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A force-free magnetic field is linear when the proportionality between its curl and itself is a spatial constant. The curl of the curl identity and zero divergence give the vector Helmholtz equation ${\nabla }^2\mathbf{B}+{\lambda }^2\mathbf{B}=0$. Here $\lambda$ has inverse-length units; specifying a constant current-to-field coefficient with another normalization changes its numerical definition.

In Einstein summation convention, the curl isApply another curl and use the contraction of two Levi-Civita symbols:Smoothness allows the partial derivatives to commute. This proves the curl of the curl identity:

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

By definition the vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. The curl of the curl identity yieldsFor incompressible flow, $\nabla \cdot \mathbf{u}=0$, so

In a charge-free and current-free vacuum, Maxwell equations giveTake the curl of Faraday's law and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. The divergence term vanishes by Gauss's law, while the time derivative of the magnetic curl supplies $-{\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$. ThusThis is the transverse electromagnetic wave equation.

Apply the curl of the curl identity $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ to Faraday's law, and use the Ampère-Maxwell equation. This givesApplying the same identity to $B$, using $\nabla \cdot B=0$, gives

For a spherically symmetric charge density, Gauss's law on a sphere of radius $r$ givesFor the unit-density ball of radius $a$,The potential is continuous at $r=a$ and tends to zero at infinity.

Let $\mathbf{p}=(1,0,0)$ and write $1_{S_n}$ for the indicator of $S_n$. The layered density decomposes asThe point $\mathbf{p}$ lies on every sphere $S_n$, whose radius is $R_n=4^{-n}$ and whose outward radial direction there is $\hat{\mathbf{x}}$. A uniform ball of density $c$ contributes $c{R}_n/3$ to the field and $c{R}_n^2/3$ to the potential at its surface. Thereforeand

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

For proper rotations the Levi-Civita symbol is invariant, so contracting the two rank-one tensors ${\partial }_j$ and $v_k$ gives $c_i={ϵ}_{ijk}{\partial }_j{v}_k$ with transformation $c_i^{\prime }=R_{i\ell }{c}_{\ell }$. Thus the curl is a vector under proper rotations. Under an orientation-reversing orthogonal change it acquires an additional determinant sign: it is an axial vector when $v$ is a polar vector.

Commuting smooth partial derivatives and using epsilon contraction gives the curl of the curl identity:Now take the curl of Faraday's law and use the vacuum Ampère-Maxwell equation: $\nabla \times \nabla \times E=-{\partial }_t(\nabla \times B)=-c^{-2}{E}_{tt}$. Since $\nabla \cdot E=0$, the identity gives $\Delta E-c^{-2}{E}_{tt}=0$. Taking the curl of the Ampère-Maxwell equation gives the same result for $B$, using $\nabla \cdot B=0$. Hence both satisfy the electromagnetic wave equation

Work in a Cartesian coordinate system, write ${\partial }_j=\partial /\partial x_j$, and use the Einstein summation convention. For a twice continuously differentiable vector field, the contraction of two Levi-Civita symbols givesHere the Kronecker deltas select components, and equality of mixed partial derivatives gives the curl of the curl identityFor the second identity, the same contraction followed by the product rule yieldsThus the curl of a cross product isThe directional derivatives in the last two terms act on the vector field immediately to their right.

Differentiate ${\mathbf{u}}^E=\frac{1}{2}\nabla \times \mathbf{E}$ with respect to time. The vacuum Maxwell equations give ${\partial }_t\mathbf{E}=c^2\nabla \times \mathbf{B}$, soThe curl of the curl identity is $\nabla \times (\nabla \times \mathbf{B})=\nabla (\nabla \cdot \mathbf{B})-{\nabla }^2\mathbf{B}$. Since $\nabla \cdot \mathbf{B}=0$,

In Cartesian coordinates, write ${\partial }_i=\partial /\partial x_i$ and use the Levi-Civita symbol. The contraction of two Levi-Civita symbols givesHere the Kronecker delta selects the indicated indices, and continuous second partial derivatives allow their order to be interchanged. Identifying the gradient, divergence and componentwise Laplacian proves the curl of the curl identity:

In source-free vacuum, Maxwell's equations readTake the curl of the third equation and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. ThenThe prescribed plane wave travels in the $+z$ direction. Its polarization is along $x$, and Faraday's law requires $B_y=(k/\omega )E_x$ with $\omega =ck$. The wave's magnetic flux density is thereforeThe direction is also fixed by $\mathbf{E}\times \mathbf{B}$ pointing along the propagation direction. A superposed static uniform magnetic field is not fixed by the oscillatory electric field; the expression above is the magnetic field belonging to the wave itself.

In a source-free vacuum, the Maxwell equations in SI arewhere ${ϵ}_0$ and ${\mu }_0$ are the vacuum permittivity and vacuum permeability. Take the curl of the last equation, substitute Faraday's law of induction, and use the curl of the curl identity:The zero divergence therefore gives the wave equationFor a propagating plane wave, write $\mathbf{B}=\mathrm{Re}[{\mathbf{B}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. The wave equation gives ${\omega }^2=c^2|\mathbf{k}{|}^2$, and Gauss's law for magnetism gives $\mathbf{k}\cdot {\mathbf{B}}_0=0$. Thus the oscillating magnetic field is a transverse wave: it is perpendicular to the propagation wavevector. A static background magnetic field, if present, is separate from this propagating component.

In SI units, the Maxwell equations in vacuum with prescribed charge density $\rho$ and current density $\mathbf{J}$ areWhen $\rho =0$ and $\mathbf{J}=0$, take the curl of the third equation and substitute the fourth. Using the curl of the curl identity and $\nabla \cdot \mathbf{E}=0$ gives the electromagnetic wave equation:For example, $\mathbf{E}=E_0\hat{\mathbf{x}}\cos (kz-\omega t)$, $\mathbf{B}=(E_0/c)\hat{\mathbf{y}}\cos (kz-\omega t)$ is a plane electromagnetic wave provided $\omega =ck$. The electric field, magnetic field and propagation direction are mutually perpendicular.

The Maxwell equations are linear, so the superposition principle applies to fields. However, the Poynting vector is quadratic:One may add the average wave intensities only when the cross terms vanish under the actual detector average. For co-propagating plane electromagnetic waves of parallel linear polarization, averaging over optical cycles but potentially resolving the difference frequency givesIf the detector integrates for $T$, the last term acquires $\mathrm{sinc}[({\omega }_1-{\omega }_2)T/2]$, with sinc function convention $\mathrm{sinc}x=\sin x/x$, evaluated at the midpoint phase. Thus $T\gg 1/|f_1-f_2|$ suppresses beats, and the infinite-time average adds the wave intensities exactly for distinct fixed frequencies. Random mutually uncorrelated wave phases also remove the cross terms in an ensemble average; orthogonal linear polarizations remove them for co-propagating waves in polarization-insensitive detection. Merely having different frequencies does not remove a resolved beat signal: two nearly equal frequencies with fixed relative wave phase are a concrete counterexample. For different propagation directions, the relevant addition is of average Poynting vectors, or their components through the chosen detector surface.

In charge-free, current-free vacuum, Maxwell's equations areTaking the curl of the third equation and using the fourth, the curl of the curl identity, and the zero divergence of the electric field givesThe magnetic flux density obeys the same wave equation. For the plane wave factor $e^{i(\omega t-kx)}$, the spatial Laplacian contributes $-k^2$ and the second time derivative contributes $-{\omega }^2$. Thusfor propagation in the positive $x$ direction. The zero-divergence conditions require $E_{0x}=B_{0x}=0$, while the curl equations require ${\mathbf{B}}_0=(k/\omega )\hat{\mathbf{x}}\times {\mathbf{E}}_0$. These conditions show that the proposed plane waves satisfy all four Maxwell's equations, rather than only their derived wave equations. Physical electric fields and magnetic fields are the real parts of the complex expressions.

The Lorenz gauge condition is $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$. Substitute the electric potential and magnetic vector potential into Gauss's law:Next the Ampère-Maxwell equation, the curl of the curl identity, and the Lorenz gauge condition giveThe gradient terms cancel. Since ${\mu }_0{ϵ}_0{c}^2=1$, the time component obeys $\square (\varphi /c)=-{\mu }_{0}c\rho$, so togetherThe four-potential can be chosen to transform as a four-vector. To make the argument rigorous, choose the retarded potential with no freely added homogeneous solution. If $\square G_{\mathrm{ret}}(x)={\delta }^{(4)}(x)$, this choice isThe retarded Green function is a Lorentz scalar under proper orthochronous Lorentz transformations, and $d^{4}y$ is invariant. The four-vector rule for $J^{\mu }$ therefore proves $A^{\prime \mu }(x^{\prime })={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }(x)$. More generally, transform any homogeneous initial data by this same rule. Both the wave equation and the Lorenz gauge condition are then preserved.

Qualification. The wave equation by itself cannot prove the transformation rule for every independently chosen gauge potential: the d'Alembert operator has a nontrivial kernel of a linear map. For a concrete counterexample, zero sources and zero fields admit $A^{\mu }=0$ in one inertial frame and $A^{\prime \mu }=(0,C,0,0)$ in another. Both satisfy the wave equation and the Lorenz gauge condition, but the latter is not the Lorentz transformation of zero. It is a residual Lorenz gauge symmetry, generated by $\chi =C{x}^{\prime }$. Thus the usual assertion is a covariant choice of four-potential, with a consistent gauge transformation and homogeneous data, rather than a uniqueness result from the displayed equation.