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Solar radiant power incident per unit area. Horizontal, beam-normal and tilted-plane irradiances are different quantities: for a direct beam on a horizontal plane, $I_h=I_n\sin \alpha$, where $\alpha$ is solar elevation.

A spark gap can rapidly discharge a charged capacitor through an inductor. After excitation, the stored energy runs down through Joule heating, radiation and other losses. A key held down can permit repeated discharges: each discharge starts a damped train of rapid oscillations. The spark repetition frequency, the electrical oscillation frequency within each train, and the rate of keyed symbols are different quantities. Repetition does not make each train a maintained sinusoid.

An original lumped model replaces the conducting spark gap and other losses by a constant total electrical resistance $R>0$ in an isolated series RLC circuit. For electric charge $q$ on the capacitor, the Kirchhoff cycle law givesFor $Q>1/2$, its characteristic roots are $-\alpha \pm \mathrm{i}{\omega }_d$, with ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$, proving the stated damped form. The electric current $\dot{q}$ has the same exponential decay scale. Its wave envelope falls by $\exp (-2\pi \alpha /{\omega }_d)$ per cycle; the energy envelope falls twice as fast in the exponent. A maintained sinusoid instead requires continuing input power balancing losses.

With $L=5\mathrm{mH}$, $C=2\mathrm{nF}$ and $R=100\Omega$, the model gives $f_0={\omega }_0/(2\pi )=50.329\mathrm{kHz}$, Q factor $Q=15.811$, amplitude time constant $1/\alpha =100\mu \mathrm{s}$, and damped frequency ${\omega }_d/(2\pi )=50.304\mathrm{kHz}$. These are chosen component values. A real spark gap switches nonlinearly, an antenna radiates, and coupling exchanges energy between circuits; one constant-$R$ free decay does not model the complete transmitter. Rapid damping also broadens the frequency spectrum, which helps explain why short trains make sharp receiver selection difficult.

The detector figure of merit $D^{\ast }=\sqrt{a\Delta f}/\mathrm{NEP}$, where NEP is noise-equivalent power, $a$ detector area and $\Delta f$ electrical noise bandwidth. Its common unit is $\mathrm{cm}\mathrm{H}{\mathrm{z}}^{1/2}{\mathrm{W}}^{-1}$. A supplied detector-noise convention must be used consistently with integration time.

Spectral radiance is radiance per unit frequency or wavelength interval. Its astronomical synonym specific intensity is usually written $I_{\nu }=L_{\nu }$ or $I_{\lambda }=L_{\lambda }$; this $I_{\nu }$ is a spectral radiance, distinct from the radiant intensity $I$ in radiometry. For positive small intervals,The two expressions refer to the same spectral slice, with the positive widths related through frequency $\nu$ and wavelength $\lambda$. For frequency measured in hertz, $\mathrm{Hz}={\mathrm{s}}^{-1}$, the units are $[L_{\nu }]=\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$. For wavelength measured in metres, $[L_{\lambda }]=\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Integrating over a spectral band gives its radiance, and then multiplying by the collecting projected area and accepted solid angle gives a small uniform beam's radiant flux. A spectral measurement therefore retains information that the band-integrated power alone loses.

For an ideal blackbody, Planck's law gives $L_{\nu }=B_{\nu }(T)$ and $L_{\lambda }=B_{\lambda }(T)$ at temperature $T$, independent of outgoing direction. Its existing frequency and wavelength peaks of Planck's law illustrate why the coordinate matters. The simpler flat-band calculation below isolates that same change of variables without the thermal physics.

Radiant power leaving a surface per surface area and per unit frequency or wavelength. Isotropic cavity radiation obeys $M_{\nu }=c{U}_{\nu }/4$; the factor $1/4$ combines the outward hemisphere and projected area.

Spectral radiant flux resolves radiant flux by wavelength or frequency:Its units are $\mathrm{W}{\mathrm{m}}^{-1}$ per metre of wavelength or $\mathrm{W}\mathrm{H}{\mathrm{z}}^{-1}$ per hertz of frequency. A density quoted per nanometre must be integrated with wavelength intervals in nanometres; its numerical value is $10^{-9}$ times the per-metre density. The frequency and wavelength spectral radiance conversion applies equally to spectral radiant flux, through the change of variables formula:with $c$ the speed of light in vacuum and ${\Phi }_{e,\lambda }$ expressed per metre. The Jacobian determinant preserves integrated power rather than the height or location of an optical spectrum's peak.

For two explicitly synthetic optical spectra, let $x=\lambda /(1\mathrm{nm})$ and quote the following densities in $\mathrm{mW}\mathrm{n}{\mathrm{m}}^{-1}$:The triangle under $f_A$ and the rectangle under $f_B$ both have area $2\mathrm{mW}$: $200\times 0.020/2=200\times 0.010=2$. Their band radiant fluxes differ: $450$–$550\mathrm{nm}$ contains $1.5\mathrm{mW}$ of $A$, while $650$–$750\mathrm{nm}$ contains $1.0\mathrm{mW}$ of $B$. These are specified functions, not measured source spectra.

The dark-subtracted electric current of a linear photodiode isusing its spectral responsivity. The figure's stipulated $R$ gives $i_A=0.60\mathrm{mA}$ and $i_B=1.00\mathrm{mA}$ despite equal total radiant flux. Symmetry places $A$'s mean wavelength at $500\mathrm{nm}$ and $B$'s at $700\mathrm{nm}$; a linear $R$ averages to its value at that mean. A single electric current therefore cannot determine an unknown broadband radiant flux without additional spectral information or a sufficiently flat calibrated response.

For a photodiode in its linear regime, spectral responsivity is the dark-subtracted electric current per incident radiant flux in a narrow band at wavelength $\lambda$, measured in $\mathrm{A}{\mathrm{W}}^{-1}$. More generally a photodetector's readout can be electrical voltage or another quantity, with corresponding units. A broad-band readout weights the incoming optical spectrum by the detector's spectral responsivity; it is not the unweighted total incident power. ^[ref].

For real negative displacement reflection amplitude $r=-a$, with $0<a<1$, useThis separates a travelling field and a standing field algebraically. Their individual powers cannot be treated as independent incident-power fractions because the two fields interfere. Reflected and transmitted power should instead be calculated from travelling-wave amplitudes with the appropriate characteristic impedance factors.

A stationary mode shape does not imply absence of energy. Write $A=2a$ and $y=A\sin (kx)\cos (\omega t)$. For the ideal wave equation on a string, the kinetic energy and elastic potential energy per unit reference length add toHere $P$ is signed power crossing a position on the string, positive to the right; $e$ is an energy density per length, not per volume. Differentiation and $\mu y_{tt}=\tau y_{xx}$ give the local conservation of energy identity ${\partial }_{t}e+{\partial }_{x}P=0$. Substitution into $P$ givesIt can be nonzero away from a node of a standing wave or an antinode, but its time average over an oscillation period is zero. Energy flows locally back and forth even though this pure standing wave transports no net energy across a section over a complete cycle. At a node of a standing wave, $y_t=0$, so kinetic energy density is zero, but $y_x$ need not vanish: elastic potential energy density can be positive there.

For a fixed-end normal mode, integration over $0\le x\le L$ givesThe last equality uses $\tau k^2=\mu {\omega }^2$. At a straight-string snapshot, $U=0$ while $K=E$; at an extreme snapshot, $K=0$ while $U=E$. This is normal-mode energy exchange. A driven real string with damping requires continuing input power to replace lost energy, so a visually stationary pattern does not establish zero average supplied power.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

The Stefan–Boltzmann law gives the total hemispherical radiant exitance of an ideal blackbody at absolute temperature $T$:Here $\sigma$ is the Stefan-Boltzmann constant. Total means integrated over all wavelengths; hemispherical means integrated over all outward directions with the projected area weighting. Thus $M_b$ is emitted power per actual emitting area, rather than directional radiance or spectral radiance. Absolute temperature must be in kelvin; raising a Celsius reading to the fourth power gives the wrong result. A uniform surface of area $A$ emits radiant flux $P_{\mathrm{emit}}=A{M}_b$. A real surface instead has $M=ϵ(T)\sigma T^4$, where $ϵ(T)$ is its total hemispherical emissivity. The older symbol $F$ in the formula above denotes this emitted radiant exitance.^[ref]

For an ideal blackbody, doubling absolute temperature multiplies radiant exitance by $16$. With the approximate Stefan-Boltzmann constant $5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, the radiant exitances at $300\mathrm{K}$ and $600\mathrm{K}$ are approximately $459.300$ and $7348.805\mathrm{W}{\mathrm{m}}^{-2}$. This steep response explains why thermal radiation becomes so consequential at high temperature. It also enables heat transfer through a vacuum, where thermal conduction through a gas and convection are absent.

A short derivation integrates Planck's law in its frequency form. For a blackbody, direction-independent spectral radiance is $B_{\nu }(T)=2h{\nu }^3/[c^2(e^{h\nu /(k_{B}T)}-1)]$, where $h$ is the Planck constant, $k_B$ the Boltzmann constant and $c$ the speed of light. The outward hemisphere contributes $\int \cos \theta d\Omega =\pi$, soExpanding $(e^x-1{)}^{-1}={\sum }_{n=1}^{\infty }{e}^{-nx}$, nonnegative terms allow termwise integration. Repeated integration by parts gives ${\int }_0^{\infty }{x}^3{e}^{-nx}dx=6/n^4$; the Riemann zeta function value $\zeta (4)={\pi }^4/90$ gives the remaining integral as ${\pi }^4/15$. Hence $M_b=\sigma T^4$ with the coefficient in Stefan-Boltzmann constant. The angular integral is essential: omitting it confuses radiance with radiant exitance.

The bolometric luminosity of a star is its total radiated power integrated over all wavelengths. A band or spectral luminosity instead specifies a restricted spectral range. The Hertzsprung-Russell diagram compares total luminosity with effective temperature.

To construct an original example rather than imitate GW150914, set $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, and adopt the rounded solar mass $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$. Compare $10+10$ and $20+20$ solar mass binaries starting at $20\mathrm{Hz}$. Stop each at the stipulated orbital compactness thresholdThis is a conservative pedagogical stopping rule, not a claimed error bound: the relative speed already reaches $\sqrt{0.10}c$. Higher-order corrections accumulate wave phase well before merger, and precision inference requires better models. The two cutoffs are approximately $102.2$ and $51.1\mathrm{Hz}$ respectively; doubling the chirp mass multiplies $\dot{f}$ at fixed $f$ by $2^{5/3}$.

For a nearby stipulated source at distance $D$, neglect redshift and define inclination $\iota$ as the angle between the orbital angular momentum and the line of sight. The leading quadrupole formula yieldsThe numerator of $A$ has units ${\mathrm{m}}^5{\mathrm{s}}^{-4}$, matching $c^{4}D$, so both polarizations are dimensionless. Its $f^{2/3}$ growth and the rising frequency produce the tightening, strengthening oscillations. For face-on viewing the polarization components have equal magnitudes in phase quadrature; for edge-on viewing the cross component vanishes, agreeing with circular-binary quadrupole radiation.

Shifting an arrival time aligns the chirp evolution, but different polarization projections need not overlay after only a sign flip. The last panel divides the ideal projected records by their known envelope factors to expose this phase difference. A physical source model must jointly explain those differences, not erase them by arbitrary rescaling. Independent integration of the separation equation, the radiated power and $d\Phi /dt=2\pi f$ checks the analytic chirp against orbital energy balance over the entire stipulated interval.

A Stirling engine repeatedly heats and cools a confined gas, whose expansion and compression drive a moving mechanical load. This makes it a particularly accessible example of a heat engine: both the repeating motion and the two regions exchanging heat can be observed.

In the published Demonstration experiments with a Stirling engine by Christopher G Deacon and others, an electrical heater warms one end of a cylinder and flowing water cools the other at $12{}^{\circ }\mathrm{C}$. A moving displacer shuttles gas through copper wool that temporarily stores and returns internal energy. A laser spot traces a closed pressure-volume path on a screen: mechanical motion controls one coordinate, pressure the other. Calibrating both scales converts the enclosed area to work per thermodynamic cycle; multiplying by rotation frequency gives gas-cycle power. At $97\mathrm{W}$ heater input, the paper reports $4.4\mathrm{W}$ gas-cycle power, while a separate Prony brake measures $3.4\mathrm{W}$ shaft output. The cold-side temperature is controlled and two higher heater settings are compared. The gas has a nonuniform temperature, and heater energy also escapes to the room: neither output/input ratio measures $W/Q_h$ for an ideal two-thermal reservoir engine. These observations demonstrate work production and losses, not a direct measurement of the bound in Carnot's theorem. The inspected apparatus, readouts and results are on pp. 180–182, Figures 1–3 and Table 1.

Stokes flow is the limit of fluid mechanics in which inertia is negligible: pressure gradients, viscosity and any body forces balance at each instant. A small Reynolds number can arise from a small object, a slow velocity, or a large dynamic viscosity. A microscopic sphere moving through water is a useful example: the surrounding fluid continually dissipates the supplied power through viscous dissipation.

The picture is a meridional slice of the three-dimensional axisymmetric flow around a sphere, with the sphere held fixed and the distant fluid moving rightward at speed $U$. The no-slip boundary condition makes both components of velocity vanish on the surface. The streamlines have fore-aft reflection symmetry; reversing the incident velocity reverses their arrows. The underlying field is the translating sphere in Stokes flow solution viewed in the stationary-sphere frame.

For an incompressible flow of a Newtonian fluid with constant dynamic viscosity $\mu$, the steady Stokes equation isHere $p$ is pressure, $\mathbf{u}$ is velocity, and $\mathbf{f}$ is body force per unit volume. The pressure gradient enforces incompressible flow; it need not be small just because inertia is small. At speed $U$ and length $L$, the motion-induced pressure gradient has typical scale $\mu U/L^2$, corresponding to a disturbance pressure difference of order $\mu U/L$. See Shreyas Mandre's fluid dynamics notes, sections 9.1.1–9.1.2.

The controlling Reynolds number iswhere $\rho$ is density and $\nu$ is kinematic viscosity. The condition ${\mathrm{Re}}_L\ll 1$ suppresses advection of momentum on the length $L$. Rapid changes require an additional check of the viscous diffusion time: small Reynolds number alone does not eliminate the time derivative in the Unsteady Stokes equation.

Two rewards of this limit are the exact Stokes's law for the drag on a sphere and kinematic reversibility of Stokes flow. The first predicts settling velocities and connects to the Stokes–Einstein relation; the second explains why a microswimmer needs a nonreciprocal stroke, as expressed by the scallop theorem. Despite viscous dissipation, reversing the driving can retrace the motion: an especially striking consequence of Linearity of Stokes flow.

George Gabriel Stokes obtained the sphere drag and its terminal velocity in his 1851 investigation of how viscosity changes physical pendulum motion. He also used the slow fall of small water droplets to discuss clouds. The paper was read on 9 December 1850; 1851 is its publication date. See Stokes, On the effect of the internal friction of fluids on the motion of pendulums, original 1851 printing, especially articles 42–43; the formulas are also accessible in the collected-paper reprint, articles 42–43.

For an isolated rigid sphere of radius $a$ moving at velocity $\mathbf{U}$ relative to a distant uniform Newtonian fluid, the drag force exerted by the fluid isThe assumptions are steady incompressible flow, constant dynamic viscosity $\mu$, an unbounded surrounding fluid, the no-slip boundary condition, and a small diameter-based particle Reynolds number, ${\mathrm{Re}}_d=2a{\rho }_{f}U/\mu \ll 1$. Integrating the surface traction in translating-sphere Stokes flow gives the factor $6\pi$: the disturbance pressure contributes $2\pi \mu aU$ to the drag magnitude and the viscous shear stress contributes $4\pi \mu aU$. See Richard Fitzpatrick, Axisymmetric Stokes Flow Around a Solid Sphere, equations 10.120–10.129.

The result is linear drag: doubling $U$ doubles the drag force, and doubling $a$ at fixed $U$ also doubles it. The required power is $6\pi \mu a{U}^2$, all supplied to viscous dissipation. Density enters the check of the particle Reynolds number, although it does not appear in the leading drag formula.

Nearby walls or other particles change the hydrodynamic resistance matrix and invalidate the isolated-sphere coefficient, even at very small Reynolds number. Finite Reynolds number introduces inertia corrections; the Oseen approximation resolves the distant region where the leading Stokes flow approximation ceases to be uniform. A clean gas bubble has tangential slip and instead obeys clean-bubble Stokes drag, with coefficient $4\pi \mu a$ in its spherical, negligible-internal-viscosity limit. See H. K. Moffatt, Six Lectures on General Fluid Dynamics, section 3.4 for the sphere solution and its small-Reynolds number limitation. The continuum mechanics and boundary assumptions must also remain valid as the particle becomes very small.

Balancing this Stokes drag law against gravity and buoyancy gives the Stokes settling velocity of an isolated rigid sphere.

The magnetic energy stored in a linear solenoid of constant self-inductance $L$ isin the long-solenoid approximation. The first formula is an inductor relation; the second identifies its storage with magnetic energy in the interior magnetic field, under the vacuum-core approximation. This agreement joins the electrical circuit description to the spatial magnetic field description. The expression $B^2/(2{\mu }_0)$ is the vacuum magnetic energy density. ^[ref].

A short derivation starts from the terminal electric potential difference $v_L=LdI/dt$, with positive electric current entering the terminal at higher potential. The absorbed power is $v_{L}I$, and integrating from zero electric current givesThe result describes recoverable storage in the inductor; Joule heating in the wire is a separate expenditure of energy. ^[ref].

For the long-coil example, $L\approx 4.9348\mathrm{mH}$ and $I=0.50\mathrm{A}$ give $U\approx 6.1685\times 10^{-4}\mathrm{J}=0.61685\mathrm{mJ}$. Substituting the ideal $B\approx 1.256637\mathrm{mT}$ and volume $A\ell \approx 9.8175\times 10^{-4}{\mathrm{m}}^3$ into $B^{2}A\ell /(2{\mu }_0)$ gives the same value. Doubling electric current quadruples stored magnetic energy. In a real coil, reducing electric current can return this energy to an electrical circuit, or release it as Joule heating in a closed RL circuit.

The 1985 experiment launched $150\mathrm{ps}$ pulses from a mode-locked laser into $1.4\mathrm{km}$ of optical fiber. Dispersion and self-phase modulation produced a broader spectrum and approximately $300\mathrm{ps}$ pulses. A Pockels cell selected a pulse for repeated amplification in neodymium glass, giving approximately $2\mathrm{mJ}$ before the double-pass diffraction grating compressor. Its transmission was 50 percent. The amplified output was reported as $1\mathrm{mJ}$ with a $2\mathrm{ps}$ width limited by the streak camera response. The compressor was adjusted on fiber-output pulses using intensity autocorrelation; that measurement gave $1.5\mathrm{ps}$. Steve Williamson was acknowledged for help developing the concept, and Philippe Bado and Marcel Bouvier for technical support.^[ref]

The crucial comparison is compression before and after amplification. It asks whether enough usable spectral phase structure survives the amplifier, rather than merely whether its output is bright. A response-limited trace cannot establish the true width below the detector's resolution or prove that every higher-order phase coefficient vanished. An intensity autocorrelation is also not a direct unique measurement of the complex electric field. These are limits on the inference, not extra controls attributed to the experiment.

Donna Strickland's 2018 lecture describes setting the compressor with the high-repetition oscillator pulses and making the amplified-pulse measurement with Steve Williamson's streak camera. It explicitly notes imperfect chirp matching.^[ref] The first apparatus therefore illustrates the strategy without establishing a universal lossless or exactly phase-matched amplifier.

Donna Strickland leads an ultrafast laser group at Waterloo investigating intense laser interactions and nonlinear optics.^[ref] Gérard Mourou's later work includes coherent combination of laser beams in the XCAN project at École polytechnique.^[ref] Their complementary research settings connect the original short-pulse experiment with the continuing problems of producing useful energy, peak power and average power together.

For two ideal strings under a common tension, the positive mechanical impedances satisfy $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ are their linear mass densities. The reflected fraction of incident power is $\mathcal{R}=[(Z_1-Z_2)/(Z_1+Z_2){]}^2$. Solving for the positive ratio $z=Z_2/Z_1$ gives $z=(1+\sqrt{\mathcal{R}})/(1-\sqrt{\mathcal{R}})$ or its reciprocal. Squaring yields the displayed linear mass density ratios. For $\mathcal{R}=1/2$, the heavier string has $17+12\sqrt{2}\simeq 33.97$ times the lighter string's linear mass density; the impedance ratio is only $3+2\sqrt{2}$.

The equivalent input noise temperature of the antenna-plus-receiver system is defined by available noise power $P_n=k_B{T}_{\mathrm{sys}}\Delta \nu$ in one matched polarization of waves mode. It includes receiver noise and the appropriate sky and instrumental contributions. It is not generally the physical temperature of a single component.