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For an isolated razor-thin disk, rescale its exterior magnetic scalar potential to satisfy $\sqrt{4\pi G/{\mu }_0}\mathbf{B}=-\nabla {\Phi }_M$. Comparing the normal magnetic boundary derivative with ${\partial }_z\Phi =2\pi G\Sigma$ identifies the displayed equivalent signed surface density. Both exterior potentials satisfy Laplace's equation. The equivalent surface density is a representation of the magnetic boundary data, rather than physical negative mass.

Use the original scanned boundary condition. The scan fixes the normal-field amplitude at the corner $x=z=0$. The supplied TeX drops the condition $x=0$ and would instead prescribe a constant normal field on the entire bottom. Those are different boundary-value problems; the solution below uses the legible original scan.

Choose a positive integer $n$, put $k=n\pi /a$, and seek a separated decaying mode. A suitable flux function isThis is a decaying linear force-free magnetic arcade. The first and third curl components then satisfy the force-free equation automatically; its second component requires $-({\partial }_x^2+{\partial }_z^2)A={\lambda }^{2}A$. ThusDirect differentiation gives $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=\lambda \mathbf{B}$. Both horizontal components vanish at $x=0,a$, and $B_z(0,0)=B_0$. The magnetic field on the bottom is $B_0\cos kx$, not a uniform $B_0$.

A strictly decaying real normal mode requires $q>0$, soFor the fundamental, single-arcade choice $n=1$, this becomes $|\lambda |<\pi /a$. Equality gives no vertical exponential decay, while a larger magnitude gives an oscillatory rather than decaying vertical dependence for that mode. The data do not explicitly fix the horizontal mode number. If arbitrarily fine higher-mode arcades are permitted, every finite real $\lambda$ admits some integer $n$ with $n\pi /a>|\lambda |$. The fundamental restriction must therefore not be advertised as a necessary bound for every possible field satisfying only the stated pointwise and wall data.

For a fundamental field line, $A$ is constant along its projection in the $(x,z)$ plane. If $0<s<1$ denotes its normalized flux-function value, its arch isThe maximum height is $q^{-1}\log (1/s)$ at $x=a/2$. In the horizontal projection,Thus the vertical projections are arches and the horizontal projections are straight sheared segments joining their paired footpoints. Positive $\lambda$ gives positive horizontal slope, negative $\lambda$ reverses the shear, and $\lambda =0$ gives an unsheared potential arcade. Higher $n$ repeats such cells across the width.

The transcription's alternative uniform-bottom condition would be inconsistent with exponential decay and zero side flux for $B_0\ne 0$: solenoidality gives $d{\int }_0^a{B}_{z}dx/dz=-B_x(a,z)+B_x(0,z)=0$, whereas a uniform nonzero bottom would give a conserved flux $a{B}_0$ that cannot decay to zero. This integral check is an independent reason not to silently solve that mistranscribed problem.

Write $b=|{\mathbf{r}}_0|$, $r=|\mathbf{r}|$, $\mathbf{P}=\mathbf{r}-{\mathbf{r}}_0$ and $P=|\mathbf{P}|$. Initially assume $0<b<a$ and a nonzero current dipole. The possible singularities of the continued formulas occur when $r=0$, $P=0$ orBy the Cauchy-Schwarz inequality, $D\ge 0$. Away from the two endpoints it vanishes exactly when $\mathbf{P}$ is antiparallel to $\mathbf{r}$, namely on the open segment $\mathbf{r}=s{\mathbf{r}}_0$, $0<s<1$. Thus both formulas are regular in the physical exterior $r>a$; the singularities requested here belong to their analytic expressions continued into the sphere.

To determine the order on the segment, let $\mathbf{e}={\mathbf{r}}_0/b$ and approach an interior point as $\mathbf{r}=z\mathbf{e}+\rho {\mathbf{e}}_{\rho }$, with $0<z<b$ and ${\mathbf{e}}_{\rho }\perp \mathbf{e}$ a unit transverse direction. Expansion givesConsequently the leading singular terms in the electric potential and magnetic scalar potential areThus a dipole with a nonzero transverse moment has first-order line singularities on the open segment: their generic growth is $1/\rho$, the inverse distance to that segment. Angular coefficients can vanish along particular approach directions without removing the line singularity.

At the source point put $\mathbf{r}={\mathbf{r}}_0+P\mathbf{n}$ with a fixed direction $\mathbf{n}\ne -\mathbf{e}$. The explicit dipole term givesThe source therefore has a second-order electric dipole singularity and, for a nonradial moment, a first-order magnetic singularity. Approaches tangent to the singular segment need not have these fixed-angle scalings.

At the origin put $\mathbf{r}=r\mathbf{n}$, with $\mathbf{n}\ne \mathbf{e}$. The leading terms areThe electric expression has a first-order point singularity there. The magnetic expression has a generally direction-dependent finite radial limit, so the origin is a nonremovable endpoint singularity of radial order zero, rather than an isolated $1/r$ pole. It can still be unbounded along approaches whose angle to the segment also tends to zero. Stating the approach matters for this nonisolated singularity.

For a radial current dipole, $\mathbf{Q}=q\mathbf{e}$, the transverse line terms cancel and the electric line singularity is removable. Its only electric singularities are the first-order term $-q/(4\pi \sigma br)$ at the origin and the second-order dipole at ${\mathbf{r}}_0$. The magnetic potential is identically zero because $\mathbf{Q}\times {\mathbf{r}}_0=0$. Finally, if ${\mathbf{r}}_0=0$, the two terms in the given exterior electric expression combine toThere is then just a second-order electric singularity at the origin. A zero dipole moment makes both potentials zero.

For this question use the normalized magnetic scalar potential $\mathcal{U}$ defined by $\mathbf{B}=({\mu }_0/4\pi )\nabla \mathcal{U}$. The previous question instead includes ${\mu }_0/4\pi$ in its scalar potential; its $U$ equals $({\mu }_0/4\pi )\mathcal{U}$. This normalization change in the paper must be kept separate from the kernel calculation.

Take the radial component of the Geselowitz formula. On the spherical boundary, ${\mathbf{n}}^{\prime }$ is parallel to ${\mathbf{r}}^{\prime }$, andThus the surface-current term contributes no radial magnetic field. The primary Biot-Savart law term gives, along a fixed direction $\hat{\mathbf{r}}$,Choose $\mathcal{U}\to 0$ at infinity. Integrating this equation fixes the full potential along every ray, givingTo express it as a source-position derivative, defineWith $\mathbf{W}=\mathbf{Q}\times {\mathbf{r}}_0$, differentiating the integrand givesso $\mathcal{U}=\mathbf{W}\cdot {\nabla }_{{\mathbf{r}}_0}\varphi$ as required.

Let $b=|{\mathbf{r}}_0|$ and $P=|\mathbf{r}-{\mathbf{r}}_0|$. An elementary antiderivative gives the spherical current-dipole logarithmic kernelFor $r>a>b$, both logarithm arguments are positive. Its radial derivative is $-1/(rP)$ and its limit at infinity is zero, which verifies the preceding integral. At $b=0$ use the continuous limit $\varphi =1/r$.

For a direct check, differentiating along $\mathbf{W}$ leaves $b$ unchanged and gives $\mathbf{W}\cdot {\nabla }_{{\mathbf{r}}_0}P=-\mathbf{W}\cdot \mathbf{r}/P$. Sinceone obtainsIts gradient is the Sarvas formula. The sphere radius and electrical conductivity disappear from this exterior magnetic result because spherical symmetry removed the radial contribution of the return currents. The original derivation and spherical-conductor assumptions are discussed in Sarvas's paper.

Integrate Poisson equation for Newtonian gravity through a narrow slab around the disk. The horizontal derivative contributions vanish as its thickness tends to zero, leaving the normal-derivative jumpThe even function symmetry of $\Phi$ makes the derivatives opposite, soIn the current-free simply connected upper half-space, Ampère's circuital law gives $\nabla \times \mathbf{B}=0$ and permits a magnetic scalar potential. Rescale it so that $\sqrt{4\pi G/{\mu }_0}\mathbf{B}=-\nabla {\Phi }_M$. The divergence-free condition makes ${\Phi }_M$ satisfy Laplace's equation, withCompare with the gravitational jump condition. Subject to the same isolated-field boundary condition at infinity, ${\Phi }_M$ is the harmonic potential of the effective surface densityThis gravity-equivalent magnetic surface density can have either sign; it is a mathematical representation of the exterior magnetic field, not physical negative mass. An imposed nondecaying field would require additional boundary data and would not be fixed by the disk surface density alone.

Because $\nabla \cdot \mathbf{B}=0$, a magnetic vector potential can be introduced with $\mathbf{B}=\nabla \times \mathbf{A}$. It is not unique: $\mathbf{A}+\nabla \chi$ gives the same field. In magnetostatics, the Coulomb gauge $\nabla \cdot \mathbf{A}=0$ gives ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{J}$ in vacuum. For localized electric currents,A magnetic scalar potential is more restricted. In a simply connected current-free region of constant permeability, write $\mathbf{B}=-\nabla \psi$; then ${\nabla }^2\psi =0$. Alternatively define the scalar potential for $\mathbf{H}$, differing by a constant permeability factor. A single-valued scalar potential generally cannot cover a domain encircling a net electric current, even where the local curl vanishes. The vector potential does not require the field itself to be curl-free.

The method of images replaces induced charges or currents on a boundary by fictitious sources outside the physical region, chosen to satisfy that boundary's conditions. For example, a point charge $q$ a height $d$ above a grounded conducting plane has an image $-q$ at the reflected point. The sum has zero potential on the plane, and the electrostatic uniqueness theorem identifies it with the physical solution in the upper half-space. Images do not describe a second physical charge below the metal.

For the superconducting plane, use the different boundary condition $B_z=0$ at its surface: magnetic flux cannot enter the ideal bulk. Choose the real ring moment $\mathbf{m}=m\hat{z}$, where $m=I\pi R^2$, at $z=h$. Its image has moment $-m\hat{z}$ at $z=-h$. This superconducting plane image dipole makes the normal fields cancel at $z=0$, while their tangential fields add. The image is a computational continuation; the physical field inside the plane is zero. The original sketch shows the dipole-scale field, appropriate to $h\gg R$; the fine field close to the ring wire is outside that approximation.

In the vacuum region the magnetic scalar potential isAt the surface $z=0$, it reduces to $-mh/[2\pi ({\rho }^2+h^2{)}^{3/2}]$. Its tangential derivative therefore givesBoth $B_z$ and $B_{\varphi }$ vanish there; the sign is inward radially for the chosen positive moment. The expression is the leading dipole result, not an exact finite-ring formula at arbitrary $R/h$.

For the force, calculate the external image field on the axis near the real dipole:Differentiate this field with the image location held fixed, then evaluate at $z=h$. The dipole levitation force above a superconducting plane isdirected upward. Differentiating a substituted separation $2h$ while moving both dipoles in the force derivative would incorrectly double the result.

To levitate ring mass $M$, balance this force against $Mg$:A light, thin wire and the smallest practical gap reduce current. At fixed gap and wire cross-section $s$, $M=2\pi Rs{\rho }_w$ gives $I_{\mathrm{lev}}\propto h^2\sqrt{s{\rho }_w}/R^{3/2}$, favouring a larger ring until the dipole-limit constraint on $R/h$ matters. There is no unique least current without geometric and material constraints. As a geometrically consistent order-of-magnitude example, take $R=1\mathrm{mm}$, $h=10\mathrm{mm}$, wire radius $50\mu \mathrm{m}$ and an assumed metallic-scale density ${\rho }_w=6.0\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Then $M=2.96\times 10^{-7}\mathrm{kg}$ and $I_{\mathrm{lev}}\simeq 2.8\times 10^2\mathrm{A}$. The large value reflects the weak $h^{-4}$ force in the stipulated far-field geometry. Uniformly shrinking this geometrical example reduces current as size to the power $3/2$, but increases required electric current density; arbitrarily thin wire is therefore not a free optimization.

This balance estimate is not a verified superconducting design. The actual wire must carry that current without quenching, its self-field counts toward the limit, the plane must remain in its assumed flux-excluding state, and leads or supports add load. Transport critical-current measurements specify temperature, applied field and the voltage criterion rather than a material-independent limiting current.^[ref]

For a normal from medium 1 to 2, the magnetic boundary conditions areWithout a free surface current, tangential $H$ is continuous. To obtain the magnetization of a permeable sphere of radius $R$ and permeability ${\mu }_0(1+\chi )$, choose magnetic scalar potentialsTangential continuity gives $H_i=H_0-C/R^3$, and normal $B$ continuity gives $(1+\chi )H_i=H_0+2C/R^3$. HenceThe fields are uniform inside. For negative susceptibility, magnetization is antiparallel to the applied field; the scalar formula retains its negative sign.

For a small weakly magnetic body, the induced magnetic energy is $U_m=-\chi V{B}^2/(2{\mu }_0)$. The factor one-half accounts for building the induced moment. The vertical force is $F_z=\chi VB{B}^{\prime }/{\mu }_0$. Balancing gravity therefore givesA diamagnet needs a field decreasing in magnitude with height for an upward force. For diamagnetic levitation in an inverse-height field,The force falls as $z^{-3}$, giving a restoring vertical derivative at equilibrium. Full lateral stability would require the transverse field geometry, which is not specified.

For the exterior of a spherical volume conductor model, put $P=|\mathbf{r}-{\mathbf{r}}_0|$, $r=|\mathbf{r}|$ and $F=r{P}^2+P\mathbf{r}\cdot (\mathbf{r}-{\mathbf{r}}_0)$. The displayed Sarvas formula follows by taking the radial component of the Geselowitz formula and integrating the magnetic scalar potential from infinity. It is independent of a spherically symmetric conductivity profile. The sign shown uses $\mathbf{B}=+{\mu }_0\nabla U/(4\pi )$; conventions defining a magnetic potential with a minus sign reverse $U$. Sarvas's original paper derives the spherical-conductor field and its consequences for the inverse problem.

For $r>b$ and $P=|\mathbf{r}-{\mathbf{r}}_0|$, the displayed kernel satisfies $\varphi ={\int }_r^{\infty }ds/[s|s\hat{\mathbf{r}}-{\mathbf{r}}_0|]$. The limit at $b=0$ is $1/r$. Differentiating along $\mathbf{Q}\times {\mathbf{r}}_0$, which preserves $b$, gives $(\mathbf{Q}\times {\mathbf{r}}_0)\cdot {\nabla }_{{\mathbf{r}}_0}\varphi =(\mathbf{Q}\times {\mathbf{r}}_0)\cdot \mathbf{r}/F$. Thus it generates the normalized magnetic scalar potential in the Sarvas formula.