After field electron emission, let $h$ denote the plate separation of the parallel-plate capacitor, which the question does not specify. The uniform capacitor electric field has magnitude $E_c=|V|/h$, and a nonrelativistic electron of mass $m_e$ has constant acceleration magnitude $a=e{E}_c/m_e$. The sign of its charge affects the direction of the radiation field but not its radiated power.

In the supplied nonrelativistic Liénard-Wiechert field, $\dot{\mathbf{\beta }}=\mathbf{a}/c$. At polar angle $\theta$ to the acceleration, $|\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})|=a\sin \theta$, henceThe outgoing Poynting vector has magnitude $S={ϵ}_{0}c{E}_{\mathrm{rad}}^2$. Its radiated power per solid angle is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ gives the Larmor formula and the requested constant power during the uniform-field part of the flight:The brackets in the given field require the acceleration at retarded time; the radiation reaches the far observer after the propagation delay. This derivation assumes that radiation reaction is a small correction to the acceleration and that the flight is nonrelativistic, for example $e|V|\ll m_e{c}^2$ for release from rest. Entry, emission and impact transients are not described by the uniform-field model. A result solely in terms of $V$ cannot be obtained without a plate spacing or a specified internal field.

First prove the two electromagnetic field invariants from the supplied three-dimensional transformations. Then distinguish the correct covariant transversality of a vacuum plane wave from the index and equivalence issues in the original tensor statement.

Treat a single weakly driven electron or an independent bound charge, so that the incident electric field is approximately uniform across its displacement. The classical calculation uses electric dipole radiation and excludes both a resonant divergence and substantial quantum recoil.

Use retarded time $t_r$ and reserve $r$ for the large observation distance. The source velocity is nonrelativistic, $|\mathbf{u}|/c\ll 1$. There are two errors in the original PDF that must be corrected before the radiation zone calculation: the vector-potential prefactor contains an extraneous $1/r$, and the displayed magnetic radiation field has the wrong sign for the stated outward unit vector $\mathbf{n}$. The extra distance factor is dimensionally wrong; the magnetic sign would make the Poynting vector point inward. The following derivation uses dimensionally consistent retarded potentials and outgoing radiation.

The transverse and longitudinal quantities characterize different correlations: spatial optical coherence across a wavefront and temporal optical coherence between delayed fields. Their numerical values require a convention for how much correlation counts as coherent.

Use a transparent reciprocal wave plate at its design wavelength. For the return calculation, assume a mirror at normal incidence with the same reflection coefficient for both transverse components; this is the usual ideal double-pass quarter-wave plate arrangement.

Take $u>0$ to mean that the mirror recedes along the incident plane electromagnetic wave. Put $\beta =u/c$, with $|\beta |<1$. The moving-mirror Doppler shift can be obtained directly by matching wave phase on the moving surface. Write the incident wave phase as $k_{i}z-{\omega }_{i}t$ and the reflected wave phase as $-k_{r}z-{\omega }_{r}t$, where $k_i={\omega }_i/c$ and $k_r={\omega }_r/c$ are positive wavenumbers. At the mirror, $z=z_0+ut$, the two phases must have the same time dependence; their constant offset incorporates the reflection phase shift. ThusConsequently the observed reflected angular frequency and ordinary frequency areThis is a redshift for a receding mirror and a blueshift for an approaching mirror. At small velocity, ${\omega }_r\simeq {\omega }_i(1-2u/c)$.

Equivalently, the relativistic Doppler effect gives ${\omega }_i^{\prime }=\gamma (1-\beta ){\omega }_i$ in the mirror rest inertial frame. Reflection there reverses propagation and preserves frequency. Transforming the backward-going reflected wave to the laboratory gives another factor $\gamma (1-\beta )$, so ${\omega }_r={\gamma }^2(1-\beta {)}^2{\omega }_i={\omega }_i(1-\beta )/(1+\beta )$. A single Doppler factor would miss the effect of the return path.

Use a boost along the wire axis when comparing frames. The current variable in a four-current must be current density, in $\mathrm{A}{\mathrm{m}}^{-2}$, rather than a scalar total current in amperes. For the wire-force calculation it is convenient to integrate charge and current over each cross-section and work with line quantities.

Use the ideal electrically short Hertzian dipole, $d\ll \lambda$, with negligible ohmic loss. Far-field statements additionally require an observation distance large compared with the wavelength and source dimensions. These assumptions distinguish its radiation from near-field coupling and from scattering by a general extended antenna structure.

The five tasks below concern a scanned Michelson interferometer. The original dashed curves are labelled visibility in arbitrary units; divide them by their maximum to obtain normalized interferometric visibility, which lies between zero and one.

First obtain the particle speed from kinetic energy rather than treating $1\mathrm{GeV}$ as its total energy. With electron rest energy $m_e{c}^2=0.510999\mathrm{MeV}$,For Cherenkov radiation, the photon propagation direction makes angle ${\theta }_C$ to the particle trajectory, with $\cos {\theta }_C=1/(n\beta )$. The speed correction is negligible at three significant digits, but the Cherenkov photon-cone and wavefront-cone angles must be distinguished.

If the given opening $\alpha$ is the full cone of photon directions, ${\theta }_C=\alpha /2=48.75^{\circ }$. The result isIf instead $\alpha$ denotes the full opening of the conical wavefront, its half-angle is complementary to the photon direction: ${\theta }_C=90^{\circ }-\alpha /2=41.25^{\circ }$. That convention givesThe original PDF supplies no cone diagram or definition distinguishing these geometries. Thus the literal full photon-cone interpretation gives $1.52$, while $1.33$ follows from the wavefront interpretation. It would be incorrect to interchange sine and cosine without stating which cone is meant.

At normal incidence, continuity of tangential wave phase makes both internal wavevectors normal to the parallel faces. For the ordinary wave in a uniaxial optical medium, the electric field is an eigenpolarization of the permittivity tensor, so $\mathbf{D}\parallel \mathbf{E}$. ThusIts energy path is also normal to the faces. The extraordinary wave differs because $\mathbf{D}\perp \mathbf{k}$ but $\mathbf{E}$ need not be perpendicular to $\mathbf{k}$. In a nonmagnetic crystal, $\mathbf{H}\parallel \mathbf{B}$ and $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ is perpendicular to $\mathbf{E}$. The vectors $\mathbf{D},\mathbf{E},\mathbf{k},\mathbf{S}$ lie in the same meridional plane, with $\mathbf{D}\perp \mathbf{k}$ and $\mathbf{E}\perp \mathbf{S}$. Therefore the electric-displacement angle from birefringent walk-off equals the angular separation between $\mathbf{S}$ and $\mathbf{k}$.

The observed lateral separation is accumulated over thickness $L$, so $\Delta =L\tan {\alpha }_e$. With $L=10\mathrm{mm}$ and $\Delta =0.1\mathrm{mm}$,The two emergent rays are parallel because their outgoing wavevectors are both normal to the faces in air; their internal energy paths can nevertheless be displaced by birefringent walk-off. The angle quoted is the positive angle between the field vectors, so the unspecified lateral direction does not affect its value. The unpolarized light excites both eigenpolarizations; it does not require their energy paths to coincide.

The London penetration depth measures the density-to-mass ratio of carriers participating in superconducting screening. With electron-counted density $n_s$ and the stipulated bare electron mass $m_e$,Using $m_e=9.109\times 10^{-31}\mathrm{kg}$, $e=1.602\times 10^{-19}\mathrm{C}$, ${\mu }_0\simeq 4\pi \times 10^{-7}\mathrm{H}{\mathrm{m}}^{-1}$ and ${\lambda }_L=22\times 10^{-9}\mathrm{m}$ givesThis is the requested conduction-electron density within the assumed local London equation model. More precisely, it is the superconducting carrier density from London screening; identifying it with all conduction electrons assumes that they participate in screening at the measurement temperature. No temperature-dependent normal-carrier fraction is supplied.

If the condensate is instead counted as Cooper pairs, the pair mass is $2m_e$ and charge magnitude is $2e$, giving pair density $n_{\mathrm{pair}}=n_s/2\simeq 2.9\times 10^{28}{\mathrm{m}}^{-3}$. Keeping electron and pair counting consistent avoids a spurious factor of two.

A charge moving uniformly through a transparent dielectric can emit Cherenkov radiation if its speed exceeds the phase velocity of light in that medium:It need not exceed the vacuum speed $c$. The passing charge polarizes the medium; the induced response produces radiation that adds coherently along selected directions. Equivalently, phase matching along the trajectory requires $\mathbf{k}\cdot \mathbf{v}=\omega$. Since $|\mathbf{k}|=n(\omega )\omega /c$,Here ${\theta }_C$ is the angle of the emitted light rays to the forward particle velocity: the rays form a cone around the track. With dispersion, both the threshold and cone angle depend on frequency; only transparent frequency bands satisfying the threshold contribute escaping radiation.

In the wavefront construction, a disturbance emitted a time $t$ earlier has radius $ct/n$, while the particle has moved $vt$. Tangents from the present particle position to these wavefronts make a backward-opening wavefront cone of half-angle $\alpha$, where $\sin \alpha =1/(\beta n)$. The light rays are normal to this cone. Consequently $\alpha +{\theta }_C=\pi /2$: the Cherenkov photon-cone and wavefront-cone angles are complementary, rather than identical.

A phased array uses fixed antennas driven coherently, with adjustable electronic delays or phases. Changing those relative phases changes the direction of constructive interference.

For $N$ identical elements at ${\mathbf{r}}_j=jd\hat{\mathbf{x}}$, take time dependence $e^{-i\omega t}$ and feed phase $j\psi$. In a far-field direction at angle $\theta$ from the array normal, the array factor isMultiply $|F{|}^2$ by the single-element antenna beam pattern to obtain the total angular intensity. All elements add in phase in the selected direction ifThus electronic control of $\psi$ steers the beam to ${\theta }_0$ without moving an element. Reciprocity gives the corresponding directionality in reception.

The aperture $Nd$ controls angular resolution: the nearest nulls obey $|\sin \theta -\sin {\theta }_0|=\lambda /(Nd)$, giving a main-beam width of order $\lambda /(Nd\cos {\theta }_0)$ away from endfire. Additional equally strong maxima can occur when $\chi =2\pi m$; these are grating lobes. Choosing $d<\lambda /2$ avoids them for all steering directions, with $d=\lambda /2$ the usual limiting spacing. Amplitude tapering reduces side lobes at the cost of a wider main beam. For a broad frequency band, true time delays preserve steering more accurately than frequency-independent phase shifts.

Molecules in clean air are much smaller than visible wavelengths. Their induced electric dipole moments undergo Rayleigh scattering. Away from molecular resonances, the induced dipole amplitude is approximately $p=\alpha E$, with only weak wavelength variation in the polarizability $\alpha$. The radiation-zone electric field is proportional to $\ddot{p}$, so the scattered intensity satisfiesThis is the Rayleigh scattering wavelength dependence. For equal incident intensities, $450\mathrm{nm}$ light scatters about $(650/450{)}^4=4.35$ times as strongly as $650\mathrm{nm}$ light. Thus light scattered out of the direct solar beam and into our line of sight is enriched in blue wavelengths.

The dipole mechanism also determines the Rayleigh scattering angular pattern. For unpolarized light, its angular factor is $1+{\cos }^2\vartheta$; near $90^{\circ }$ scattering, the scattered light has strong linear polarization. The sky's perceived colour depends additionally on the solar spectrum, atmospheric transmission and the eye's spectral response, so the ${\lambda }^{-4}$ law alone does not predict a violet sky. Over a long near-horizontal path the direct beam loses more short-wavelength light, explaining red sunsets. Large aerosol particles instead produce Mie scattering, which is less strongly wavelength-selective and can make haze look whitish.