Michael Faraday's induction ring separates two questions: can one electrical circuit affect another without an electrical connection, and does that effect continue under steady conditions? The surviving apparatus uses insulated copper windings around an iron ring; the Royal Institution dates the successful experiment to 29 August 1831.^[ref]

Faraday connected one winding group to a battery, the other to a galvanometer. Making battery contact deflected the needle; maintaining contact let it return to rest; breaking contact reversed the response. Reversing the battery reversed the deflections. His separate cylinder comparison enhanced induction with iron, whereas replacing it with copper gave no enhancement over the windings alone.^[ref]

The contrast between a switching response and a quiet steady state makes mutual induction tangible. A steady primary electric current can maintain magnetic flux without sustaining a secondary electromotive force. Changes in magnetic flux linkage matter, as expressed by Faraday's law of induction. This is why a driven electrical transformer uses repeated changes in magnetic flux to supply secondary power.

The galvanometer readout of an induction pulse also separates the measured motion from the inferred electric current. Its moment of inertia and damping mean that a deflection is not an instantaneous trace of electric current. The following coupled RL induction model calculates currents in a stipulated circuit, rather than fitting Faraday's needle motion or reconstructing his switching times.

Choose the positive loop normal and its positive circulation by the right-hand rule. A spatially uniform external magnetic field of fixed direction and signed magnitude $B_0\cos \omega t$, at angle $\theta$ to that normal, gives magnetic flux $\Phi =\pi r^2{B}_0\cos \theta \cos \omega t$. Neglecting the loop's self-inductance, Faraday's law of induction and Ohm's law give the displayed electric current. At a quarter-cycle the external field is zero but its normal component is decreasing most rapidly, so the current produces a field in the direction of the initial normal flux, as required by Lenz's law. With appreciable self-inductance the circuit instead obeys $L\dot{I}+RI=-{\dot{\Phi }}_{\mathrm{ext}}$.

A uniform steady film of thickness $d$ on a plane inclined by $\theta$ above horizontal has downslope coordinate $s$ and outward normal coordinate $x$. Neglecting air shear, the pressure is $p_{\mathrm{atm}}+\rho g\cos \theta (d-x)$ and its downslope gradient vanishes. The Newtonian fluid momentum equation is $\eta v^{\prime \prime }+\rho g\sin \theta =0$. The no-slip boundary condition $v(0)=0$ and shear-free free surface $v^{\prime }(d)=0$ give the displayed parabolic profile. Integrating over thickness gives volume flux per unit width $q$. The wall shear $\rho gd\sin \theta$ balances the film's downslope weight, and the vertical-wall case is $\theta =\pi /2$.

The polynomial identityfollows by multiplying out the right side. For real $a,b$, the quadratic factor is $a^2+ab+b^2=(a+b/2{)}^2+3b^2/4$, so it is nonnegative and vanishes only when $a=b=0$. The identity is a special case of the finite geometric series factorisation $a^n-b^n=(a-b){\sum }_{j=0}^{n-1}{a}^{n-1-j}{b}^j$.

Replacing $b$ by $-b$ in the difference of two cubes givesFor real cube roots $u=a^{1/3}$ and $v=b^{1/3}$, this simplifies $(a^{1/3}+b^{1/3})(a^{2/3}-(ab{)}^{1/3}+b^{2/3})$ to $a+b$. Specifying real cube roots matters when using such identities with negative real inputs; unconstrained complex exponentiation has logarithm-branch choices.

The distinction appears at a smooth join between a straight segment and a curved boundary. Consider the convex hullThe upper tangent from $P$ meets the unit circle at $T=(1/2,\sqrt{3}/2)$. The supporting line isEvery point of the unit disk has $T\cdot z\le 1$, with equality only at $T$. Hence the convex face of $C$ where equality holds is exactly the line segment $TP$.

The point $T$ is extreme. If a nontrivial convex combination of two points of $C$ equals $T$, both must belong to the convex face $TP$. Since $T$ is an endpoint of that line segment, both must equal $T$.

Yet $T$ is not exposed. A nonzero linear functional maximizing at $T$ on $C$ must also maximize there on the included unit disk. Its vector must be a positive multiple of $T$, by the equality condition in the Cauchy-Schwarz inequality. It then gives $P$ the same value as $T$. No linear functional can select $T$ alone. Being an extreme point is a property of possible mixtures; being an exposed point adds the ability to distinguish it by one linear objective.

An exposed point is a point uniquely selected by maximizing a linear functional. In ${\mathbb{R}}^d$, a point $p$ of a convex set $C$ is exposed if some vector $c$ satisfiesGeometrically, a supporting hyperplane touches the convex set only at $p$. Every exposed point is an extreme point: a convex combination that attains the maximum can put positive weight only on maximizers, and there is only one.^[ref]

The unit disk calculation in extreme point exposes every point of the unit circle. For a corner $(a,b)$ of the square, $c=(a,b)$ exposes it: $ax+by$ attains two only at $(a,b)$. However, maximizing $x$ exposes no single point on the right edge, since all of them have the same objective value. A linear functional may have several optimal points, even when some of them are extreme.

The extensional strain rate is the rate of relative stretching along a specified material direction. If a short material line of length $\ell$ follows a principal stretching direction, $\dot{ϵ}=d\log \ell /dt$ is the corresponding eigenvalue of the rate-of-strain tensor. For the planar model in extensional flow, $\dot{ϵ}=\partial u_x/\partial x$; the two transverse rates are $-\dot{ϵ}$ and zero. For uniaxial extensional flow, each transverse rate is $-\dot{ϵ}/2$.

Consider an illustrative water–air mixture, initially saturated at $T_0=293.15\mathrm{K}$. Let $\lambda =V/V_0$ be the expansion ratio and approximate the carrier gas by a uniform ideal gas with constant heat capacity ratio $\gamma =1.40$. Before significant condensation, assume fixed numbers of carrier and vapor molecules, negligible incoming heat, and the reversible adiabatic expansion relation. The first law of thermodynamics, $n{C}_{V}dT=-pdV$, with the ideal gas law, givesThe vapor follows the same temperature, so its partial pressure scales as $T/V$. A rapid real piston motion need not be reversible; the formula is an idealized bulk estimate after neglecting flow and losses.

Use a physically motivated Integrated Clausius-Clapeyron relation for the equilibrium water vapor pressure, with stipulated rounded constants $p_0=2339\mathrm{Pa}$ at $T_0$, molar latent heat $L_m=45000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$:These constants define the example, without fitting any historical measurements. Throughout the plotted interval $1\le \lambda \le 1.17$, the temperature stays above the freezing point of water. InitiallyThus the decrease in vapor pressure initially beats dilution. At $\lambda =1.15$, the model gives $T=277.21\mathrm{K}$, $p_v=1923.33\mathrm{Pa}$, $p_{\mathrm{sat}}=809.07\mathrm{Pa}$ and $S=2.3772$.

Why does $S>1$ not instantly make every molecule a visible drop? In the capillarity approximation of classical nucleation theory, a neutral spherical drop of radius $r$ has a positive surface energy cost and a negative bulk Gibbs free energy contribution. For molecular liquid volume $v_l$, surface tension $\sigma$ and Boltzmann constant $k_B$,Differentiating $\Delta G$ shows that the critical nucleus at $r_{\ast }$ is a maximum: smaller neutral drops tend to shrink, whereas larger ones can grow. With stipulated $\sigma =0.075\mathrm{J}{\mathrm{m}}^{-2}$, $v_l=2.99\times 10^{-29}{\mathrm{m}}^3$ and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, the $\lambda =1.15$ example gives $r_{\ast }=1.353\mathrm{nm}$ and $\Delta G_{\ast }/(k_{B}T)=150.3$. A substantial neutral-drop barrier can coexist with supersaturated vapor.

This is deliberately a neutral-drop comparison. An ion changes the cluster free energy through electrical interactions and molecular hydration; the formula does not calculate the ion-assisted nucleation threshold. At nanometre radii, using bulk surface tension is itself approximate. There is no predicted droplet count, growth time or detection efficiency here. Once many drops grow, latent heat release, vapor depletion and heat transfer from the chamber walls invalidate the dry adiabatic expansion trajectory. The example therefore explains the direction of the effect and the need for nucleation, without extrapolating its curves to Wilson's cloud limits.

Charles Thomson Rees Wilson's 1927 lecture recalls electron cloud trails produced with X-rays in 1911, then alpha particle trails from a radium-tipped source. The improved 1912 chamber had a wide, shallow space with a suddenly falling floor to limit stirring. A roof–floor clearing electric field removed unwanted ions. Sharp photography required expansion, then the ionizing particle, then illumination.^[ref]

The piston mechanism connects the space beneath the moving floor to an evacuated vessel: opening the valve lets the floor drop. Water in a trough supplies the vapor. The Smithsonian's surviving demonstration object is explicitly a 1961 replica of the 1912 apparatus.^[ref]

After dust removal, X-rays increased droplet formation above a lower expansion threshold; larger expansions also produced dense background cloud.^[ref]

The control logic separates a localized ionization trail from fog. Delays permit ion displacement and drop growth; stirring deforms the trail. A photograph alone supplies neither an absolute event time nor a calibrated particle energy.

A triggered cloud chamber expands in response to an event indicated by other detectors, rather than hoping that a particle arrives during a randomly chosen sensitive interval. The ionization trail is produced before expansion. The apparatus must respond while the ions remain sufficiently localized to make a useful photograph.

A biexciton is a two-electron, two-electron hole excitation. If $E_X$ and $E_{XX}$ are total excitation energies relative to the empty emitter, the cascade $XX\to X\to 0$ emits photons of energies $E_{XX}-E_X$ and $E_X$. Define its binding energy by $B_{XX}=2E_X-E_{XX}$. The first cascade photon then has energy $E_X-B_{XX}$. Positive binding shifts that line below the single-exciton line; confinement can also produce negative binding, shifting it above. The cascade does not emit a single photon of twice the exciton energy.

Direct exchange comes from Coulomb matrix elements between overlapping localized electronic orbitals together with the antisymmetry required by the Pauli exclusion principle. It produces a spin-dependent energy without being a classical dipole interaction. Its competition with kinetic processes and orbital occupation helps determine whether neighbouring moments prefer parallel or antiparallel alignment.

Superexchange couples localized moments through virtual hopping via intermediate, often nonmagnetic, ligand orbitals. Half-filled orbitals sharing an effective hopping amplitude $t$ give an antiferromagnetic scale $J\simeq 4|t{|}^2/U$ in the repulsive one-band limit. Ligand geometry, orbital occupation and exchange within intermediate states can change the balance and allow ferromagnetic coupling. Real charge motion can remain gapped while these virtual processes couple an insulating magnet's spins.

Let $a,b$ be distinct orthonormal spatial orbitals occupied by two identical spin one-half particles, with a weak repulsive Coulomb interaction. The symmetric and antisymmetric spatial combinations have first-order energy shifts $J+K$ and $J-K$, where $J$ is the direct Coulomb integral andThe Coulomb kernel has positive Fourier transform $4\pi /|\mathbf{k}{|}^2$, so $K\ge 0$. Fermionic exchange symmetry pairs the symmetric spatial function with the spin-one-half singlet state and the antisymmetric one with the spin-one-half triplet state. For these fixed orbitals the triplet lies lower by $2K$ when $K>0$. This does not compare arbitrary singlets and triplets with different orbital occupations; for negligible orbital overlap the splitting can vanish.

For an equatorial null geodesic in Schwarzschild spacetime, let $k=(1-2\mu /r)dt/d\lambda$ and $h=r^{2}d\varphi /d\lambda$ be the conserved quantities for an affine parameter $\lambda$. The null metric condition gives ${\dot{r}}^2/2+V_{\mathrm{eff}}=c^2{k}^2/2$. For $h\ne 0$, the effective potential is zero at the Schwarzschild event horizon, positive outside, and tends to zero at infinity. Its unique exterior maximum is at $r=3\mu$, with value $h^2/(54{\mu }^2)$ and second derivative $-h^2/(81{\mu }^4)$. This maximum gives the unstable photon sphere.

The impact parameter of an incoming ray is $b=|h|/(ck)$ for $k>0$. Capture requires its radial energy to exceed the maximum, namely $b<3\sqrt{3}\mu$. At equality the incoming ray approaches the circular orbit asymptotically; it does not cross the horizon. For $h=0$, the potential is identically zero and the ingoing radial ray is captured.

Two physically distinguishable noninteracting species with fixed particle numbers and the same single-particle partition function have the displayed canonical partition function. If $Z_1\propto V{T}^{3/2}$, the leading large-particle-number entropy is $S=k_B{\sum }_{i=A,B}{N}_i[\ln (Z_1/N_i)+5/2]$. With $N=N_A+N_B$ and $x_i=N_i/N$, this equals the one-species expression at total number $N$ plus $-N{k}_B{\sum }_i{x}_i\ln x_i$. Merely attaching labels to otherwise identical indistinguishable atoms does not justify this mixing term.

Residual entropy is entropy remaining in the zero-temperature limit. An equilibrium ground-state degeneracy $g_0$ gives $S_0=k_B\ln g_0$ for the complete system; an extensive residual entropy requires a correspondingly large state count. Frozen configurational disorder can also leave an experimentally inferred residual entropy when a sample fails to equilibrate. The zero-entropy perfect-crystal form of the third law of thermodynamics assumes a unique, fully ordered equilibrium ground state.

For a body of constant heat capacity $C$ cooling from $T_i$ to a thermal reservoir at $T_0>0$, the body's entropy change is $C\ln (T_0/T_i)$, while the reservoir gains $C(T_i-T_0)/T_0$. Their sum is nonnegative because $r-1-\ln r\ge 0$ for $r=T_i/T_0>0$. The logarithmic body contribution is computed along a hypothetical reversible path; the actual finite-temperature-difference heat transfer is irreversible.