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Locally a U(1) connection whose gauge curvature has no $(2,0)$ or $(0,2)$ part can be written $A=\partial u+\bar{\partial }v$ by the Dolbeault-Poincaré lemma. Unitarity permits $v=-\bar{u}$. Writing $u=p+iq$, a gauge transformation removes $idq$ and leaves $A=\partial p-\bar{\partial }p$. Setting $f=-2p$ gives the displayed unitary form, whose gauge curvature is $\partial \bar{\partial }f$. Its remaining anti-self-duality of gauge curvature condition is the scalar Laplace equation. Alternatively a complex gauge transformation by $e^{-u}$ gives $A=\bar{\partial }(v-u)$; this second gauge generally does not preserve unitarity.

// Destination: complex-geometry.bigb

Antiplane shear has displacement normal to a plane and independent of that normal coordinate. In an isotropic linear elastic body, ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with their symmetric partners, are the only nonzero stress tensor entries. Static balance without body force reduces to the Laplace equation for $w$.

Consider a uniform-density cylinder with an axisymmetric radial Fourier mode $\eta =ϵe^{ikz}$. Its displaced boundary gives a first-order surface mass density $\rho \eta$, although the interior density perturbation vanishes. The gravitational Poisson equation reduces to Laplace equation away from the interface. Regularity internally and decay externally give ${\psi }_{\mathrm{in}}=A{I}_0(kr)e^{ikz}$ and ${\psi }_{\mathrm{out}}=B{K}_0(kr)e^{ikz}$. Continuity at $R$ and the derivative jump ${\psi }_{\mathrm{out}}^{\prime }-{\psi }_{\mathrm{in}}^{\prime }=4\pi G\rho ϵe^{ikz}$, together with $I_0(\kappa )K_0^{\prime }(\kappa )-I_0^{\prime }(\kappa )K_0(\kappa )=-1/\kappa$, give the title amplitude. The prime differentiates the modified Bessel function argument. This axial mode is different from an azimuthal mode on a gravitational potential of a displaced cylindrical interface.

For boundary data $f(u)$ on the sphere $r=a$, with $u=\cos \theta$, define its Legendre polynomial coefficients by $f_{\ell }=(2\ell +1){\int }_{-1}^{1}f(u)P_{\ell }(u)du/2$. Separation of variables in the Laplace equation gives radial factors $r^{\ell }$ and $r^{-\ell -1}$. Regularity at the origin selects the first for the interior Dirichlet problem; decay at infinity selects the second for the exterior problem. Matching each boundary coefficient gives the displayed series. For continuous boundary data the resulting harmonic functions attain those data, with convergence interpreted through their harmonic extension rather than requiring uniform convergence of the unweighted boundary series. The maximum principle for harmonic functions proves uniqueness in the bounded interior and decaying exterior classes.

The diagrams solve an explicitly bounded Dirichlet problem in a uniform charge-free medium of constant permittivity. Take a width scale $W=1$ and terminal electric potential difference $V=1$. Two zero-thickness ideal plate segments occupy $|x|\le W/2$ at $y=\pm d/2$, with electric potentials $\pm V/2$. Fix $\varphi =0$ on the entire rectangle $x=\pm 2W$, $y=\pm 1.5W$. This boundary is an additional enclosing equipotential, rather than a condition at infinity. The model is invariant along the omitted third direction. The connection between the electric potential, its gradient, and the charge-free Laplace equation follows from electrostatics.^[ref]

On a square mesh of spacing $h=W/160$, the finite difference method imposesat free nodes, while plate and outer nodes retain their prescribed electric potentials. There are $641\times 481$ nodes. Each plotted gap aligns exactly with the mesh; the smallest gap has 16 mesh intervals. The generators solve the resulting linear system by the conjugate gradient method, stopping when the maximum dimensionless five-point residual is below $2\times 10^{-10}$, and use central finite differences to approximate $\mathbf{E}=-\nabla \varphi$ away from plate nodes.

The discrete maximum principle keeps the computed electric potential between $-V/2$ and $+V/2$. Reflection checks require evenness in $x$ and oddness in $y$; halving $h$ tests discretization effects, while enlarging the outer rectangle tests sensitivity to its artificial boundary conditions. These are different checks: a small residual only establishes that the discrete equations were solved. Differentiating near zero-thickness endpoints is especially sensitive to the mesh, so the plots do not report a converged edge peak. No total electric charge, exact open-domain three-dimensional capacitance, or measured material property is extracted from this model.

For a bounded numerical check, compare $h=W/80$ with the plotted $h=W/160$ at midplane positions $x/W=0,0.25,0.40,0.50,0.60,0.80$. At $d/W=0.10$, the largest change in $-E_{y}d/V$ is approximately $0.0265$, at $x/W=0.50$; the other five sampled changes are below $0.00472$. At $d/W=0.25$, moving the outer rectangle from $\pm 2W\times \pm 1.5W$ to $\pm 3W\times \pm 2W$ while retaining $h=W/80$ changes these samples by at most approximately $0.00622$. These calculated sensitivities support the qualitative comparison and expose its limits; they are not experimental uncertainties or evidence that the artificial boundary has converged to infinity.

The capacitor gap aspect ratio compares the gap $d$ with a lateral plate width $W$. For rectangular plates of widths $W_x,W_z$, require both $d/W_x\ll 1$ and $d/W_z\ll 1$ for the leading parallel-plate capacitor approximation. Large area alone is insufficient: a long narrow strip can have large area and still have $d$ comparable with its smaller width. In addition, the observation point must be many gap widths from an edge. Reducing $d/W$ improves the interior approximation without making the edge itself disappear.

In a two-dimensional straight gap, write the electric potential as the infinite-plate solution plus a perturbation:The perturbation satisfies the Laplace equation and vanishes on the two ideal plates. Separation of variables in a semi-infinite strip, with $s$ the distance inward from its entrance, gives transverse modesThus edge disturbances have a natural penetration scale set by $d$. Modes forbidden by a particular symmetry have zero coefficients; this expression is an explanation of localization in an ideal strip, not a universal numerical correction to finite-plate capacitance. The outside boundary conditions determine the coefficients.

There is no single aspect-ratio cutoff guaranteeing an assigned accuracy: the requested observable, plate shape and thickness, surroundings, dielectric distribution, and guard geometry all matter. The small-gap approximation is local and is not uniform across an edge. It also does not justify extrapolating $C\simeq ϵA/d$ to arbitrarily large $d$, where finite electrical conductors and their external electric field dominate. At fixed $V$, formally shrinking an ideal gap makes the interior electric field grow like $1/d$; real electric currents across an imperfectly insulating gap, surface roughness and contact eventually invalidate that ideal limit.

For a static D3-brane with one scalar and no magnetic field, taking the Legendre transform of the DBI action gives the displayed identity. It bounds energy by the electric scalar charge and is saturated at $\mathbf{d}=\nabla X$. The opposite orientation uses the analogous identity with reversed signs. On saturation, the constitutive relation gives $\mathbf{E}=\mathbf{d}$ and the Gauss law constraint in gauge theory reduces to the Laplace equation for $X$.

A quasi-static chemical potential outside a fixed-shape droplet solves the same Laplace equation as an electrostatic potential outside an equipotential conductor. If $\mu ={\mu }_s$ on the droplet and zero on the surrounding reservoirs, its outward diffusive current iswhere $C$ is the corresponding electrostatic capacitance in a medium of permittivity $ϵ$. The arbitrary $ϵ$ cancels against the capacitance's proportionality to $ϵ$. Thus electrostatic geometry directly determines evaporation rates.

For a three-dimensional droplet with concentration jump $\Delta \varphi$, constant mobility $M$, and quasi-static exterior chemical potential,The first formula solves the exterior Laplace equation. The second follows from conservation: the total outward flux $4\pi MR({\mu }_s-{\mu }_{\infty })$ removes excess composition at the rate $-4\pi R^2\Delta \varphi \dot{R}$. For a symmetric mixture, $\Delta \varphi =2{\varphi }_B$ and the Gibbs--Thomson relation gives ${\mu }_s=\gamma /({\varphi }_{B}R)$.

For the gap $a<r<b$, put $u_{\ell }(r)=r^{\ell }-a^{2\ell +1}{r}^{-\ell -1}$, $v_{\ell }(r)=r^{-\ell -1}-r^{\ell }/b^{2\ell +1}$ and $d_{\ell }=(2\ell +1)[1-(a/b{)}^{2\ell +1}]$. With orthonormal spherical harmonics, the Dirichlet Green function isFor a short verification, $u_{\ell }(a)=v_{\ell }(b)=0$ and both solve the radial Laplace equation. Their Wronskian is $u_{\ell }{v}_{\ell }^{\prime }-u_{\ell }^{\prime }{v}_{\ell }=-d_{\ell }/r^2$. Thus the radial derivative jump is $-1/r^{\prime 2}$; completeness of the spherical harmonics gives the stated Dirac delta function normalization. A point charge has electric potential $qG/(4\pi {ϵ}_0)$, and the inner-wall surface charge density is $-q{\partial }_{r}G{|}_{a^{+}}/(4\pi )$, with the angular source coordinates held fixed.

For two incompressible inviscid layers at rest between rigid horizontal boundaries, with the heavier fluid below and no surface tension, the displayed dispersion relation follows from Laplace equation for the velocity potentials, no normal flow at the walls, equal normal velocities at the interface, and pressure continuity. The positive density contrast supplies the restoring force. In the long-wave limit the squared wave speed is $g({\rho }_2-{\rho }_1)/({\rho }_1/h_1+{\rho }_2/h_2)$.

In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

For two resting incompressible inviscid layers between impermeable horizontal walls, the upper layer has density ${\rho }_1$ and depth $L_1$, and the lower layer has ${\rho }_2<{\rho }_1$ and depth $L_2$. Solving the Laplace equation for the velocity potential and imposing the linearized kinematic and capillary pressure conditions gives the displayed dispersion relation. With positive surface tension $\gamma$, exponential Rayleigh-Taylor instability occurs for $0<k<k_c$, where $k_c=\sqrt{g({\rho }_1-{\rho }_2)/\gamma }$. The depths change the growth rate, but not this cutoff.

For the Laplace equation on $0<y<1$, prescribe bottom data $g$ and zero top data. A horizontal Fourier transform reduces the equation to ${\hat{u}}_{yy}-s^2\hat{u}=0$, giving the displayed multiplier and its limit $1-y$ at $s=0$. Integrable transformed data give a bounded inverse-transform solution. The bounded solution is unique by the maximum principle for harmonic functions, using barriers $\epsilon \cosh (\delta x)\cos (\delta (y-1/2))$ for $0<\delta <\pi$ on expanding rectangles. Without a growth condition, one can add $e^{\pi x}\sin (\pi y)$, so the two boundary traces alone do not determine an arbitrary harmonic function on an infinite strip.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

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 planar parabolic coordinates and $g(u,v)=f(x(u,v),y(u,v))$, the multivariable chain rule gives $g_u=u{f}_x+v{f}_y$ and $g_v=-v{f}_x+u{f}_y$. Differentiating once more and adding cancels the first partial derivatives and the mixed partial derivatives:Consequently, on every local coordinate patch away from the origin, $f$ is a harmonic function if and only if $g$ is a harmonic function. The Laplacian changes by a positive scale factor, so solutions of the Laplace equation are preserved. The restriction away from the origin is needed to divide by the Jacobian determinant.

The Field Mapper Kit (PK-9023), Product Manual 012-04346D by PASCO scientific (p. 3, Parallel plate capacitor) proposes a direct investigation of the plate-length-to-gap ratio using conductive-paper electric potential mapping. Its finite straight electrodes represent a cross-section of the plates. The apparatus is a gridded resistive sheet on a corkboard, silver-ink electrodes, pushpin electrical contacts, a $5$–$20\mathrm{V}$ direct supply capable of $25\mathrm{mA}$, and a voltmeter with input electrical resistance at least $10\mathrm{M\Omega }$. The manual specifies waiting 20 minutes for the ink to reach its maximum electrical conductivity, then checking each electrode: its internal voltage drop should be no more than $1\%$ of the applied electric potential difference.^[ref]

For the readout, keep one voltmeter lead on the reference electrode and scan the other across the sheet. Mark points having the same reading to trace an equipotential. A second method holds the two probe tips at a fixed separation and pivots them to find the largest electric potential difference; a suggested separation is one centimetre. Avoid printed grid marks when making contact.^[ref]

The inference follows from the measured electric potential, rather than assuming straight electric field lines. With probe separation $\Delta s$ and direction $\hat{\mathbf{t}}$, a local finite difference givesAn arrow toward rising electric potential follows $+\nabla \varphi$, so it must be reversed to show $\mathbf{E}$. To investigate the fringing field of a capacitor, compare contours well inside the gap with contours near and beyond the ends, then repeat at a changed gap while keeping the sheet boundary and plate length fixed. Curving contours and an outside gradient establish nonuniformity in this planar setup. Compare $-E_{y}d/V$ at recorded positions to distinguish simple voltage rescaling from geometric changes; no measured values are supplied here.

Additional useful controls are repeat reference readings to detect drift, reversal of supply polarity, and repeating selected readings with reduced probe spacing. Keep contact quality and sheet edges fixed during the aspect-ratio comparison. Finite probe separation averages the gradient; reducing it improves spatial resolution but magnifies the effect of voltage uncertainty. Voltmeter loading, imperfect ink electrodes, and inhomogeneous sheet electrical conductivity can mimic distortions. These proposed checks supplement the manufacturer's electrode check, rather than claiming a completed experiment.

The sheet carries a steady electric current. For uniform electrical conductivity, Ohm's law and electric charge conservation give the two-dimensional Laplace equation away from electrodes, as explained in conductive-paper electric potential mapping. Uncontacted sheet edges have zero normal electric current, whereas the diagram model fixes zero outer electric potential. These distinct boundary conditions must be matched before a quantitative comparison. The apparatus reveals a planar electric potential map; it does not directly measure the free-space three-dimensional capacitance of metal plates.

A second-order partial differential equation is of mixed type when the type of its principal part changes across its domain. For $A{u}_{xx}+2B{u}_{xy}+C{u}_{yy}=0$, the sign of $B^2-AC$ distinguishes a hyperbolic partial differential equation, parabolic type and an elliptic partial differential equation. For example, $u_{xx}+\mathrm{sign}(y)u_{yy}=0$ is elliptic above $y=0$, hyperbolic below it, and degenerate on the interface. On its hyperbolic side, the change $\xi =x+y$, $\eta =x-y$ reduces it to $u_{\xi \eta }=0$; above it is the Laplace equation. Type classification alone supplies no matching rule for the normal derivative.

The modified Helmholtz equation isIt reduces to the Laplace equation when $m=0$.