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In SI units the Maxwell equations in free space are

The trace of the electromagnetic stress-energy tensor in four spacetime dimensions isFor its covariant divergence, the source-free Maxwell equations eliminate the derivative of the first factor. Contracting the Bianchi identity ${\nabla }_{[a}{F}_{bc]}=0$ with $F^{ac}$ giveswhich cancels the derivative of the trace term. HenceFor $J^a=T^{ab}{V}_b$, stress-energy conservation and symmetry of $T^{ab}$ now imply

Differentiating $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, substituting Maxwell equations, and using the stated vector identity gives the Maxwell stress tensor conservation lawMoving the divergence to the left identifies

Split the microscopic sources into free and bound parts and defineSubstitution of $\rho ={\rho }_{\mathrm{f}}-\nabla \cdot P$ into Gauss's law gives $\nabla \cdot D={\rho }_{\mathrm{f}}$. Similarly, using $J=J_{\mathrm{f}}+{\partial }_{t}P+\nabla \times M$ in the Ampère-Maxwell equation gives $\nabla \times H=J_{\mathrm{f}}+{\partial }_{t}D$. The homogeneous Maxwell equations are unchanged. Thus

Maxwell equations areTaking the divergence of the Ampère-Maxwell equation and using Gauss's law givesThereforebecause $J$ vanishes on the boundary. Componentwise integration by parts also giveswith the boundary term again zero.

Varying $\bar{\psi }$ and $\psi$ independently givesand its adjointwhere $D_{\mu }\bar{\psi }={\partial }_{\mu }\bar{\psi }-ie{A}_{\mu }\bar{\psi }$. Varying $A_{\nu }$ and integrating the Pauli term by parts gives the modified Maxwell equations

Varying $A_{\nu }$ gives the Maxwell equationsOn the vacuum manifold, choose unitary gauge so that $\varphi =v$. Then $\mathbf{J}=-2e^2{v}^2\mathbf{A}$ and the London equation isFor a static configuration, $\nabla \times \mathbf{B}=\mathbf{J}$ and $\nabla \cdot \mathbf{B}=0$, soThe magnetic field therefore decays exponentially inside the condensate. This is the Meissner effect, with penetration depth equal to the inverse gauge-boson mass.

The first vacuum Maxwell equations equation is preserved for every vector field because part a givesOn two-forms in four dimensions, the mixed volume tensor $T_{ab}{}^{cd}$ represents twice the Hodge star operator. Part b therefore says that a conformal Killing field commutes with the Hodge star:It follows thatHence $\tilde{F}={\mathcal{L}}_{X}F$ satisfies both vacuum Maxwell equations whenever $F$ does.

Let $\hat{n}$ be the unit normal pointing from the minus medium to the plus medium. Integrating the Maxwell equations across an infinitesimal pillbox or loop gives the dielectric-interface boundary conditionsandThus tangential $E$ and $H$ are continuous, as are normal $D$ and $B$.

Because $D=ϵE$, the normal electric field generally jumps according toThe tangential displacement field generally jumps because ${ϵ}_{+}\ne {ϵ}_{-}$. Since the permeability is common, continuity of tangential $H$ is continuity of tangential $B$; together with normal-$B$ continuity, every component of $B$ is continuous. The same common $\mu$ also makes the normal component of $H$ continuous.

The same Maxwell equations givewhere the last equality uses $\nabla \cdot \mathbf{E}=0$. Combining with part (ii),The divergences of the symmetric gradients satisfyThus the local identity is ${\partial }_t(u_i^E+u_i^B)={\partial }_j(S_{ij}^E-c^2{S}_{ij}^B)$. Integrate over the fixed volume and apply the divergence theorem componentwise:Here $d{A}_j=n_{j}dA$ uses the outward normal vector.

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 $\nabla \cdot \mathbf{B}=0$,

Apply the symmetric gradient and curl decomposition separately to the electric field and the magnetic field:The corresponding axial parts areThe second equalities use the vacuum Maxwell equations. Their divergence constraints also show that both symmetric gradient tensors are traceless.

The divergence of a cross product identity and the vacuum Maxwell equations implyThe divergence theorem therefore gives the required balance withAn additive time-independent constant in $\lambda$ is immaterial. This is the vacuum Poynting theorem in the normalization of the given equations: equivalently ${\partial }_t\lambda +\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. The order of the cross product accounts for the plus sign in the stated integral identity.

A photonic crystal has a spatially periodic dielectric response on the scale of the optical wavelength. For a nonmagnetic, lossless, nondispersive dielectric, the source-free Maxwell equations can be written as the eigenproblemThe Bloch theorem gives ${\mathbf{H}}_{n\mathbf{k}}=e^{i\mathbf{k}\cdot \mathbf{r}}{\mathbf{u}}_{n\mathbf{k}}(\mathbf{r})$ with periodic $\mathbf{u}$. The allowed frequencies form photonic band structures. Coupling of counterpropagating waves near a Brillouin zone boundary opens a stop band. A one-dimensional dielectric multilayer is a simple example; two-dimensional hole arrays and three-dimensional dielectric networks allow more extensive control.

A complete photonic band gap excludes bulk propagating modes for every direction and polarization in a frequency interval. Periodicity alone does not guarantee such a gap: dimensionality, symmetry and refractive-index contrast matter. A missing or altered element can support a localized defect state in the gap, making a cavity; a line defect can guide light. Finite crystals have leakage and wavelength-dependent transmission rather than an infinitely sharp exclusion. Applications include selective mirrors, optical filters, small resonators and integrated waveguides.^[ref]

Write a nonzero vacuum plane electromagnetic wave with wavevector $\mathbf{k}=(\omega /c)\mathbf{s}$, where $\mathbf{s}$ is a unit propagation direction. Its propagation Lorentz four-vector is $K^{\mu }=(\omega /c,\mathbf{k})$, or, up to a scalar normalization in any one frame, $n^{\mu }=(1,\mathbf{s})$. It is null because $K_{\mu }{K}^{\mu }=0$.

Choose coordinates $x^{\mu }=(ct,\mathbf{x})$ and Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. With the supplied three-dimensional boost formulas, a consistent convention assigns the printed array to upper-index components,The index conventions for the electromagnetic field tensor then require $F_{0i}=+E_i/c$ and $F_{ij}=F^{ij}$ when both indices are lowered. The printed lower-index label therefore cannot be kept together with its electric signs and a forward propagation vector in this convention.

This is not merely an arbitrary overall sign. For a valid forward wave with $\mathbf{s}=\hat{\mathbf{z}}$, $\mathbf{E}=E_0\hat{\mathbf{x}}$ and $\mathbf{B}=(E_0/c)\hat{\mathbf{y}}$, multiplying the printed array, unchanged, by $n^{\mu }=(1,0,0,1)$ gives an $x$ component $-2E_0/c$, rather than zero. One consistent repair is to retain the printed array as $F^{\mu \nu }$ and contract with $n_{\mu }=(1,-\mathbf{s})$. Equivalently, lower both tensor indices before forming the contraction asked for in the question.

Using that corrected lower-index tensor givesTo establish both electric and magnetic transversality, include the dual electromagnetic field tensor. Choose ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; lowering both indices then givesThe source-free plane-wave Maxwell equations are precisely the pair of covariant conditionsTheir temporal components give $\mathbf{k}\cdot \mathbf{E}=0$ and $\mathbf{k}\cdot \mathbf{B}=0$. Their spatial components give the associated amplitude and orientation relations, $\mathbf{E}/c=-\mathbf{s}\times \mathbf{B}$ and $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$.

To see the Lorentz covariance explicitly, let $V_{\nu }=K^{\mu }{F}_{\mu \nu }$. It is a covector, so if $V_{\nu }=0$ in one inertial frame, its Lorentz transformation is also zero in every other inertial frame. The same is true of $K^{\mu }{\tilde{F}}_{\mu \nu }$. Since the transformed wave still has $K^{\prime 0}>0$ and $|{\mathbf{k}}^{\prime }|={\omega }^{\prime }/c$, their temporal components implyHere $K^{\mu }$, rather than a vector renormalized to have temporal component one in every frame, is the actual Lorentz four-vector; multiplying a zero contraction by a nonzero normalization does not affect the argument.

Finally, the two scalar perpendicularity statements alone are not equivalent to one tensor contraction. They do not enforce the amplitude relation between $\mathbf{E}$ and $\mathbf{B}$. Conversely $n^{\mu }{F}_{\mu \nu }=0$ alone cannot exclude a longitudinal magnetic field: adding $B_{\parallel }\mathbf{s}$ leaves its cross product unchanged. For a genuine source-free plane electromagnetic wave, the remaining Maxwell equations supply that missing constraint, and both tensor contractions vanish. This qualification and the corrected index placement are necessary for the intended invariance argument.

Use the inverse coordinate transformation,Substitution into the wave phase $\mathbf{k}\cdot \mathbf{r}-\omega t$ and comparison of the coefficients of ${\mathbf{r}}^{\prime }$ and $t^{\prime }$ givesThe first is the relativistic Doppler effect; the changed direction of the wavevector describes relativistic aberration. Equivalently,The electric-field wave amplitude follows from part (b), after rotating the boost axis to $\mathbf{v}$:For a vacuum plane wave, Maxwell equations give ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$ and $\mathbf{k}\cdot {\mathbf{E}}_0=0$. Thus an expression using only the given electric wave amplitude and wavevector isEquivalently, ${\mathbf{E}}_0^{\prime }=\gamma ({\mathbf{E}}_0+\mathbf{v}\times {\mathbf{B}}_0)-[{\gamma }^2/(\gamma +1)]\mathbf{v}(\mathbf{v}\cdot {\mathbf{E}}_0)/c^2$. The wave amplitude is not a scalar carried unchanged by the wave phase transformation; its direction must remain transverse to the transformed wavevector.

In vacuum, the source-free Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-{\partial }_{t}B$ and $\nabla \times B=c^{-2}{\partial }_{t}E$. Put $\vartheta =\omega (t-z/c)$ and denote the field amplitude by $E_0$. The corresponding travelling-wave magnetic field isBoth divergences vanish because the fields have no longitudinal component and no dependence on the transverse coordinates. Direct differentiation gives $\nabla \times E=(\omega E_0/c)(\cos \vartheta ,\sin \vartheta ,0)=-{\partial }_{t}B$ and $\nabla \times B=(\omega E_0/c^2)(-\sin \vartheta ,\cos \vartheta ,0)=c^{-2}{\partial }_{t}E$.

Introducing potentials by $B=\nabla \times A$ and $E=-\nabla \varphi -{\partial }_{t}A$ automatically satisfies the magnetic-divergence and Faraday equations. The remaining Maxwell equations become equations for these potentials, with the usual gauge freedom. For $\omega \ne 0$ take $\varphi =0$ andDifferentiation verifies both potential relations.

The electromagnetic force is $Q(E+v\times B)$. Confinement also supplies a normal reaction force. For motion in a plane of constant $z$, both $v$ and $B$ are tangent to the plane, so their cross product is normal and is balanced by that reaction. It does not enter the in-plane equation $m\ddot{r}=Q{E}_0(\cos \vartheta ,\sin \vartheta )$.

At the fixed plane, the solution with no uniform drift isIt is circular, withOther initial data can add a constant drift velocity; the question only asks for the possibility of circular motion.

The source-free Maxwell equations in vacuum areTheir real and imaginary parts are exactlyPut $\theta =\omega t-K\cdot x$. Differentiating gives $\nabla \cdot W=(\Omega \cdot K)\sin \theta$ and $\nabla \times W=-(\Omega \times K)\sin \theta$, while $(i/c){\partial }_{t}W=-i\omega \Omega \sin \theta /c$. Thus the necessary and sufficient conditions areHere the complex dot product is bilinear, without complex conjugation. Taking the dot product of the second equation with $\Omega$ gives $\Omega \cdot \Omega =0$. Crossing it with $K$ givesso, since $\Omega \ne 0$, $|K{|}^2={\omega }^2/c^2$. Writing $\Omega =e+icb$, the real conditions areBoth physical fields are linearly polarized: $E=e\cos \theta$ and $B=b\cos \theta$ each oscillate along a fixed real direction. Combining two fields as $E+icB$ does not turn the electric field into a circularly polarized field.

The source-free Maxwell equations in vacuum areTheir real and imaginary parts are exactlyPut $\theta =\omega t-K\cdot x$. Differentiating gives $\nabla \cdot W=(\Omega \cdot K)\sin \theta$ and $\nabla \times W=-(\Omega \times K)\sin \theta$, while $(i/c){\partial }_{t}W=-i\omega \Omega \sin \theta /c$. Thus the necessary and sufficient conditions areHere the complex dot product is bilinear, without complex conjugation. Taking the dot product of the second equation with $\Omega$ gives $\Omega \cdot \Omega =0$. Crossing it with $K$ givesso, since $\Omega \ne 0$, $|K{|}^2={\omega }^2/c^2$. Writing $\Omega =e+icb$, the real conditions areBoth physical fields are linearly polarized: $E=e\cos \theta$ and $B=b\cos \theta$ each oscillate along a fixed real direction. Combining two fields as $E+icB$ does not turn the electric field into a circularly polarized field.

Historically, electromagnetism was central to Albert Einstein's 1905 argument. His opening example compares a moving magnet and a moving conductor. The same relative motion produces the observed induction, but a description using electric fields in one frame and the magnetic part of the Lorentz force in another seemed to divide the explanation according to which object was called stationary. The unsuccessful search for an observable preferred state of rest also motivated treating the laws of mechanics, optics and electromagnetism on a common footing. That establishes a strong historical connection, not a deduction that relativity applies only to electromagnetic systems.^[ref]

The logical starting points of the kinematical argument are Einstein's principle of relativity and an invariant speed of light in vacuum, together with the assumptions needed to compare homogeneous inertial frames and physical measuring standards. Einstein synchronisation assigns distant clock readings by light exchange. In standard coordinates, the resulting Lorentz transformation for a boost along $x$ isNeither the derivation of this transformation nor the deductions of relativity of simultaneity and time dilation require solving Maxwell equations for every particular material system. They constrain the relation between measurements made in different inertial frames.^[ref]

There is nevertheless an empirical commitment in identifying $c$ with the speed of vacuum light. It is not a freely chosen numerical label: the theory says different inertial observers, using the prescribed standards, obtain the same measured value. More generally, $c$ specifies the causal structure of Minkowski spacetime. A theory could contain a massive scalar field with a Lorentz invariant equation and no electromagnetic field at all. Its particles, proper times and causal propagation would still be governed by relativistic geometry. This gives a direct sense in which the framework is logically more general than electromagnetism.

Conversely, the presence of Maxwell equations alone would not establish the full relativity principle for all matter. One could append a law involving a physically preferred timelike vector field, giving a nonelectromagnetic sector a preferred frame while retaining the usual electromagnetic equations. The covariant equation with a preferred timelike vector is an explicit model of this distinction. Requiring every relevant interaction to respect Lorentz invariance adds content; writing its components in four-dimensional notation does not supply that content automatically.

The magnet–conductor puzzle is resolved because electric fields and magnetic fields are related parts of one electromagnetic field, with the Lorentz force transforming consistently. The measured current must be compared for the same physical arrangement, including the conductor's state of motion, rather than by demanding the same electric and magnetic components separately. This is an application of the general frame transformations to a particular interaction, not the definition of every possible relativistic interaction.

Albert Einstein's later distinction between a principle theory and a constructive theory clarifies what the 1905 argument provides. It places universal restrictions on admissible laws without constructing every rod and clock from a particular microscopic model. A constructive explanation can additionally investigate why actual material standards approximate the required behavior. Harvey R. Brown and Oliver Pooley emphasize this dynamical question; it does not make the kinematical transformations meaningless or confine them to electromagnetic phenomena.^[ref][ref]

The strongest account therefore distinguishes historical motivation, the operational use of light, and the theory's universal scope. The 1905 paper addresses an electromagnetic problem through a framework whose restrictions extend to every physical process claimed to obey special relativity.