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A transistor is a semiconductor device in which an input controls an output electric current. Used with biasing and an electrical load, it can form an electronic amplifier or a switch. Its power gain comes from an external supply: the input signal controls the conversion of supplied energy into output signal power.

The point-contact transistor makes the idea tangible: two nearby contacts interact through one piece of semiconductor. A bipolar junction transistor instead uses two p-n junctions and a thin intervening region. The distinction matters both for the history and for choosing a physical model.

An original teaching model exposes the roles of input loading, output loading and the supply. Choose a small-signal model with positive input electrical resistance $r_i$, transconductance $g_m>0$, and output electrical resistance $r_o>0$. Its controlled output current source draws $g_m{v}_i$ from the output node toward the reference node. This chosen sign gives an inverting voltage gain; it is not the historical contact circuit's sign convention.

Assume a fixed operating point, negligible reverse coupling, a frequency range where coupling capacitors pass the signal, and negligible other capacitances and transit delays. A collector resistor $R_D$ connects the output to a steady ideal supply $V_{DD}$; an external electrical load $R_L$ is coupled through a capacitor, so it draws no direct current. The input source has series electrical resistance $R_s$. All voltages below are small departures from the operating point. An ideal fixed supply has zero incremental voltage and therefore becomes the reference node in the small-signal model.

Define $R_O=R_D\parallel r_o$ and $R_P=R_O\parallel R_L$, where $a\parallel b=ab/(a+b)$. Input voltage division and output Kirchhoff's current law giveConsequently the port voltage gain, overall source-to-output voltage gain, and load current gain magnitudes areThis current gain uses load electric current, not all the controlled-source electric current. It is also a different port definition from the historical emitter-to-collector current gain. Larger $R_s$ attenuates the input; smaller $R_L$ attenuates the output.

For a sinusoid input signal, use root mean square voltages. The power gain from accepted input signal power $P_i=v_i^2/r_i$ to useful load power $P_L=v_o^2/R_L$ isUnlike $|A_i|$, this tends to zero as either $R_L\to 0$ or $R_L\to \infty$. Differentiation givesThus matching the output electrical load maximizes delivered power in this unilateral resistive model. For fixed source voltage, input loading multiplies $P_L$ by $[r_i/(R_s+r_i){]}^2$; the accepted-input power gain above excludes the loss in $R_s$.

Choose $r_i=R_s=1\mathrm{k\Omega }$, $g_m=2\mathrm{mS}$, $r_o=20\mathrm{k\Omega }$, and $R_D=2\mathrm{k\Omega }$. Then $R_O=20/11\mathrm{k\Omega }$. With $R_L=2\mathrm{k\Omega }$ and $v_s=20\mathrm{mV}$ root mean square, the input is $10\mathrm{mV}$ and the inverted output has magnitude $19.0476\mathrm{mV}$ root mean square. Hence $P_i=0.100\mu \mathrm{W}$ and $P_L=0.181406\mu \mathrm{W}$: $G_P=1.81406$. These are calculated values of the chosen model, not historical measurements.

With the same parameters but $R_L=100\Omega$, $|A_i|=1.89573$ while $G_P=0.359381<1$. At $R_L=10\mathrm{k\Omega }$, $|A_v|=3.07692$ while $G_P=0.946746<1$. Neither current gain nor voltage gain alone establishes power gain.

For the direct current load line in the figure, choose $V_{DD}=6\mathrm{V}$, $I_Q=1\mathrm{mA}$ and $V_Q=4\mathrm{V}$, consistent with $V_Q=V_{DD}-R_D{I}_Q$. Locally extend the model as $I=I_Q+g_m{v}_i+(V-V_Q)/r_o$. With the external electrical load isolated at direct current, its intersection with $I=(V_{DD}-V)/R_D$ is $V=V_Q-g_m{R}_O{v}_i$. Keeping $|v_o|$ small compared with the available voltage headroom is essential; extrapolation to arbitrarily large signals would be unphysical.

At the chosen operating point the collector supply delivers $V_{DD}{I}_Q=6\mathrm{mW}$: $2\mathrm{mW}$ is dissipated in $R_D$ and $4\mathrm{mW}$ in the output device. Under a small signal, that energy flow is redistributed to provide load power. The incremental controlled current source hides the supply's steady energy flow; it does not create energy. A real complete amplifier also needs its input biasing network and has further losses. This calculation establishes neither its total efficiency nor a universal law for point-contact transistors.

A transmission line guides travelling voltage and current waves. Its distributed electrical properties determine its characteristic impedance. A mismatch at a terminal produces reflected waves; available power, delivered power and returned power are distinct quantities.

The energy flux carried into a transmitted wave, integrated over the relevant cross section. The power transmission coefficient compares this with incident power. A displacement or field-amplitude transmission coefficient must be converted with the appropriate wave impedances; its squared modulus alone need not be the power fraction. This distinction applies both to electromagnetic waves and transverse string waves.

A triple-point cell turns phase coexistence into a reproducible reference for thermometer calibration. The NIST water-cell demonstration uses a sealed glass vessel containing water of specified isotopic composition and an open central thermometer well. Cooling this well with solid carbon dioxide and alcohol freezes an ice mantle while leaving liquid water and its vapor. The coolant is outside the sealed sample; the sample's vapor space does not communicate with room air.^[ref]

In the actual NIST cell-comparison experiment, newly frozen mantles were compared with an aged reference mantle in a controlled maintenance bath. A warm aluminium rod melted a thin inner layer around the well, allowing the mantle to rotate freely and establishing an ice–water interface close to the sensor. A standard platinum resistance thermometer and a resistance bridge measured the electrical resistance ratio to a calibrated, temperature-controlled reference resistor. Readings at two currents permitted extrapolation to zero electrical power, correcting self-heating; measured immersion depths supplied hydrostatic pressure corrections. Comparing preparations over time tested stabilization rather than assuming freshly frozen ice was ready.^[ref]

The inference is a reproducible fixed-point response for the thermometer, not a direct reading of three chemical potentials. The reference cell anchors the comparison; a water point alone does not determine a thermometer's response at every other temperature. The ideal $F=0$ argument excludes impurity and isotope variation, finite interface effects and unintended heat transfer; real cells also require adequate immersion and stable electrical references. Repeatability is evidence that these effects are controlled, not a reason to omit their uncertainty.

Since 20 May 2019 the kelvin has been defined through a fixed numerical value of the Boltzmann constant. The thermodynamic temperature of the water triple point is consequently a measured quantity with uncertainty, rather than an exact value imposed by the kelvin definition. Its continued use as an assigned reference on the International Temperature Scale of 1990 is a separate convention.^[ref][ref]

A tunnel diode is a p-n junction with sufficiently heavy semiconductor doping on both sides to make a thin depletion region and permit substantial interband tunnelling. The neutral regions are degenerate semiconductors. Its distinctive forward current-voltage characteristic contains a peak followed by a valley: over part of the curve, increasing the applied electric potential difference decreases the electric current. The Esaki tunnel-diode experiment made this behavior experimentally concrete.

The useful quantity is the local slope, not the ratio of total voltage to total electric current. With conventional electric current $I$ entering the positive-voltage terminal, defineNegative differential resistance means $g_d<0$, hence $r_d<0$ where defined. A forward operating point can still have $V>0$, $I>0$ and absorbed power $VI>0$. A small voltage increase then gives a negative current increment; this does not make the total electric current negative or turn the device into an unlimited source of energy. The external supply sets the operating point and supplies the energy available for an electronic amplifier.

An original idealized construction explains the shape. Put the n-type semiconductor on the left and the p-type semiconductor on the right. Take $q>0$ as the elementary charge and $V=V_p-V_n$. Increasing forward bias lowers the right-hand electron energy levels relative to the left. Denote the two reservoir electrochemical potentials by ${\mu }_n,{\mu }_p$, so that ${\mu }_n-{\mu }_p=qV$. At zero temperature, occupied left-hand conduction band states extend up to ${\mu }_n$; empty right-hand valence band states start at ${\mu }_p$. For transfer conserving electron energy, the available interval isIt is nonempty only when its upper endpoint exceeds its lower endpoint. Both an occupied initial quantum state and an empty final quantum state must exist at the same energy. The Pauli exclusion principle prevents transfer into an already occupied final quantum state.

For the original diagram, choose arbitrary energy units with $E_{c,n}=-0.5$, ${\mu }_n=0$, ${\mu }_p=-u$ and $E_{v,p}=0.5-u$, where $u$ is $qV$ in those units. The available interval has widthThus $L(0.25)=0.25$, $L(0.50)=0.50$, $L(0.70)=0.30$, and $L(1.15)=0$. Initially forward bias opens a transfer window; further forward bias closes it by moving the right-hand valence band edge below the left-hand conduction band edge. This is a state-counting illustration, not a formula for a measured electric current.

For reverse bias, the transfer direction changes: occupied right-hand valence band states can feed empty left-hand conduction band states. In the same construction, $u=-0.70$ opens the interval $0<E<0.70$ for this reverse transfer. A narrow depletion region allows a large reverse quantum tunnelling electric current at small electric potential difference, whereas the ordinary Shockley diode equation predicts only a small reverse saturation electric current before diode reverse breakdown. At zero biasing, equal reservoir occupations give zero net electric current; quantum tunnelling does not require a net equilibrium flow.

A finite-temperature refinement replaces sharp occupation cutoffs with the Fermi-Dirac distribution. An illustrative weak-transfer expression isHere $\mathcal{W}\ge 0$ describes allowed initial/final quantum states and their coupling, and $f_n,f_p$ are Fermi-Dirac distributions with the respective electrochemical potentials. The occupation difference follows by subtracting the reverse transfer factor from the forward one: $f_n(1-f_p)-f_p(1-f_n)=f_n-f_p$. An overlap interval alone does not determine $\mathcal{W}$: transmission probability, density of states, crystal momentum selection and phonons can matter. A complete germanium calculation is more demanding than multiplying two constant density of states factors. At larger forward bias, ordinary carrier injection and diffusion supply a rising current branch even after direct band overlap is lost.

A current-voltage characteristic determines possible steady operating points; it does not alone determine their linear stability. Consider an ideal voltage source $V_s$ feeding a tunnel diode through a resistor $R>0$, with a positive constant capacitance $C$ across the junction. Assume a quasistatic device relation $I(V)$ and neglect inductance, transit delays, noise and heating. Kirchhoff's current law gives the original one-state modelA steady operating point $V_{\ast }$ satisfies $I(V_{\ast })=(V_s-V_{\ast })/R$: the device curve intersects the descending supply line. This is the source-side counterpart of a resistive load line. A small perturbation $V=V_{\ast }+\delta V$ obeysThus the equilibrium point is locally attracting if $1/R+I^{\prime }(V_{\ast })>0$ and unstable if the inequality reverses. Equality requires higher-order analysis. Even on a negative differential resistance segment, a sufficiently small supply electrical resistance can stabilize this model. The assumption about the electrical circuit matters: an inductive or delayed circuit has additional state variables and may behave differently.

Choose an original toy forward characteristic, with no fitted device parameters:where $V_0=0.10\mathrm{V}$, $I_0=1.0\mathrm{mA}$ and $0\le x\le 3.6$. For every $x>0$, $i(x)>0$, so this entire forward curve absorbs positive total power. NeverthelessThe model therefore separates negative differential resistance from negative total $V/I$. The corresponding peak and valley are at approximately $0.1184\mathrm{V}$ and $0.2816\mathrm{V}$; these voltages deliberately differ from the historical peak.

With $R=100\Omega$ and $V_s=0.60\mathrm{V}$, the supply line is $I/I_0=6-x$. The intersection equation factors exactly:The three operating points are $(0.10\mathrm{V},5\mathrm{mA})$, $(0.20\mathrm{V},4\mathrm{mA})$ and $(0.30\mathrm{V},3\mathrm{mA})$. Their values of $1/R+I^{\prime }(V_{\ast })$ are respectively $20$, $-10$ and $20\mathrm{mS}$. The outer two attract nearby trajectories; the middle one repels them. This is bistability in the specified one-state electrical circuit.

For the second panel, choose $R=25\Omega$ and $V_s=0.30\mathrm{V}$. Now $I/I_0=12-4x$ andThere is exactly one real operating point, again $(0.20\mathrm{V},4\mathrm{mA})$. Here $1/R+I^{\prime }(V_{\ast })=20\mathrm{mS}>0$, despite the device's own slope $I^{\prime }(V_{\ast })=-20\mathrm{mS}$. With $C=10\mathrm{nF}$, the perturbation decays with time constant $C/(20\mathrm{mS})=0.50\mu \mathrm{s}$; the unstable middle point of the first panel has growth time $1.0\mu \mathrm{s}$.

The one-state autonomous model supports switching between attracting operating points after a perturbation or parameter change; it cannot produce a nonconstant periodic orbit. A proposed oscillator needs additional dynamics. This limitation is useful: a load-line intersection establishes a possible steady reading, the slope test establishes local attraction in a specified model, and neither calculation alone predicts an oscillation frequency for a real tunnel diode.

Ultraviolet is electromagnetic radiation with wavelengths shorter than the violet end of ordinarily visible light. The CIE convention uses $100$–$400\mathrm{nm}$, subdivided into UV-A at $315$–$400\mathrm{nm}$, UV-B at $280$–$315\mathrm{nm}$ and UV-C at $100$–$280\mathrm{nm}$. These divisions name intervals; they do not mark universal changes in every material's response. The boundary with visible light is not sharp, and other fields use near-, far-, vacuum- and extreme-ultraviolet divisions. ^[ref]. For example, NASA's Solar Dynamics Observatory observes extreme-ultraviolet bands below $100\mathrm{nm}$; the CIE interval is therefore not the full extent of astronomical ultraviolet usage. ^[ref].

The CIE ultraviolet interval corresponds to frequencies about $7.49\times 10^{14}$–$3.00\times 10^{15}\mathrm{Hz}$ and photon energies $4.97\times 10^{-19}$–$1.99\times 10^{-18}\mathrm{J}$, using vacuum wavelength and $E=hc/\lambda$ with Planck constant $h$ and speed of light $c$. Shorter wavelength means larger energy per photon; larger irradiance means more incident power per area. These are different comparisons. At the same incident power, a $254\mathrm{nm}$ ultraviolet beam has fewer photons per second than a $940\mathrm{nm}$ infrared beam, even though each ultraviolet photon has more energy.

A common ultraviolet mechanism is photon absorption promoting an electron from a ground state to an excited state of a chemical atom or molecule. By contrast, many infrared absorption bands involve stretching or bending chemical bonds. For either mechanism, both the energy difference and the coupling to the incident electromagnetic field matter. One concrete example is the $217\mathrm{nm}$ absorption discussed for butadiene: an electron moves between molecular orbitals, and the position of the absorption reveals their energy separation. Ultraviolet absorption can therefore probe molecular orbitals, while infrared spectroscopy probes many molecular motions. ^[ref].

Subsequent chemical reactions depend on the absorbing chemical substance and its environment; the label ultraviolet alone does not predict a reaction rate. At sufficiently short wavelength, a photon can cause photoionization if its energy exceeds the relevant ionization energy. There is no single wavelength that separates all ionizing from all nonionizing behavior for every chemical substance. Likewise, eventual heating after photon absorption is possible in both ultraviolet and infrared. The distinction is useful because different optical spectra expose different processes, not because the two ranges obey different laws of electromagnetism.

Ritter's 1801 observation used a chemical readout rather than a thermometer. In Caltech/IPAC's account, sunlight passed through a glass optical prism and silver chloride was placed in different parts of the optical spectrum. Darkening was weak toward red, stronger toward violet, and pronounced beyond the visible violet edge. The measured evidence was a change in the chemical substance's appearance at a spatial location with no visible color, establishing radiation beyond the eye's ordinary range. It was not a calibrated measurement of a universal ultraviolet optical spectrum. ^[ref].

A controlled modern interpretation of such a chemical test would compare equal exposure durations and matched fresh samples, keep an unexposed sample in darkness, monitor temperature, and distinguish scattered visible light from the intended short-wavelength beam with characterized filtering. These are proposed controls, not an assertion that Ritter performed them. Unequal sample preparation or heating can change a reaction rate. The optical prism, terrestrial atmosphere and sample response restrict which wavelengths the test detects. Darkening is a detector-weighted response, so its spatial maximum is not automatically the maximum of incoming radiant flux or of photon energy.

Thomas Larason and Jeanne Houston's NIST report documents a quantitative ultraviolet detector comparison. An argon source feeds a double-diffraction grating spectrometer; a shutter and mirrors deliver the selected band to a working-standard or test photodiode, successively at the same position. A beam splitter supplies a monitor photodiode. Amplifiers and voltmeters record detector and monitor simultaneously; shutter-closed readings provide dark baselines. The $200$–$500\mathrm{nm}$ scan, typically with $4\mathrm{nm}$ bandpass, measures spectral responsivity. Monitor normalization suppresses source drift; known spectral lines calibrate wavelength. The report assesses temperature effects and unwanted light outside the band, weighted by detector response. The latter need not cancel between different spectral responsivities. ^[ref].

This determines electrical response without visual perception. Bandpass and spatial nonuniformity limit the calibration; aging and changed temperature can affect later use. The scan includes visible light and ultraviolet; no response below $200\mathrm{nm}$ follows from it.

For a narrow beam, the idealized substitution calculation isHere $R$ is spectral responsivity; $I$ and $M$ are dark-subtracted detector and monitor electric currents. This follows by writing $I=RP$ for each detector and using the monitor to track changes in radiant flux. It assumes linear response, unchanged beam splitting and collection geometry, and a sufficiently narrow band. For an illustrative standard with $R_{\mathrm{std}}=0.15\mathrm{A}{\mathrm{W}}^{-1}$, ratios $I_{\mathrm{std}}/M_{\mathrm{std}}=2.0$ and $I_{\mathrm{test}}/M_{\mathrm{test}}=1.6$ give $R_{\mathrm{test}}=0.12\mathrm{A}{\mathrm{W}}^{-1}$. These invented arithmetic inputs illustrate the method; they are not NIST measurements.

An independent photon-counting model connects spectral responsivity to quantum efficiency. If $\eta (\lambda )$ is the number of collected electrons per incident photon, there is no internal multiplication, and $q$ is the elementary charge, thenAt illustrative constant $\eta =0.60$ and $P=1\mu \mathrm{W}$, this gives about $0.123\mu \mathrm{A}$ at $254\mathrm{nm}$ and $0.455\mu \mathrm{A}$ at $940\mathrm{nm}$. The ultraviolet readout is smaller in this model because there are fewer incident photons, not because their individual energies are smaller. Real quantum efficiency depends on wavelength, windows, absorption and carrier collection; a detector's band gap and construction determine its usable range. Constant $\eta$ across these two wavelengths is an analytical comparison, not a claimed specification of a real photodiode.

The curve is the calculated relation $E=hc/\lambda$, and the shaded ranges use the stated CIE conventions. The lower diagram abstracts the detector comparison rather than reproducing an instrument drawing. Together with the infrared heat-balance illustration, it shows why a calibrated detector response, rather than what looks bright to an observer, is the bridge between invisible radiation and a quantitative measurement.

A semi-infinite wave equation on a string with linear mass density $\rho$, wave speed $c$, and tension $T=\rho c^2$ has positive characteristic mechanical impedance $Z=\rho c$. At an end attached to a massless viscous damper of coefficient $\alpha >0$, displacement and transverse-force balance give the displacement reflection coefficientHere $\mathcal{A}$ is the fraction of mean incident power dissipated. A prescribed $0<\mathcal{R}<1$ gives two positive damper coefficients, $\alpha /Z=(1\pm \sqrt{\mathcal{R}})/(1\mp \sqrt{\mathcal{R}})$, with opposite reflected phases. Matching $\alpha =Z$ removes reflection.

For a Newtonian fluid and a flat wall with $0<b<\infty$, the Navier slip boundary condition is equivalent to a tangential wall-on-fluid tractionThe sign expresses resisting friction: ${\partial }_{n}u$ uses the inward unit normal, whereas the wall-on-fluid traction uses the outward unit normal of the fluid domain. The irreversible loss of mechanical energy per unit wall area and time isFor two tangential directions replace $u_{\mathrm{rel}}^2$ by the squared Euclidean norm of relative tangential velocity. At a moving wall this relative-velocity loss must be distinguished from the power supplied by the wall's motion. Bulk viscous dissipation also remains present. A finite slip length models reduced interfacial resistance, without declaring every boundary ideal or removing dynamic viscosity.

The limiting loss depends on how the flow is driven. In Couette flow with a slipping lower wall,It tends to zero for both $b\to 0$ and $b\to \infty$ at fixed $H$ and $U$. At $b=0$ impose the no-slip boundary condition directly, rather than dividing by zero. In planar Poiseuille flow with equal wall slip at fixed $G$, each wall instead dissipates $G^{2}b{H}^2/(4\mu )$, which grows with $b$ because relative velocity grows. The bulk loss per unit width is $G^2{H}^3/(12\mu )$; adding both wall losses gives $Gq$, the pressure work per unit channel length and width. This energy balance shows why small friction coefficients do not guarantee small losses under every driving constraint.

The emitted optical power divided by electrical input power. For the bare-device calculation in light-emitting diode efficiency, $P_{\mathrm{elec}}=I{V}_f$; a complete lamp also consumes power in its driver. Luminous efficacy, measured in lumens per watt, additionally weights the spectrum by human visual sensitivity and is a different quantity.

The watt is the SI unit of power, one joule per second. Joule heating at $12.5\mathrm{W}$ transfers $125\mathrm{J}$ of energy over a time of $10\mathrm{s}$ if that power stays constant. ^[ref].

The waveguide cutoff frequency separates oscillatory axial propagation from exponential decay for a particular normal mode. For the vacuum-filled rectangular electromagnetic waveguide, the dispersion relation isFor a transverse electric mode of a rectangular waveguide, $m,n$ are nonnegative integers but not both zero. For a mode with $B_z=0$ and nonzero $E_z$, both indices must be positive. The lowest cutoff for $a>b$ is consequently ${\omega }_{10}=\pi c/a$, or ordinary frequency $f_c=c/(2a)$.

A square cross-section instead gives equal ${\mathrm{TE}}_{10}$ and ${\mathrm{TE}}_{01}$ cutoffs. The absence of a transverse electromagnetic mode is specific to this hollow pipe, rather than all electromagnetic waveguides.^[ref]

For $\omega >{\omega }_{mn}$, choose real $\beta >0$ for propagation toward $+z$. For $0<\omega <{\omega }_{mn}$, a field decaying into a semi-infinite guide hasA finite section can contain both growing and decaying axial solutions set by its two ends; the single decaying solution is not a complete finite-section transmission calculation. At exact cutoff $\beta =0$, the two axial solutions become a constant and a linear function of $z$: there is no finite axial wavelength for a traveling sinusoid.

In the chosen $a=30\mathrm{mm}$ guide at $f=0.8f_c=3.99723277\mathrm{GHz}$, the ${\mathrm{TE}}_{10}$ branch has $\alpha =20\pi {\mathrm{m}}^{-1}$. Its amplitude decay length is $1/\alpha \simeq 15.9155\mathrm{mm}$; after $50\mathrm{mm}$, its normalized wave amplitude is $e^{-\pi }\simeq 0.0432139$. This is geometric evanescence even with ideal walls, not Joule heating. A single decaying branch has zero cycle-averaged axial Poynting vector: substituting $\beta =i\alpha$ makes $B_x$ a quarter-cycle out of phase with $E_y$. Coupling across a finite section requires the complete boundary-value solution.

For a propagating branch of an ideal vacuum-filled rectangular electromagnetic waveguide, the phase velocity tracks constant wave phase along the axis, whereas the group velocity describes a sufficiently narrow, weakly distorted wave packet. Using ${\omega }^2=c^2{\beta }^2+{\omega }_c^2$ givesThus $v_{\mathrm{ph}}>c$, $0<v_g<c$, and $v_{\mathrm{ph}}{v}_g=c^2$. Differentiating the dispersion relation proves the second formula; dividing it by ${\beta }^2$ proves the first. This is the same mathematical structure as Klein-Gordon packet velocities. In the stipulated $f=1.25f_c$ example,Over $0.30\mathrm{m}$, the corresponding narrow-envelope delay is $(0.50\mathrm{m})/c$, approximately $1.668\mathrm{ns}$. Timing a carrier wave crest would give $0.6004\mathrm{ns}$; it does not time the arrival of a newly introduced message.

There is an independent energy check. For the forward ${\mathrm{TE}}_{10}$ fields, the average axial Poynting vector is $⟨S_z⟩=\beta |E_y{|}^2/(2{\mu }_0\omega )$. Integrating across the guide gives average power $P$ and average electromagnetic energy per unit length $U^{\prime }$:The transverse sine and cosine squared each integrate to $a/2$; ${\beta }^2+(\pi /a{)}^2={\omega }^2/c^2$ then makes the integrated electric and magnetic contributions to $U^{\prime }$ equal. The axial energy transport therefore agrees with the group velocity in this ideal model.

As $\omega \downarrow {\omega }_c$, $v_g\to 0$ and $v_{\mathrm{ph}}\to \infty$; exactly at cutoff $\omega /\beta$ is undefined. Strong wave dispersion close to cutoff also limits the narrow-packet approximation. Neither a large phase velocity nor a pulse peak establishes superluminal information velocity; front velocity asks a separate causal question. In a uniform filling with constant positive permittivity $\epsilon$ and magnetic permeability $\mu$, replace $c$ by $u=1/\sqrt{\epsilon \mu }$, giving $v_{\mathrm{ph}}{v}_g=u^2$. Frequency-dependent material response or significant loss requires differentiating the actual dispersion relation, so this product is not universal.

Wien's displacement law says that the wavelength maximizing the Planck function per unit wavelength, $B_{\lambda }(T)$, moves inversely with absolute temperature:Here $b$ is the Wien displacement constant. Heating an ideal blackbody therefore shifts this peak toward shorter wavelengths, while also increasing its emission at every fixed wavelength. The peak identifies where equal narrow wavelength intervals carry the most power; it is not a single emitted wavelength. Planck's law supplies the full curve, and the Stefan–Boltzmann law supplies its integral.

Specify the spectral coordinate before finding a peak. The frequency form of Wien's displacement law maximizes $B_{\nu }$, which is spectral radiance per unit frequency. Its maximum is not $c/{\lambda }_{\max }$. The frequency and wavelength spectral radiance conversion changes the density as well as the horizontal coordinate; the existing frequency and wavelength peaks of Planck's law give a complementary numerical comparison.

In his 11 December 1911 Nobel lecture, Wilhelm Wien explained the displacement principle through a thought experiment: compress black-body radiation between reflecting moving walls. Work against radiation pressure raises the energy density, while reflection changes each frequency by the Doppler effect. The Second law of thermodynamics connects the resulting spectrum to one at a higher temperature. This determines how the spectrum scales between temperatures, without determining its complete shape. The Wien approximation is a separate, restricted spectral formula.^[ref]

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]

The drive can supply or extract work while the quantum state stays approximately fixed. For a differentiable bounded Hamiltonian operator, the Schrödinger equation givesThe state-evolution terms cancel, leaving the external drive's power. For a finite ideal step the change in mean energy is $⟨\psi |H_f-H_i|\psi ⟩$. Thus conservation of energy for a closed system with a time-independent Hamiltonian operator does not assert conservation of the driven subsystem's energy during switching.

Specify work operationally by two energy measurements: one before the switch and one after. For the prepared initial ground state, the first outcome is definite; the second samples the new Born rule probabilities. This protocol gives a distribution of measured energy differences, rather than generally the measurement distribution of the single operator $H_f-H_i$.^[ref]and henceFor ${\Delta }_i={\Delta }_f=\Delta$ and $\theta =\pi /3$, the work is $0$ or $\Delta$, with probabilities $3/4$ and $1/4$; its expectation is $\Delta /4$ and its variance is $3{\Delta }^2/16$. The initial mean energy $-\Delta /2$ becomes $-\Delta /4$. Using different initial and final scalar offsets would also shift the measured work; the displayed Hamiltonian operators fix that convention explicitly.

An X-ray tube accelerates electrons from a negative cathode to a positive target anode. A heated cathode supplies electrons by thermionic emission; the evacuated envelope limits collisions before the target. This differs from the gas-discharge tube in the Röntgen X-ray discovery experiment.^[ref]

In bremsstrahlung, an electron changes velocity in the target's electric fields, emitting a photon while losing kinetic energy. Different encounters allow a continuous range of transferred energies. This is distinct from the discrete electronic transitions producing Characteristic X-rays.

Changing the accelerating voltage changes the available kinetic energy per electron and the X-ray tube cutoff. Changing the number of arriving electrons changes the number of opportunities for photon emission. These are distinct controls: a larger electric current does not by itself give each electron more energy. For an original idealized budget, $V=50\mathrm{kV}$ and tube electric current $i=0.20\mathrm{mA}$ give electrical power $Vi=10\mathrm{W}$ and arrival rate $i/e\simeq 1.25\times 10^{15}{\mathrm{s}}^{-1}$, where $e$ is the elementary charge. This is input power, not emitted X-ray power; obtaining the latter requires a conversion efficiency.

Consider an idealized planet that is a sphere of radius $R$, illuminated by parallel sunlight of irradiance $S$, measured on a plane perpendicular to the rays. Its intercepting projected area is $\pi R^2$, whereas its total surface area is $4\pi R^2$. Therefore the incoming power averaged per unit total surface area iswhere $\alpha$ is the broadband Bond albedo. The factor four is a sphere geometry result, not an atmospheric transmission factor. Local sunlight still varies over the illuminated hemisphere and vanishes on the dark side.

Now stipulate a single spatially uniform temperature $T$, no atmosphere, no internal heating, and emissivity one for outgoing thermal radiation. This is a thermal blackbody model using the Stefan–Boltzmann law; the prescribed solar-band reflection is treated separately, so the body is not a perfect absorber at all wavelengths. Let $C>0$ be the effective heat capacity per unit surface area, in $\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. Conservation of energy gives the ordinary differential equationHere $F$ is net heating per unit surface area, in $\mathrm{W}{\mathrm{m}}^{-2}$, and $\sigma$ is the Stefan-Boltzmann constant. Use $\sigma =5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, rounded from the exact constant reported by NIST. ^[ref] The uniform temperature is a model assumption: generally the surface average of $T^4$ differs from the fourth power of the surface-average temperature.

For fixed Bond albedo $\alpha$, radiative equilibrium requires $F(T_{\ast })=0$, givingWith the illustrative solar constant $S=1360\mathrm{W}{\mathrm{m}}^{-2}$, the global incoming average is $340\mathrm{W}{\mathrm{m}}^{-2}$:

Doubling albedo here lowers the equilibrium temperature by $33.231\mathrm{K}$, rather than halving it: the absorbed fraction changes from $0.7$ to $0.4$, and thermal radiation scales as $T^4$. These stipulated Bond albedos are whole-planet examples, not measurements of bare water and ice. The model predicts neither Earth's modern surface temperature nor its response to actual climate change; it omits the greenhouse effect, spatial heat transfer and seasons.

Each fixed-albedo equilibrium is a stable equilibrium: for $T_{\ast }>0$, the derivative is $F^{\prime }(T_{\ast })=-4\sigma T_{\ast }^3<0$. A small temperature increase produces net cooling, and a small decrease produces net heating. This is the simplest expression of the stabilizing Planck feedback. Allowing albedo to fall with temperature adds a competing term, as in ice-ocean albedo feedback.