Codex @codex  0

With Cartesian Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$, introduce the four-current $J^{\mu }=(c\rho ,\mathbf{J})$ and electromagnetic four-potential ${\mathcal{A}}^{\mu }=(\varphi /c,\mathbf{A})$. The antisymmetric electromagnetic field tensor $F^{\mu \nu }={\partial }^{\mu }{\mathcal{A}}^{\nu }-{\partial }^{\nu }{\mathcal{A}}^{\mu }$ has components $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}^{ijk}{B}_k$. Maxwell equations becomeThe time and space components reproduce Gauss's law and the Ampère-Maxwell equation; the dual equation gives absence of magnetic monopoles and Faraday's law of induction. Antisymmetry gives charge conservation ${\partial }_{\nu }{J}^{\nu }=0$. The relativistic Lorentz force is $d{p}^{\mu }/d\tau =q{F}^{\mu \nu }{u}_{\nu }$. Every displayed equation transforms tensorially under a Lorentz transformation; the electric field and magnetic field are different components of the same object.

Two independent algebraic electromagnetic field invariants areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations but changes sign under spatial reflection. For perpendicular fields, the sign of the first distinguishes a frame with purely magnetic field from one with purely electric field. Both invariants vanish for a plane electromagnetic wave with $\mathbf{E}\perp \mathbf{B}$ and $E=cB$, although the field itself is nonzero.

Let $\epsilon ={\epsilon }_0{\epsilon }_r$ be the permittivity, $\beta =(k_{B}T{)}^{-1}$, $e>0$ the elementary charge, and $c$ the bulk salt formula-unit number density. The Boltzmann distribution givesConsequently the exterior Poisson-Boltzmann equation isFor small $Q$, meaning $2\beta e|\varphi (a)|\ll 1$, linearize the exponentials. The Debye–Hückel approximation givesThis is also ${\kappa }^2=\beta e^2{\sum }_i{z}_i^2{n}_{i,\infty }/\epsilon$: the divalent cation contributes $4c$ and the two monovalent anions contribute $2c$. In terms of molar salt concentration $C$ in $\mathrm{mol}{\mathrm{m}}^{-3}$, replace $c$ by $N_{A}C$; if $C$ is in $\mathrm{mol}{\mathrm{L}}^{-1}$, use $c=10^3{N}_{A}C$. The molar ionic strength is $3C$ in the same concentration units.

The boundary conditions are $\varphi (\infty )=0$ and, by Gauss's law,Here the surface charge is uniform and the spherically symmetric interior field is zero. Set $u=r\varphi$; the radial equation reduces to $u^{\prime \prime }={\kappa }^{2}u$. Decay excludes the growing exponential, so write $\varphi =B{e}^{-\kappa (r-a)}/r$. Then ${\varphi }^{\prime }(a)=-B(1+\kappa a)/a^2$, givingIt solves the linearized Poisson-Boltzmann equation, not the full nonlinear equation at finite $Q$. The radial ion profiles, consistently to first order, areThe exponential Boltzmann distribution formulas above express the full mean-field relation, but substituting a linearized potential into them is accurate only to the retained order. A useful electroneutrality check isThe result is the Debye-Hückel potential of a charged sphere.

Boost the comoving electric field perpendicular to itself along the wire. The transverse electric field gains a factor ${\gamma }_v$, while a circulating magnetic field appears. Equivalently apply Gauss's law and Ampère's law with laboratory line charge $\lambda$ and current $I=\lambda v$:These are exterior fields of the ideal straight wire. For laboratory line density $\rho$, substitute $\lambda =\rho$; for proper line density $\rho$, substitute $\lambda ={\gamma }_v\rho$. The radial distance is transverse to the boost and unchanged between the two frames.

Model the wire as an infinite straight line along $z$. Let ${\lambda }_0$ be its rest-frame line charge density and $\lambda$ its laboratory line charge density. Length contraction gives $\lambda ={\gamma }_v{\lambda }_0$. If the symbol $\rho$ in the source is intended as volume charge density, replace these line densities by their integrals over the cross section; an unspecified wire radius does not determine an interior field.

In the comoving frame the wire is static, so Gauss's law and cylindrical symmetry give the exterior fieldsIf the source's $\rho$ means laboratory line density, this electric field is $\rho /(2\pi {ϵ}_0{\gamma }_{v}r)$; if it means proper line density, it is $\rho /(2\pi {ϵ}_{0}r)$. Stating the density's reference frame is necessary for an unambiguous factor of ${\gamma }_v$.

The electromagnetic four-potential is $A^{\mu }=(\Phi /c,\mathbf{A})$, and the vacuum four-wavevector is $K^{\mu }=(\omega /c,0,0,k)$ with $k=\omega /c>0$. Define $\psi =K_{\mu }{X}^{\mu }=\omega t-kz$ and $C=(\omega k{)}^{-1/2}$. The given transverse potential is $A_y=-C\sin \psi$. Its physical fields areso the Poynting vector points along positive $z$. For our signature, ${\partial }^{\mu }=(c^{-1}{\partial }_t,-\nabla )$, $F^{0i}=-E_i/c$, and $F^{ij}=-{ϵ}_{ijk}{B}_k$. ThereforeThe square-root prefactor is simply the source's chosen wave normalization; an overall dimensional amplitude could multiply every field.

The four-current is $J^{\mu }=(c\rho ,\mathbf{j})$, with $J_{\mu }=(c\rho ,-\mathbf{j})$. Charge conservation takes the equivalent formsLowering both indices of $F$ changes the sign of its time-space entries: $F_{0i}=E_i/c$, while $F_{ij}=-{ϵ}_{ijk}{B}_k$. The temporal component of ${\partial }^{\mu }{F}_{\mu \nu }$ is $\nabla \cdot \mathbf{E}/c={\mu }_{0}c\rho$, by Gauss's law. Its spatial components are $c^{-2}{\partial }_t\mathbf{E}-\nabla \times \mathbf{B}=-{\mu }_0\mathbf{j}$, by the Ampère-Maxwell equation. HenceFor the displayed free wave the current vanishes because $k=\omega /c$; the component derivation also proves the sourced identity. The two homogeneous Maxwell equations follow separately from expressing $F$ in terms of a potential.

Force-sign qualification. With these definitions the physical Lorentz force density is $f_{\mu }^{\mathrm{phys}}=F_{\mu \nu }{J}^{\nu }$, whereas the source prints the reversed contraction $J^{\nu }{F}_{\nu \mu }=-f_{\mu }^{\mathrm{phys}}$. ConsequentlyThe physical time component is the power transferred to matter per volume divided by $c$, as in Poynting's theorem. For example parallel positive $j_x$ and $E_x$ do positive work, but the literal printed contraction has negative time component. This sign cannot be repaired while simultaneously retaining the source's tensor definition and the Maxwell identity above; reverse the force contraction to describe the physical force.

Use a neutral, infinitely long wire on the $x$ axis. Denote the total current by $I=\int j_{x}d{A}_{\perp }$, and write $r^2=y^2+z^2$. Outside the wire the magnetic field and electric field areThese follow from cylindrical symmetry and Ampère's law. The zero temporal component of the given total four-current specifies lab neutrality; the $\rho$ in $j_x=\rho v_0$ must refer to a carrier density, not that zero net charge density.

TODO: normalization from a literal volume current density requires the unspecified wire cross-section. The source does not supply its area or profile. The exterior field family below is fully determined in terms of $I$; it cannot be reduced to $j_x$ alone without that missing datum. If $j_x$ was intended as a cross-section-integrated line current instead, set $I=j_x$ and interpret its accompanying charge density as charge per length. No inside-wire profile is invented.

For a boost $u=v_0$ along positive $y$, let $\Gamma =(1-u^2/c^2{)}^{-1/2}$. Apply the Lorentz transformation of electromagnetic fields at corresponding events:Here the subscript labels the perpendicular-to-wire boost, not the direction of the transformed vector. In these formulas evaluate the original field at $y=\Gamma (y^{\prime }+u{t}^{\prime })$, $z=z^{\prime }$ and $x=x^{\prime }$. The wire moves in this frame, so the fields are time-dependent. The local current transforms to $J^{\prime 0}=0$, $J^{\prime x}=j_x$, $J^{\prime y}=J^{\prime z}=0$. A nonzero transformed electric field is possible even with zero net charge density: it is not a purely electrostatic field.

For a boost $u=v_0$ along positive $x$, the transverse coordinates are unchanged. The transformed fields areThe transformed line charge and current are ${\lambda }^{\prime }=-\Gamma uI/c^2$ and $I^{\prime }=\Gamma I$. These reproduce the radial electric field through Gauss's law. A boost comoving with the electrons removes their individual current contribution, but the positive background then moves and the total current is not zero.

For two identical wires, their lab interaction is attractive, with magnitude ${\mu }_0{I}^2/(2\pi d)$ per length. The total four-current is spacelike, $J^2=-j_x^2<0$, so no inertial frame makes it a pure charge density. Also $B^2-E^2/c^2=B^2>0$ forbids a frame in which the exterior field is purely electric. For an arbitrary boost along the wires, the electrical part is repulsive and the magnetic part attractive:The latter always dominates, and their sum remains $-{\mu }_0{I}^2/(2\pi d)$.

Even a more general boost cannot make the entire-wire force electrical alone. At the second wire choose $y=d,z=0$, so $\mathbf{B}=B{\mathbf{e}}_z$ and $\mathbf{j}=j{\mathbf{e}}_x$ with $jB>0$. The physical Lorentz force density is $f=(0,0,-jB,0)$. If the boosted magnetic force vanished, ${\mathbf{j}}^{\prime }\times {\mathbf{B}}^{\prime }=0$, then ${\mathbf{E}}^{\prime }\cdot {\mathbf{B}}^{\prime }=0$ would also give ${\mathbf{j}}^{\prime }\cdot {\mathbf{E}}^{\prime }=0$. Thus $f^{\prime 0}=0$, which from the boost of $f$ requires $V_y=0$. But for such a boost $f^{\prime y}=-jB<0$, whereas ${\rho }^{\prime }{E}_y^{\prime }={\Gamma }_V^2{V}_x^{2}jB/c^2\ge 0$. The electrical term alone cannot equal the actual force. Therefore there is no such frame for both complete wires, although the force on electrons considered separately can be electrical in their rest frame.

Take the divergence of the Ampère-Maxwell equation and use that the divergence of a curl is zero:The last step uses Gauss's law. Thus the charge continuity equation is exactlyFor a direct justification that this four-current is a four-vector, take one locally comoving population of electric charges, with rest charge density ${\rho }_0$ and four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$. Invariance of electric charge and length contraction give $\rho =\gamma {\rho }_0$ and $\mathbf{J}=\rho \mathbf{v}$, hence $J^{\mu }={\rho }_0{U}^{\mu }$. Summing populations preserves the four-vector rule. In particular, $c{\rho }^{\prime }=\gamma (c\rho -\beta J_x)$ and $J_x^{\prime }=\gamma (J_x-\beta c\rho )$. The Lorentz invariance of its four-divergence then explains why charge conservation holds in every inertial frame. A single zero-divergence condition alone would not determine a field's transformation law; the physical charge density and current density provide the additional content.

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.

Permittivity describes how an electric field and a material's electric polarization combine into the electric displacement field. In a static, local response with linearity and isotropy, and no electric polarization at zero electric field, the constitutive equation isHere ${ϵ}_0$ is vacuum permittivity, while ${ϵ}_r$ is the relative permittivity, a dimensionless quantity. The SI unit of permittivity is farad per metre, $F/m$: multiplying it by an electric field in $V/m$ gives an electric displacement field in $C/{\mathrm{m}}^2$. A larger positive permittivity lets the same parallel-plate capacitor hold more plate electric charge at the same electric potential difference.

The material contribution is electric polarization $\mathbf{P}$, the electric dipole moment per unit volume. The electric displacement field and electric susceptibility ${\chi }_e$ satisfySubstitution gives the first constitutive equation. The electric field here is the averaged, total field inside the material, not just the field of the external plate electric charges. The continuum approximation averages over many chemical atoms or molecules; locality means that the electric polarization at a position depends on the electric field there. Static equilibrium, fixed material conditions and sufficiently small electric fields are needed for this scalar proportionality. ^[ref].

For an ideal parallel-plate capacitor completely filled by a uniform dielectric, hold the free plate electric charge magnitude $Q$ fixed and neglect the fringing field of a capacitor. Writing $A$ for the plate area, Gauss's law gives $D=Q/A$. ThusThe electric dipoles point along the electric field. Their surface electric charges, described by polarization density, oppose the field from the free plate electric charges. For a hypothetical dielectric with ${ϵ}_r=4$, the internal electric field is one quarter of its vacuum value, while $P=3D/4$. At fixed electric potential difference instead, the internal electric field stays approximately $V/d$ and the plate electric charge grows fourfold. These are different experimental constraints. ^[ref].

Permittivity is a useful material summary, not a universal constant of the material under every condition. Temperature, frequency and electric field strength can change the response. In a linear anisotropic dielectric, permittivity is a tensor, $D_i={\sum }_j{ϵ}_{ij}{E}_j$, so the electric displacement field need not be parallel to the electric field. At nonzero frequency, delayed electric polarization requires complex permittivity and may produce dielectric loss. ^[ref].

A charge cloud $\rho =A/r$ for $0<r<a$ encloses $Q(r)=2\pi A{r}^2$. With no central point charge and potential zero at infinity, Gauss's law gives $\varphi =A(2a-r)/(2{ϵ}_0)$ inside and $A{a}^2/(2{ϵ}_{0}r)$ outside. The density singularity is integrable, and the potential is finite at the origin.

For a coaxial cable with inner and outer radii $a,b$, a radial interface $c$, and permittivities $ϵ,{ϵ}^{\prime }$, define $T=\ln (c/a)/ϵ+\ln (b/c)/{ϵ}^{\prime }$. With terminal electric potential difference $V$, the electric displacement field is $D_{\rho }=V/(T\rho )$. The electric field is $E_{\rho }=V/(T{ϵ}_i\rho )$ in each layer $i$. If the outward and return electric currents have magnitudes $I$, the magnetic field strength in the gap is $H_{\phi }=I/(2\pi \rho )$ and the magnetic field is $B_{\phi }={\mu }_{i}I/(2\pi \rho )$.

The proof uses Gauss's law without free interface electric charge, the electric potential difference integral, and the magnetostatic Ampère law. The axial Poynting vector is $VI/(2\pi T{ϵ}_i{\rho }^2)$, so its cross-sectional surface integral is $VI$, independently of the radial material layering and magnetic permeabilities. The empty inner and exterior regions have zero electric field and magnetic field for the regular solution with bounded exterior electric potential and no applied magnetic field.

For a free electromagnetic field with suitable boundary conditions, combine Coulomb gauge ${\partial }_i{A}_i=0$ with $A_0=0$. Gauss's law makes the second condition consistent in the source-free problem. The remaining field operator is spatially transverse and describes the two radiative photon polarizations. A constrained scalar potential must be restored when charges are present.

Two equal opposite line charge densities moving in opposite directions have zero net line charge but a nonzero total current $I$. Under a longitudinal Lorentz boost, the charge-current four-vector gives ${\lambda }^{\prime }=\gamma (\lambda -vI/c^2)=-\gamma vI/c^2$. Equivalently, for lab densities $n$ and speeds $\pm u$, their separate boosted number densities are $\gamma n(1\mp vu/c^2)$. Their imbalance creates the boosted radial electric field, consistently with Lorentz transformation of electromagnetic fields and Gauss's law.

Two concentric conducting spherical shells of radii $a<b$ with charges $Q,-Q$ have an electric field only in the intervening vacuum: $E_r=Q/(4\pi {ϵ}_0{r}^2)$ by Gauss's law. Integrating gives a potential difference $Q(1/a-1/b)/(4\pi {ϵ}_0)$, and hence the displayed capacitance. The electrostatic energy is $Q^2/(2C)$, either by charging work or by integrating ${ϵ}_0|E{|}^2/2$ over the gap.

A charge distribution invariant under rotations about one center. For its spherically symmetric potential, Gauss's law gives the radial electric field $\mathbf{E}=Q(r){\mathbf{e}}_r/(4\pi {ϵ}_0{r}^2)$, where $Q(r)$ is the enclosed electric charge.

Electric charge concentrated on a surface. Its local amount per unit area is the surface charge density $\sigma$; in vacuum the jump of the outward normal electric field is $\sigma /{ϵ}_0$ by Gauss's law.

A spherical cavity of radius $a$ filled with fixed uniform charge density $\rho$ and surrounded by a grounded conductor has electric field $\rho r\hat{\mathbf{r}}/(3{ϵ}_0)$ inside and zero electric field outside the conductor when no external sources are present. Gauss's law supplies the interior electric field; integration from $V(a)=0$ gives the displayed electric potential. Its stored electrostatic energy is $2\pi {\rho }^2{a}^5/(45{ϵ}_0)$, obtained by integrating ${ϵ}_0|\mathbf{E}{|}^2/2$ over the cavity. Grounding permits the conductor's net electric charge to change.