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This second-order linear differential equation extends the Legendre differential equation by its order parameter $m$. Its solutions are associated Legendre functions. In a degree-one potential problem, with $x=\tanh \eta$, two explicit solutions are $e^{m\eta }(m-\tanh \eta )$ and $e^{-m\eta }(m+\tanh \eta )$. Their Wronskian with respect to $\eta$ is $2m(1-m^2)$, so the two formulas are not a fundamental pair at $m=0$ or $m=\pm 1$.

Let ${\rho }_q=3q/(4\pi a^3)$ be the uniform charge density and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Rigid rotation produces current density $\mathbf{J}={\rho }_q\mathbf{\omega }\times \mathbf{r}$. The magnetic dipole moment isSpherical symmetry gives $\int r_i{r}_j{d}^{3}r={\delta }_{ij}\int r^2{d}^{3}r/3$, while $\int r^2{d}^{3}r=4\pi a^5/5$. ThereforeIt is parallel to the angular velocity for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|{\mathbf{m}}_s|=|q||\omega |a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.

Axisymmetry permits a purely azimuthal magnetic vector potential, $\mathbf{A}=f(r)\sin \theta \hat{\mathbf{\phi }}$, with $A_r=A_{\theta }=0$. For the source's associated Legendre function convention, $P_1^1(\cos \theta )=+\sin \theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces toInside the sphere, $J_{\phi }={\rho }_q\omega r\sin \theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is ${\alpha }_0=-{\mu }_0{\rho }_q\omega /10$. Regularity at the origin and decay at infinity giveThere is no surface sheet electric current, so both normal and tangential magnetic field components are continuous at $r=a$. Usingthis is equivalent to continuity of $f$ and $f^{\prime }$. The two equations give $A={\mu }_0{\rho }_q\omega a^2/6$ and $C={\mu }_0{\rho }_q\omega a^5/15$. ThusIn the homogeneous expansion, ${\alpha }_1=A$, ${\beta }_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^{\ell }$ and decaying $r^{-\ell -1}$ terms with zero source at $\ell >1$ forces both coefficients to zero. The outside magnetic field is exactly the magnetic dipole field of ${\mathbf{m}}_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_{\phi }={\mu }_0{m}_s\sin \theta /(4\pi r^2)$, which is the printed cross-product magnitude expression.

For the flux coupling of a rotating charged sphere to a circular loop, take a loop of radius $b>a$ in the equatorial plane and use Stokes theorem on the regular full vector potential, evaluated around the rim. Its linked magnetic flux isWith $m_s=\pi a^2{I}_0$, the mutual inductance defined using this effective electric current isIntegrating an ideal magnetic dipole field through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.

For the prescribed time dependence, the quasistatic Faraday's law of induction result and the purely resistive loop approximation givePositive loop electric current is defined to produce a positive $z$ magnetic dipole moment, so the minus sign is Lenz's law. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's self-inductance. If that is not negligible, the governing equation is instead $L_b{\dot{I}}_b+R{I}_b=-{\dot{\Phi }}_b$; $R$ alone is insufficient for the exact dynamic response.

For the scalar wave separation in Kerr spacetime, continue to use $A=r^2+a^2$ and $s=\sin \theta$. First verify the determinant in the hint. Direct multiplication of the covariant $t,\varphi$ components givesHence the block determinant is $-\Delta {\sin }^2\theta$. Inverting this block givesThe other inverse components are $g^{rr}=\Delta /\Sigma$, $g^{\theta \theta }=1/\Sigma$, and $\sqrt{-g}=\Sigma s$. The covariant wave operator on a scalar consequently has the divergence formLet $m_{\varphi }$ denote the azimuthal mode number, to distinguish it from the axial vector $m$. Insert the mode $\Psi =e^{-i\omega t+i{m}_{\varphi }\varphi }R(r)\Theta (\theta )$ in the massless Klein-Gordon equation. Single-valuedness makes $m_{\varphi }$ an integer. The $t,\varphi$ derivatives giveWith $K(r)=A\omega -a{m}_{\varphi }$, division by the mode factor gives, on patches where $R\Theta \ne 0$,The radial and angular expressions must be opposite constants. Defining the separation constant as $\Lambda$, we obtain the two ordinary differential equationsThese equations also hold at zeros of a mode by continuity, without dividing there. Regular angular solutions are scalar spheroidal harmonics, with discrete $\Lambda ={\Lambda }_{\ell m_{\varphi }}(a\omega )$. For $a\omega =0$ the angular equation becomes the associated Legendre function equation, with $\Lambda =\ell (\ell +1)$ and $\ell \ge |m_{\varphi }|$, providing a useful check of the signs and normalization. The radial function here is exactly $R$ in the chosen ansatz, without an additional factor of $1/r$.

Here $\mu =\cos \theta$ and $a_{\ell ,m}=\sqrt{({\ell }^2-m^2)/[(2\ell +1)(2\ell -1)]}$. This follows by combining the associated Legendre function recurrence $(2\ell +1)\mu P_{\ell }^m=(\ell -m+1)P_{\ell +1}^m+(\ell +m)P_{\ell -1}^m$ with the normalized spherical harmonic constants. It gives the adjacent-multipole couplings in a photon Boltzmann hierarchy.