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Quasi-adiabatic continuation implements a gapped path of local Hamiltonians by a quasi-local unitary. It dresses local operators with exponentially decaying tails while preserving their action within the corresponding low-energy spaces.

A large collar translating slowly at speed $c$ over a film $h_{+}$ and leaving a thinner film $h_{-}$ must grow: its leading excess volume increases at rate $c(h_{+}-h_{-})$. An exact travelling wave connecting unequal stationary films is impossible, since integration of the thin-film equation would give a common constant $q-ch=-c{h}_{+}=-c{h}_{-}$. In a slowly evolving collar core the leading pressure gradient vanishes. Therefore $h^{\prime }+h^{\prime \prime \prime }=0$, and a centred, reflection-symmetric collar has $h=A(1+\cos x)+B$ on $-\pi <x<\pi$. Its leading volume is $2\pi A$, so for constant small $B$, $2\pi \dot{A}\simeq c(h_{+}-h_{-})$. The speed is supplied by collar-to-film capillary matching.

Let $-1<s<1$ and let $S_s$ be the outward-oriented boundary of $x^2+y^2+z^2\le 1$, $z\le s$. For a continuous function $h:[-1,s]\to \mathbb{R}$, the vector field $\mathbf{F}=h(z)(x,y,z{)}^T$ has outward fluxOn the curved unit sphere, the outward unit normal is $(x,y,z{)}^T$, so the normal component is $h(z)$. A horizontal spherical band has area $2\pi dz$, producing the first term. On the top disk, the outward unit normal is $(0,0,1{)}^T$, the normal component is $sh(s)$, and the area is $\pi (1-s^2)$, producing the second term.

For $h=f^{\prime }$ with a continuous first derivative, the fundamental theorem of calculus givesThis direct surface integral proof needs no second derivative of $f$. For $h=1$, it reduces to $\pi (2+3s-s^3)$; the divergence theorem identifies this with three times the enclosed volume.

For uniform probability per volume in a sphere of radius $R$, the radial probability density function is $3r^2/R^3$. Its root-mean-square radius is $\sqrt{3/5}R$, whereas its scalar-radius standard deviation is $\sqrt{3/80}R$. These are distinct from a fully specified quantum momentum uncertainty.

For a homogeneous Poisson point process of intensity $\lambda$ in ${\mathbb{R}}^d$, mapping $x$ to $v_d|x{|}^d$, where $v_d$ is the unit-ball volume, gives a rate-$\lambda$ Poisson process on $[0,\infty )$. Mapping to $|x{|}^d$ instead gives rate $\lambda v_d$.

For $u=A{T}^4$, $P=u/3$ and physical cell volume $R^3$, the first law integrates to $S=4u{R}^3/(3T)$ with zero chemical potentials and a fixed relativistic species set. Entropy conservation through a species threshold instead requires tracking $g_{\ast s}{T}^3{R}^3$.

If an open aortic valve equilibrates pressure much faster than peripheral outflow, the arteries and ventricle share a common pressure while their combined blood volume is conserved during the short adjustment. Let arterial vascular compliance be $C_a$ and systolic ventricular elastance be $E_s$. Equating the initial and final arterial-plus-ventricular stored volumes gives the displayed pressure. Subsequent peripheral drainage sees the combined compliance $C_a+1/E_s$; during filling a closed aortic valve removes the ventricular contribution.

Rate-zonal centrifugation resolves an initially narrow sample zone by differences in sedimentation coefficient during a finite interval. Its bands remain transient: extending the interval can carry material into a centrifugation pellet. Thus a density gradient need not imply an equilibrium buoyant density measurement.^[ref]

A typical preparation layers a narrow sample above a preformed density gradient, often containing sucrose. The initial zone must be narrow enough that differences in starting position do not overwhelm the later separation. The run ends while useful bands remain inside the liquid. In the simple case where all particles are denser than every part of the column, none has an interior isopycnic centrifugation position.^[ref][ref]

The sedimentation coefficient $s=(m-{\rho }_{f}V)/f$ explains why separation is not an ordering by mass alone. At fixed local centrifugal acceleration, increasing displaced volume can increase buoyancy, while a less compact shape can increase the linear friction coefficient and reduce velocity. Even for a sphere, Stokes's law gives $s=2b^2({\rho }_p-{\rho }_f)/(9\eta )$, so both the density difference and the dynamic viscosity matter. The resulting sedimentation coefficient can change along a density gradient.

Rayleigh scattering is elastic scattering of electromagnetic radiation by particles much smaller than the incident wavelength. In the stationary particle approximation, the scattered electromagnetic radiation keeps the incident frequency: its direction changes, rather than its color. The incident electric field induces an oscillating electric dipole moment, which produces electric dipole radiation. This is a useful example of how a microscopic response can explain a familiar large-scale sight, why the sky is blue. ^[ref].

The smallness condition for Rayleigh scattering by a small sphere of radius $a$ is $ka\ll 1$, where the wavenumber is $k=2\pi /\lambda$ and $\lambda$ is the surrounding medium's wavelength. For ordinary materials with modest refractive index, the incident electric field then varies little across the particle, making the electric-dipole approximation for radiation useful. For very large refractive index, also check the internal size $|m|ka\ll 1$ before applying the simple electric dipole model. Once these conditions fail, use Mie scattering for a sphere instead of extrapolating the Rayleigh scattering wavelength dependence.

Historically, the 1871 paper ^[ref] connected the color and polarization of an electromagnetic wave of scattered skylight with particles smaller than the wavelength. Its dimensional analysis anticipates scattered light intensity proportional to particle volume squared and to ${\lambda }^{-4}$. The original mechanical picture is historical; the modern explanation uses induced electric dipole radiation.

For one chemical reaction in a closed system of fixed volume $V$, with uniform concentrations and no competing reactions or appreciable accumulation of unlisted species, choose the chemical equationBrackets denote amount of substance per volume, for example $[\mathrm{A}]=n_{\mathrm{A}}/V$. The stoichiometry gives one instantaneous reaction ratewhere $\xi$ is the existing extent of reaction. The minus signs make forward consumption positive; dividing by the coefficients makes measurements of different chemical species agree. Units can be $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. See IUPAC Gold Book: rate of reaction and Flowers et al., Chemistry 2e, §12.1: chemical reaction rates.

For the illustrative chemical equation $2\mathrm{A}+\mathrm{B}⟶3\mathrm{C}$, a measured $d[\mathrm{C}]/dt=0.060\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ implies $v=0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $d[\mathrm{A}]/dt=-0.040\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, and $d[\mathrm{B}]/dt=-0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. These numbers express stoichiometry, without supplying a rate law.

Each derivative is the slope of a concentration curve at one instant. Over a finite interval, $-\Delta [\mathrm{A}]/(a\Delta t)$ instead gives an average reaction rate. For A converting reversibly to B, the net reaction rate is $v=d[\mathrm{B}]/dt=J_f-J_r$; both directional reaction rates can remain positive when $v=0$ at chemical equilibrium.

A real gas departs from the ideal gas law because interactions and molecular volume matter. Nonideal equations of state, including the Dieterici equation, model such departures.

A rectangular parallelepiped is a parallelepiped whose three edge directions at each vertex are mutually perpendicular. Its faces are rectangles; if its edge lengths are $a,b,c$, its volume is $abc$. A cube is the special case $a=b=c$.

The derivative $V_0^{-1}dV/dT$ uses a fixed reference volume $V_0$. It differs from the instantaneous fractional coefficient $V^{-1}dV/dT$ by the factor $V/V_0$. The volume convention must be specified before comparing coefficients.

At fixed particle number and volume, a classical monatomic ideal gas has $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$. The canonical energy fluctuation identity $\mathrm{Var}(E)=k_B{T}^2{C}_V$ therefore gives ${\sigma }_U/U=\sqrt{2/(3N)}$. Absolute fluctuations grow as $\sqrt{N}$, while relative fluctuations decrease as $N^{-1/2}$. A particle reservoir changes the fluctuation problem because energy and particle-number fluctuations are then correlated.

Primary official report republished in Archives parlementaires, volume XC, pp. 74–80. Documents the administrative and financial prosecution and referral; accusations do not establish the truth of every alleged fraud.

For an original idealized mathematical model, let $n_{{\mathrm{O}}_2}(t)$ be the amount of substance of molecular oxygen consumed since the initial reading and $n_{{\mathrm{CO}}_2}(t)$ the carbon dioxide produced. Suppose a fraction $f$ of that carbon dioxide is removed from the gas, all other gas inventories remain fixed, and temperature $T$ and gas pressure $P$ stay constant. The ideal gas law gives the positive gas contractionwhere $R$ is the molar gas constant and $V$ the gas volume. With complete absorption, $f=1$, the contraction reflects molecular oxygen uptake regardless of the amount of carbon dioxide made. For a uniform pipette bore of area $A$, an inward boundary displacement $x$ gives $D=Ax$. A graduated pipette can instead supply the displaced volume directly.

Let $D_s$ denote sample contraction and $D_b$ the matched blank contraction. Define $D_{\mathrm{corr}}=D_s-D_b$ with both taken relative to their own initial readings. Under the preceding assumptions,The slope of corrected volume against time estimates uptake over the chosen interval. The approximate sign acknowledges that a liquid boundary also responds to hydrostatic and capillary pressure, while the gas can exchange water vapour. A fixed-volume apparatus would instead show a pressure decrease; its ideal conversion is $n_{{\mathrm{O}}_2}\simeq V(P(0)-P(t))/(RT)$ after complete absorption and appropriate blank correction.

The ideal gas law also exposes drift without any biology: for fixed gas inventory, $V\propto T/P$. A one-percent rise in absolute temperature at fixed pressure produces a one-percent expansion. Matching the initial gas spaces makes such common changes easier to subtract. Subtracting unmatched blanks can leave a physical signal that looks biological.