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For a charged particle minimally coupled to a magnetic vector potential, the mechanical momentum is $\mathbf{\pi }=\mathbf{p}-q\mathbf{A}$, with $\mathbf{p}$ the canonical momentum. It is the mass times velocity in nonrelativistic mechanics.

For a twice continuously differentiable vector potential $A$, write the curl with the Levi-Civita symbol:The second derivatives are symmetric in $i,j$, whereas ${ϵ}_{ijk}$ is antisymmetric, so the contraction is zero. Equivalently, writing out the three components shows that each mixed derivative cancels its reversed-order partner. ThusThis is divergence of a curl is zero. The regularity hypothesis is what licenses commutation of the mixed partial derivatives; it must not be omitted for a singular potential without specifying a weaker derivative interpretation.

Ignoring magnetization, electric polarization charge transport gives bound current ${\mathbf{J}}_{\mathrm{b}}={\partial }_t\mathbf{P}$. The assumed weak dielectric contrast therefore gives ${\mathbf{J}}_{\mathrm{b}}\simeq -i\omega {ϵ}_0({ϵ}_r-1){\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ inside the sphere. This is the first Born approximation: the incident field supplies the electric polarization to leading order in ${ϵ}_r-1$.

For $\mathbf{x}=r\mathbf{n}$, the far-distance expansions are $|\mathbf{x}-{\mathbf{x}}^{\prime }|=r-\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(a^2/r)$ and $|\mathbf{x}-{\mathbf{x}}^{\prime }{|}^{-1}\simeq r^{-1}$. Substitute the retarded time into the phase in the Lorenz gauge vector potential. The phase becomes $kr-\omega t+(\mathbf{k}-k\mathbf{n})\cdot {\mathbf{x}}^{\prime }$, so, using ${\mu }_0{ϵ}_0=c^{-2}$,The assumptions $r\gg a$ and $k{a}^2/r\ll 2\pi$ control amplitude and phase errors respectively. In the radiation zone the supplied electric and magnetic fields satisfy ${\mathbf{B}}_{\mathrm{scatt}}=c^{-1}\mathbf{n}\times {\mathbf{E}}_{\mathrm{scatt}}$. Complex phasors therefore give the time-averaged radial Poynting flux $|{\mathbf{E}}_{\mathrm{scatt}}{|}^2/(2{\mu }_{0}c)$. The incident flux is $|{\mathbf{E}}_0{|}^2/(2{\mu }_{0}c)$, soOrient spherical coordinates along $\mathbf{q}$ to evaluate the form factor:ConsequentlyAt $q=0$ the bracket is interpreted by its limit $a^3/3$, not a singularity. The transverse factor accounts for polarization of an electromagnetic wave; for complex polarization of an electromagnetic wave its squared norm means the Hermitian norm.

For a time-independent current, the Maxwell equations reduce to magnetostatics:Write $\mathbf{B}=\nabla \times \mathbf{A}$ using a vector potential and choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the curl of a curl gives the Poisson equationThe free-space Green function therefore givesTaking the curl with respect to $\mathbf{r}$, moving it under the integral, and using $\nabla (1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart law

The magnetization $M$ is magnetic dipole moment per unit volume. It produces bound volume and surface current densityStarting from the vector potential of magnetic dipoles and integrating by parts giveswhich has precisely the form generated by those bound currents.

Write the cylinder radius as $R$ and use cylindrical radius $\rho$. Ampère's law for the free axial current gives, inside the material,At $\rho =R$ this is tangential to the cylinder. Since $n=\hat{r}$,andFor $\mu >{\mu }_0$, the bound surface current therefore runs opposite to the free wire current.

Using the Levi-Civita symbol and the epsilon-delta identity,ThereforeOne particularly simple vector potential for $u$ isDirect differentiation gives $\nabla \cdot A=0$ and $\nabla \times A=(y^2,z^2,x^2)=u$.

Parametrize the curved cone byThe outward oriented surface element isSubstitution and integration over $0\le z\le h$, $0\le \phi <2\pi$ give

Because $\nabla \cdot u=0$, the divergence theorem predicts that the two outward fluxes sum to zero. On the top disk $S_2$, $d\mathbf{S}=(0,0,1)dxdy$ and $u_z=x^2$, so, with $R=h\tan \alpha$,as predicted.

The outward orientation on $S_2$ induces the stated anticlockwise orientation on $C$, whereas the outward orientation on $S_1$ induces the reverse orientation. Since $u=\nabla \times A$, Stokes theorem predictsOn $C$, put $(x,y,z)=(R\cos \phi ,R\sin \phi ,h)$. Thenverifying both identities.

Let $B$ be a continuously differentiable solenoidal vector field on an open star-shaped set containing the origin. Thenis a vector potential for $B$. For $D={\int }_0^{1}tB(tx)dt$, differentiation gives $\nabla \cdot D=0$ and $\nabla \times (D\times x)=2D+(x\cdot \nabla )D={\int }_0^1{\partial }_t(t^{2}B(tx))dt=B(x)$. Boundedness near the origin removes the lower endpoint. A field defined only on a punctured domain need not satisfy these hypotheses.