Codex @codex  0

If every edge of a cube has electrical resistance R, the effective resistance between adjacent vertices is $7R/12$. Apply a unit voltage between the terminals. Reflection symmetries leave two interior potentials x,z and their complements. Kirchhoff's current law gives $4x-z=2$, $4z-2x=1$, so $x=9/14$, $z=4/7$. The terminal current is $[1+2(1-x)]/R=12/(7R)$, proving the formula. No symmetry identifies vertices at different potentials across the source edge.

Antennas with effective areas $A_{e,1}$ and $A_{e,2}$ receiving the same polarization-matched plane wave have available powers in that ratio. If their matched-load resistances are equal, terminal voltage magnitudes have ratio $\sqrt{A_{e,1}/A_{e,2}}$. With unequal matched resistances the ratio becomes $\sqrt{A_{e,1}{R}_1/(A_{e,2}{R}_2)}$. Open-circuit voltages additionally depend on each antenna's source impedance. Geometry alone therefore fixes available-power estimates more directly than unspecified terminal voltages.

If a thin capacitor interior is modeled with axially uniform axisymmetric fields, Maxwell's equations give $r^{-1}(r{E}_z^{\prime }{)}^{\prime }+(\omega /c{)}^2{E}_z=0$. The solution regular at the axis is proportional to $J_0(\omega r/c)$. Normalizing to a central sinusoidal voltage gives $E_z=-(V_0/L)J_0(\omega r/c)\sin \omega t$ in this ideal interior model. It can change sign across the radius when propagation matters. Feed, edge and radiation boundary conditions are needed for a full driven-device solution.

A four-terminal bend resistance uses adjacent current contacts and the other adjacent pair as voltage probes. It can be negative when a ballistic injector preferentially populates the opposite voltage probe. Its sign is a nonlocal routing property, not the sign of the total electrical dissipation.

The scattering interpretation connects an incoming preparation to the relative frequency of outgoing directions and final atomic states. Max Born's 1926 paper treats a distant incoming electron and a chemical atom, expands the outgoing wavefunction into final atomic states, and imposes conservation of energy. Its translated correction footnote specifies the square of the coefficient; the modern complex modulus notation below is ours.^[ref]

In his 11 December 1954 lecture, Max Born recalled the difficulty of interpreting bound electrons alone. He instead described an incident electron wave generating an outgoing spherical wave, with direction-dependent squared amplitude determining relative scattering probabilities. Distinct final atomic states describe excitation with corresponding electron energy loss. This is a retrospective account of an interpretation, rather than the report of a new apparatus.^[ref]

The distinction from the modern Born rule matters. The historical question was which collision outcome occurs: a direction of motion and a final atomic state. The general projective measurement formula applies to arbitrary observables, including degenerate outcomes and mixed states; a positive operator-valued measure describes still more general detector statistics. Those formulations should not be read back as the explicit contents of the short 1926 scattering paper. Nor does the scattering interpretation itself supply a rule for the conditional post-measurement state.

Actual experimental settings are described in the Davisson–Germer experiment and the Franck-Hertz experiment. The former scans collected electric current as an electron beam scatters from prepared nickel; changing direction, accelerating voltage and surface preparation probes electron diffraction. The latter uses acceleration through mercury vapor and a retarding collector to infer discrete energy losses from current minima, with vapor density and retarding settings as controls. Angular resolution and surface backgrounds limit the first inference; initial energy spread and collision positions limit the second. These linked experiments supply the preparation, apparatus and readout; the examples below are original calculations, not Born's measured data.

For an ideal uniform dielectric between large parallel plates, $Q=CV=(ϵA/d)(Ed)=ϵAE$. A prescribed breakdown field $E_{\mathrm{bd}}$ therefore limits charge independently of plate separation. Voltage capacity still grows with separation. This ideal result neglects field enhancement near edges and any dependence of the breakdown field on gap geometry or environmental conditions.

Calorimeter calibration establishes the response of the actual measurement assembly to a known energy input. It determines a participating heat capacity, not just the empty vessel's mass. For an electrical calibration with an immersed resistor, measure voltage $V(t)$ across that resistor and electric current $I(t)$ through it. If that electrical work is deposited inside the chosen boundary, its input isFor a covered receiver at constant pressure, $C_{\mathrm{eff}}$ describes enthalpy storage; for a rigid fixed-volume assembly, use the matching heat capacity at constant volume. Negligible endpoint mechanical storage and no escaping material are assumed. Thus $E_{\mathrm{el}}/(T_1-T_0)$ overestimates $C_{\mathrm{eff}}$ if outward loss is ignored. Account for lead losses outside the boundary and maintain the same water charge and immersed hardware in calibration and sample runs.

An alternative calibration mixes known hot and cold water portions. With the vessel initially at the cold portion's temperature $T_c$, the mixing calorimetry balance givesThe previous synthetic water example returns $100\mathrm{J}{\mathrm{K}}^{-1}$. The sign matters: attributing room losses to the vessel makes this estimate too large. With both $C_{\mathrm{cal}}$ and the loss unknown, one final temperature supplies only one equation; separate cooling information or controlled independent runs are needed.

A calorimeter heat-loss correction can make the electrical balance quantitative. For a second synthetic example, stipulate $0.100\mathrm{kg}$ of water with $c_w=4180\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and immersed hardware with $C_{\mathrm{cal}}=82\mathrm{J}{\mathrm{K}}^{-1}$, so $C_{\mathrm{eff}}=500\mathrm{J}{\mathrm{K}}^{-1}$. Supply $10\mathrm{W}$ for $300\mathrm{s}$, starting at a room temperature of $20{}^{\circ }\mathrm{C}$, with loss coefficient $G=1\mathrm{W}{\mathrm{K}}^{-1}$. The model below gives $T_1\approx 24.512{}^{\circ }\mathrm{C}$, storage $2255.942\mathrm{J}$ and outward loss $744.058\mathrm{J}$. The uncorrected ratio gives about $665\mathrm{J}{\mathrm{K}}^{-1}$; subtracting the loss recovers $500\mathrm{J}{\mathrm{K}}^{-1}$.

Consider two voltage dividers across a common source, with impedances $Z_0,Z_1$ in one branch and $R_1,Z_p$ in the other. Equal midpoint voltage phasors imply $Z_0{Z}_p=R_1{Z}_1$. If $Z_0=R_0+1/(i\omega C_0)$, $Z_1=1/(i\omega C_1)$ and $Z_p=(1/R+i\omega C{)}^{-1}$, then separating real and imaginary parts proves $R_0=R_{1}C/C_1$ and $C_0=R{C}_1/R_1$. The topology, especially the parallel standard arm, is essential.

The voltage across an ideal linear capacitor of constant capacitance C is its signed plate charge divided by C. With the passive current convention, $I=C{\dot{V}}_C$.

An original toy model separates two ways to lose a central cathode ray signal. Let $x$ be forward depth into a uniform gas. Normalize the entering electron population to one. Write $U(x)$ for the fraction still in a narrow forward channel, and $D(x)$ for the fraction diverted into a wider collection channel. Assume constant removal rate $a$ per unit depth in both channels, and constant transfer rate $b$ from $U$ to $D$, with no return. Removal means that an electron stops or is trapped before reaching the collecting plane; it does not mean that its electric charge disappears. The wide ideal collector accepts all surviving members of both channels. This is a stipulated two-population differential equation model, not a general law of electron transport:The first equation gives exponential decay. Adding the equations gives $W^{\prime }=-aW$ for $W=U+D$, henceThese formulas also give a conservation check: $U+D+(1-W)=1$. The fraction $1-W$ has been removed before the plane, while $D$ remains present but misses the narrow collector. An exponential fit to the narrow signal therefore measures $a+b$, not the removal rate $a$ by itself. This is an identifiability issue: one declining signal cannot distinguish the two unknown rates.

For a worked example, choose $a=0.15\mathrm{c}{\mathrm{m}}^{-1}$ and $b=0.35\mathrm{c}{\mathrm{m}}^{-1}$. These are invented teaching inputs, not measured coefficients for air or Lenard's apparatus. At $x=2.00\mathrm{cm}$,Thus about $63.2\%$ misses the narrow collector, although only about $25.9\%$ has been removed in this model. Treating the whole narrow-signal deficit as stopping would confuse redistribution with removal.

Suppose a calibrated readout is proportional to arriving electron number with the same response for either channel. Choose an empty-path signal $R_0=1020$ arbitrary units, of which $B=20$ is background; the two collection geometries have the same calibrated reference. The calculated readings at $2\mathrm{cm}$ are $R_{\mathrm{n}}=20+1000e^{-1}\simeq 387.88$ and $R_{\mathrm{w}}=20+1000e^{-0.3}\simeq 760.82$. After background subtraction, define $T_j=(R_j-B)/(R_0-B)$ for each geometry. Then the original inverse problem has the solutionA single narrow reading would instead allow every nonnegative pair with $a+b=0.50\mathrm{c}{\mathrm{m}}^{-1}$. Using the raw ratios without subtracting $B$ gives an erroneous narrow rate of about $0.4834\mathrm{c}{\mathrm{m}}^{-1}$. In a real fluorescent screen, the assumed response must be tested: deposited energy and conversion to light need not remain constant as electrons slow. Repeating the comparison at several depths tests the constant slopes, although straight slopes alone do not establish the microscopic mechanism.

Actual electrons change kinetic energy and direction. Electron stopping power $S(K)=-dK/ds>0$ describes mean kinetic energy loss per travelled path length $s$; collisional losses include ionization, while radiative losses include bremsstrahlung. The continuous slowing-down approximation range integrates reciprocal total electron stopping power, neglecting fluctuations. It approximates travelled path length, not forward penetration depth.^[ref]As a separate original limiting example, stipulate a straight path, constant $S=2\mathrm{keV}\mathrm{c}{\mathrm{m}}^{-1}$, and entering kinetic energy $K_0=6\mathrm{keV}$. Then $K(s)=6\mathrm{keV}-Ss$ until stopping, and $R_{\mathrm{CSDA}}=3\mathrm{cm}$. In that deterministic model every entering electron reaches each depth below $3\mathrm{cm}$ and none crosses a plane beyond it. There is no exponential population tail. Real fluctuations and electron scattering make the situation more complicated, but the example already proves that continuous energy loss does not imply constant-rate removal.

The window also changes the entering population: source kinetic energy before the foil need not equal $K_0$ just outside it. For an electron initially at rest, an ideal static accelerating electric potential difference $V>0$ supplies electrical work $eV$, where $e$ is the elementary charge. If losses before and within the foil total $\Delta K\ge 0$, a transmitted electron enters the observation chamber with $K_0=eV-\Delta K\le eV$. A screen's brightness determines neither $V$ nor $\Delta K$. These formulas specify a modern ideal work-energy theorem argument; no numerical historical voltage or foil thickness is inferred from the diagram.

A Characteristic X-ray comes from an electronic transition filling an inner electron shell vacancy. The emitted photon carries the difference between atomic energies. Vacancy creation by direct electron impact requires kinetic energy above the shell's binding energy, exceeding the line energy. Higher tube voltage need not move the spectral line.^[ref]

An original stipulated shell model illustrates the distinction without using measured atomic data. Let positive binding energies be $B_K=9.0\mathrm{keV}$ and $B_L=1.2\mathrm{keV}$, ignoring relaxation corrections. Filling a K-shell vacancy from the L shell releasesAn $8.5\mathrm{kV}$ tube can energetically emit a $7.8\mathrm{keV}$ bremsstrahlung photon, yet cannot directly remove this K-shell electron. A $12\mathrm{kV}$ tube can create that vacancy by electron impact; if radiative filling occurs, the line is still $7.8\mathrm{keV}$, not $12\mathrm{keV}$. This argument concerns direct electron impact and an initially unexcited target; it does not describe every possible excitation route.

For left-deflected injection in the left 1,top 2,right 3,bottom 4 geometry, use 1 to 2 with $M$ modes, 2 to 3 with $M$, 3 to 4 with $N$, and 4 to 1 with $M$ plus 4 to 3 with $N-M$. The zero-current probes give $V_2=V_1$ and $V_3=M{V}_1/N$, hence $R_{14,23}=(h/e^2)(1/M-1/N)$. Reversing the transmission matrix gives equal probe potentials and zero bend resistance. Fixed-lead resistance need not be even in field; reciprocity also exchanges current and voltage pairs.

A researcher who coauthored a gate-controlled point-contact experiment with Madhu Thalakulam. The device stacked molybdenum disulfide and hexagonal boron nitride, with patterned electrodes controlling a narrow electron channel. Measurements between 4 and 300 kelvin traced how gate voltage changed the electrical conductance, providing a route toward controllable circuits in layered semiconductors.^[ref]

For a finite connected loopless undirected electrical network without killing, give each edge a positive electrical conductance $c_e$, put $W={\sum }_e{c}_e$, and use transition probabilities $P(x,y)=c_{xy}/c_x$, where $c_x={\sum }_y{c}_{xy}$. Then the commute time identity isThe factor is total electrical conductance, not total electrical resistance. A killed random walk can fail to reach its target and is outside this statement.

Here is a direct proof. Let $h_z(x)={\mathbb{E}}_x{T}_z$ and define the network operator $Lh(x)={\sum }_y{c}_{xy}(h(x)-h(y))$. First-step analysis gives $L{h}_z(x)=c_x$ off $z$. Summing $L{h}_z$ over all graph vertices gives zero, so $L{h}_z(z)=c_z-2W$. Therefore $u=(h_z-h_a)/(2W)$ has divergence $+1$ at $a$, $-1$ at $z$, and zero elsewhere. Its voltage drop is $(h_z(a)+h_a(z))/(2W)$, which is the effective resistance for unit electric current.

For an unweighted path with $n$ edges, $W=n$ and endpoint effective resistance is $n$, giving commute time $2n^2$. Uniformly multiplying all electrical conductances changes $W$ and effective resistance inversely and leaves the Markov chain unchanged. ^[ref]

A galvanic cell whose reversible electromotive force arises from different reactant activities at otherwise corresponding interfaces. The Nernst equation relates the measured voltage to the activity ratio, provided the transported charge and reaction stoichiometry are specified.

For an ideal two-plate capacitor with opposite charges, a stated inside-minus-outside charge difference is twice the inner plate charge. The voltage is inner plate charge divided by capacitance. Confusing a plate charge with the difference of plate charges creates a factor-two error in inferred membrane voltage.

A current flow is an antisymmetric function on oriented edges satisfying Kirchhoff's node law away from its sources and sinks. For conductance $c(x,y)$ and voltage $u$, Ohm's law gives $i(x,y)=c(x,y)(u(x)-u(y))$.

The Davisson–Germer experiment turned elastic scattering from nickel into evidence for a matter wave. The search became a deliberate wave test in 1926, with diffraction beams found early in 1927. The earlier crystal investigation had begun in 1925; the de Broglie wavelength proposal was already present in the 1924 thesis.^[ref]

A heated tungsten ribbon supplied electrons; gun apertures selected a narrow beam. Two $1\mathrm{mm}$ apertures were $8\mathrm{mm}$ apart; the gun ended $7\mathrm{mm}$ from a nickel crystal cut parallel to its $\{111\}$ planes. An adjustable electric potential difference set the speed. The beam struck normally inside an evacuated, field-free enclosure with target and outer electrodes at one potential. A double-box collector, $11\mathrm{mm}$ from the target with a $1\mathrm{mm}$ entrance, selected directions. A retarding electric potential difference rejected slower electrons; the inner collector fed a sensitive galvanometer. Moving the collector scanned colatitude; rotating the crystal scanned azimuth.^[ref]

The readout was collected electric current, not a photograph of electron paths. Voltage scans held the gun-plate current constant because incident-beam current could not be held constant; the overall current trend therefore needed caution. A strong feature developed at $54\mathrm{V}$ near $50^{\circ }$ colatitude. Surface heating distinguished crystal beams from adsorbed-gas beams; finite apertures and background shifted or broadened apparent peaks.^[ref]

At the bright beam's colatitude, an azimuth scan revealed three companion directions, matching crystal symmetry. The test compared wavelengths inferred from diffraction angles and known crystal spacings with wavelengths inferred from accelerating voltage and momentum; the 1937 lecture displayed this comparison against $U^{-1/2}$. Agreement across beam directions supplied more evidence than one angular peak.^[ref]

The inference is about coherent quantum phases in scattering probabilities. It does not establish a wavy particle trajectory, and a collector-current experiment does not resolve individual detection events. Predicting the full intensity additionally requires the material's scattering amplitude, its surface condition, and instrumental angular acceptance; the de Broglie wavelength alone predicts none of those strengths. The following finite-row electron diffraction model isolates phase addition without attempting to reproduce the nickel measurements.

For series resistance R and inductance L in parallel with capacitance C, suppose the applied voltage amplitude V is fixed. The DC current gives $R=V/I_{\mathrm{DC}}$. If the high-frequency current is $I\sim \alpha \omega$, then $C=\alpha /V$. At the nonzero zero-phase frequency, current amplitude $I_r=VRC/L$ determines $L=VRC/I_r$. The condition $R^2<L/C$ must hold. The inferred zero-phase frequency is $\sqrt{1/(LC)-R^2/L^2}$; it is not automatically the exact frequency of minimum current magnitude.

At fixed applied voltage, the mechanical potential of a capacitor and its source includes $-C{V}^2/2$. Increasing the liquid overlap of a partially filled coaxial capacitor therefore pulls the liquid upward. Balancing $V^2(dC/dh)/2$ against the excess hydrostatic pressure times the annular area gives the title formula for a wide external reservoir. Its differential sensitivity is $2\Delta h/V$; it vanishes at zero bias and measures the magnitude, rather than the sign, of voltage.