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

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

Consider an original hypothetical atom with a lower angular momentum multiplet $F=1$, $m=-1,0,+1$, and one opposite-parity excited level $F^{\prime }=0,m^{\prime }=0$. Stipulate that all spontaneous emission returns to these three lower quantum states, with no other decay channel. This is an isolated four-state model, not the actual sodium or rubidium D-line structure. In particular, the physical rubidium cell above contains multiple hyperfine structure levels.

The electric-dipole selection rule permits $\Delta F=-1$ here; ${\sigma }^{+}$ absorption with $\Delta m=+1$ excites only $m=-1$. Decay from $F^{\prime }=0$ has three equal integrated branching probabilities $b_{-}=b_0=b_{+}=1/3$: rotational symmetry gives equal squared Clebsch-Gordan coefficients. The emitted photons carry the three permitted angular momentum projections; equal integrated branching does not imply equal intensity in every observation direction. Thus $m=0$ and $m=+1$ are both dark. The chosen beam cannot prepare a unique pure $m=+1$ state.

Let $R\ge 0$ be the absorption rate of an atom in $m=-1$, and let $\Gamma$ be the excited-state spontaneous emission rate, both in inverse seconds. Assume $R,\gamma \ll \Gamma$, negligible stimulated emission, and no maintained quantum coherence between sublevels. Eliminating the short-lived excited population gives normalized lower populations $p_{-}+p_0+p_{+}=1$ to this approximation. At each scattering event the atom leaves $m=-1$ with probability $2/3$; the remaining $1/3$ simply returns it there.

Model depolarizing collisions and other population relaxation by resets to the unpolarized distribution at rate $\gamma \ge 0$. The rate equations areThis relaxation prescription assumes that Zeeman splitting is small compared with thermal energy, so the unpolarized distribution approximates thermal equilibrium. It is not an energy-independent relaxation law at arbitrarily low temperature.

Every transfer has a nonnegative rate: optical transfers $-\to 0$ and $-\to +$ each have rate $R/3$, and relaxation transfers between each distinct pair have rate $\gamma /3$. These define a continuous-time Markov chain. The three derivatives sum to zero, and at $p_i=0$ the corresponding derivative cannot be negative. Hence the model conserves population and preserves nonnegative populations. Optical cycles that return to $m=-1$ contribute to scattering even though they cause no net population transfer.

Independent stochastic, approximately fixed-noise and fractional-response contributions add in variance. The relative resolution is $A/\sqrt{E}\oplus B/E\oplus C$. Signal-counting and sampling fluctuations enter $A$; fixed equivalent energy noise enters $B$; residual response nonuniformity enters $C$. The powers approximate a fitted energy range rather than describing every leakage or Pile-up effect exactly.

For $f(y)=y-y^2+\gamma y^3/3$ with $0<\gamma <1$, the fold points of a critical manifold are $y_{\pm }=(1\pm \sqrt{1-\gamma })/\gamma$. When $\delta \ll ϵ\ll 1$, the fastest variable satisfies $z\simeq f(y)$ even during jumps of the intermediate variable. At fixed $x$, the outer branches of $x=f(y)$ attract the fast $y$ dynamics because $f^{\prime }(y)>0$; its middle branch repels. If $y_{-}<\alpha <y_{+}$, the slow equation $\dot{x}=\alpha -y$ drives the low branch to its fold point of a critical manifold and the high branch back to the other fold point of a critical manifold. Fast jumps at almost fixed $x$ close a relaxation cycle. This explains why a reduction justified on the slow arcs must also be checked on the jump time scale. For $\gamma >1$, the cubic is strictly increasing; the reduced energy $V=[x-f(\alpha ){]}^2/2+ϵ(y-\alpha {)}^2/2$ has $\dot{V}=-(y-\alpha )[f(y)-f(\alpha )]\le 0$, precluding this oscillation mechanism.

For a time-independent Hamiltonian operator $H$, an energy eigenstate satisfiesFor a one-dimensional particle of mass $m$ in a real potential energy $U(x)$,Its full time-dependent wavefunction is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$: substitution in the Schrödinger equation checks both the minus sign and the factor $\hslash$. Thus an energy eigenstate has stationary position probability density, even though its quantum phase evolves. The boundary conditions and operator domain are part of this eigenvalue problem, selecting physically allowed solutions; the infinite square well and Radial Schrodinger equation for the hydrogen atom give complementary confined examples.

For a complete discrete orthonormal basis of energy eigenstates, linearity gives $\Psi (x,t)={\sum }_n{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$, with ${\sum }_n|c_n{|}^2=1$. Cross terms in $|\Psi {|}^2$ oscillate at angular frequencies $(E_n-E_m)/\hslash$. A continuous spectrum requires the corresponding spectral integrals; for a free particle on the whole line, individual plane waves are generalized eigenfunctions, and a normalized Gaussian wave packet is their integral quantum superposition. Adding a spatially constant real $C$ to $U$ shifts every energy eigenvalue by $C$ and multiplies the full state by $e^{-iCt/\hslash }$, a global phase; energy differences, position probabilities and probability current are unchanged.

Invariance under a shift of the time coordinate, with other required background structures held fixed. A continuous time-translation symmetry gives an energy conservation law through Noether theorem. A generic expanding cosmological background is not stationary and need not supply a corresponding globally conserved total energy.

An effective low-energy lattice model obtained from a strongly repulsive Hubbard model near one electron per site. The retained Hilbert space excludes double occupancy. Its usual nearest-neighbor Hamiltonian has projected hopping $-t{\sum }_{⟨ij⟩,\sigma }({\tilde{c}}_{i\sigma }^{†}{\tilde{c}}_{j\sigma }+h.c.)$ and antiferromagnetic superexchange $J{\sum }_{⟨ij⟩}({\mathbf{S}}_i\cdot {\mathbf{S}}_j-n_i{n}_j/4)$. Virtual intermediate states with one double occupancy cost energy $U$, giving $J=4t^2/U$ to leading order. Three-site hopping terms of the same expansion order are omitted in this simplified version.

At half filling the projected hopping is blocked. Doped holes restore charge motion coupled to spin correlations. Nearest-neighbor singlet pairing can support a d-wave bond order ansatz on a square lattice, but the Hamiltonian alone is not a proof of a unique superconducting ground state for all parameters.

A topological insulator has an insulating bulk electronic spectrum and boundary electronic behavior constrained by a nontrivial bulk band topology and the relevant protecting symmetries. In common three-dimensional time-reversal-invariant examples, the conducting surface can be approximated at low energy by a Dirac cone. The number of surface cones and the range of the linear approximation depend on the material; boundary conduction and thermal bulk phonons therefore need separate state counts.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

For steady rotation at nonzero current, the electromagnetic braking torque is $C{I}^2$ and balances applied torque $G$. Thus $I^2=G/C$, while the circuit requires $\omega =R/C$. With $H=NI/l$, these give the displayed field and $G\omega =I^{2}R$. This is a steady energy balance, not a general nonlinear stability claim.

For a steady Keplerian accretion disk with inward accretion rate $\dot{M}$ and vanishing inner torque, the outward angular-momentum torque is $\mathcal{G}=\dot{M}(\sqrt{GMr}-\sqrt{GM{r}_{\mathrm{in}}})$. It transports energy outward at rate $\Omega \mathcal{G}$. The binding-energy acquisition rate between radii $b$ and $a$ is $B_{a,b}=GM\dot{M}(1/a-1/b)/2$. Integrating the vertical viscous dissipation and then the annulus givesSince $\Omega \mathcal{G}=GM\dot{M}[r^{-1}-\sqrt{r_{\mathrm{in}}}{r}^{-3/2}]$, subtraction proves the displayed energy balance. Far from the inner edge $D_{a,b}\simeq 3B_{a,b}$, whereas the boundary torque powers vanish for the complete disk and $D=B=GM\dot{M}/(2r_{\mathrm{in}})$. Local dissipation is therefore not supplied solely by local changes in orbital binding energy.

An ideal fluid issuing from a small hole a vertical depth $h$ below a free surface has speed

Here $h$ is the vertical distance from the free surface to the centre of the orifice, and $g$ is the positive magnitude of gravitational acceleration. The ideal prediction assumes incompressible flow, the same ambient pressure at the free surface and emerging jet, negligible velocity of the reservoir's free surface, and negligible viscous dissipation and surface tension effects. For a falling free surface, a quasi-steady approximation additionally assumes that the local outflow adjusts much faster than $h$ changes. It applies the Bernoulli equation at each instant, rather than claiming that the whole drainage is steady flow.^[ref]

A short proof uses a streamline connecting the reservoir's free surface at elevation $z_s$ to the emerging jet at elevation $z_o$. With constant mass density $\rho$, negligible surface speed and no mechanical energy losses, the Bernoulli equation givesEqual pressures cancel, leaving $v^2/2=gh$: a unit mass gains the same kinetic energy as a body in free fall through height $h$. This makes Torricelli's law a useful first estimate before the detailed velocity field is known. At one quarter of the original head, ideal jet speed is one half as large; it is not proportional to head.

Far from the accelerating outlet region, the reservoir has approximately hydrostatic pressure $p_{\mathrm{ambient}}+\rho gh$ at the outlet's elevation. The emerging jet instead has approximately ambient pressure when surface tension is negligible. Do not assign that bulk hydrostatic pressure to the emerging jet as well as adding the height difference: that would count the available mechanical energy twice. A sealed reservoir, a submerged outlet or significant surface tension changes the pressure conditions and requires a different balance.

Actual volumetric flow rate is commonly represented by a discharge coefficient:where $a$ is the geometric area of the orifice. This definition compares volumetric flow rates, not just speeds. If a nearly uniform jet at its vena contracta has area $a_j$ and speed $v_j$, define the two separate ratios $C_c=a_j/a$ and $C_v=v_j/\sqrt{2gh}$. Then $Q=a_j{v}_j$ gives $C_d=C_c{C}_v$. A contracted jet can consequently have a speed close to $\sqrt{2gh}$ while $C_d$ is appreciably below one. No particular $C_d$ is universal: calibrate it for the opening and operating range.

A circular shaft twisted uniformly through small end angle $\varphi$ over length $L$ has local engineering shear strain $r\varphi /L$ at radius $r$. Its shear stress is $Gr\varphi /L$, so the integrated torque is $C={\int }_{A}r(Gr\varphi /L)dA=G{J}_p\varphi /L$, using the polar second moment of area. The restoring torque has the opposite sign to the imposed twist. The stored elastic energy is $C\varphi /2=G{J}_p{\varphi }^2/(2L)$. The circular cross-section is important: general noncircular sections can warp and have a different torsion constant.

Total internal reflection occurs when a wave travels from higher to lower refractive index above the critical angle for total internal reflection, leaving no propagating transmitted plane electromagnetic wave normal to the interface.

At an infinite planar interface, total internal reflection returns all incident normal power in the reflected plane electromagnetic wave. Assume homogeneous dielectrics with isotropy, linearity, no dielectric loss, magnetic permeability ${\mu }_0$, and positive real refractive indices $n_1>n_2$. Above the critical angle for total internal reflection, both TE polarization and TM polarization have unit optical reflectance despite the external evanescent wave.^[ref]

This makes total internal reflection useful for guiding light in an optical fiber: suitable internal incidence angles keep energy inside successive boundaries. Its more surprising feature is that perfect optical reflectance coexists with an external electric field. The evanescent field in total internal reflection can interact with a nearby object, as the frustrated total internal reflection experiment demonstrates.

For an original idealized example, choose $n_1=1.5$, $n_2=1$ at vacuum wavelength ${\lambda }_0=633\mathrm{nm}$. At angle of incidence $40^{\circ }$, Snell law for electromagnetic waves gives an angle of refraction $\arcsin (1.5\sin 40^{\circ })\approx 74.6186^{\circ }$, so a transmitted light ray still emerges. At ${\theta }_c\approx 41.8103^{\circ }$, its direction becomes tangent to the boundary. At $50^{\circ }$, the proposed sine of a real angle of refraction would be $1.5\sin 50^{\circ }\approx 1.14907>1$: the transmitted wavevector must instead have an imaginary normal component. These are predictions for the stipulated media, not measured properties of a particular block.

A real semicircular-block experiment isolates this threshold. The University of Maryland laboratory instructions place a laser, motor-driven rotation stage and photodetector together; an aperture limits the detector voltage. The semicircular block is centered on the rotation axis, with the photodetector aligned along its flat face. The instructions identify a strong grazing-beam signal and a weaker signal just before the critical angle for total internal reflection. The readout is detector voltage versus stage angle. Retracing the incident laser beam identifies normal incidence for alignment.^[ref]

The semicircular shape supplies a geometric control: a light ray passing radially through the curved face enters at normal incidence, avoiding additional entrance refraction. The weaker signal near the flat face locates a limiting angle of refraction; a vanishing remote-detector signal alone would not establish that the external electric field vanishes. Beam width and alignment limit angular resolution; no measured threshold or decay depth is supplied.^[ref]

The track grain density is the number $N$ of resolved developed grains along reconstructed track length $L$, divided by $L$. It is a response variable, not a universal conversion from darkness to energy. Overlapping developed grains, exposure fading, processing and optical spatial resolution can all change what is counted. Calibration must use comparable material and viewing conditions.

Cesare Lattes specifically recalls distinguishing meson tracks from proton tracks using their greater deflection by scattering and changing track grain density toward the stopping point. He also recounts a secondary track that left the coating, requiring extrapolation rather than a directly measured stopping distance.^[ref]

For a candidate joined track, the useful questions are whether both branches actually meet in depth, whether the incoming branch is consistent with stopping there, and whether the outgoing branch has the expected particle range and track grain density. A visible change of direction alone cannot distinguish a decay from scattering or an accidental crossing. Examining successive focal depths tests a projected crossing; comparison with ordinary stopping tracks tests the identification. These are measurement checks, not a claim that the original plates provided time-resolved movies.

A nuclear emulsion integrates exposure over time. It does not by itself supply the time between two apparently joined branches. Finite spatial resolution also limits how tightly one can establish their common origin: developed marks and the microscope response are much larger than a subatomic particle. There is no single resolution number applicable to every emulsion, developer and optical arrangement.

An electrical transformer transfers energy between windings through electromagnetic induction. Its operation depends on changing magnetic flux, rather than a static magnetic field alone.

The ratio of the witness accelerating field to the driver decelerating field, with peak or loaded-field conventions stated. Under ideal common-length depletion, it also relates witness energy gain to driver energy loss.

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 transition-edge sensor detects deposited energy using the strong change in electrical resistance near a superconducting transition temperature.