.

The two quadratic roots have product $a^2$. For $z\notin S$, neither root lies on $|\zeta |=a$, since that circle maps into $S$. Consequently exactly one root lies outside the circle, and it is the branch $g$ chosen at infinity. Also, if two exterior points have the same image, thenwhose second factor cannot vanish. Thus the Joukowski map is a conformal map from $|\zeta |>a$ onto $D$, and $g$ is conformal throughout $D$. The two endpoints are branch points and are excluded.

The complex potential in the circle plane is the sum of the uniform flow term $U{e}^{-i\alpha }\zeta$, its circle-image term $U{a}^2{e}^{i\alpha }/\zeta$, and the point vortex term $-i\Gamma \log \zeta /(2\pi )$. On $\zeta =a{e}^{i\theta }$, the first two sum to $2Ua\cos (\theta -\alpha )$, which is real, while the imaginary part of the vortex term is the constant $-\Gamma \log a/(2\pi )$. Hence the circle is a streamline, giving no normal flow. At infinity the first term gives the required uniform flow; the logarithmic term supplies circulation $\Gamma$. The potential may be multivalued by a real constant, but its velocity is single-valued.

Conformal transport therefore givesIn the frame where the plate is horizontal, its length is $4a$ and the stream has angle $\alpha$; rotating the frame gives the equivalent incidence description. The complex velocity isAt the trailing edge $z=2a$, corresponding to $\zeta =a$, the denominator has a simple zero. Finite velocity requiresso the Kutta condition isWith this choice, cancellation of the common zero givesThus the limiting velocity has zero transverse component and is tangent to the plate.

Use the convention that positive vortex strength is counterclockwise and complex velocity is $u-iv$. A complex potential of a two-dimensional potential flow for the superposed uniform flow and two point vortices isLocal choices of logarithm branch change no velocity. Differentiation givesFor nonzero vortex strength and $U>0$, its two zeros areThe number under the square root is nonzero, so these are distinct stagnation points, neither at a vortex.

A free point vortex moves with the velocity induced by the other vortex and the background, omitting its own singular field. At either vortex, the other induces vertical velocity $-\Gamma /(4\pi a)$, while the background contributes $(U,0)$. Their velocities are therefore equal. Their separation stays $(2a,0)$, making this induced velocity constant for all time. Both travel along parallel straight lines with

In a sufficiently uniform flow the Stokes force is $F=\gamma v$ toward trap 2. Its work over separation $d$ lowers the second effective well by $Fd$. Equal effective free energies require $\gamma vd=k_{B}T\log 2$. Henceat 298 K. The temperature was not specified, so the numerical result scales as $T/298\mathrm{K}$. This approximation treats a weak uniform drag as a constant tilt and neglects changes in barrier shape and hydrodynamic interactions with boundaries.

For a fixed impermeable sphere of radius $a$ and incident uniform flow $U{\mathbf{e}}_z$, the velocity potential is the displayed expression and the Stokes streamfunction is $\Psi =(U/2)(r^2-a^3/r){\sin }^2\theta$. Differentiation gives $u_r=U(1-a^3/r^3)\cos \theta$ and $u_{\theta }=-U(1+a^3/(2r^3))\sin \theta$. Thus the no-penetration boundary condition holds, the poles are stagnation points, and the equatorial surface speed is $3U/2$. The disturbance is a potential dipole decaying as $r^{-3}$ in velocity. The sphere theorem for a Stokes streamfunction constructs this solution from ${\Psi }_0=U{r}^2{\sin }^2\theta /2$.

For $0<a<1$, the Joukowski map $z=\zeta +a^2/\zeta$ maps the exterior unit circle to an ellipse of horizontal and vertical semiaxes $1+a^2$ and $1-a^2$. The Milne-Thomson circle theorem gives $W(\zeta )=U(e^{-i\alpha }\zeta +e^{i\alpha }/\zeta )$ for a uniform flow inclined by $\alpha$. The chain rule produces the displayed complex velocity in the physical plane. Since $z\sim \zeta$, the far-field flow retains its speed and direction. Its two stagnation points areobtained from $\zeta =\pm e^{i\alpha }$. The map derivative is nonzero on and outside the unit circle, so these zeros are preserved without introducing a boundary singularity.

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

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

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