Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 76 (Brain Imaging Methods via Electro-and Magneto- Encephalography)

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.

Put $\lambda =(r/a{)}^2$, so the inverted observation point is $\overline{\mathbf{r}}=\mathbf{r}/\lambda$ and $\overline{r}=r/\lambda$. Let $\mathbf{R}=\overline{\mathbf{r}}-{\mathbf{r}}_0$. The Kelvin transform correction in the first bracket is transformed usingThus its dipole contribution is exactlyFor the remaining correction use the dipole line-integral identity. Writing ${\mathbf{P}}_c=\mathbf{r}-c{\mathbf{r}}_0$ and $P_c=|{\mathbf{P}}_c|$, it isIt can be proved by differentiating the right side with respect to $c$ and checking its zero value at $c=0$. Alternatively resolve into directions parallel and perpendicular to ${\mathbf{r}}_0=b\mathbf{e}$: if $\mathbf{r}=\rho {\mathbf{e}}_{\rho }+z\mathbf{e}$, the two integrals are $(z/r-(z-bc)/P_c)/(b\rho )$ and $(1/P_c-1/r)/b$, respectively, which combine to the displayed expression. The limiting cases follow by continuity.

Set $c=\lambda$. Then ${\mathbf{P}}_c=\lambda \mathbf{R}$, $P_c=\lambda R$, $\mathbf{r}=\lambda \overline{\mathbf{r}}$ and $r=\lambda \overline{r}$. Substitution givesDotting with $\mathbf{Q}/(4\pi \sigma )$ reproduces the second correction. Including the primary dipole therefore proves the Spherical Neumann dipole images representation,The upper limit is $(r/a{)}^2$ in the PDF; extracting the small stacked fraction as text can incorrectly invert it. The extra terms are one scaled dipole and a continuous dipole line in this parametrized method of images. Their positions and strengths depend on the observation radius, so they should not be treated as a fixed set of independent physical sources. For an interior point $r<a$, the entire image segment satisfies $t|{\mathbf{r}}_0|\le (r/a{)}^2|{\mathbf{r}}_0|<r$, so it introduces no new interior singularity. The only physical interior dipole singularity remains at ${\mathbf{r}}_0$.

Let $\overline{\mathbf{r}}=a^2\mathbf{r}/r^2$ and $\hat{\mathbf{r}}=\mathbf{r}/r$. Its Jacobian isThe chain rule therefore yieldsTaking its cross product with $\mathbf{r}$ kills the radial term and proves that spherical inversion preserves angular derivatives:This is also the invariance of the differential expression underlying the angular momentum operator.

A scalar spherical harmonic $Y$ depends only on direction, so $Y(\hat{\mathbf{r}})=Y(\widehat{\overline{\mathbf{r}}})$ and its gradient is tangential. Thus $\hat{\mathbf{r}}\cdot {\nabla }_{\overline{\mathbf{r}}}Y=0$, andEach of the three normalized vector spherical harmonic families is consequently unchanged as an angular basis function:The normalization requires $n\ge 1$ for $\mathbf{B}$ and $\mathbf{C}$; at $n=0$ the gradient vanishes and only $\mathbf{P}$ remains.

This substitution into the basis functions is distinct from the pushforward of a vector field of a physical vector field by inversion. Under that Jacobian map the radial vector picks up $-a^2/r^2$ and the two tangential vectors pick up $+a^2/r^2$. Likewise, the weighted Kelvin transform of a scalar includes the extra factor $a/r$. Neither operation changes the angular differential identity just proved.

For a degree-$n$ homogeneous polynomial, the Euler homogeneous function theorem gives $\mathbf{r}\cdot \nabla H_n=n{H}_n$. On $r>0$ in three-dimensional Euclidean space,Apply the product rule for the Laplacian:For $s=-(2n+1)$ the second term vanishes. Hence homogeneous harmonic inversion givesThe nonzero factor $r^{-(2n+1)}$ proves both directions: the transformed function is harmonic exactly when $\Delta H_n=0$ away from the origin. Since $\Delta H_n$ is a polynomial, vanishing on the punctured space makes it identically zero everywhere.

Equivalently, homogeneity shows that the Kelvin transform isIt sends a regular solid harmonic to an irregular solid harmonic, exchanging the radial powers $r^n$ and $r^{-n-1}$. The origin is excluded from the harmonicity statement: for example, the $n=0$ transform $1/r$ is harmonic only away from zero and has a distributional Dirac delta function source there.