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A non-polarizing beam splitter divides the incident spectrum into a fixed arm and an arm with a translating mirror. Recombining their returned fields gives a detector signal as a function of mirror position $x$. The round-trip optical path difference is $D=2x$ relative to the equal-arm position, so the delay is $\tau =2x/c$ in air approximated as vacuum. Record both detector signal and calibrated position, subtract the constant background, and take a Fourier transform of the interferogram.

For matched arm polarizations and balanced throughput, write the ideal signal aswhere $S(\omega )$ is the incident spectral power density and $C$ is a constant set by the beam splitter and detector. This is the cosine-transform form of Fourier transform spectroscopy, or equivalently the Wiener-Khinchin theorem applied to delayed self-interference.

The delay-range resolution in Fourier transform spectroscopy follows because recording a finite delay range multiplies the correlation by a window. Its transform broadens each spectral line by an instrumental width of order the reciprocal delay range. A mirror travel range $L$ creates a delay range of order $2L/c$, soAt fixed wavelength, absolute wavelength resolution therefore improves inversely with travel range. The numerical coefficient depends on whether $L$ is a one-sided or total travel, on the window or apodization, and on whether resolution means first zero, half width or another criterion. The proportionality is the requested convention-independent result. Position sampling must also be fine enough to sample the fastest fringe variation without aliasing.

Birefringence gives two orthogonal linear eigenpolarizations with different refractive indices, $n_f$ and $n_s$. They accumulate different wave phases through the same physical thickness $d$. Their relative phase retardation isHere ${\lambda }_0$ is the vacuum wavelength. For the thinnest positive-retardance plate, $d={\lambda }_0/[4(n_s-n_f)]$. The plate uses a difference in phase velocity, rather than differential wave absorption. Equal wave amplitudes along the two axes, obtained from input linear polarization at $45^{\circ }$ to them, acquire phase quadrature and become circular polarization. Reversing that conversion gives linear polarization from the corresponding circular input. A different input angle generally gives elliptical polarization on one pass.

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.

The given spatial and carrier factors combine into retarded copies of the same random field:On the $y$ axis, $t_1=ky/(\sqrt{2}\omega )$ and $t_2=-ky/(\sqrt{2}\omega )$. Expand the squared electric-field wave amplitude. Each self term contributes $I_0/2$, and the two cross terms together contribute the real part of the full-field correlation. Stationarity and $\Gamma (-\tau )=\Gamma (\tau {)}^{\ast }$ giveIn vacuum, $k=\omega /c$ and $\tau (y)=\sqrt{2}y/c$. Reversing the delay convention replaces $\tau$ by $-\tau$ without changing this real-part result. This $I$ is the electric-field light intensity convention of the question. For two nonparallel waves it should not automatically be identified with the magnitude of the net Poynting vector; part (d) gives that geometric conversion explicitly.

A line with lifetime broadening has a Lorentzian line shape. Write its half-width at half-maximum in angular-frequency units as $\kappa >0$. The Wiener-Khinchin theorem, or the Fourier transform of the Lorentzian, givesTherefore in vacuumThe sketch is a cosine fringe pattern about baseline $I_0$, bounded by $I_0[1\pm e^{-\sqrt{2}\kappa |y|/c}]$. Its central value is $2I_0$, its spatial interference fringe spacing is $\lambda /\sqrt{2}$, and its visibility decays on length $c/(\sqrt{2}\kappa )$. For an isolated transition whose population lifetime is $T_1$ and which has no extra dephasing, $\kappa =1/(2T_1)$, so the field coherence time is $2T_1$, not $T_1$. Introducing $\kappa$ avoids confusing population, wave amplitude and spectral-width conventions when no numerical linewidth is given.

For each beam in vacuum, its magnetic field is ${\mathbf{B}}_j={\hat{\mathbf{k}}}_j\times {\mathbf{E}}_j/c$. Write the two complex scalar contributions, including the beam splitter factor, as $u_1=f(t-t_1)/\sqrt{2}$ and $u_2=f(t-t_2)/\sqrt{2}$. Since their electric fields are both along $z$,The optical-cycle and ensemble-averaged Poynting vector isThe mean individual light intensities are equal, soThus energy flows along the bisector of the two beam directions, crossing a detector plane normal to $x$. Such a sensor can in principle record the variation across its $y$ coordinate. Its actual normal energy flux is $I(y)/\sqrt{2}$ in the normalization used above; this fixed factor does not change interferometric visibility.

There is a practical resolution condition in addition to the direction of energy flow. The interference fringe spacing $\lambda /\sqrt{2}$ is very small, so direct pixels would need spacing no larger than $\lambda /(2\sqrt{2})$ as required by the Nyquist–Shannon sampling theorem, or optical magnification and sufficient angular aperture to preserve both incident directions. An ordinary optical camera with much larger effective pixels averages the interference fringes. The Poynting-vector argument establishes the allowed measurement geometry, not guaranteed resolution with any optical camera.

A physical delay is a time, whereas the stated $\lambda /(4\sqrt{2})$ is a length. First interpret it literally as an additional optical path length $\ell =\lambda /(4\sqrt{2})$ in arm 1, with the other arm unchanged. In vacuum this delays its full field by ${\tau }_d=\ell /c$. The exact stationarity argument from part (b) then givesDelaying arm 2 changes the plus sign before ${\tau }_d$ to a minus sign. As a function of $y$, the arm-1 result is the old pattern evaluated at $y+\ell /\sqrt{2}=y+\lambda /8$.

For a quasi-monochromatic source, ${\tau }_d$ is much shorter than the coherence time. The envelope correlation can then be treated as unchanged, while the carrier wave phase cannot:HenceThis literal optical-path delay is not a quarter-cycle wave phase shift. If the intended length instead specifies a displacement of the entire recorded fringe pattern by one quarter of its period, $\Delta y=\lambda /(4\sqrt{2})$, the equivalent delay is $\sqrt{2}\Delta y/c=\lambda /(4c)$ and the wave phase shift is $\pi /2$. Under that different interpretation the narrow-band result is $I_0+\mathrm{Im}\Gamma (\tau (y))$, with its sign reversed for the opposite displacement. The exact expression is still a shifted argument of $\Gamma$, not merely a wave phase multiplier when the envelope varies. The original wording does not specify such a fringe-pattern displacement, so these interpretations must be distinguished.

The emission mechanism is Cherenkov radiation. An electron moving faster than the phase velocity $c/n$ of light in the material excites electric polarization disturbances whose radiation adds coherently on a cone. This does not require the electron to exceed the speed of light in vacuum. Taking $n=1.4$ as constant over the detected optical band, the Cherenkov cone angle satisfiesEmission requires $\beta >1/n$. The detector accepts ${\theta }_C<20^{\circ }$, soNumerically,The lower endpoint has vanishing Cherenkov radiation yield, and the upper endpoint is excluded by the stated strict angular acceptance. Faster electrons emit at larger cone angles: at $v\simeq c$, ${\theta }_C\simeq {\cos }^{-1}(1/1.4)=44.4^{\circ }$, outside this detector's acceptance. The angular condition gives the accepted speed interval conditional on rays reaching the detector; no collection geometry or total signal yield is specified.

Birefringence means that two independent polarization eigenmodes have different refractive indices at the same frequency and propagation direction. The rotatory-power formula relevant here requires circular birefringence: the eigenmodes are opposite circular polarizations. General linear birefringence instead introduces phase retardation between linearly polarized components and need not rotate every incident linear polarization through one common angle.

Use the complex amplitude convention $\mathrm{Re}[\mathbf{E}{e}^{-i\omega t}]$ and transverse unit vectors ${\mathbf{e}}_1,{\mathbf{e}}_2$. DefinePropagation through thickness $d$ adds phases ${\varphi }_{\pm }=(2\pi /{\lambda }_0)n_{\pm }d$, where ${\lambda }_0$ is the vacuum wavelength. Recombining the two components givesThus the rotation and the specific rotatory power per unit length areThe units of $\rho$ are $\mathrm{rad}{\mathrm{m}}^{-1}$. The sign follows the specified circular basis and viewing convention. Equal attenuation of the two eigenmodes has been assumed; unequal attenuation also produces elliptical polarization.

Only the $1/r$ fields transport a nonzero outgoing power to infinity. Retain those terms from the Hertzian dipole fields:The brackets mean evaluation at the retarded time,for a point source at the origin in vacuum. They are not a time average. The radial Poynting vector in the radiation zone isIntegrating over a sphere usesThereforeNear-field terms either give inverse powers of $r$ after the surface integral or describe reversible exchange with the near field. The emitted power labelled by source time $t_{\mathrm{src}}$ is ${\mu }_0\ddot{p}(t_{\mathrm{src}}{)}^2/(6\pi c)$; the enclosing surface receives it one light-travel time later. For a changing dipole orientation, the general form is $P={\mu }_0|\ddot{\mathbf{p}}{|}^2/(6\pi c)$.

The given static induced surface charge and the dipole integral from part (b) giveHere $ϵ$ is the dimensionless relative permittivity, as required by the stated $ϵ\pm 1,2$ expression; ${\alpha }_E$ is the SI electric polarizability. Take ${\mathbf{E}}_{\mathrm{inc}}(t)=E_0\cos \omega t\hat{\mathbf{e}}$ and $k=\omega /c=2\pi /\lambda$. In the Rayleigh scattering regime the field is approximately uniform across the sphere. Its induced dipole radiates mean powerThe incident time-averaged light intensity in vacuum is $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$. Dividing power by intensity gives the total cross section,The units are ${\mathrm{m}}^2$; the comparison is with the geometric projected area $\pi R^2$, not with a length:The inequality holds for an ordinary nonabsorbing dielectric, with finite contrast factor and $kR\ll 1$. It explains why a sufficiently small transparent sphere scatters weakly and why its scattered power has the characteristic ${\lambda }^{-4}$ dependence. Near a material resonance where $ϵ+2$ is small, this simple inequality and static-polarizability approximation need reconsideration, including loss and radiative corrections. For the stated low-frequency ordinary dielectric the supplied polarizability is the intended approximation.

In SI units, Maxwell's equations areDotting the last equation with $\mathbf{E}/{\mu }_0$, and using the curl equation for $\mathbf{E}$, givesIntegrate over the fixed region and apply the divergence theorem. This proves Poynting's theorem: the rate of increase of electromagnetic energy stored inside, plus the outward Poynting flux, plus the rate of work on charges, is zero. The last term need not be positive for arbitrary externally driven currents; when positive it is power transferred from the field to matter.

For the travelling wave put $\phi =kz-\omega t$. A purely oscillatory magnetic field satisfying Faraday's law isBoth divergences vanish. The vacuum Ampère-Maxwell law then requires $k^2={\mu }_0{ϵ}_0{\omega }^2$, so with the positive constants in the question,A superposed static background is not part of the pure plane-wave solution. The electric and magnetic energy contributions are equal, givingConsequently $\boxed{|\mathbf{P}|=c\mathcal{U}}$ pointwise, including the nodes where both sides vanish.

A gravitational wave produces a time-dependent tidal distortion. For a normally incident plus-polarized wave in the long-wavelength limit, two perpendicular freely suspended arms change by $\delta L_x/L=h_{+}/2$ and $\delta L_y/L=-h_{+}/2$. Their differential length is thereforefor a simple round-trip Michelson interferometer. LIGO uses kilometre-scale orthogonal arms with optical cavities that enhance the phase response, a stabilized laser and suspended mirrors in vacuum. As an illustrative scale, $h=10^{-21}$ across a 4 km arm corresponds to differential length $4\times 10^{-18}\mathrm{m}$. The measured quantity is returning light phase, not mirror displacement against a rigid external ruler. Optical storage and control loops give a frequency-dependent calibrated response.

Seismic and terrestrial disturbances dominate low frequencies, thermal motion of suspensions and mirror coatings contributes at intermediate frequencies, and photon shot noise limits high-frequency readout. Isolation and suspension systems reduce ground coupling; low-loss materials reduce thermal noise; vacuum suppresses gas scattering and forces. Greater optical power reduces shot noise but increases radiation-pressure fluctuations and thermal demands. Stable optical alignment and feedback are essential, yet their own sensor and actuator noise must be controlled. These are engineering requirements because the desired strain is extraordinarily small.^[ref]

A candidate should appear coherently at separated observatories with consistent propagation delay and antenna response. Auxiliary environmental monitors, calibration injections and time-shifted coincidence backgrounds test instrumental explanations. Matched filtering is powerful for modelled compact-binary chirps; coherent burst searches allow less predictable signals. Binary-pulsar orbital energy loss already provided indirect evidence for gravitational radiation before 2013, but it is distinct from directly measuring passing waves.^[ref] At the examination date, direct detection remained a search objective. Reliable detection requires both sufficient strain sensitivity and a demonstrated astrophysical signal rather than correlated or accidental noise.

For a homogeneous canonical scalar field, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Potential-energy domination gives $P_{\varphi }\simeq -{\rho }_{\varphi }$, accelerated expansion and a decreasing comoving Hubble radius. Hubble friction can permit slow roll down a sufficiently flat potential. Enough inflation stretches an initially small causally connected region to encompass our observable universe, suppresses the relative spatial-curvature term and dilutes unwanted massive relics. quantum fluctuations supply possible seeds for cosmological structure; a successful model must also end inflation and create the hot post-inflationary universe. These statements do not imply that arbitrary initial conditions inevitably inflate.

Use the stated reduced-Planck units throughout the field calculation. Differentiating the Friedmann constraint and substituting the field equation givesDivision assumes the nontrivial expanding branch $H>0$; the vacuum endpoint follows continuously.

During rapid oscillations, require $m\gg H$ and $|\dot{A}|\ll mA$. To leading order, $\dot{\varphi }\simeq -mA\sin mt$, while the Friedmann constraint gives $H_m^2\simeq m^2{A}^2/6$. ThusAveraging over many oscillations on a timescale still short compared with $H^{-1}$ givesThe quadratic-inflaton Hubble oscillations obey, to leading order in $H_m/m$,Differentiating this expression reproduces the cosine term, with smaller corrections from the slowly varying amplitude. Instantaneously $H$ remains nonincreasing because its exact derivative is $-{\dot{\varphi }}^2/2$; an oscillatory correction does not mean oscillatory growth of the total Hubble parameter.

The mean equation integrates toThe shift $t_{\ast }$ is set by matching to the earlier evolution, not necessarily the Big Bang origin of the full solution. Averaged kinetic and potential energies are equal, so $⟨P_{\varphi }⟩\simeq 0$, $⟨{\rho }_{\varphi }⟩\propto a^{-3}$ and $A\propto a^{-3/2}$. The coherent quadratic oscillations behave as pressureless matter, not radiation, even though they are a scalar field rather than a gas of cold particles. radiation would instead give $a\propto t^{1/2}$. The averaged approximation is controlled only after oscillations are fast compared with expansion.

reheating requires interactions beyond the isolated field equations above. The oscillating inflaton can decay into other particles, whose scattering and further decays establish a thermal bath. The perturbative inflaton reheating equations areExpansion redshifts the components differently; transfer becomes efficient around $H\sim \Gamma$, and radiation domination follows when the transferred relativistic energy dominates. For suitable couplings, nonperturbative preheating by parametric resonance can precede thermalization, with backreaction and rescattering affecting its completion. Neither the decay rate nor thermalization is specified by $m$ and $V$ alone. Instantaneous reheating idealizes all this as immediate conversion of the end-of-inflation energy to radiation.^[ref]

For instantaneous-reheating horizon matching, specify the comparison: take the initial patch to be of order $H_i^{-1}$, compare the same-shaped inflationary history with approximately fixed $H_i/H_{\mathrm{end}}$, and assume instantaneous reheating followed by adiabatic expansion. Ignoring changes in relativistic degrees of freedom, $a_0/a_{\mathrm{reh}}\propto T_{\mathrm{reh}}$, while energy conservation and the Friedmann equation give $H_{\mathrm{end}}\propto T_{\mathrm{reh}}^2$. The present size of this inflated patch therefore scales asAt fixed present Hubble radius, a higher instantaneous reheating scale requires more e-folds in this standard comparison, approximately $\Delta N\simeq \ln (T_2/T_1)$, up to slow scale and entropy-degree corrections. Hotter reheating gives more subsequent expansion, but the associated initial Hubble patch is smaller by a stronger power of temperature. If instead one artificially holds the initial physical patch or $H_i$ fixed while varying the end energy independently, the post-inflationary geometric effect alone would give the opposite trend. temperature without an inflation-scale comparison convention does not fix that alternative problem; this is why the matching assumptions must be stated.^[ref]

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.

The electromagnetic energy density in vacuum is $u_E={ϵ}_0{E}^2/2$. For a length ${\ell }_z$ of the coaxial capacitor, integrate over annular shells:Since $C=C^{\prime }{\ell }_z=2\pi {ϵ}_0{\ell }_z/\eta$, this becomesThe integral is over the inter-cylinder electric field, and gives the stored energy, independently of which electrode is assigned zero electric potential.

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, Maxwell's equations in vacuum, allowing charge density $\rho$ and electric current density $\mathbf{J}$, areThe source-free specialization sets $\rho =0$ and $\mathbf{J}=0$; it cannot describe the charged objects at their locations. Applying the divergence theorem to the first equation gives Gauss's law:For an isolated point charge $q_1$, spherical symmetry makes the electric field radial and constant in magnitude on a concentric sphere. Therefore $4\pi r^2{E}_r=q_1/{ϵ}_0$. Multiplying by a second point charge gives Coulomb's law,The force points away from the first charge for like signs and toward it for opposite signs.

For the spherical capacitor, a Gaussian sphere between the shells encloses $q$, so $E_r=q/(4\pi {ϵ}_0{r}^2)$ for $a<r<b$. Integrating the electric field inward from the outer shell yieldsAn arbitrary constant $V(b)$ remains unless a reference is specified. The capacitance is determined by the potential difference:For the subsequent charging operation, take the initial outer shell to be earthed, or interpret the stated $V$ as the inner-to-outer voltage with the outer reference zero. Isolating the inner shell fixes its electric charge at $q=CV$. Removing the outer shell leaves an isolated conducting sphere. Its electric field and electric potential, now with zero at infinity, areConsequentlyThese changes of electric potential are compatible with constant electric charge. Without the initial reference convention the absolute stated $V$ does not determine $q=CV$: for example, if the outer shell were initially isolated and neutral, the absolute inner electric potential would instead be $q/(4\pi {ϵ}_{0}a)$.

Finally take the connected spheres to be sufficiently far apart that the long wire's capacitance and their mutual influence can be neglected. Their common electric potential requiresso $q_a=q/3$ and $q_{2a}=2q/3$. The magnitudes of the surface electric fields areThis is the usual widely separated-sphere approximation. A wire enforces an equipotential, but alone does not make this ratio exact for arbitrary arrangements. As a concrete counterexample, if the second shell enclosed the first concentrically and they were connected, all net electric charge would reside on the exterior and the inner sphere's surface electric field would be zero.

In vacuum, Maxwell's equations in differential form areHere $\rho$ is charge density, $\mathbf{J}$ is electric current density, $\mathbf{E}$ is electric field, and $\mathbf{B}$ is magnetic flux density. In a source-free region, taking the curl of the third equation and using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$ gives the wave equationThe same argument gives the wave equation for $\mathbf{B}$. A concrete family satisfying all four Maxwell's equations isThe function $F$ may be any twice differentiable profile. These are transverse travelling waves; reverse the magnetic field direction for propagation in the opposite direction. Because the source-free Maxwell's equations are homogeneous linear differential equations, any finite linear combination of such paired electric fields and magnetic fields is again a solution. Infinite superpositions also work whenever the required derivatives converge. A solution of the component wave equations alone must still satisfy the divergence and mutual-curl constraints.

In vacuum, Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:Take opposite electric charges $\pm Q$ on the two electrical conductors, with surface charge densities $\pm \sigma =\pm Q/A$. A Gaussian pillbox crossing one isolated large sheet gives $2ES=\sigma S/{ϵ}_0$, so each sheet contributes $\sigma /(2{ϵ}_0)$ on either side. By the superposition principle, the electric fields add between the sheets and cancel outside. Ignoring the fringing field of a capacitor, the gap electric field is $E=\sigma /{ϵ}_0$. The electric potential difference and capacitance of this parallel-plate capacitor are consequentlyThe approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with permittivity $ϵ$ replaces ${ϵ}_0$ by $ϵ$.