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Use peak phasors with time dependence $e^{-i\omega t}$, real positive electrical conductivity $\sigma$, and absolute permeability ${\mu }_a=\mu {\mu }_0$. Ampère-Maxwell equation and Faraday equations giveso ${ϵ}^{\prime }=ϵ+i\sigma /(\omega {ϵ}_0)$. The sign of the imaginary term follows from the stated time convention. For a transverse plane wave in a homogeneous conductor,In the good-conductor limit, choose the root that decays into $z>0$:The electric phasor and physical field are consequently ${\tilde{E}}_x=E_0{e}^{(i-1)z/\delta }$ and $E_x=\mathrm{Re}[E_0{e}^{i(z/\delta -\omega t)-z/\delta }]$, as required. Tangential boundary matching sets $E_0$; it need not equal the incident amplitude.

Faraday's law fixes the magnetic field, including its phase:For real $E_0$ and $\psi =z/\delta -\omega t$, this means $H_y=E_0{e}^{-z/\delta }(\cos \psi -\sin \psi )/(\omega {\mu }_a\delta )$, with other components zero. The time-averaged Poynting vector isThe energy entering a thin slab exceeds that leaving it by $-d⟨N_z⟩/dz$ per unit volume, givingDirectly, $\mathbf{J}=\sigma \mathbf{E}$ and $\mathbf{J}\cdot \mathbf{E}=J^2/\sigma$ at each time. Therefore $⟨J^2/\sigma ⟩=\sigma |{\tilde{E}}_x{|}^2/2$ is exactly this loss. The factors one half use peak rather than root mean square phasors; amplitude decays over $\delta$, while intensity decays over $\delta /2$.

For a cold collisionless unmagnetized plasma, electron inertia gives $\tilde{\sigma }=i{ϵ}_0{\omega }_p^2/\omega$ and real ${ϵ}^{\prime }=1-{\omega }_p^2/{\omega }^2$. Above the plasma frequency the wave propagates without this resistive attenuation. Below it, a semi-infinite plasma supports an evanescent field with electric and magnetic components in quadrature and zero mean inward energy flux: the incident energy is reflected, not absorbed as Joule heating. electron kinetic energy is reversibly stored, so the nondispersive dielectric field-energy formula alone is incomplete. Collisions introduce a real dissipative electrical conductivity and can produce absorption.^[ref]

For a linear nondispersive medium with real absolute permittivity ${ϵ}_a=ϵ{ϵ}_0$ and permeability ${\mu }_0$, instantaneous field energy density and energy flux arewith local balance ${\partial }_{t}u+\nabla \cdot \mathbf{N}=-\mathbf{J}\cdot \mathbf{E}$. energy stored locally and energy passing through a surface are distinct: integrating the balance relates their rates and the power given to charges.

Between the coaxial conductors, cylindrical Gauss and Ampère laws giveThe inner line charge is $\lambda =2\pi {ϵ}_{a}V/\ln (b/a)$, so the capacitance per length is $C=2\pi {ϵ}_a/\ln (b/a)$. Integrating magnetic energy over the annulus gives $\int B^2(2\pi rdr)/(2{\mu }_0)={\mu }_0{I}^2\ln (b/a)/(4\pi )=L{I}^2/2$, hence $L={\mu }_0\ln (b/a)/(2\pi )$. ThusThese are per-length parameters for ideal conductors, excluding internal magnetic penetration.

Now let the voltage and current amplitudes be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular dielectric. The two nontrivial curl equations reduce toConsequently both $k^2={\omega }^2{\mu }_0{ϵ}_a$ and $V_0=Z_0{I}_0$ are required for a forward wave. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy Maxwell's equations. This TEM travelling-wave voltage-current relation and the dispersion constraint identify the transverse electromagnetic mode and its speed $v_p=1/\sqrt{LC}$.

For the travelling wave, $⟨N_z⟩=\mathrm{Re}(V_0{I}_0^{\ast })/[4\pi r^2\ln (b/a)]$. Integrating over the dielectric annulus yieldsThis is the familiar circuit power, carried here through the dielectric rather than through the ideal metal. In root mean square notation it is $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ for the forward lossless wave; the corresponding static fields give $VI$ without a time-average factor. The average line energy is $(C|V_0{|}^2+L|I_0{|}^2)/4=C|V_0{|}^2/2$, whose product with wave speed agrees with the power.

Finite conductor resistance requires a small longitudinal electric field to drive current. Its cross product with the azimuthal magnetic field produces a radial Poynting component into each conductor: inward radially at the inner conductor and outward radially into the outer conductor. This Poynting flux into lossy coaxial conductors explains why forward power decreases along the cable as energy becomes heat in the metal. The mode is then only approximately TEM; strictly zero longitudinal electric field would be inconsistent with resistive current.

The dimensions are $[\rho ]=M{L}^{-3}$, $[\sigma ]=M{T}^{-2}$ and $[\lambda ]=L$. Equating exponents in $L{T}^{-1}=M^{\alpha +\beta }{L}^{-3\alpha +\gamma }{T}^{-2\beta }$ yieldsThus the model has $v_{\mathrm{ph}}\propto \sqrt{\sigma /(\rho \lambda )}$. Its common dimensionless factor cancels in the ratio:They are nearly equal within this capillary-wave model. The assumed power law omits gravitational restoring forces and finite-depth effects; dimensional analysis cannot establish those omissions for an arbitrary real surface wave.

Here $V(x)$ is potential energy, $E$ is the stationary state's total energy, and $\psi (x)$ is its spatial probability amplitude. The wave function has time factor $e^{-iEt/\hslash }$; $|\psi {|}^2$ is a position probability density for a normalized state. Stationary plane-wave scattering solutions are idealizations described by probability-current ratios, or components of a normalizable wave packet.

Above an upward step, a classical particle is fully transmitted with reduced speed. Quantum mechanically the abrupt wavelength change generally produces both reflected and transmitted amplitudes even though both regions are classically allowed. With unit incident amplitude and no wave arriving from the right,The reflected component has signed wavevector $-k_1$ and the same wavenumber magnitude as the incident component. A finite potential jump with constant mass and no delta interaction requires continuity of $\psi$ and ${\psi }^{\prime }$:The probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ gives probability reflection and transmission $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(k_2/k_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The amplitude $t$ may exceed one without violating conservation of total probability, because the transmitted group speed is smaller.

For $0<E<V_0$, let $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The bounded right-hand solution is $t{e}^{-\kappa x}$, an evanescent wave tail with amplitude penetration scale $1/\kappa$. It carries no outgoing probability current in an infinitely long forbidden half-space, so reflection is total. This differs from transmission through a finite potential barrier, where a second boundary can support quantum tunnelling.

For a downward step, replace $k_2$ by $\sqrt{2m(E+V_0)}/\hslash >k_1$. The same boundary conditions give a nonzero negative $r$, corresponding to a reflected amplitude with phase reversal. If reflection were forbidden, $r=0$ would require both $t=1$ and $k_1=k_2$, impossible for a nonzero step. Reflection therefore does not require a classical potential barrier; the discontinuity of wavenumber mismatch is enough.

For the specified electron energies, $k_2/k_1=\sqrt{40/100}=\sqrt{0.4}$, so the requested amplitude coefficients areThe corresponding probability coefficients are approximately $\mathcal{R}=0.05069$ and $\mathcal{T}=0.94931$. electron mass cancels from these ratios; there is no need to invent an additional material effective mass.

For a flexible uniform string with linear mass density $\rho$ and tension $T$, dimensional analysis gives $[T/\rho ]=L^2{T}^{-2}$ and hence $c\propto \sqrt{T/\rho }$. The small-slope force balance $\rho {\psi }_{tt}=T{\psi }_{xx}$ fixes the factor to one:Here $\rho$ has units kg/m, not the volume-density units used for a bulk fluid. The phase moves along the string at $c$, while a material point moves transversely with ${\psi }_t=A\omega \sin (kx-\omega t)$. Its maximum speed is $A\omega$, which is not the phase velocity. Small slope $Ak\ll 1$ makes $A\omega \ll c$ in this model.

The kinetic energy and extra elastic energy of a segment areThe second expression follows from extra length $\sqrt{1+{\psi }_x^2}dx-dx\simeq {\psi }_x^{2}dx/2$ against the pre-existing constant tension. Time averaging a harmonic wave gives $⟨{\mathcal{E}}_K⟩=\rho A^2{\omega }^2/4$ and $⟨{\mathcal{E}}_U⟩=T{A}^2{k}^2/4=\rho A^2{\omega }^2/4$. Therefore

At the driven endpoint the string pulls along its local tangent into $x>0$, with vertical component $T{\psi }_x$ for small slope. The point exerts the opposite force on the string, so the driven-string endpoint power isFor the outgoing wave $\psi =A\cos (kx-\omega t)$, at $x=0$ the derivatives are ${\psi }_x=Ak\sin \omega t$ and ${\psi }_t=-A\omega \sin \omega t$. Thus $P=T{A}^{2}k\omega {\sin }^2\omega t$ andThe energy flux of a transverse string wave follows by combining the energy density $\mathcal{E}=(\rho {\psi }_t^2+T{\psi }_x^2)/2$ and flux $P=-T{\psi }_x{\psi }_t$, which satisfy ${\partial }_t\mathcal{E}+{\partial }_{x}P=0$. For a pure right-going nondispersive wave, $P=c\mathcal{E}$ even instantaneously. energy is transported with the wave; a material string element merely oscillates about its unperturbed position.

Coastal Kelvin-wave radiation of long Rossby waves requires a common phase convention to match the two waves. The supplied Rossby wave formula conventionally multiplies $e^{i(k_{R}x+{\ell }_{R}y-{\omega }_{R}t)}$, while the coastal expression uses $e^{i({\omega }_{K}t-{\ell }_{K}y)}$. Since physical fields are real, matching may use the complex-conjugate representative; in either convention the boundary forcing fixes the absolute frequency and alongshore wavenumber. Thus $|{\omega }_R|=|{\omega }_K|$ and $|{\ell }_R|=|{\ell }_K|$.

For $k_R^2+{\ell }_R^2\ll L_d^{-2}$, the shallow-water Rossby-wave dispersion relation reduces toChoosing $k_R>0$ recovers the printed magnitude relation. Both the zonal phase velocity and zonal group velocity are approximately $-\beta L_d^2$, so the long Rossby waves carry the coastal signal westwards into the basin.

The coastal condition $|{\ell }_K|L_d\ll 1$ alone does not guarantee a long zonal wavelength. Since $|{\omega }_K|=c|{\ell }_K|$, matching requires $|k_R|=c|{\ell }_K|/(\beta L_d^2)$ and hence the stronger consistency conditionFor example, at $10^{\circ }$ a coastal wavenumber with $|{\ell }_K|L_d=0.05$ gives $|k_R|L_d\simeq 0.56$, which is not asymptotically small even though the meridional wavelength is long. Thus the source's proposed excitation is possible at sufficiently low frequency; it is not an automatic consequence of its meridional long-wave assumption.

At $10^{\circ }$ N or S the beta plane parameter isTaking a representative Pacific basin width $W\simeq 1.5\times 10^7$ m givesA width of $1.4$–$1.7\times 10^7$ m gives roughly $2.0$–$2.4$ years. This is propagation near $10^{\circ }$ latitude, not an equatorial $f=0$ calculation. Slow Rossby wave adjustment supplies a basin-scale delay relevant to El Niño–Southern Oscillation: changing thermocline depth can affect later sea surface temperature and coupled wind forcing. Reflection and coupling may help produce interannual memory, but this travel-time estimate alone does not establish a climate oscillation or its period.

Differentiate the approximate dispersion: $v_g=3k{v}_{th}^2/\omega$, while $v_{\varphi }=\omega /k$. Therefore $\boxed{v_g{v}_{\varphi }=3v_{th}^2}$. In the high-phase-speed regime that justified the approximation, the group speed is correspondingly small. Trying to make $v_g$ much larger than $v_{th}$ would put the phase speed into the thermally populated distribution, where resonant Landau damping is strong and the undamped-fluid estimate becomes unreliable. This explains the stated tendency for these electron electrostatic waves, not a universal prohibition on fast electromagnetic waves in plasma.

For a 1 micrometre laser, $n_c\simeq 1.1\times 10^{27}$ m${}^{-3}$ and a cold underdense wave has ${\gamma }_{\varphi }^2\simeq n_c/n_e$. At peak wave scale $E_0$, optimal passage through an accelerating half-cycle gives ideal stage gain $\Delta \mathcal{E}\sim 4(n_c/n_e)m_e{c}^2$; crest-to-zero injection gives half that. With the illustrative $n_e\simeq 7\times 10^{22}$ m${}^{-3}$, the optimal ideal gain is about 32 GeV and a 9.9 TeV increment needs order 300 stages, or order 600 with the crest convention.

TODO — no unique stage count follows without the missing density and laser/wake amplitude assumptions. Depletion, diffraction, guiding, beam loading, injection, energy spread, driver evolution and interstage losses can all increase or otherwise alter the requirement. The parametric count is $N\gtrsim \Delta {\mathcal{E}}_{\mathrm{total}}/[4\eta (n_c/n_e)m_e{c}^2]$, with the phase and field conventions explicitly retained.

Put $\beta =m\alpha /{\hslash }^2$. The delta interaction reduces the integral equation to ${\psi }_s=e^{ikx}+i\beta e^{ik|x|}{\psi }_s(0)/k$. At zero, ${\psi }_s(0)=(1-i\beta /k{)}^{-1}$. HenceTheir sum is one, as required for a real potential. The explicit wave is the indicated incident-plus-reflected form for $x<0$ and $t{e}^{ikx}$ for $x>0$; it is continuous and has derivative jump $-2\beta \psi (0)$.

The first Born approximation substitutes the incident wave for the exact wave inside the scattering integral. ThusFor the attractive delta this is ${\beta }^2/k^2$, the leading small-$\beta /k$ expansion of the exact result. Physically one neglects the distortion of the incident state by repeated interaction with the potential. At low $k$ the Born result can exceed one, while the exact reflection remains bounded and approaches one; it is not valid in that limit.

The full equation rearranges to $L\psi =-U\psi$. Since a bound state has no incoming free wave, convolution with the decaying inverse gives $\boxed{\psi (x)=-m\int e^{-\kappa |x-x^{\prime }|}U(x^{\prime })\psi (x^{\prime })d{x}^{\prime }/(\kappa {\hslash }^2)}$. Applying $L$ to this identity recovers the differential equation and the decay boundary conditions.

The supplied density is $4\times 10^{24}$ m${}^{-3}$. ThereforeThe cold relativistic field scale is $E_0=m_{e}c{\omega }_p/e\simeq 1.92\times 10^{11}$ V m${}^{-1}$, corresponding to an energy gradient $\boxed{e{E}_0\simeq 192\mathrm{GeV}{\mathrm{m}}^{-1}}$. It is an upper scale for the stipulated large-amplitude wave, not proof that the particular weak driver produces that field.

The accumulated phase along a one-dimensional ray. In an oscillatory WKB approximation, the two travelling waves have phases $\pm \int kdx$. In an evanescent region, the positive local decay rate $\kappa =\sqrt{-k^2}$ replaces $k$, producing exponents $\pm \int \kappa dx$.

The study of physical properties and processes in the oceans, including fluid flow, waves and transport of heat.

Physical optics describes light as a wave, including wave interference and diffraction.

A point on a plasma wave branch where the local wavenumber tends to zero at finite frequency. The mode may change between propagation and evanescence. For a finite magnetized cold-plasma dielectric tensor, candidate cutoffs satisfy $PRL=0$; special propagation angles require checking which factor belongs to the chosen branch.

A limiting point of a plasma wave branch with divergent refractive index. At finite coefficients of the magnetized cold-plasma dielectric tensor, the generic condition is $S{\sin }^2\theta +P{\cos }^2\theta =0$. Poles of that magnetized cold-plasma dielectric tensor require a separate limiting analysis. Collisions, thermal dispersion or mode conversion can regularize a singular cold prediction.

Polarization of waves describes the direction and time-dependent geometry of a transverse wave's vector during its oscillation at a fixed position. In three spatial dimensions, a transverse wave can have two independent directions of oscillation, both perpendicular to its direction of propagation. For polarization of an electromagnetic wave, the relevant vector is the transverse electric field; for a mechanical transverse wave, it can be the transverse displacement. A purely longitudinal wave has no choice between these transverse directions, and the orientation of a static electric field is not polarization of waves.

Choose transverse axes $x,y$ in a Cartesian coordinate system. A monochromatic wave with a fixed phase difference has the real electric fieldHere $A_x,A_y$ are wave amplitudes, $\omega$ is the angular frequency, and $\delta$ is the phase difference of the $y$ oscillation relative to the $x$ oscillation. The vector's tip traces its path during the wave period $T=2\pi /\omega$. Both plotted coordinates are real electric field components; neither axis represents the imaginary part of a complex amplitude.

For nonzero wave amplitudes, $\delta =0$ or $\pi$ modulo $2\pi$ gives linear polarization: the vector moves back and forth along a fixed line. Equal nonzero wave amplitudes and $\delta =\pm \pi /2$ modulo $2\pi$ give circular polarization: the tip traces a circle of constant radius. Unequal nonzero wave amplitudes with $\delta =\pm \pi /2$ give a noncircular ellipse whose axes coincide with $x,y$. More generally, nonzero wave amplitudes and $\sin \delta \ne 0$ give elliptical polarization, with circular polarization as a special case. If exactly one wave amplitude vanishes, the nonzero wave has linear polarization for every $\delta$; if both vanish, there is no nonzero wave to assign a polarization of waves. ^[ref].

For example, take $\omega =2\pi {\mathrm{s}}^{-1}$, so the wave period is $T=1\mathrm{s}$. The choices $(A_x,A_y,\delta )=(2,1,0)$, $(1,1,\pi /2)$ and $(2,1,\pi /2)$, with electric field wave amplitudes in $V/m$, produce respectively the line segment $E_y=E_x/2$, the circle $E_x^2+E_y^2=1$ and the ellipse $E_x^2/4+E_y^2=1$, when coordinates are measured in $V/m$. At $t=T/4$, their tips are respectively $(0,0)$, $(0,-1)$ and $(0,-1)V/m$. The last two share a point without having the same path.

Resolving the electric field into two transverse components makes polarization of waves useful: the same frequency and wave amplitudes can describe different paths when their phase difference changes. An ideal linear polarizer selects the component along its pass axis, letting its orientation probe the direction of linear polarization through Malus's law. A fixed pair of wave amplitudes and a fixed phase difference describe one stable path; unpolarized light requires a statistical description rather than one fixed ellipse.

Refraction changes the direction of a transmitted wave when it crosses between media with different phase velocities. For light at a planar interface, Snell's law relates the angles measured from the normal vector through the refractive indices. At normal incidence the direction remains unchanged even though the wavelength and phase velocity can change. The worked example under geometrical optics distinguishes transmission from total internal reflection.

For light, refraction explains why a submerged object looks nearer the surface and why an optical lens can form an image. A wavefront crossing a boundary must retain a consistent wave phase along that boundary: its normal spacing changes with phase velocity, and oblique light rays change direction. Normal incidence separates these two effects especially clearly.

Take stationary, transparent, homogeneous media with isotropy, linearity of their optical response, and real positive refractive indices, at one chosen frequency $f$. With $c$ the vacuum speed of light and ${\lambda }_0=c/f$ the vacuum wavelength, medium $j$ hasThe frequency is unchanged at a stationary boundary; the phase velocity, wavelength and wavenumber can change. Here $n_j$ is a phase refractive index, not the group refractive index. If there is optical dispersion, evaluate $n_j$ at this same frequency on both sides.^[ref]

For an original idealized air–water example, stipulate $n_{\mathrm{a}}=1$ and $n_{\mathrm{w}}=4/3$, with vacuum wavelength $600\mathrm{nm}$. Then $f=c/(600\mathrm{nm})\approx 4.99654\times 10^{14}\mathrm{Hz}$, the water phase velocity is $3c/4$, and its wavelength is $450\mathrm{nm}$. At an air angle of incidence $30^{\circ }$, Snell's law gives a water angle of refraction $\arcsin (3/8)\approx 22.0243^{\circ }$. These stipulated values illustrate the relations; actual refractive indices depend on material conditions and frequency. By reversibility of an optical ray, the same path emerging from water has outgoing air angle $30^{\circ }$.

An interaction that changes the state or propagation of an incident particle or wave. Distinguish elastic scattering from inelastic scattering, and the general concept from its specifically quantum treatment in quantum scattering.