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The Lorentz force isFor $\mathbf{E}=0$, the magnetic part is perpendicular to the velocity, so its instantaneous power $\mathbf{F}\cdot \mathbf{v}$ vanishes. The speed and kinetic energy therefore remain constant. Decompose the initial velocity into $v_{\parallel }$ along the uniform magnetic field and $v_{\perp }$ perpendicular to it. The parallel component is unchanged, while the perpendicular component rotates uniformly.

For a nonrelativistic particle with $qB\ne 0$, the resulting helical motion hasHere $r_L$ is the Larmor radius, ${\omega }_c$ the cyclotron frequency in angular units, and $h$ the axial advance per turn. Opposite signs of electric charge give opposite senses of rotation. Pure perpendicular motion gives a circle; pure parallel motion gives a straight line. If $q=0$ or $B=0$, there is no magnetic deflection. At relativistic speeds, replace $m$ in the radius and period by $\gamma m$; $\gamma$ stays constant because the magnetic force does no work.

The magnetic dipole moment is $\mathbf{\mu }=IA\hat{\mathbf{n}}$ and its magnetic torque is ${\mathbf{\tau }}_{\mathrm{mag}}=\mathbf{\mu }\times \mathbf{B}$. Its component along the rotation axis opposes the imposed rotation. For positive $\omega$, the external driving torque needed to keep the angular velocity constant isAveraging over a complete rotation, with $⟨{\sin }^2\omega t⟩=1/2$, givesAs an independent sign and dimension check, $\omega ⟨{\tau }_{\mathrm{drive}}⟩=R⟨I^2⟩$. Thus the mechanical input power equals the mean Joule heating; a passive loop cannot supply a sustained driving torque. This average assumes the same negligible-self-inductance approximation as the current expression.

During switch-on, the Joule heating in $R_2$ has instantaneous power $P_2=R_2[V_0/(R_1+R_2){]}^2{e}^{-2t/\tau }$. Integrating the finite transient givesAlthough the source continues supplying power to $R_1$ indefinitely, $R_2$ is eventually shorted and its heat integral converges. It would be incorrect to identify this heat with the final stored magnetic energy, because the source and $R_1$ also participate during switch-on.

The intensity of an isotropic radiator at distance $r$ is the radiated power divided by spherical area: $\mathcal{I}=P/(4\pi r^2)$. For a locally plane far-field electromagnetic wave in air, its mean Poynting flux is $\mathcal{I}=E_{\mathrm{rms}}^2/Z_0=E_{\mathrm{peak}}^2/(2Z_0)$, where the vacuum wave impedance is $Z_0\simeq 376.7\Omega$. ThusandBoth conventions are given because “field strength” alone does not specify RMS or peak. This estimate treats the stated transmitter power as radiated power and ignores ground reflection and absorption.

Use phasors with time dependence $e^{i\omega t}$ and electrical impedance $Z=\tilde{V}/\tilde{I}$. The element laws $V=IR$, $I=CdV/dt$ and $V=LdI/dt$ giveA complex number records both amplitude and phase, turning the linear differential equations of a sinusoidally driven electrical circuit into algebraic equations. A resistor has in-phase voltage and electric current and dissipates average power. For an ideal capacitor, the electric current leads the voltage by $90^{\circ }$; for an ideal inductor, it lags by $90^{\circ }$. Their imaginary electrical impedances express reversible energy storage and return, with zero net average power over a period. Using $e^{-i\omega t}$ would conjugate the displayed reactive signs without changing predictions.

In the illustrated Maxwell bridge, the left and right vertices are the supply terminals and the upper and lower vertices are detector terminals. Insert a sensitive AC null detector between the latter and drive the former with a sinusoidal source. The upper-left arm is $Z_x=R_x+i\omega L_x$, the upper-right arm is $R_3$, the lower-left arm is $R_2$, and the lower-right arm is the parallel combinationAt a null no detector electric current flows, so the two voltage dividers giveMultiplying by the parallel-arm electrical admittance yieldsEquating real and imaginary parts therefore givesTwo independent adjustments are needed to annul the in-phase and quadrature detector signals. For example, change $R_2$ to set $L_x$ and $R_1$ to set $R_x$, iterating until the detector reads zero. Alternatively, with fixed $R_2,R_3$, independently vary $R_1$ and $C$.

The Maxwell bridge compares an unknown lossy inductor with a stable standard capacitor, avoiding the need for an accurately known reference inductor. The ideal balance equations contain no frequency, and a null measurement need not rely on absolute detector calibration. In practice choose a suitable frequency below the coil's self-resonance, where the series $R_x,L_x$ model and the standard capacitor remain accurate.

If the initial setting is far from balance, the bridge as drawn needs adjustable components and a detector, not merely a more precise readout. Fit resistance boxes or switched ranges for $R_1$ and a ratio arm, and a selectable standard $C$. Choose ranges aroundThen approach the null with coarse adjustments followed by finer ones. A large required $R_1$ can be brought into range by increasing $C$ and decreasing $R_2{R}_3$ proportionally. Start with reduced excitation or reduced detector sensitivity to avoid overload, then increase sensitivity near the null. Changing frequency alone cannot repair an impossible component range, since the ideal balance conditions are frequency-independent. Stray capacitance and lead electrical resistance set practical accuracy limits and may require shielding and short connections.

The component of magnetic force along the slope has magnitude $IBs\cos \varphi$. The rate at which the sliding bar does work against this magnetic braking isBy Ohm's law, $\mathcal{E}=IR$, soThis exactly equals the Joule heating in the resistor in the negligible-self-inductance approximation. The normal force does no work because it is perpendicular to the velocity.

The gravitational power is instead $mgv\sin \varphi$. By the work-energy theorem,It equals the Joule heating only at constant speed. For a concrete counterexample, during the initial acceleration the kinetic energy increases, so some gravitational power remains mechanical.

For the imposed electric current $I=I_0\cos \omega t$, Faraday's law gives the self-induced electromotive forceThe externally applied terminal voltage, with the passive sign convention, has the opposite sign:Thus the terminal voltage leads the electric current by $\pi /2$. The instantaneous input power isAn ideal inductor alternately stores magnetic energy and returns it to the source. Real winding electrical resistance would add mean Joule heating $I_0^2{R}_w/2$; the ideal zero result assumes that loss is neglected.

The single secondary turn links the primary magnetic flux ${\mu }_{0}n\pi a^2{I}_p$, so its mutual inductance with the primary isChoose positive winding directions so that positive secondary electric current would reinforce the primary magnetic flux. Faraday's law and Kirchhoff's voltage law then giveAfter the transient proportional to $e^{-Rt/L_s}$ has decayed, put $\alpha ={\tan }^{-1}(\omega L_s/R)$. The steady electric current isIts wave amplitude is $\omega M{I}_0/\sqrt{R^2+(\omega L_s{)}^2}$ and it lags the induced electromotive force by $\alpha$. Relative to the primary $I_0\cos \omega t$, its phase shift is $-\pi /2-\alpha$, modulo $2\pi$. Reversing the chosen secondary winding direction adds $\pi$ to this signed phase shift and leaves all power predictions unchanged.

The lamp's mean Joule heating isFor $\omega L_s\ll R$ the response is nearly in phase with the induced electromotive force; for $\omega L_s\gg R$ the secondary electric current nearly opposes the primary electric current in the stated winding convention.

The positive lamp power comes from extra work by the primary current source. The secondary electric current creates a changing magnetic flux through the primary, so maintaining the same prescribed primary electric current requires an additional terminal voltage. With the preceding winding convention,Using phasors, ${\tilde{I}}_s=-i\omega M{\tilde{I}}_p/(R+i\omega L_s)$, hence the primary sees the complex impedanceIt follows directly thatThis is the reflected impedance of the secondary load. The self-inductance and mutual inductance terms exchange stored magnetic energy, while the lamp dissipates it. Multiplying the two coupled equations from Kirchhoff's voltage law by their respective electric currents also gives the instantaneous conservation of energy identityThe derivative averages to zero in periodic steady state. A fixed primary current source supplies the required extra voltage; a fixed voltage source would generally change its primary electric current when the lamp is connected.

The restoring, damping and electric forces on the electron give $m_e\ddot{x}=-kx-b\dot{x}-eE(t)$. Dividing by $m_e$ gives a Lorentz oscillator model withFor the $e^{i\omega t}$ convention, substitute $x=\mathrm{Re}(x_0{e}^{i\omega t})$. Then $x_0=-e{E}_0/[m_e({\omega }_0^2-{\omega }^2+i\gamma \omega )]$ andThe time-averaged dissipated power isFor positive frequency the squared term is minimized to zero only at $\omega ={\omega }_0$. Thus the power maximum is exactly at the undamped natural frequency, even though the displacement maximum is slightly shifted. Here $P_{\max }=e^2|E_0{|}^2/(2b)$. The half-power equation is $|{\omega }_0^2-{\omega }^2|=\gamma \omega$, whose two positive roots areThis is the absorbed-power bandwidth of a viscously damped oscillator, exact within the model. Near resonance the profile is approximately a Lorentzian line shape. With $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$,For this wavelength linewidth of a damped-oscillator power resonance we used ${\omega }_{+}{\omega }_{-}={\omega }_0^2$. The half-power wavelength endpoints are not exactly symmetric around ${\lambda }_{\mathrm{peak}}$, although that distinction is tiny here.

In a constant field the settled displacement is $x_s=-e{E}_{\mathrm{dc}}/k$. On sudden removal, with zero initial velocity, the free motion isThe electron oscillates about its original equilibrium, losing the previously stored spring energy. The period is $3.52\mathrm{fs}$, the amplitude decay time is $2/\gamma =0.319\mathrm{ps}$, and the cycle-averaged energy decay time is $1/\gamma =0.160\mathrm{ps}$. This is the effective classical model supplied, not a prediction of the full hydrogen line radiation spectrum.

In Fourier transform spectroscopy, a scanning Michelson interferometer measures intensity against optical path difference $\ell$ rather than selecting one narrow wavelength at a time. After subtraction of the constant background, an ideal stationary source gives an interferogramA cosine Fourier transform recovers the spectral power density $S(\nu )$ after instrumental calibration. A mirror displacement $x$ changes the round-trip path by $2x$; the zero-path-difference burst aligns the scan, and a reference laser can calibrate the displacement.

Finite maximum path difference $L$ multiplies the interferogram by a window and convolves the recovered radiation spectrum with that window's Fourier transform. Thus $\delta \nu$ is of order $c/L$; precise factors depend on scan range and resolution convention. Uniform path samples of separation $\Delta \ell$ need ${\nu }_{\max }<c/(2\Delta \ell )$ to avoid aliasing without bandpass information. interferogram apodization reduces sidelobes at the price of broader lines. The broad accepted optical throughput can be advantageous; the multiplex signal-to-noise advantage depends on detector noise and is not universal in photon-noise-limited measurements. Source drift during a scan, finite beam divergence, imperfect sampling and wave phase errors require control. The finite-window reasoning is the delay-range resolution in Fourier transform spectroscopy.^[ref]

Joining p-type and n-type regions initially allows majority carriers to diffuse across and recombine. Exposed negative acceptor charge on the p side and positive donor charge on the n side form a depletion region. Its electric field points from n to p and balances diffusion at equilibrium. The Fermi level is then spatially constant, while the conduction and valence band edges bend across the built-in potential; this equilibrium barrier cannot supply continuous power to a load.

forward bias makes p positive relative to n, reducing the barrier from $q{V}_{bi}$ to $q(V_{bi}-V)$. electrons enter p material and holes enter n material, becoming minority carriers whose diffusion and recombination sustain the current. reverse bias enlarges the barrier and leaves a much smaller minority-carrier current until breakdown. The ideal diffusion model gives $I=I_0[e^{qV/k_{B}T}-1]$, the Shockley diode equation; depletion-region recombination, series resistance and breakdown limit this approximation. Minority-carrier injection on both sides is essential to the junction theory.^[ref]

Absorption of photons with $h\nu \gtrsim E_g$ creates electron-hole pairs, the upward band-diagram arrow. In the depletion region the built-in field drives electrons towards n material and holes towards p material; diffusion also delivers carriers generated within a diffusion length outside that region. Selective contacts collect the two signs. Illumination maintains a nonequilibrium quasi-Fermi-level splitting, allowing a voltage and useful current through an external load; it is the light, rather than the equilibrium barrier, that supplies the energy.

Taking forward diode current as positive, an ideal illuminated junction haspower is delivered where $VI<0$ in this convention. Reflection, below-gap transmission, carrier thermalization, recombination and resistive losses limit real conversion; even an ideal single junction has radiative detailed-balance limits.^[ref]

For an experimental-noise interpretation of Nyquist's theorem, a resistor $R$ in thermal equilibrium at temperature $T$ has one-sided open-circuit voltage-noise spectral densityin the classical range $hf\ll k_{B}T$, over an approximately rectangular bandwidth $\Delta f$. Across a matched load, half the open-circuit voltage appears, giving available noise power $k_{B}T\Delta f$. This is thermal agitation, present without a mean current, and is distinct from shot noise. A real readout weights the spectrum with its transfer function: $⟨v_{\mathrm{out}}^2⟩={\int }_0^{\infty }{S}_V(f)|H(f){|}^{2}df$. Cooling, reduced bandwidth and suitable impedance/noise matching reduce measured noise; the resistor's noise does not simply disappear in an ideal amplifier model.^[ref]

The bare name also has a sampling interpretation. The Nyquist–Shannon sampling theorem permits exact reconstruction of a bandlimited signal from sufficiently fast uniform samples. The practical condition is $f_s>2f_{\max }$, with an anti-aliasing filter before sampling; otherwise distinct continuous frequencies produce identical samples. Ideal reconstruction uses shifted sinc functions. Critical-rate equality needs an appropriate signal class and can fail for unrestricted edge-frequency sinusoids with unknown phase. Sampling fast enough prevents aliasing, but does not remove thermal noise. This distinction covers the source's unspecified use of “Nyquist's theorem”.^[ref]

The angle in the source ray sketch is inside the gap. Two extra reflections add a full round trip. Comparing adjacent emerging rays on a common outgoing phase plane subtracts the lateral-path contribution, so the optical path difference is $2\mu d\cos \theta$, rather than $2\mu d/\cos \theta$. Thusat transmission maxima. The two internal reflection phases add to an integer full cycle in the stated ideal symmetric-coating model. Snell's law, $\mu \sin \theta ={\mu }_0\sin {\theta }_{\mathrm{out}}$, converts to the external direction. A fixed internal angle defines a cone and hence an equal-inclination ring in a focal plane.

For identical lossless mirrors the first transmitted amplitude has magnitude $(1-R)$ times the incident amplitude. Each additional round trip multiplies it by $R{e}^{i\delta }$, because $R$ is an intensity reflection coefficient for one mirror but the product of the two amplitude reflection magnitudes. Summing the geometric series givesThe maximum is one in this symmetric lossless limit; absorption, unequal mirrors or additional incoherent reflections change it. At fixed angle and negligible material dispersion, $m\lambda =2\mu d\cos \theta$ and adjacent orders differ by one. For $m\gg 1$,

order-safe angular scanning of a Fabry–Pérot spectrum should cover the approximately $38\mathrm{nm}$ spectral span within one order while making the instrumental width smaller than $1\mathrm{nm}$. Near normal incidence, requiring the smallest approximate FSR to exceed the span gives $\mu d\lesssim 5.3\mu \mathrm{m}$, and $\mathcal{F}$ should exceed roughly $40$. These are design bounds, not a unique optimum. One suitable choice isThe physical spacing is $d=5.0\mu \mathrm{m}/\mu$; an air gap gives $d\simeq 5\mu \mathrm{m}$. Approximate FSRs near normal incidence are $40.1\mathrm{nm}$ at $633\mathrm{nm}$ and $44.9\mathrm{nm}$ at $670\mathrm{nm}$, with widths about $0.65$ and $0.73\mathrm{nm}$. For a concrete angular scan choose order $m=14$. The $670\mathrm{nm}$ pair lies near internal $\theta =20.3^{\circ }$ and the $633\mathrm{nm}$ line near $27.6^{\circ }$; their widths within this scan remain below $1\mathrm{nm}$. Restricting to this calibrated order window avoids the other-order $633\mathrm{nm}$ ring near $18.3^{\circ }$. Index dispersion and unknown extra source lines would need calibration or order-sorting filters.

The resolving power is approximately $m\mathcal{F}$: many coherent cavity round trips create a much longer effective optical delay than one traversal. A Michelson instrument with the same available physical delay has less resolving power unless its mirror travel provides a correspondingly long maximum optical path difference. Mirror separation errors spread the round-trip phase; keeping them below a linewidth requires roughlyFor this design the scale is about $6\mathrm{nm}$ in an air gap. Both surface flatness and parallelism over the illuminated area, as well as angular collimation and coating uniformity, must support the intended finesse.

The string speed is $c=\sqrt{T/\rho }$. The characteristic impedance is the ratio of the forward-wave transverse force to transverse velocity, $Z=T/c=\sqrt{T\rho }$. With peak displacement amplitude $A$, the mean power isDisplacement and transverse force are continuous at a massless junction. Since $Z_2=2Z_1$, the displacement amplitudes are $r=(Z_1-Z_2)/(Z_1+Z_2)=-1/3$ and $t=2Z_1/(Z_1+Z_2)=2/3$. Taking the incident amplitude real, the light-string motion isIn this standing-and-travelling decomposition of a reflected string wave, the first term travels and the second stands; the decomposition is not a separation into independent incident-power fractions because these fields interfere. The reflected wave has a phase reversal. power coefficients and standing-wave ratio are

For a uniform inserted matching section, unequal exterior impedances require a quarter-wave impedance transformer: $Z_M=\sqrt{Z_1{Z}_2}$ and $k_{M}a=(2j+1)\pi /2$. At equal tension this means middle density $2{\rho }_0$ and an odd-quarter-wavelength length in that section. Reflections at its two ends then cancel at the design frequency. An arbitrary density or an arbitrary half-wavelength segment cannot perfectly match these unequal exterior strings; the ideal match is frequency-selective.

amplitude scaling along a slowly tapered string instead follows from matching over an adiabatic frequency range. The source fixes frequency, while conserved power and the final characteristic impedance determine amplitude. In the heavy string,Its wave speed halves. The no-reflection approximation needs small fractional characteristic impedance change over each local wavelength, rather than length alone for a taper containing abrupt features.

For $m\ddot{x}+b\dot{x}+kx=F_0\cos \omega t$, define ${\omega }_0=\sqrt{k/m}$ and $\beta =b/(2m)$. The parameter relating quality factor and viscous damping regimes isAt weak damping it also equals $2\pi$ times stored energy divided by energy lost per cycle. The stored-energy convention and frequency matter: this cycle-based interpretation should not be extended to free motions that never complete a cycle.

For $\beta <{\omega }_0$ the free motion is $e^{-\beta t}[A\cos {\omega }_{d}t+B\sin {\omega }_{d}t]$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. Thus underdamping means $Q>1/2$. High $Q$ gives many oscillations per amplitude-decay time, narrow resonance and strong frequency selectivity; a lightly loaded tuning fork, optical cavity or low-loss electrical resonator provides examples. The amplitude decay time is ${\tau }_A=1/\beta =2Q/{\omega }_0$, while the slowly averaged energy decays with ${\tau }_E=1/(2\beta )=Q/{\omega }_0$. These times differ by a factor of two.

At critical damping, $\beta ={\omega }_0$ and $Q=1/2$, with $x=(A+Bt)e^{-{\omega }_{0}t}$. There is no ringing; a simple critically damped position indicator can settle without oscillation. For $\beta >{\omega }_0$ and $Q<1/2$, there are two real negative decay rates $-\beta \pm \sqrt{{\beta }^2-{\omega }_0^2}$. Increasing very strong drag slows the long-time return, whose rate approaches ${\omega }_0^2/(2\beta )$. A strongly viscously damped mechanical indicator illustrates this regime. These classifications concern the linear homogeneous equation, not every possible nonlinear damping law.

The driven displacement amplitude isThe displacement-resonance threshold of a viscous oscillator requires $\beta <{\omega }_0/\sqrt{2}$, or $Q>1/\sqrt{2}$; it lies at $\sqrt{{\omega }_0^2-2{\beta }^2}$. Some underdamped systems therefore have no such maximum. Average absorbed power instead peaks at ${\omega }_0$ for this viscous model. Its half-power frequencies are $\sqrt{{\omega }_0^2+{\beta }^2}\pm \beta$, so their separation is $2\beta$ and $Q={\omega }_0/\Delta \omega$. For broad response this ratio remains a useful parameter but not a claim of many free oscillations. The familiar displacement-linewidth version is a high-$Q$ approximation.

A series RLC circuit has ${\omega }_0=1/\sqrt{LC}$ and $Q={\omega }_{0}L/R$ for the charge oscillation. Resistance dissipates energy, inductance provides the inertial analogue and capacitance the restoring analogue. Ringdown and a calibrated half-power scan independently estimate loss; parasitic resistance, loading and radiation change the measured value. A large $Q$ increases frequency selectivity but also lengthens settling time, so it is not universally the best instrumental choice.

A carrier in a locally parabolic isotropic band obeys $m^{\ast }\dot{\mathbf{v}}=q\mathbf{v}\times \mathbf{B}$, soIlluminate a high-mobility crystal with a transverse alternating electric field, sweep a calibrated static magnetic field at known microwave or far-infrared frequency, and detect resonant absorbed power. Cooling lengthens the relaxation time; resolving the resonance generally needs ${\omega }_c\tau \gg 1$. Varying field direction measures band anisotropy, while circular polarization distinguishes the two senses of rotation for opposite carrier charges. This is orbital motion rather than electron spin resonance.

The classic silicon/germanium measurements used single crystals near $4\mathrm{K}$, light modulation and circularly polarized microwaves, with directional scans to infer anisotropic carrier-energy surfaces.^[ref] Carrier generation, calibration, scattering linewidth and inhomogeneity must be separated from a genuine mass shift.

In a general band the cyclotron mass is an orbit property, $m_c={\hslash }^2(\partial A_k/\partial E)/(2\pi )$ for the appropriate field-normal orbit. It need not equal every component of the band-curvature mass tensor. For an ellipsoidal parabolic band with field along a principal $z$ direction it is $\sqrt{m_x{m}_y}$. Thus a single resonance does not determine a whole effective-mass tensor without angular information.

At the zero-slope boundary, the rightgoing and leftgoing slopes cancel, so $A_r=A_i$. ConsequentlyThis free-end reflection of a transverse string wave uses a Neumann boundary condition: the displacement has an antinode and is not forced to vanish. The instantaneous flux $-T{\psi }_x{\psi }_t$ through the boundary is zero, confirming that it transmits no power.

time average of harmonic power. Expand the two real parts:The terms proportional to $e^{\pm 2i\omega t}$ average to zero over complete periods, while $F{u}^{\ast }$ is time-independent. For real peak amplitudes this is $F_0{u}_0\cos \varphi /2$.

String dynamics. A short element of length $dx$ experiences transverse force $T[{\psi }_x(x+dx)-{\psi }_x(x)]$. Newton's second law therefore gives $\rho dx{\psi }_{tt}=Tdx{\psi }_{xx}$, orThe approximation is a flexible, lossless string with small slope and constant tension; $\rho$ is its linear mass density.

energy transport. Across a cut toward increasing $x$, the transverse force is $F_y=-T{\psi }_x$ and the transverse velocity is $u={\psi }_t$. For a rightgoing wave these have complex amplitudes $ikTA$ and $i\omega A$. HenceThis real mechanical impedance is force divided by transverse velocity for a travelling wave, with units $\mathrm{kg}{\mathrm{s}}^{-1}$.

Matching and power ratios. At an ordinary massless junction without an external transverse force, displacement is continuous and transverse tension is balanced: ${\psi }_1={\psi }_2$, $T_1{\psi }_{1x}=T_2{\psi }_{2x}$. For the same tension, the slope is therefore continuous even if density changes. With real positive impedances,Their sum equals $P_i$; the transmitted power includes the impedance ratio and is not simply $t_{\psi }^2{P}_i$.