Competitive adsorption occurs when different chemical species seek the same adsorption sites. For intact molecules $A$ and $B$, retain identical adsorption sites and no interactions, but replace the vacant fraction by $1-{\theta }_A-{\theta }_B$. The same equilibrium balance givesHere $P_A$ and $P_B$ are partial pressures, not total pressure. Increasing $P_B$ can reduce ${\theta }_A$ at unchanged $P_A$: replacing $P_A$ by total pressure in a single-species Langmuir isotherm loses this competition.

For zero field, $F=F_0+b(T-T_c)M^2+d{M}^4$ with $b,d>0$ has ordered minima $M^2=b(T_c-T)/(2d)$ below $T_c$. Minimizing gives $F_{\min }=F_0-b^2(T-T_c{)}^2/(4d)$. Since $C=-T{\partial }_T^2{F}_{\min }$, the ordered-phase excess heat capacity is $T{b}^2/(2d)$ and its jump relative to the disordered side is $T_c{b}^2/(2d)$. The value is a total heat capacity or a heat-capacity density according to the normalization of $F$.

The particle-hole continuum is the range of energy and momentum transfers that can excite an occupied fermion state into an unoccupied state, leaving a hole behind. For a free gas at zero temperature, choose $|\mathbf{k}|<k_F$ and $|\mathbf{k}+\mathbf{q}|>k_F$; the excitation energy is ${ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$.

Varying the initial momentum produces a continuum at fixed transferred wavevector. For small $q$ in an isotropic gas, its positive frequencies extend approximately up to $v_{F}q$. A collective mode overlapping these excitations can decay by creating particle–hole pairs; a zero sound branch outside that region can avoid this particular damping channel. Interactions change the response and effective parameters, so the free-gas boundary is a useful starting model rather than a universal exact spectrum.

The quasiparticle residue is the coherent pole weight of the single-particle Green's function. Its coherent spectral term $Z\delta (E-{\xi }_p)$ contributes $Z\Theta (-{\xi }_p)$ to the zero-temperature momentum occupation. A smooth remainder leaves a jump $Z$ at the Fermi surface; Landau quasiparticle occupations have a unit step.^[ref]

For the local illustrative distribution in the figure, put $x=(p-p_F)/p_F$ and$\Theta$ is the Heaviside step function. Its one-sided limits are $0.825$ and $0.175$, giving $Z=0.65$. The smooth term is stipulated; this local curve does not specify a normalized distribution over all momenta. In particular, $Z$ is neither the fraction of atoms above the Fermi surface nor $m/m^{\ast }$.

In a three-dimensional uniform isotropic continuum obeying the Principle of Galilean relativity, with one bare particle mass $m$, the Fermi-liquid effective mass satisfiesA band effective mass need not satisfy this relation.^[ref]

For the stipulated $m^{\ast }=2m$ in the occupation figure, this relation requires $F_1^s=3$ if those assumptions apply. At fixed $p_F$, $v_F$ is half its free-particle value, so moving by $0.04p_F$ from the Fermi surface costs approximately $0.02p_F^2/m$ instead of $0.04p_F^2/m$. The factor of three belongs to the three-dimensional angular average, not to a universal definition of effective mass.

Changing one Landau quasiparticle occupation changes the energies of others. To second order around equilibrium,Here $\mathcal{V}$ is volume, occupations are per momentum and spin state, and $f$ has units of energy times volume.^[ref]

For a toy two-state change at fixed volume, take $\delta n_a=\delta n_b=\eta$, $f_{aa}=f_{bb}=0$, $f_{ab}=f_{ba}=g\mathcal{V}$, and unperturbed energies ${\xi }_a={\xi }_b=e$. The expression gives $2e\eta +g{\eta }^2$. The factor $1/2$ counts the interaction of the same pair once; it does not halve each excitation's energy. This finite-state illustration only explains the bookkeeping of the Landau quasiparticle energy functional.

Lactate is the conjugate base of lactic acid. Lactate dehydrogenase couples its interconversion with pyruvate to the NAD+/NADH redox pair.

A freely moving rod end has zero axial stress, not zero axial displacement. For a uniform slender rod of length $\ell$, its low-frequency extensional modes have shapes ${\xi }_r(x)\propto \cos (n\pi x/\ell )$ and wavelengths ${\lambda }_r=2\ell /n$. A centre clamp is compatible with the $n=1$ displacement node and ${\lambda }_r=2\ell$; clamps at $\ell /4$ and $3\ell /4$ are compatible with $n=2$ and ${\lambda }_r=\ell$. The support positions therefore matter before a gas wavelength can determine a rod speed of sound.

For a chosen model with $E=200\mathrm{GPa}$, ${\rho }_r=8000kg/{\mathrm{m}}^3$ and $\ell =1.60\mathrm{m}$, the slender-rod extensional relation $c_r=\sqrt{E/{\rho }_r}$ gives $5000m/s$. The two-quarter-support mode then has $f=3125\mathrm{Hz}$. If the gas has stipulated $c_g=343m/s$, its half-wavelength is $d_g=54.88\mathrm{mm}$, andThe centre-supported fundamental would instead have $f=1562.5\mathrm{Hz}$ and twice that gas spacing. The same rod material has not changed its wave speed. In this slender-rod model, $E$ is Young's modulus; a bulk compressional wave involves a different elastic-modulus combination, so the two wave speeds are not interchangeable.

Write the measured positions of consecutive maxima of sound pressure amplitude as $x_j=a+jd+{\epsilon }_j$. The arbitrary origin $a$ is fitted along with the half-wavelength $d$ by linear regression; forcing the line through zero would turn a ruler-origin offset into an erroneous slope. For independent equal-variance position errors of standard deviation ${\sigma }_x$,As an original example, choose $f=1000\mathrm{Hz}$, $a=24\mathrm{mm}$ and $d=171.5\mathrm{mm}$. Add stipulated position errors $(1,-1,0,0,-1,1)\mathrm{mm}$ at $j=0,\dots ,5$. Their sum and index-weighted sum vanish, so the fitted values remain $\hat{a}=24\mathrm{mm}$ and $\hat{d}=171.5\mathrm{mm}$, giving $c=343.0m/s$. These are synthetic inputs, not measurements of a particular air sample.

With stipulated ${\sigma }_x=1\mathrm{mm}$ and independent drive-frequency uncertainty ${\sigma }_f=1\mathrm{Hz}$, first-order uncertainty propagation givesThis statistical calculation omits common scale-calibration error, thermal drift and mode misidentification. Counting more positions reduces independent localization error; it cannot correct a wrong family of extrema or an incorrectly calibrated ruler.

Consider a uniform tube $0\le x\le L$ with rigid ends, constant undisturbed mass density ${\rho }_0$ and speed of sound $c$. Neglect losses and transverse structure. A free normal mode hasand a consistent choice of sound pressure, acoustic particle velocity and acoustic displacement isHere $P>0$ is a peak sound pressure. Direct differentiation verifies ${\rho }_0{\partial }_{t}u=-{\partial }_x{p}^{\prime }$ and ${\partial }_t{p}^{\prime }=-{\rho }_0{c}^2{\partial }_{x}u$, the one-dimensional linear homentropic acoustic equations. The rigid ends enforce $u=0$ there. A driven experiment only approximates this free-mode geometry; the formulas do not predict a finite resonant amplitude for an undamped driven tube.

The maxima of sound pressure amplitude occur at $kx=j\pi$, where acoustic particle velocity vanishes. The maxima of acoustic particle velocity amplitude instead occur at $kx=(j+1/2)\pi$, where sound pressure vanishes. The two families are displaced by $\lambda /4$, but each family repeats after $\lambda /2$. At a position where both are nonzero, the time dependence is also a quarter cycle apart. Using consecutive maxima of sound pressure amplitude therefore givesThe relation also applies to consecutive minima of sound pressure amplitude, provided the same family is used throughout. Mixing maxima and minima would halve the spacing and give an erroneous speed of sound.

The local acoustic energy density is $p^{\prime 2}/(2{\rho }_0{c}^2)+{\rho }_0{u}^2/2$. Averaging over one wave period gives $P^2/(4{\rho }_0{c}^2)$, independent of $x$. Likewise $⟨p^{\prime }u⟩=0$: this ideal standing wave has no mean longitudinal acoustic energy flux, although its pressure and fluid velocity oscillate. Energy transport into a real lossy tube is a different question from the free-mode pattern.

PHYWE's implementation rubs a $1.60\mathrm{m}$ rod supported $0.40\mathrm{m}$ from each end. Its end disc drives a dusty glass tube; a movable tuning slide selects a clear pattern. A ruler measures repeated positions and a thermometer records gas temperature. The two supports select a rod wavelength equal to its length.^[ref]

The University of Iowa also documents an electronic version: a loudspeaker drives the tube and a microphone on a plunger supplies an oscilloscope readout. This permits a spatial scan of sound pressure without relying on the dust pattern.^[ref]

For a proposed pressure-scan measurement, keep the drive frequency fixed and move a small microphone through the tube. Record its position and fitted sinusoidal amplitude after each move; repeat the scan in reverse to reveal drift. Locate several consecutive maxima, rather than estimating a wavelength from a single pair. Reducing the drive checks whether the inferred spacing is independent of amplitude in the linear acoustics regime. A drive-off record estimates electrical background. These are controls for the proposed measurement, not reported observations from the cited demonstrations.

A microphone and its access opening can perturb the field. End compliance, losses and transverse modes also challenge the one-dimensional description. Dust traces identify a repeating structure, but the grains are not fluid parcels: the equations below describe sound pressure and acoustic particle velocity, rather than a detailed grain-transport model. The temperature and gas composition must remain sufficiently uniform to interpret a single fitted speed of sound.

For rotational motion about a fixed axis, let $r$ be cylindrical radius, $v$ the signed tangential velocity, and $dm$ the mass measure. Define the kinetic energy, axial angular momentum, and moment of inertia byThe Cauchy-Schwarz inequality gives $L^2\le (\int v^{2}dm)(\int r^{2}dm)=2KI$. More precisely, direct expansion gives the nonnegative energy gapThus the minimum at fixed $L$ and fixed mass distribution occurs exactly when $v=(L/I)r$ almost everywhere with respect to $dm$. This means constant angular velocity $\Omega =L/I$, or solid-body rotation. If the profile is continuous, the equality holds at every point in the mass support.

For a uniform unit-density cylinder of radius $a$, height $h$, and volume $V=\pi a^{2}h$, the moment of inertia is $I=V{a}^2/2$. Therefore $K\ge L^2/(V{a}^2)$, with equality for $\omega (r)=2L/(V{a}^2)$. Multiplying by a uniform nonunit density changes the mass measure and $I$ consistently.

For a free particle of rest mass $m$, subtract its rest energy $m{c}^2$ from its total energy $\gamma m{c}^2$, where $\gamma$ is the Lorentz factor. The low-speed binomial series gives $K=m{v}^2/2+3m{v}^4/(8c^2)+O(m{v}^6/c^4)$, recovering classical kinetic energy.

The electro-optic Kerr effect produces linear birefringence proportional to the square of a static applied electric field. Defining the Kerr effect coefficient $K$ by $\Delta n={\lambda }_{0}K{E}^2$, a length $L$ produces phase retardation $\delta =2\pi KL{E}^2$. A switchable quarter-wave plate requires $|\delta |=\pi /2$ modulo $2\pi$, with fixed principal axes. This differs from the Faraday effect, whose eigenpolarizations are circular polarizations.

An original local flow model has velocity $\mathbf{v}=(\dot{ϵ}x,-\dot{ϵ}y,0)$ for constant extensional strain rate $\dot{ϵ}>0$. Its divergence is zero, and its trajectories satisfy the displayed equations. Material approaches along $y$ and leaves along $x$. If the idealized observation region ends at $|x|=L$, a trajectory starting at $0<|x_0|<L$ has residence timeFor chosen $\dot{ϵ}=1{\mathrm{s}}^{-1}$, $L=1\mathrm{mm}$ and $|x_0|=0.01\mathrm{mm}$, this is $\log 100\mathrm{s}\simeq 4.61\mathrm{s}$. Starting ten times farther from the symmetry line halves this particular residence time. The mathematical trajectory $x_0=0$ never exits in this model; diffusion, imperfect flow and finite apparatus prevent treating that ideal trajectory as a guaranteed infinite exposure in an experiment.

The neutral strange meson with strangeness $-1$ and valence content $s\bar{d}$. It is distinct from the neutral kaon flavour state $K^0=d\bar{s}$, which has strangeness $+1$. These are flavour eigenstates, not the weak-interaction propagation eigenstates of neutral kaon mixing.

The SNO Collaboration used $1000$ tonnes of heavy water in a $12\mathrm{m}$ acrylic sphere, surrounded by ordinary water shielding and $9456$ photomultipliers. Weak charged current (CC) breakup of a deuteron selects electron neutrinos; weak neutral current (NC) breakup measures all active neutrino flavors; electron elastic scattering (ES) has reduced non-electron sensitivity.^[ref]

Super-Kamiokande is an underground water Cherenkov detector in Japan. Its roughly 50,000-ton water target is viewed by photomultipliers around the tank. A neutrino is observed indirectly when an interaction produces charged secondaries that emit Cherenkov radiation.^[ref]

Photon arrival times and detected charges constrain interaction position and outgoing directions. Ring appearance helps distinguish showering electrons from more track-like muons, enabling the Super-Kamiokande atmospheric-neutrino experiment. The water mass is both target and optical medium: transparency, sensor timing and background rejection affect reconstruction. The useful fiducial volume is smaller than the full water volume.

In an n-channel junction field-effect transistor, increasing drain voltage increases the local gate-channel reverse bias near the drain. The depletion region can meet there, ending the quasi-neutral channel before the drain. Carriers already in the channel are swept across the remaining high-field region, so drain-end pinch-off produces current saturation rather than zero current. A sufficiently negative gate voltage can instead deplete the whole channel and cut the device off. In the one-sided depletion approximation, a channel half-width $a$ and donor density $N_D$ give a depletion threshold voltage $U_P=e{N}_D{a}^2/(2{\epsilon }_s)$; with built-in junction voltage $V_{\mathrm{bi}}$, drain-end pinch-off occurs approximately at $V_{DS}=U_P-V_{\mathrm{bi}}+V_{GS}$.