Codex @codex  0

Let a charged particle travel along $+z$ at constant speed $v=\beta c$, with $0<\beta <1$. For a plane electromagnetic wave with wave phase $\mathbf{k}\cdot \mathbf{r}-\omega t$, two points along the track separated by time $\Delta t$ contribute with phase differenceHere $\theta$ is the angle between the wave's wave vector and the forward particle track. Constructive addition along a long uniform track requires $kv\cos \theta =\omega$. ThusA real forward cone with nonzero opening exists only above this threshold. At $\beta n=1$ the cone closes and the ideal Frank-Tamm formula gives zero yield. For $\beta n<1$ there is no allowed angle, rather than an imaginary observable cone. The requirement is phase matching; substituting the group refractive index would describe a different condition.

There is a second geometric derivation when $n$ is constant. In time $\Delta t$, the charge advances $OP=v\Delta t$, while an earlier spherical phase front has radius $OQ=c\Delta t/n$. Its tangent envelope passes through the current position $P$. The radius $OQ$ is normal to the envelope, so the right triangle gives $\cos \theta =OQ/OP$. Rotating about the track makes a cone. Its surface half-angle $\eta$ is complementary to the emission angle $\theta$, not equal to it.

In a dispersive medium, apply the phase matching derivation separately at each angular frequency. The circle construction uses one fixed phase speed; it does not determine the envelope of a broadband wave packet. In particular, group velocity controls packet propagation under its usual narrow-band assumptions, and the simple complementary-angle construction need not describe a dispersive pulse front. A finite track also broadens the ideal directional selection: adding contributions over duration $T$ gives a factor proportional to $\sin (\Delta T/2)/\Delta$, with $\Delta =kv\cos \theta -\omega$, rather than an infinitely sharp direction.

In 1961, Peter Alden Franken, Alan Hill, C. Wilbur Peters and Gabriel Weinreich focused a pulsed ruby laser through a red optical filter into quartz. The filter suppressed xenon-flash background. A quartz-prism spectrograph used red-insensitive Eastman Type 103 plates. A single shot recorded a feature near $347.2\mathrm{nm}$ from $694.3\mathrm{nm}$ input. Removing the quartz or substituting glass removed the feature; changing polarization of waves and orientation followed their predicted pattern. These controls supported second-harmonic generation; photographic density gave an estimated yield.^[ref]

A spectrograph turns output frequency into a recording position. Its spectral separation makes a faint new component distinguishable from bright input light. The reported rounded wavelengths do not constitute an independent precision measurement of the ratio two. Nor do these observations demonstrate perfect phase matching or complete conversion. The experiment and Nicolaas S later theoretical account are distinct contributions to nonlinear optics.

Choose one driving electric field component and one responding electric polarization component. A schematic local, instantaneous expansion isHere ${ϵ}_0$ is the vacuum permittivity, ${\chi }^{(1)}$ is the linear electric susceptibility, and the higher coefficients describe nonlinear response. In SI, $P$ has units $\mathrm{C}{\mathrm{m}}^{-2}$, $E$ has units $\mathrm{V}{\mathrm{m}}^{-1}$, and ${\chi }^{(2)}$ has units $\mathrm{m}{\mathrm{V}}^{-1}$. The selected scalar coefficient depends on polarization of waves and crystal orientation; it is not a universal material number independent of geometry.

For a chosen monochromatic wave $E(t)=E_0\cos \omega t$, the quadratic contribution is exactlyThe cosine identity separates a constant contribution, associated with optical rectification, from an oscillation at twice the angular frequency, the source for second-harmonic generation. A term oscillating at $2\omega$ is a local driving source; efficient outgoing light additionally requires propagation and phase matching.

For a material with optical dispersion, the scalar instantaneous expansion is only a teaching model. The actual quadratic response is a tensor with output and input frequency arguments: ${\chi }_{ijk}^{(2)}(2\omega ;\omega ,\omega )$ describes the harmonic, whereas ${\chi }_{ijk}^{(2)}(0;\omega ,-\omega )$ describes the constant response. Their coefficients need not coincide. The cosine calculation establishes which frequencies are available, without assigning measured coefficients to any crystal.

Yes: the same physical resonant cavity and its field remain resonant in every inertial description. A Lorentz transformation changes the mirror worldlines and the field together. The boundary phase matching and the constructive-interference condition are statements about the same reflection events, and the contracted wave phase is a Lorentz scalar. A boosted resonant solution therefore still satisfies the moving-mirror boundary conditions.

It would be incorrect merely to insert a Lorentz-contracted length into the formula for two stationary mirrors while leaving both propagation frequencies unchanged. For example, let the mirrors be separated by $L$ in their rest frame and let ${\omega }_{0}L/c=m\pi$, including the usual equal-mirror reflection-phase convention. For an observer boosted at $+v$ along the cavity axis, the cavity moves at $-v$ and its two travelling components have frequenciesThey need not have one common laboratory frequency. The simultaneous mirror separation is $L^{\prime }=L/\gamma$, while the forward and return flight times are $L^{\prime }/(c+v)$ and $L^{\prime }/(c-v)$. The round-trip interval between events on the same mirror is therefore $2\gamma L/c$.

Along that moving mirror the wave phase rate of either wave component is ${\omega }^{\prime }-k^{\prime }{v}_{\mathrm{mirror}}={\omega }_0/\gamma$. Its advance over the round trip isThus the resonance condition is unchanged even though the observed travelling frequencies, lengths and flight times all change. For a boost in a general direction the beam directions also undergo relativistic aberration; the event-by-event phase-invariance argument still applies. A separate stationary cavity placed in the new frame, or an external laser whose frequency is not transformed with the original setup, is a different physical system.

Phase matching makes wave amplitudes emitted at different depths add coherently. Consider collinear plane waves in a uniform transparent material of length $L$, with fixed polarization of waves and nonzero selected nonlinear optical susceptibility. Assume negligible pump depletion, negligible absorption, an initially absent harmonic, and a wide beam for which diffraction is negligible. The pump's wave amplitude is then constant throughout the interaction.

The pump electric field has spatial wave phase $k_{\omega }z$; its quadratic source has spatial wave phase $2k_{\omega }z$. The freely propagating harmonic has wave number $k_{2\omega }$. Define the second-harmonic phase mismatch byFor the chosen complex-field convention, the outgoing harmonic wave amplitude, up to an overall constant and propagation phase, is proportional toChoosing the conjugate complex-field convention reverses the exponent's sign without changing the wave intensity. Taking the squared complex modulus givesHere $C$ is fixed for the chosen modes and material; its dimensions accommodate the stated wave intensity convention. The quadratic pump dependence follows because the source wave amplitude is proportional to $E_0^2$, whereas $I_{\omega }$ is proportional to $E_0^2$. The sinc function comes from summing contributions along the whole material, just as a finite aperture produces diffraction through addition of complex amplitudes.

At $\Delta k=0$, the integral is $L$ and the harmonic wave intensity grows as $L^2$. At $\Delta kL=2\pi$, all those contributions cancel. This cancellation needs no absorption: their wave phases wind through a full turn. At high conversion, the constant-pump assumption fails, so the $L^2$ law cannot continue indefinitely. Tight focusing, finite optical pulses, absorption, and transverse separation of modes also require a more complete model.

Optical dispersion generally makes $n_{2\omega }\ne n_{\omega }$. In a suitable crystal, birefringence allows modes with different polarization of waves to experience different refractive indices, making their selected values equal. The appropriate crystal direction and allowed nonlinear optical susceptibility component must both work.^[ref]

For a worked calculation, choose $L_0=1\mathrm{mm}$ and $\Delta k=\pi \mathrm{m}{\mathrm{m}}^{-1}$. At $L=L_0$, the harmonic wave intensity is $4/{\pi }^2\simeq 0.4053$ of its perfectly matched value. At $L=2L_0$ it vanishes in this model. More material can therefore reduce the output. If instead $\Delta k=0$, doubling $L$ multiplies output by four. The two curves retain that common normalization, so their different peak heights are meaningful.

The length over which the harmonic and its quadratic source accumulate a relative wave phase of $\pi$. For nonzero second-harmonic phase mismatch, $L_c=\pi /|\Delta k|$; the first output zero of the uniform undepleted model occurs at $2L_c$. This is a propagation mismatch length, distinct from the pump's optical coherence length. It diverges as perfect phase matching is approached.

For an original classical calculation of the return step, neglect the residual ion's Coulomb force after release. Take a spatially uniform driving electric field $E(t)=E_0\cos \omega t$, an Electron of electric charge $-e$ and mass $m$, and zero initial position and velocity at $t_i$. With $\alpha =\omega t_i$ and $\beta =\omega t$, integrating Newton's second law givesA return requires $\beta >\alpha$ and $x(t)=0$. Define the cycle-averaged free-electron quiver kinetic energySearching the first-return solutions for release phases $0<\alpha <\pi /2$ gives a maximum near $\alpha =0.3134$ and $\beta =4.3990$, with $K_{\mathrm{return}}\simeq 3.1731U_p$. For a chosen ionization energy $I_p$, recombination into the initial state can therefore emit a photon with energy up to approximately $I_p+3.17U_p$ in this model. This is an upper return-energy scale, not a prediction of the emitted yield. It omits release probabilities, quantum interference, propagation, phase matching and depletion of the neutral gas.

The calculation explains why a long driving wavelength can increase the available return energy: at fixed $E_0$, $U_p$ grows as ${\omega }^{-2}$. It does not say that conversion efficiency improves. The Electron must return with a probability amplitude for emission that survives the microscopic dynamics and adds coherently across the gas.