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Use a representative conduction-electron density $n\sim 10^{29}{\mathrm{m}}^{-3}$, roughly one mobile electron per atom in a dense metal, and the bare electron mass for this estimate. The plasma frequency is thenThis is an order-of-magnitude answer because no particular metal or carrier density is specified. It corresponds to an ordinary frequency $f_p\simeq 2.8\times 10^{15}\mathrm{Hz}$, excitation energy $\hslash {\omega }_p\simeq 12\mathrm{eV}$, and free-space wavelength near $1.1\times 10^{-7}\mathrm{m}$.

The physical oscillation is collective motion of the electron gas against the positive ionic background. Displacing the electrons relative to the ions produces a restoring electric field. In the simplest uniform-displacement picture, $\ddot{x}+{\omega }_p^{2}x=0$: the frequency increases with the density of mobile electric charge and decreases with carrier inertia. Quantizing this longitudinal collective oscillation gives a plasmon.

The same scale controls the ideal collisionless Drude model optical response. Its cold collisionless plasma dielectric response is ${ϵ}_r(\omega )=1-{\omega }_p^2/{\omega }^2$. Below ${\omega }_p$, the negative relative permittivity makes a transverse electromagnetic wave evanescent, producing strong reflection; above it, the ideal model admits propagation. Real-metal collisions, interband photon absorption and background polarization modify this simple optical edge. Thus ${\omega }_p$ is a collective charge-response scale, not a single-electron scattering rate or a sharp transparency threshold for every real metal.

Nondegenerate carrier product. Assume three-dimensional parabolic energy bands near their extrema, spin degeneracy two, and no additional uncounted valley multiplicity. Near the conduction edge, a Boltzmann approximation to the Fermi-Dirac distribution givesAngular integration and the Gaussian radial integral yieldFor valence electron holes, the empty-state probability is $1-f(E)\simeq e^{(E-\mu )/(k_{B}T)}$. Measuring electron hole kinetic energy downwards from $E_v$ similarly givesThe factors two count spin once in each band. Multiplying cancels the chemical potential and proves the semiconductor mass-action law:In an intrinsic neutral semiconductor, $n=p=n_i$. In a doped equilibrium semiconductor the same product holds under the stated nondegenerate assumptions, even when $n\ne p$. Extra valleys or several valence branches must be included in the effective density-of-states factors rather than counted inconsistently.

Equilibrium junction. Put the p region on the left and the n region on the right. Before contact their majority carriers are electron holes and electrons, respectively. Diffusion initially sends electrons from n to p and electron holes from p to n. This leaves negative ionized acceptors on the p side and positive ionized donors on the n side, creating a depletion region and an electric field directed from n to p. At equilibrium this field balances diffusion, so the net electric current vanishes.

For a homojunction the gap remains constant while the conduction band and valence band edges bend downwards from p to n by $e{V}_{\mathrm{bi}}$. The equilibrium chemical potential is flat. Since $E_{c,v}(x)=E_{c,v}^{\mathrm{ref}}-e\varphi (x)$, the electrostatic potential rises from p to n. The carrier profiles satisfyWith fully ionized nondegenerate dopants, the neutral p side has $p_p\simeq N_A$, $n_p\simeq n_i^2/N_A$, while the neutral n side has $n_n\simeq N_D$, $p_n\simeq n_i^2/N_D$. Thus the electron concentration rises across the junction and the electron hole concentration falls, with both small relative to the far-side majority densities in much of the depleted region. The ideal depletion approximation neglects these mobile populations when calculating the fixed space electric charge, but the actual equilibrium profiles are smooth.

The built-in potential follows from these Boltzmann factors:There is no applied terminal bias; the internal electrostatic potential variation does not contradict a constant electrochemical potential.

Opposing carrier fluxes under reverse bias. Use $V>0$ for the magnitude of reverse bias and $I>0$ for reverse electric current, as in the PDF. The junction barrier increases by $eV$. The Shockley diode equation in that orientation can be writtenThe reverse-directed contribution is supplied by minority electrons on the p side and minority electron holes on the n side. They reach the depletion edge by diffusion and are swept across by the built-in field. The opposing contribution is associated with majority carriers thermally crossing the barrier in the forward direction; reverse bias suppresses it by the Boltzmann factor. At zero bias detailed balance makes the two equal, while at large enough reverse bias the opposing flux becomes negligible.

In the ideal neutral-region diffusion model,Here $D_n,D_p$ are minority-carrier diffusion coefficients, $L_n,L_p$ their diffusion lengths, and $\mathcal{A}$ is junction area. The two terms inside this positive prefactor are the electron and electron hole contributions; both flow in the same reverse-current direction. This separates the two carrier-species contributions from the two opposing fluxes in the Shockley diode equation. Calling the saturation term a drift current refers to its passage across the depleted region, and does not eliminate the minority-carrier diffusion needed to supply it.

Reading the measured curves. At $eV\ll k_{B}T$, the ideal equation gives $I\simeq I_{\mathrm{sat}}eV/(k_{B}T)$, initially proportional to reverse electric potential difference. At $eV\gg k_{B}T$, it approaches a plateau $I\simeq I_{\mathrm{sat}}$. The thermal electric potential differences at the three temperatures are about $21$–$26\mathrm{mV}$, consistent with a rise at small reverse biases and near saturation at larger ones. Higher temperature produces a much larger plateau because thermal electron-hole excitation raises $n_i^2$ exponentially.

The lower-temperature curve has a noticeable residual electric potential difference dependence rather than a perfectly flat plateau. Depletion-region generation, surface leakage or other departures from the ideal diffusion model can contribute. The sharp upturn at the largest reverse electric potential differences is reverse breakdown, outside the unextended Shockley diode equation; the plot alone does not identify a unique microscopic breakdown mechanism. The original horizontal axis duplicates the final decade label, so a precise breakdown electric potential difference cannot be read reliably. Neither issue prevents estimating the gap from the intermediate-bias saturation region.

Band-gap estimate. Use plateau electric currents at a common intermediate reverse bias, away from both the small-bias rise and the breakdown upturn. Approximate readings are $I_{\mathrm{sat}}(298\mathrm{K})\sim 2.5\times 10^{-6}\mathrm{A}$, $I_{\mathrm{sat}}(273\mathrm{K})\sim 1.5$–$2\times 10^{-7}\mathrm{A}$, and $I_{\mathrm{sat}}(248\mathrm{K})\sim 10^{-8}\mathrm{A}$. Kelvin temperatures are required in the activation law.

For fixed effective masses, $n_i^2\propto T^3{e}^{-E_g/(k_{B}T)}$. If the remaining transport prefactor is approximately constant, extracting a semiconductor band gap from diode saturation current therefore usesUsing the widest temperature separation and the rounded readings givesNeglecting all algebraic prefactors and fitting $\ln I_{\mathrm{sat}}$ instead gives about $0.70\mathrm{eV}$. These differ by roughly $0.07\mathrm{eV}$, a material correction on the graph's precision scale. Thus an honest graph-based conclusion isThe nonflat low-temperature plateau, rough electric current readings, unprovided temperature dependence of $D/L$, and possible gap variation prevent a more precise extraction. If depletion-region generation dominated rather than the specified $I_{\mathrm{sat}}\propto n_i^2$ mechanism, its activation slope could involve a different energy scale; the displayed estimate explicitly uses the diffusion-dominated saturation model requested here.

For an electron wavepacket in one isolated band under a uniform electric field,Because $ϵ(k+2\pi /a)=ϵ(k)$, acceleration through successive Brillouin zones makes the velocity periodic rather than increasing indefinitely. For $ϵ={ϵ}_0-2J\cos (ka)$,for a positive field along the chosen axis. Integration givesThe excursion amplitude of a Bloch-oscillating wavepacket is $2|J|/(e|E|)$ about its mean and peak-to-peak extent $4|J|/(e|E|)$. The turning of velocity corresponds semiclassically to Bragg reflection near a zone boundary.

Coherent Bloch oscillations require scattering/dephasing times long compared with $T_B$ and a field weak enough to avoid substantial interband tunnelling. Ordinary metals usually scatter too rapidly; semiconductor superlattices and cold atoms in periodic optical potentials provide cleaner realizations. A completely filled band has canceling velocities, so its net current does not oscillate simply because individual wavepackets can.^[ref]

Relaxation-time model. The Drude model treats mobile charge carriers as independent particles accelerated by the Lorentz force between scattering events. Collisions randomize the drift momentum toward local equilibrium and are represented by a uniform, field-independent momentum relaxation time. For the present calculation the carrier density, isotropic mass and relaxation time are fixed, the fields are uniform and stationary, and quantum orbital quantization and intercarrier correlations are neglected. The steady drift velocity obeysHere $n$ is the number of carriers of that species per unit volume, $q$ their signed charge, $m>0$ their inertial or isotropic band mass, $\tau$ the momentum relaxation time, and ${\sigma }_0$ their single-species electrical conductivity at zero field.

One-carrier tensor. With $\mathbf{B}=B\hat{\mathbf{z}}$ and $\mathbf{J}=nq\mathbf{v}$, solving for the electric field givesSince $(\mathbf{J}\times \mathbf{B}{)}_x=B{J}_y$ and $(\mathbf{J}\times \mathbf{B}{)}_y=-B{J}_x$,Thus the resistivity tensor, defined by $\mathbf{E}=\mathbf{\rho }\mathbf{J}$, isIts off-diagonal signs follow from the stated axes and signed $q$. For zero transverse electric current, $E_y/J_x=B/(nq)$ is the Hall electrical resistivity; reversing charge reverses it. The single-species longitudinal electrical resistivity is independent of $B$ in this ideal isotropic model.

Adding electrons and electron holes. Let both species have density $n$ and charges $q=\pm e$, with $e>0$. The stipulated equal zero-field electrical conductivity means equal positive charge carrier mobility $\mu ={\sigma }_0/(ne)$, even if their masses and momentum relaxation times separately differ. Put $r=1/{\sigma }_0$ and $h=B/(ne)$. Inverting each electrical resistivity matrix givesBoth species experience the same electric field, so their electric current densities add. Therefore it is their conductivity tensors, rather than their resistivity tensors, that must be summed. The Hall parts cancel andThis is compensated two-carrier magnetoresistance. Although the net Hall field vanishes, each carrier species is bent in its own direction; maintaining the total longitudinal electric current requires an increasing longitudinal field. Equal carrier numbers alone do not ensure zero Hall response if charge carrier mobilities differ; both equalities given here are used.

Large-field limit and quantum oscillations. Expanding the exact diagonal entry,For $\mu B\gg 1$,Here the last equality applies to either species with its own mass and transport time; the stipulated charge carrier mobilities give the same threshold. This is the classical strong-field condition ${\omega }_c{\tau }_{\mathrm{tr}}\gg 1$: carriers turn substantially before their drift momentum relaxes.

Observable quantum oscillations additionally require sufficiently sharp Landau levels and a sufficiently cold Fermi-Dirac distribution, approximatelyWhen ${\tau }_q\simeq {\tau }_{\mathrm{tr}}$, the disorder field scale is comparable to the onset scale for large classical magnetoresistance. The thermal condition can demand a much larger field, and when ${\tau }_q\ll {\tau }_{\mathrm{tr}}$ even the disorder criterion is stronger. The quantitative damping factors in Q2(c) show that these are crossover scales rather than sharp visibility boundaries. A large smooth $B^2$ magnetoresistance therefore does not itself demonstrate Landau level; it can exist while thermal smearing suppresses oscillations. Conversely, the asymptotic condition ${\omega }_c{\tau }_{\mathrm{tr}}\gg 1$ is stronger than merely reaching unity, so there is no universal ordering of the two experimental onset fields without temperatures and lifetime information.

Scaling function. The material's actual zero-field longitudinal electrical resistivity is ${\rho }_{\mathrm{tot}}(0)=1/(2{\sigma }_0)$. Dividing the exact answer by this value givesThe assumption that $n$ stays constant is essential to eliminating further temperature-dependent parameters. $B{\sigma }_0$ is a dimensionful scaling variable; $B{\sigma }_0/(ne)=\mu B$ is its dimensionless version. The corresponding fractional increase is $[\rho (B)-\rho (0)]/\rho (0)=(\mu B{)}^2$.

What the supplied data test. In the original PDF's left panel, the field dependence is strongly temperature dependent. In the right panel, plotting the fractional magnetoresistance against the squared rescaled field makes the displayed 125, 150, 175, 200 and 225 K curves collapse approximately onto one straight line, over the field intervals where they overlap. This supports the proposed one-parameter scaling and, on that range, the quadratic field law. The 100 K curve does not collapse: it bends away and has a different slope in the rescaled plot. The indicated scaling is therefore supported for the displayed 125–225 K data, not for the entire displayed temperature range. The graph alone does not locate a sharp transition between 100 and 125 K or establish behavior above 225 K.

This empirical collapse is Kohler's rule. It is consistent with a common temperature-dependent charge carrier mobility or relaxation scale and a fixed carrier population; it is not a unique proof of exact compensation or of one microscopic scattering mechanism. A change in relative scattering times, carrier population or Fermi-surface structure could spoil the collapse, and this figure alone does not select the cause of the 100 K departure.

There is a factor-of-two normalization convention in the printed text. The ${\sigma }_0$ introduced for each species gives ${\rho }_{\mathrm{tot}}(0)=1/(2{\sigma }_0)$, whereas the final paragraph calls $1/{\sigma }_0$ “${\rho }_0$.” If ${\rho }_0$ denotes the experimentally measured zero-field electrical resistivity, the consistent slope isIf one instead keeps the literal single-species definition ${\rho }_s=1/{\sigma }_0$, the same model has $[\rho (B)-{\rho }_s]/{\rho }_s=-1/2+[B/{\rho }_s{]}^2/(2n^2{e}^2)$. The choice rescales and offsets the axes by temperature-independent constants, so it does not change which temperature curves collapse, but it must be resolved before inferring a carrier density from the slope.

Equilibrium junction. Put the p-type side on the left and n-type side on the right. When they are joined, electrons diffuse from n to p and electron holes diffuse from p to n. This leaves negative ionized acceptors on the p-side and positive ionized donors on the n-side of the depletion region. The resulting built-in electric field points from n to p, opposing further diffusion. At equilibrium there is one constant Fermi level, and drift and diffusion currents cancel for each carrier species.

In the nondegenerate, fully ionized limit, the neutral-region carrier densities areThe conduction band and valence band lie higher on the p-side and bend down towards the n-side. The gap between them stays fixed for a homojunction. If $\psi$ increases towards the n-side, both edges shift by $-e\psi$, whileThus $n$ rises and $p$ falls continuously across the junction; neither mobile population is an exact step function. Their depletion exposes fixed dopant charge density, rather than an absence of every kind of charge. The built-in voltage is

Light emission. Forward bias lowers the barrier and injects minority carriers across the p-n junction. In a light-emitting diode, electron-hole recombination converts some of their energy into photons. A direct band gap permits efficient radiative transitions with nearly conserved crystal momentum; nonradiative channels compete. Near-edge emission has $\hslash \omega \simeq E_g$ and $\lambda \simeq hc/E_g$. Material composition and confinement select the emission colour. Reverse-bias saturation is not the ordinary LED operating mode.

Reverse current. Take $V=V_R>0$ and $I_R>0$ to mean reverse voltage and reverse current magnitudes. The Shockley diode equation in this convention isThe bias-independent $I_S$ term is the supply of thermally excited minority carriers from the neutral regions, swept across the junction by its electric field. The opposing injection current is reduced by the extra barrier energy $e{V}_R$ and is $I_S{e}^{-e{V}_R/(k_{B}T)}$. Their equality gives zero net current at zero bias; for $e{V}_R\gg k_{B}T$, the reverse current approaches $I_S$ until leakage or breakdown invalidates this ideal law. The collecting field need not create the minority carriers; their thermal supply limits the diffusion-dominated current.

For an ideal long junction of area $\mathcal{A}$, steady diffusion with recombination gives the diode minority-carrier diffusion currentIt depends on doping, geometry, carrier diffusion coefficients and lifetimes, and especially the intrinsic carrier concentration. Depletion-region generation current and surface leakage can add nonideal contributions to the measured reverse current.

Band-gap estimate from the data. Read a nearly flat region well above the small-bias knee and below the high-bias upturn, rather than estimating $I_S$ in the breakdown region. For a diffusion-dominated junction with nearly temperature-independent gap and effective masses,so $I_S=C(T)T^3{e}^{-E_g/(k_{B}T)}$, where $C(T)$ includes transport and doping factors. Extracting a semiconductor band gap from diode saturation current amounts to measuring the slope of $\ln [I_S/(C(T)T^3)]$ against $1/T$: it is $-E_g/k_B$. If the prefactor is taken approximately constant over a short interval,Approximate readings from the original PDF plateaus areUsing the two warmer curves and $k_B=8.6173\times 10^{-5}eV/K$ givesKeeping the $T^3$ factor but taking the residual transport factor constant changes this estimate to approximately $0.57\mathrm{eV}$. The justified graphical estimate is thereforeThe colder curve is a genuine limitation of the supplied data: the $0$ to $-45{}^{\circ }\mathrm{C}$ pair gives only about $0.32\mathrm{eV}$ with a constant prefactor, and the endpoint pair gives about $0.42\mathrm{eV}$. Thus all three cannot support one precise diffusion-only activation slope. A depletion-region generation current, approximately proportional to $n_i$ rather than $n_i^2$, can have a roughly half-gap activation energy; other leakage and temperature-dependent transport can also alter the slope. The PDF alone does not identify their relative contributions. A reliable precision gap measurement would use more temperatures and verify the transport regime instead of forcing one fit through these three readings.

At equilibrium, majority electrons undergo diffusion from n to p, while majority electron holes diffuse from p to n. Their departure exposes positive donors on the n side and negative acceptors on the p side. The resulting electric field points from n to p and opposes further diffusion. Equilibrium is reached when the chemical potential is constant across the junction. The built-in potential is this electrostatic adjustment, not a discontinuity of the material's band gap.

Since electronic band energies shift by $-eV(x)$, both the conduction band and valence band undergo band bending down from p to n by $e{V}_b$ at zero applied bias. Their separation remains the same gap. In the nondegenerate thermal equilibrium limit the built-in voltage can also be expressed using neutral-edge densities as $e{V}_b=k_{B}T\log (n_n/n_p)=k_{B}T\log (n_n{p}_p/n_i^2)$.

Take the n-contact potential and its electrochemical potential as reference. Applying $V_{\mathrm{ext}}$ to the p side changes the total depletion-layer drop toThus $V_{\mathrm{ext}}<0$ is reverse bias: it increases the barrier and widens the depletion region. The bands have horizontal tangents at the neutral edges in this approximation and a smooth cubic bend within the linearly graded layer.

A biased junction is not in global thermal equilibrium. Its contact electron electrochemical potentials obeyUnder reverse bias the p-contact level is higher by $e|V_{\mathrm{ext}}|$. A single horizontal equilibrium chemical potential cannot describe this state; electron and hole quasi-Fermi levels describe the nonequilibrium populations. The sketch marks the contact levels. Detailed spatial quasi-Fermi profiles require transport and recombination information beyond the electrostatic doping profile.

At equilibrium the diffusion currents and drift currents cancel separately for each carrier species. Electron diffusion is from n to p, while the electric field drives electrons from p to n; hole diffusion is from p to n, while hole drift is from n to p. The net current is zero, although these opposing microscopic fluxes are present. In conventional-current signs, both majority diffusion contributions point p to n and the balancing drift contributions point n to p.

Forward bias, $V_{\mathrm{ext}}>0$, reduces the barrier and enhances majority-carrier diffusion and minority carrier injection. The ideal low-injection Shockley diode equation giveswhere positive current is p to n. For reverse bias, majority diffusion is suppressed and the field sweeps thermally supplied minority carriers across the junction, giving approximately $J=-J_s$. This saturation is limited by carrier supply rather than a current proportional to the field. Depletion-layer generation, recombination and eventual diode reverse breakdown modify the ideal result.

Let the neutral edges be $-x_p$ and $x_n$ and take the exterior electric field to vanish. Integrating $d{E}_x/dx=\rho /{\epsilon }_s=e\alpha x/{\epsilon }_s$ between the edges gives $x_n^2=x_p^2$, so $x_n=x_p=w/2$. This is also the condition of equal and opposite integrated depletion charges.

A second integration with $E_x(w/2)=0$ yieldsIt is negative inside the layer, pointing from n to p, and zero at both edges. Its largest magnitude occurs at the metallurgical junction:For reference, choosing $V(-w/2)=0$ gives the cubic electric potential

The drift velocity is the small ensemble-average velocity acquired by charge carriers in an applied electric field. It is distinct from their much larger random thermal or Fermi velocity. For conduction electrons of charge $-e$, the isotropic Drude model giveswhere $\tau$ is the momentum relaxation time.

The positive electron mobility ${\mu }_e=|{\mathbf{v}}_d|/|\mathbf{E}|$ measures the ease of this drift in linear response. In the same approximation,Its units are ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$; in an anisotropic crystal the mobility is a tensor.

The band effective mass describes the acceleration implied by the curvature of the dispersion relation. The group velocity is $\mathbf{v}={\hslash }^{-1}{\nabla }_{\mathbf{k}}E$, while $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$. ThusFor an isotropic parabolic minimum this reduces to $E=E_c+{\hslash }^2{k}^2/(2m^{\ast })$ and $m^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}$. It incorporates the periodic crystal potential and need not equal the bare electron mass. Negative electronic curvature near a band maximum is conveniently described by positive-mass electron holes.

Use the shock frame, with upstream and downstream fluid velocities $u_1=u$ and $u_2=u/4$ pointing in the same direction. These are the Strong-shock Rankine-Hugoniot conditions for a nonrelativistic monatomic gas with adiabatic index $5/3$. The two scattering media approach one another at relative speed $U=u_1-u_2=3u/4$, to first order in $u/c$. An energetic electron repeatedly crosses the shock wave, scatters from magnetic turbulence carried by the local plasma, and returns. Each scattering approximately conserves its energy in the local fluid rest frame while changing its direction; comparing successive fluid frames produces a systematic energy and momentum gain. The energy comes from the converging bulk flow, through the moving magnetic field irregularities and their associated electric fields. This is diffusive shock acceleration, a first-order Fermi acceleration process.

Let $\eta$ be the positive cosine of the crossing angle with the shock wave's normal. The Lorentz transformation between fluid frames gives, for either crossing direction with the corresponding sign of the normal reversed,For an almost isotropic population, the number crossing a unit area is weighted by its normal velocity. Thus the normalized crossing distribution is $2\eta d\eta$, rather than a uniform distribution in $\eta$, andFor an ultrarelativistic electron, $E\simeq pc$ and its speed $v\simeq c$, soMore generally $dE/dp=v$ gives $⟨\Delta p⟩/p=2U/(3v)$, while a nonrelativistic kinetic energy has $⟨\Delta K⟩/K=4U/(3v)$. A complete upstream–downstream–upstream cycle contains two crossings and has ultrarelativistic fractional energy gain $u/c$.

The initially cold electrons need an injection stage: shock wave heating and small-scale plasma processes must first produce particles with speed much larger than $u$ that can return across the front. The $v\simeq c$ result is a description of that accelerated population; a cold electron cannot simply be assigned $E\simeq pc$ at its first encounter.

Thermal bremsstrahlung is radiation from the acceleration of electrons in the Coulomb force of ions. At fixed temperature, the encounter rate per electron is proportional to ion number density $n_i$, and the number of radiating electrons per unit volume is $n_e$. Therefore $j_{\nu }\propto n_e{n}_i$ in a dilute, optically thin plasma. This is a binary-encounter rate, not a one-particle emission process.

For a fixed encounter geometry, the ionic electric field and hence the electron acceleration are proportional to ionic charge $Ze$. The Larmor formula is quadratic in acceleration, giving $j_{\nu }\propto Z^2$ at fixed species number densities. The dimensionless Gaunt factor $g(\nu ,T)$ corrects the simple classical result for the detailed quantum scattering and radiation calculation, after averaging over the thermal electron distribution. It depends on frequency and temperature and, in a more complete notation, ionic charge.

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.

Use coordinates $x^{\mu }=(ct,\mathbf{x})$, signature $(+---)$, and the convention $F^{0i}=-E_i/c$, $F^{ij}=-{ϵ}_{ijk}{B}_k$. The antisymmetric electromagnetic field tensor combines six field components:It transforms as $F^{\prime \mu \nu }={\Lambda }^{\mu }{}_{\alpha }{\Lambda }^{\nu }{}_{\beta }{F}^{\alpha \beta }$. Lowering indices changes the time–space signs: $F_{0i}=E_i/c$, while $F_{ij}=F^{ij}$. The covariant tensor instead uses the inverse coordinate Jacobian on each lower index.

For a frame moving with velocity $\mathbf{v}$ relative to the original one,Electric and magnetic fields therefore mix under a boost; they are frame-dependent parts of one tensor. For example, a purely transverse electric field acquires a magnetic field in a boosted frame. The electromagnetic field invariants $B^2-E^2/c^2$ and $\mathbf{E}\cdot \mathbf{B}$ restrict which fields can be eliminated by a frame change. The covariant force law $d{p}^{\mu }/d\tau =q{F}^{\mu }{}_{\nu }{U}^{\nu }$ guarantees consistent particle dynamics in all inertial frames.

Use Cartesian coordinates $x^a=(t,x,y,z)$ with $c=1$ and ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$. Spatial dot and cross products below use the Euclidean three-metric. The alternating tensor is given the orientation ${ϵ}^{0123}=+1$.

Vector and covector transformation. The vector $A=A^b{\partial }_b$ is independent of coordinates, and the chain rule gives ${\partial }_b=(\partial x^{\prime a}/\partial x^b){\partial }_a^{\prime }$. Likewise the one-form $A_{b}d{x}^b$ is independent of coordinates. Comparing coefficients provesTheir contraction $A_a{B}^a$ is unchanged because the forward and inverse Jacobians cancel. Here lowering an index uses the metric, not an identification of the numerical upper and lower components.

Why antisymmetrization matters. Write $J^c{}_a=\partial x^c/\partial x^{\prime a}$. Differentiating the covector law givesThe last term obstructs the general tensor transformation law. It is symmetric in $a,b$, so it cancels from ${\partial }_a^{\prime }{A}_b^{\prime }-{\partial }_b^{\prime }{A}_a^{\prime }$. ThereforeEquivalently $F=dA$ is the exterior derivative of the electromagnetic four-potential one-form. The statement that ${\partial }_a{A}_b$ is not a tensor needs a qualification: under affine Cartesian Lorentz transformations it does transform tensorially, because the second derivatives vanish. The failure concerns arbitrary nonlinear coordinate changes, including curvilinear coordinates in flat spacetime. This distinction is precisely the difference between general covariance and covariance under the restricted inertial-frame transformations.^[ref]

Electric and magnetic components. Choose $A^a=(\Phi ,\mathbf{A})$, so $A_a=(\Phi ,-\mathbf{A})$, $\mathbf{E}=-\nabla \Phi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$. ThenThus the lower-index field tensor isThis is a derived component representation of the field, with the source's sign convention.

Raised indices and the first invariant. Raising both indices changes the time-space signs and leaves the spatial block unchanged:In the scalar contraction, each independent antisymmetric pair is counted twice, givingUnder a Lorentz transformation each inverse transformation from a lower index cancels its matching transformation from an upper index, leaving the same scalar. In particular a plane electromagnetic wave has $E=B$ in these units and makes this invariant vanish, while a pure electric field makes it negative.

The printed dual convention. The PDF explicitly defines ${}^{\ast }{F}^{ab}={ϵ}^{abcd}{F}_{cd}$, with no $1/2$. Each independent pair therefore contributes twice: for example ${}^{\ast }{F}^{01}=F_{23}-F_{32}=-2B_x$. The complete result isThe time-space pairs contribute $-4\mathbf{E}\cdot \mathbf{B}$ and the spatial pairs contribute another $-4\mathbf{E}\cdot \mathbf{B}$. The conventional Hodge dual includes $1/2$ and would instead give the contraction $-4\mathbf{E}\cdot \mathbf{B}$; it must not silently replace this paper's definition. With fixed orientation the latter contraction is a pseudoscalar: invariant under proper Lorentz transformations but odd under orientation reversal if compared using a fixed alternating symbol.

Energy density and flux. Substitution into the given mixed stress-energy tensor yieldsFor the first formula, $F^{0c}{F}_{0c}=-E^2$ and $F_{cd}{F}^{cd}/4=(B^2-E^2)/2$. For the second, $F^{0j}{F}_{ij}=(\mathbf{E}\times \mathbf{B}{)}_i$. The minus sign on $T^0{}_i$ arises from its lowered spatial index; it does not reverse the physical energy flow. Raising that index gives $T^{0i}=S_i$, and symmetry gives $T^{i0}=S_i$, with $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$ the Poynting vector in $c=1$ units. Thus $T^{00}$ is energy density and $T^{i0}$ represents energy flux. In SI coordinates $x^0=ct$, restore $T^{0i}=S_i/c$ and $T^{00}=({ϵ}_0{E}_{\mathrm{SI}}^2+B^2/{\mu }_0)/2$, with ${ϵ}_0{\mu }_0{c}^2=1$. This keeps the mixed-index signs and the units distinct.

The required invariant is obtained by tensor contraction, rather than by an ordinary sum of squares of the entries of the electromagnetic field tensor. With ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, raising both indices givesThe electromagnetic field tensor is an antisymmetric second-rank tensor, so every independent off-diagonal entry occurs twice in its contraction. The three electric contributions each acquire a minus sign, whereas the magnetic contributions retain their sign. ConsequentlyAll indices are contracted, so this quantity is a Lorentz scalar. Since the speed of light is invariant,is Lorentz invariant. Both terms have the units of squared electric field, ${\mathrm{V}}^2{\mathrm{m}}^{-2}$ in SI units.