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Lenz's law expresses the minus sign in Faraday's law of induction: the induced electromotive force drives a passive resistive loop's electric current in the direction whose own magnetic field opposes the imposed change in magnetic flux. It opposes the change of flux, not invariably the external magnetic field. Choose an oriented surface and its positive boundary direction first, find the sign of $d{\Phi }_{\mathrm{ext}}/dt$, and then use the right-hand rule to translate the signed electric current into a physical direction. ^[ref].

In the fixed-loop example in Faraday's law of induction, out-of-page magnetic flux increases at $0.0060Wb/s$. With counterclockwise positive, $I=-0.0030\mathrm{A}$ is clockwise, producing an in-page magnetic field through the loop. Reverse the ramp to $B(t)=0.40\mathrm{T}-(0.30T/s)t$ on the same one-second interval: now $\mathcal{E}=+0.0060\mathrm{V}$ and $I=+0.0030\mathrm{A}$. The counterclockwise electric current produces an out-of-page magnetic field, supporting the decreasing external magnetic flux. A steady out-of-page magnetic field with fixed loop geometry gives no induced electromotive force in either direction.

For the increasing-area example in motional electromotive force, clockwise electric current runs down the right-hand rod. Its magnetic force on a current-carrying wire points left, opposing the imposed motion to the right. Writing $I_{\mathrm{mag}}=B\ell v/R$ for the current magnitude, the power required to maintain constant speed isThus the sign agrees with conservation of energy: the external drive supplies the Joule heating. The magnetic forces do no work on individual charges; mechanical work by the drive supplies the electrical energy. This also explains the usefulness of electromagnetic braking of a sliding rod and the terminal speed of a falling conducting loop. ^[ref].

The sign of induced electromotive force alone does not determine an arbitrary circuit's instantaneous electric current. Retaining constant self-inductance $L$ gives $LdI/dt+RI=-d{\Phi }_{\mathrm{ext}}/dt$, so a pre-existing electric current can persist while the induced drive opposes its change. The distinction matters in the lossless limit of magnetic loop braking. The resistive toy examples use $L=0$ as an approximation, not a general consequence of Lenz's law.

A light-emitting diode is a semiconductor diode producing electroluminescence under electric current. Forward bias injects electrons and electron holes; radiative electron-hole recombination emits photons. A double heterostructure uses band offsets to retain both charge carriers in an active layer. A sufficiently thin layer is a quantum well.^[ref]

The optical and electrical questions are different: photon energy sets a color scale, while light-emitting diode efficiency determines how much supplied energy escapes as light. The blue light-emitting diode made compact white light-emitting diodes and full-color displays practical. A modulated light-emitting diode can also supply the optical signal in a chopped-light lock-in experiment.

Efficiency has several denominators. Specify a steady electric current $I>0$, positive terminal electric potential difference $V_f$, and a simplified narrow emission spectrum with photon energy $E_{\gamma }$. Assume at most one radiative photon per injected electron and electron hole pair. Let ${\eta }_{\mathrm{inj}}$ be the fraction of the electrical pair-supply rate $I/q$ reaching the active layer, ${\eta }_{\mathrm{rad}}$ the fraction of those pairs recombining radiatively, and ${\eta }_{\mathrm{ext}}$ the fraction of internally emitted photons that escape. Then counting the successive events givesDefine internal quantum efficiency here conditionally on pairs reaching the active layer: ${\eta }_{\mathrm{IQE}}={\eta }_{\mathrm{rad}}$. The internal photon yield per electrical pair is instead ${\eta }_{\mathrm{inj}}{\eta }_{\mathrm{IQE}}$. Stating this convention matters when another definition folds injection into internal quantum efficiency. The external quantum efficiency and wall-plug efficiency areThese are photon-count and energy-flow identities for the declared model. For a broad spectrum, replace $E_{\gamma }$ by the mean escaping photon energy, weighted by photon number, when computing optical power.

A stipulated example chooses $I=20\mathrm{mA}$, $V_f=3.20\mathrm{V}$, $\lambda =450\mathrm{nm}$, ${\eta }_{\mathrm{inj}}=0.90$, ${\eta }_{\mathrm{IQE}}=0.80$ and ${\eta }_{\mathrm{ext}}=0.50$. For every 100 electrical pairs, 90 reach the active layer, 72 emit internally, and 36 photons escape. Thus ${\eta }_{\mathrm{EQE}}=0.36$, not $0.80$. Using $q=1.602176634\times 10^{-19}\mathrm{C}$ givesThe dimensional analysis is transparent: $A/C={\mathrm{s}}^{-1}$, $J/s=\mathrm{W}$, and $\mathrm{CV}=\mathrm{J}$. The optical and electrical efficiencies differ because $E_{\gamma }/q=2.75520\mathrm{V}$ differs from the chosen $V_f$.

The factors suggest different improvements. Better confinement can reduce injection leakage; suppressing nonradiative recombination raises the radiative fraction; improving light extraction efficiency helps already-emitted photons escape. A photon trapped by total internal reflection has already passed the radiative step and can still be lost through absorption. Improving extraction therefore need not change the internal radiative fraction.

Shuji Nakamura's lecture cautions that why indium gallium nitride emits efficiently despite many defects was not completely settled in that account.^[ref]

As an independent consistency check on the 1994 reported values, treating its broad spectrum as monochromatic at the $450\mathrm{nm}$ peak givesThis agrees at the quoted precision with the reported $2.7\%$. Its optical/electrical power ratio is instead $1.5\mathrm{mW}/(20\mathrm{mA}\times 3.6\mathrm{V})=2.0833\%$. The first estimate uses a peak in place of a photon-number-weighted mean, so agreement is a consistency check rather than an exact reanalysis of the measured spectrum.

Each LIGO instrument is a Michelson interferometer with approximately $4\mathrm{km}$ arms containing Fabry–Pérot interferometers. A stabilized $1064\mathrm{nm}$ laser illuminates suspended $40\mathrm{kg}$ fused-silica mirrors in vacuum. Four-stage suspensions and active isolation suppress ground motion; low-loss materials reduce thermal noise. Power recycling increases circulating power; signal recycling shapes the response to the gravitational wave.^[ref]

Why measure two perpendicular arms? In the long-wavelength limit, take a gravitational wave with plus polarization normally incident on ideal perpendicular arms. Its transverse-traceless tensor has $h_{xx}=h_{+}$ and $h_{yy}=-h_{+}$. Expanding each proper length to first order givesHere $h_{+}$ is dimensionless, whereas $\Delta L$ is a length. A stipulated $h_{+}=10^{-21}$ and $L=4000\mathrm{m}$ imply $\Delta L=4\times 10^{-18}\mathrm{m}$, with individual arm changes of opposite sign and magnitude $2\times 10^{-18}\mathrm{m}$. These are equivalent differential arm-length changes, not a photograph of mirrors moving against a rigid external ruler. The observable compares returning light phases; coordinate systems can distribute the description between mirror coordinates and light propagation.

For unit arm directions $\mathbf{u},\mathbf{v}$, define the detector tensor $D^{ij}=(u^i{u}^j-v^i{v}^j)/2$. If $e_{ij}^{+},e_{ij}^{\times }$ are the transverse polarization tensors, thenThe dimensionless factors $F_A$ encode sky direction and the polarization basis. Thus even identical sources can have different observed amplitudes and wave phases. Motion changing both arms equally cancels in this ideal differential observable, although real asymmetries can couple common disturbances into it.

Suspension supplies an inertial reference above its resonances. For example, a single ideal harmonic oscillator with support motion $y$ obeys $\ddot{x}+{\omega }_0^2(x-y)=0$. Its response at angular frequency $\omega$ is $x/y={\omega }_0^2/({\omega }_0^2-{\omega }^2)$, which decreases in magnitude as $({\omega }_0/\omega {)}^2$ well above resonance. The mirror follows slow support drift but approximately free motion in the high-frequency limit. Real multistage suspensions also require damping and alignment control; resonant frequencies are not regions of good isolation.

Rainer Weiss's 1972 report proposed suspended masses, multiple-pass optical delay lines, and a servo holding an interference fringe. Its noise analysis separately treats photon-counting fluctuations, laser frequency instability, mechanical thermal noise, radiation pressure and ground motion. Those delay lines should be distinguished from the resonant arm cavities of Advanced LIGO.^[ref]

An ideal single-round-trip Michelson interferometer converts differential length into phase difference $\delta \varphi =4\pi \Delta L/\lambda$. If its dark-port power scale is $P_0$ and its static phase offset is ${\varphi }_0$, thenExactly at a dark fringe, ${\varphi }_0=0$, power is quadratic in small displacement and loses its sign. A deliberate offset supplies an optical reference field: interference with the small signal field produces a linear term. Equivalently, $|E_{\mathrm{ref}}+E_{\mathrm{sig}}{|}^2$ contains $2\mathrm{Re}(E_{\mathrm{ref}}^{\ast }{E}_{\mathrm{sig}})$. The photodetector measures power, not a signed displacement directly.

Advanced LIGO uses this DC homodyne principle with a small differential arm offset and an output mode cleaner. Its feedback control maintains optical resonance and alignment, using actuation distributed over the suspension stages.^[ref] The simple formula above omits resonant storage: arm and recycling cavities give a frequency-dependent optical gain and delay, so no constant pass-count multiplier describes the entire observing band.

To see why a locked detector can still measure a disturbance, use a linear feedback control model. Let $x_{\mathrm{ext}}$ be the external differential displacement, $d_{\mathrm{err}}$ the residual readout, $d_{\mathrm{ctrl}}$ the recorded control command, $C(f)$ the sensing response and $A(f)$ the actuation response. With the convention $x_{\mathrm{res}}=x_{\mathrm{ext}}-A{d}_{\mathrm{ctrl}}$, reconstruction givesHere tildes denote Fourier transforms; $C$ has readout units per metre and $A$ metres per command unit. Feedback suppresses residual motion but records the compensating command. Omitting that command would suppress the reconstructed low-frequency signal too. In the actual GW150914 experiment, a modulated auxiliary laser exerted calibrated radiation pressure on a mirror to measure the response; periodic excitations monitored calibration.^[ref]

Linear drag opposes the velocity relative to the surrounding medium with magnitude proportional to its speed, where $\alpha >0$. Its power is ${\mathbf{F}}_d\cdot \mathbf{v}=-\alpha |\mathbf{v}{|}^2$, so it dissipates kinetic energy. Under a constant external force, the resulting linear ordinary differential equation relaxes exponentially on the time scale $m/\alpha$.

The electric field does work on matter at local power per volumeHere $\mathbf{J}$ is current density and the dot product includes the relative direction. Positive $\mathbf{J}\cdot \mathbf{E}$ means energy passes from the field to matter; negative means matter supplies energy to the field. In general this is a work rate, not automatically an irreversible heating rate.^[ref]

For a stationary isotropic Ohmic conductor with real scalar electrical conductivity $\sigma >0$, local Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Thus the local Joule heating rate isThe electrical resistivity ${\rho }_{\mathrm{el}}$ has units $\Omega \mathrm{m}$, and electrical conductivity has units $(\Omega \mathrm{m}{)}^{-1}$. These formulas use the material's rest frame and dissipative conduction current density. Moving media, complex conductivity, and additional storage or transport processes require their appropriate energy balance rather than this scalar substitution.

For uniform current in a straight homogeneous wire of length $L$, wire cross-sectional area $A$ and constant electrical conductivity, $E=V/L$, $J=I/A$ and $R=L/(\sigma A)$. The volume integral recovers terminal power:As a numerical check, the wire already used in electrical resistance has $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$, $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$ and $V=5\mathrm{V}$. It gives $J=2.5\times 10^{6}A/{\mathrm{m}}^2$, $p_{\mathrm{vol}}=1.25\times 10^{7}W/{\mathrm{m}}^3$ and $P=12.5\mathrm{W}$, agreeing with $I=2.5\mathrm{A}$ and $R=2\Omega$. Concentrating the same electric current in a smaller wire cross-sectional area therefore raises the local heating rate if electrical resistivity stays fixed.

For real $m>-1$ and integer $n\ge 0$, start from ${\int }_0^1{x}^{m}dx=1/(m+1)$ and differentiate $n$ times with respect to $m$. Each derivative adds one factor $\log x$, givingTo justify differentiation under the integral sign, restrict $m$ to a compact interval with lower endpoint $m_0>-1$. The derivatives are dominated by $x^{m_0}|\log x{|}^n$, which is integrable: $x=e^{-t}$ turns its integral into ${\int }_0^{\infty }{e}^{-(m_0+1)t}{t}^{n}dt$. A power of $t$ is integrable against this decaying exponential. This also explains why $m=-1$ is the convergence threshold.

The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are

Low-temperature calorimetry measures a small sample's heat capacity by resolving a small temperature change after a known electrical energy input. The experimental challenge is that the support, heater and thermometer also warm up. Their calorimeter addenda must be removed before interpreting the sample's phonons or electronic heat capacity. The Debye T-cubed test turns the corrected response into a test of the low-temperature vibrational spectrum.

Manuel Brando's 2009 experiment used a sapphire platform suspended by four alloyed platinum wires from a copper ring fixed to a cryostat stage. Grease attached the sample; a chip heater and ruthenium-oxide resistance thermometer served platform PLS-2. A controller stabilized the bath; a lock-in amplifier supplied heating pulses and read the calibrated sensor. The steady rise calibrated thermal conductance as $G=P/\Delta T$. A separate $59.1\mathrm{mg}$, $99.99\%$ pure gold sample, compared with reference heat capacities, established the calorimeter addenda. Small pulses limited property variation; imperfect sample contact required a second relaxation time. The platform contribution rose below $1.5\mathrm{K}$, making simple extrapolation unsafe.^[ref]

For an original idealization, assume sample and platform share one temperature $T$, the bath remains at $T_b$, and both the total heat capacity and the weak-link thermal conductance $G$ are approximately constant across one small pulse. Ignore other heat transfer and any phase transition. The first law of thermodynamics becomesHere $C_s$ is sample heat capacity, $C_a$ is calorimeter addenda, $P$ is deposited electrical power from Joule heating, and $\tau$ is a time constant. This is the same balance as calorimeter heat-loss correction, used here with a deliberately weak cryogenic link and a separately subtracted platform. For a constant pulse $P_0$ beginning at bath temperature, then switched off at $t_h$,The cooling exponential decay determines $\tau$, while the heating plateau determines $G=P_0/(T_{\infty }-T_b)$. Together with calorimeter addenda, they give $C_s=G\tau -C_a$. Cooling alone determines a ratio, not an absolute heat capacity. With finite $t_h$, the observed end rise is smaller than the plateau; fit the full response or apply the displayed correction rather than identifying the two.

For the stipulated figure parameters, $G=5.00\times 10^{-8}\mathrm{W}{\mathrm{K}}^{-1}$, $\tau =20.0\mathrm{s}$, and $C_a=2.00\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$ give $C_s=8.00\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$. The asymptotic rise is $0.0400\mathrm{K}$, only $2\%$ of bath temperature. At $100\mathrm{s}$, the rise is $0.0397305\mathrm{K}$. Of $2.00\times 10^{-7}\mathrm{J}$ input, only $3.97305\times 10^{-8}\mathrm{J}$ remains stored: the other $1.60270\times 10^{-7}\mathrm{J}$ has flowed to the bath. Treating this long pulse as an isolated $Q/C$ experiment would overestimate total heat capacity by a factor of $5.034$.

Suppose independent standard uncertainties are $u(G)=0.05\times 10^{-8}\mathrm{W}{\mathrm{K}}^{-1}$, $u(\tau )=0.2\mathrm{s}$ and $u(C_a)=0.10\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$. First-order propagation of uncertainty givesThe uncertainty is one standard deviation under those assumptions, about $2.2\%$ of sample heat capacity. Shared thermometer calibration errors can correlate fitted $G$ and $\tau$; then their covariance contributes too. Smaller samples amplify the relative influence of calorimeter addenda. Proposed checks include varying pulse amplitude, repeating with a reference sample, checking sensor self-heating, and comparing single- and two-time-constant fits. A drifting bath, slow internal equilibration or a transition invalidates this simple reduction even if a cooling curve looks smooth.

Luminosity is emitted radiative power: the energy carried away by electromagnetic radiation per unit time, measured in watts. Bolometric luminosity includes all wavelengths; a measurement through one photometric passband captures only part of it. Spectral luminosity records how that power is distributed over frequency. Received observed radiative flux has different units, watts per square metre, because it measures what crosses an observer's collecting area. ^[ref].

For example, an ideal source with luminosity $10^{26}\mathrm{W}$ emits $10^{26}\mathrm{J}$ each second regardless of whether it is viewed from $20$ or $40$ parsecs. Its observed radiative flux changes with distance, while its luminosity remains fixed. Combining stellar parallax with observed radiative flux therefore lets us compare the actual outputs of stars rather than their apparent brightnesses.

The luminous efficacy of radiation is luminous flux divided by total radiant flux, in $\mathrm{lm}{\mathrm{W}}^{-1}$, for the specified visual conditions. For nonzero radiant flux, it is a spectral weighted mean of spectral luminous efficacy. Specify the denominator: a lamp's luminous efficacy relative to electrical input power also includes conversion losses and is a different ratio. ^[ref]

For laboratory velocity $v\hat{r}+w\hat{\theta }$ and field $B\hat{z}$, $F_{B,r}=qBw$ and $F_{B,\theta }=-qBv$. Their powers are opposite. The field-dependent guide-reaction power is $-qBvw$; the total ideal radial reaction for bead mass $m_b$ and constant expansion speed additionally contributes the ordinary centripetal term $-m_{b}v{w}^2/R$.

For thin parallel circular plates of radius $a$ and separation $L$, the quasistatic axial field $E_z(t)$ induces $H_{\varphi }={ϵ}_{0}r{\dot{E}}_z/2$. Integrating cycle-averaged vacuum energy densities for sinusoidal drive gives ${\overline{U}}_B/{\overline{U}}_E=(\omega a/c{)}^2/8$. The same fields have radial Poynting flux $S_r=-{ϵ}_{0}r{E}_z{\dot{E}}_z/2$, whose leading net inward power equals the time derivative of stored electric energy. The small ratio diagnoses the quasistatic regime.

A sputter-deposition method using a magnetic field near a target to confine electrons and sustain ionization efficiently. Multiple targets permit co-deposition. A local flux proportional to target power is a useful calibration model over a restricted range, rather than a universal linear law for composition or thickness.

The maser cavity stores a resonant electromagnetic wave so that an arriving molecule encounters a field left by earlier molecules. An output coupling lets part of that stored energy reach a detector. The same coupling also removes energy, so increasing extraction can prevent startup. The cavity provides feedback; the selected molecular beam supplies the useful internal energy.

Let $U$ be stored field energy, $\omega =2\pi {\nu }_0$, and $\kappa ={\kappa }_{\mathrm{int}}+{\kappa }_{\mathrm{out}}$ the total passive energy decay rate. Internal dissipation and output extraction contribute ${\kappa }_{\mathrm{int}}U$ and ${\kappa }_{\mathrm{out}}U$ to lost power. With this energy convention, the loaded quality factor isThe field amplitude decays at $\kappa /2$; confusing amplitude and energy decay introduces a factor of two into a maser oscillation threshold calculation.

Ammonia gas entered a vacuum chamber through fine channels; the selected beam entered a cavity with input and output waveguides.^[ref] With external microwaves swept across a molecular resonance, the 1954 experiment read an increase in transmitted power; with sufficient beam emission, it sustained oscillation.^[ref]

The 1955 readout mixed two maser signals with a local oscillator in a crystal detector, amplified them at an intermediate frequency, and combined them in a diode detector. An oscilloscope showed their beat frequency. Changing cavity tuning, source pressure or focuser field tested sensitivity; excessive detuning stopped oscillation.^[ref] The beat tests relative frequency stability; it does not by itself establish an absolute frequency or an intrinsic linewidth independent of detector noise. Detuning also changes the field–molecule interaction, so the threshold model's constant $G$ cannot describe arbitrary tuning scans.

Multiplying the upper-level and photon equations by $h{\nu }_0$ makes the cancellation of internal transfer visible:Incoming excitation becomes unused excitation carried away, internal dissipation, useful output, or a change in stored energy. Translational kinetic energy, the source's other internal states and preparation losses lie outside this two-level budget. The output cannot exceed the incoming excitation power $h{\nu }_{0}F$.

Choose $G=10^{-4}{\mathrm{s}}^{-1}$ per unit inversion, $\gamma =10^4{\mathrm{s}}^{-1}$, ${\kappa }_{\mathrm{int}}={\kappa }_{\mathrm{out}}=10^6{\mathrm{s}}^{-1}$, and $F=4\times 10^{14}{\mathrm{s}}^{-1}$. These are stipulated teaching parameters. Then $N_0=4\times 10^{10}$, $F_{\mathrm{th}}=2\times 10^{14}{\mathrm{s}}^{-1}$, $N_{\mathrm{ss}}=2\times 10^{10}$ and $n_{\mathrm{ss}}=5\times 10^7$. Using the rounded ammonia inversion transition frequency, the loaded quality factor is approximately $7.50\times 10^4$. The incoming excitation power is $6.327\mathrm{nW}$: $4.745\mathrm{nW}$ leaves unused, $0.791\mathrm{nW}$ becomes internal dissipation and $0.791\mathrm{nW}$ is useful output, with rounding in these displayed values.

There is an instructive extraction optimum within this model. With the same $F$, $G$ and $\gamma$, put $\rho ={\kappa }_{\mathrm{out}}/{\kappa }_{\mathrm{int}}$ and $p=FG/(\gamma {\kappa }_{\mathrm{int}})=4$. For $0\le \rho <p-1$,Differentiation gives ${\rho }_{\mathrm{opt}}=\sqrt{p}-1=1$. At zero coupling no power is extracted; at $\rho =3$ the device reaches onset and the nonzero branch disappears. The optimum balances extraction against the inversion required to compensate that extraction. Equal residence rates and loss of molecular phase memory are simplifying assumptions; the calculated efficiency is not a universal bound for beam masers.

An ideal string of negligible mass and fixed length, transmitting tension when taut. In an ideal guide with no friction, the tension magnitude is the same on both sides: otherwise the force balance on an arbitrarily short massless segment would fail. Its fixed length gives a mechanical constraint relating the endpoint velocities. The two endpoint tension powers sum to zero when the guide is stationary; the string can transfer energy between connected particles without adding net energy.

A body moving at constant velocity $V$ incorporates material at rate $\dot{M}$ with zero initial momentum along its motion. The required external force is $F=\dot{M}V$, so the supplied power is $P=\dot{M}{V}^2$. Only $\dot{M}{V}^2/2$ becomes the added material's translational kinetic energy; the other half is dissipated during inelastic incorporation. This conclusion concerns the component along the body's motion; an incoming transverse velocity can add further dissipation.