A cleavage furrow is the inward indentation of the cell membrane during cytokinesis. It is the visible deformation; the actomyosin ring is machinery beneath that boundary.^[ref]

An actomyosin ring is a contractile assembly containing actin filaments and myosin II. In typical animal cytokinesis, it forms in the cell cortex beneath the cell membrane. Its contraction draws the boundary inward to produce a cleavage furrow; an inward groove is not yet complete separation of the daughters.^[ref]

Define $\Delta V={\Phi }_{\mathrm{entry}}-{\Phi }_{\mathrm{exit}}$ as the electric potential drop along the particle's passage. If the electric field changes negligibly during that short crossing, the work givesThe sign matters: a positive ion gains kinetic energy while going to lower electric potential; a negative ion needs the opposite drop. A fixed-sign field would accelerate on one crossing and decelerate on the next. The alternating field must reverse before the particle comes back.

For a fixed electric charge sign and circulation direction, absorb this alternating crossing direction into the definition of the arrival phase ${\varphi }_n$. If $V_0$ is the peak electric potential difference across the whole gap, rather than the amplitude of either electrode relative to ground, a thin-gap model isArrival at ${\varphi }_n=0$ gives the largest accelerating kick; $|{\varphi }_n|<\pi /2$ gives a positive kick. Arrival at ${\varphi }_n=\pi$ gives deceleration. For an ideal classical particle initially on crest, the cyclotron resonance condition keeps successive arrival phases equal. It does not automatically bring an initially mistimed particle onto crest.

Cyclotron resonance is resonant absorption by charged particles in a magnetic field when an applied oscillating electric field drives their cyclotron motion. For a parabolic band, the resonance frequency is ${\omega }_c=|q|B/m^{\ast }$, so measuring it determines the cyclotron effective mass. In the quantum description, transverse electric dipole transitions connect adjacent Landau levels, whose separation is $\hslash {\omega }_c$. For an electron and a field along $z$, only the circular polarization rotating in the electron's orbital sense drives the positive-frequency resonance. Scattering broadens the absorption peak; resolving it requires an orbital period short compared with the relaxation time.

The Euler spiral is the planar curve parametrized by the Fresnel integrals. In the optical convention,The tangent has unit length and argument $\pi u^2/2$, so its signed curvature is $\pi u$, proportional to signed arc length. In Fresnel diffraction, an interval integral is the directed chord joining its endpoint parameters on the Cornu spiral. Chord length gives the interval factor's modulus, while chord direction gives its complex argument.

An interstitial defect is an atom occupying a position between the regular atomic positions in a crystal structure. Small solute atoms or an extra host atom can occupy such positions. Different interstitial orientations can induce anisotropic local strain; an applied stress changes their relative energies. Thermally activated hopping among those positions can then produce delayed elastic strain and the Snoek effect.

A missing atom at a lattice site that would be occupied in the reference crystal. A vacancy defect is a point defect and can contribute configurational entropy because many choices of empty sites can have the same macroscopic energy. Formation energy and pressure-volume work oppose this entropy gain. Vacancies can be present in thermal equilibrium; they need not all be preparation mistakes. Their allowed number and creation mechanism depend on whether the lattice size, atom number and surface reservoirs are fixed.

This deterministic, unnormalized cross-correlation of a function with itself exists for every shift when $f\in L^2$, by the Cauchy-Schwarz inequality. It satisfies $h(0)=\Vert f{\Vert }_2^2$ and $h(-x)=\overline{h(x)}$. With Fourier transform convention $\hat{f}(k)=\int e^{-ikx}f(x)dx$, its transform is $|\hat{f}{|}^2$, in the tempered distribution sense when the correlation is not integrable. Fourier inversion at zero gives the Parseval identity, first for Schwartz functions and then by approximation in L2 space. Unlike the normalized statistical autocorrelation, its value at zero is not generally one.

Direct CP violation concerns the decay transition itself. At leading order, the charged and neutral two-pion decay amplitude ratios separate a common $ϵ$ from channel-dependent ${ϵ}^{\prime }$.^[ref]

For the corresponding double ratio of partial decay widths,The expansion follows from $(1-2x)/(1+x)=1-3x+O(x^2)$. A single nonzero long-lived two-pion yield does not isolate $x$: it can already occur with ${ϵ}^{\prime }=0$. Thus the 1964 evidence for CP violation must be distinguished from the later experimental establishment of direct CP violation through ${ϵ}^{\prime }$.

For a coherently prepared state proportional to $\rho |K_S⟩+|K_L⟩$, evolve each component with ${\lambda }_j=m_j-i{\Gamma }_j/2$. The proper time is $t$, and $\hslash =1$. Since both components reach the same final state, their decay amplitudes add before squaring. Factoring out $|A_S{|}^2$ and a common beam normalization gives the two-pion intensityThe last term is quantum interference. It vanishes upon averaging over an independently random preparation phase, but cannot be discarded from a coherent preparation. Summing the two positive contributions alone would lose the phase-sensitive information.

At late proper time, the short-lived contribution and the quantum interference term die away, leaving the long-lived tail $|{\eta }_{\pi \pi }{|}^2{e}^{-{\Gamma }_{L}t}$. Such a tail is easier to identify after the initially abundant short-lived component has disappeared. The useful experimental balance is between eliminating that component and retaining enough long-lived decays to observe a rare decay channel.

Use the kaon CP eigenstates convention $CP|K^0⟩=-|{\bar{K}}^0⟩$, matching kaon CP mixing parameter. Then $|K_1⟩=(|K^0⟩-|{\bar{K}}^0⟩)/\sqrt{2}$ is CP even, while $|K_2⟩=(|K^0⟩+|{\bar{K}}^0⟩)/\sqrt{2}$ is CP odd. With CPT symmetry and a common small mixing parameter, the physical states can be writtenHere $S,L$ mean short and long mean lifetimes; the kaon CP mixing parameter $ϵ$ depends on the flavor-state phase convention. A positive flavor-exchange convention instead makes the symmetric flavor combination CP even, as explained in exchange-positive CP convention for neutral kaon mixing. Mixing these conventions would reverse the interpretation of the written sums.

Why does the two-pion final state test CP symmetry? A spin-zero kaon decaying into two spin-zero pions has relative orbital angular momentum $\ell =0$ by angular momentum conservation. Two negative intrinsic parities multiply to plus, and $(-1{)}^{\ell }=1$. Charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$; the $\ell =0$ state is even under this exchange. Hence this two-pion state has CP eigenvalue $+1$. If a decay transition operator $T$ preserves CP symmetry, inserting CP on both sides of $⟨\pi \pi |T|K_2⟩$ changes its sign without changing its value: that decay amplitude must vanish.

Assume for illustration that CP violation occurs only through the state admixture and the decay transition preserves CP symmetry. Put $A_1=⟨\pi \pi |T|K_1⟩$. ThenNeglecting the tiny difference in two-body Lorentz-invariant phase space between the physical masses gives ${\Gamma }_L^{\pi \pi }/{\Gamma }_S^{\pi \pi }\simeq |ϵ{|}^2$. A branching fraction, however, divides a partial decay width by the total decay width. With $\hslash =1$, ${\tau }_i=1/{\Gamma }_i$, soFor chosen values $|ϵ|=0.01$ and ${\tau }_L/{\tau }_S=100$, the branching fraction ratio is $0.01$, whereas the partial decay width ratio is $0.0001$. This is original arithmetic, not a historical fit. The longer observation time amplifies the integrated fraction relative to the instantaneous partial decay rate.

Neutral kaon regeneration is the production of a short-lived component when a neutral kaon beam passes through matter. In the phase convention of kaon CP eigenstates, let the forward transmission amplitudes of $K^0$ and ${\bar{K}}^0$ be $t$ and $\bar{t}$. Expanding $t|K^0⟩+\bar{t}|{\bar{K}}^0⟩$ in the kaon CP eigenstates gives the displayed equation. Different forward scattering amplitudes, including absorption, can therefore generate a two-pion signal even with CP symmetry. Ordinary matter is not a CP-symmetric preparation.

Tungsten regeneration calibrated the signal; scanning its position sampled acceptance. Helium regeneration was insufficient, bag walls lay outside acceptance, vertex distributions excluded collimator regeneration, and hydrogen data cross-checked the effect.^[ref] These controls are central: the question is whether the long-lived component itself reaches two pions, rather than whether material has prepared a fresh short-lived component.

A two-track invariant mass and a direction test answer different questions. In units $c=1$, assign each track the pion mass $m_{\pi }$ and reconstruct its momentum before magnetic deflection. With $p_{\pm }=|{\mathbf{p}}_{\pm }|$ and opening angle $\alpha$,This follows by expanding the squared sum of the two four-momenta. The angle $\theta$ between the reconstructed parent momentum $\mathbf{P}={\mathbf{p}}_{+}+{\mathbf{p}}_{-}$ and the neutral particle beam direction $\hat{\mathbf{n}}$ satisfiesA fully reconstructed two-body decay channel gives the parent mass and $\theta \simeq 0$, within experimental resolution. A three-body decay channel with an unseen neutrino generally misses both constraints; an accidental correct invariant mass alone is insufficient. Notice that the track opening angle $\alpha$ need not be small even when the parent direction angle $\theta$ vanishes.

The 1964 analysis selected $490<M_{\pi \pi }<510$ MeV and remeasured events with $\cos \theta >0.9995$, comparing adjacent mass bands.^[ref]

For an original dimensionless example, take $m_{\pi }=1$, $\hat{\mathbf{n}}=(0,0,1)$ and ${\mathbf{p}}_{\pm }=(\pm \sqrt{3},0,3)$. Each energy is $\sqrt{13}$; the opening angle is $60$ degrees since $\cos \alpha =(9-3)/(9+3)=1/2$. Yet $\mathbf{P}=(0,0,6)$ points exactly along the beam, and $M_{\pi \pi }^2=4(13)-36=16$. Thus a parent of mass $4$ passes both ideal tests. These chosen numbers illustrate four-momentum conservation, not detector performance.

James H. Christenson, James Watson Cronin, Val Logsdon Fitch and René Turlay reported the result in 1964. At Brookhaven's AGS, 30 GeV protons struck beryllium; collimators selected a neutral particle beam at 30 degrees. A downstream helium decay volume reduced interactions.^[ref]

The flight distance suppressed the short-lived kaon component. Each arm of a two-arm magnetic spectrometer had spark chambers before and after its magnet. Cherenkov detectors and scintillation counters triggered photographic recording in coincidence.^[ref]

The helium sample yielded $45\pm 9$ signal events; $(2.0\pm 0.4)\times 10^{-3}$ was the ratio to all charged decay channels.^[ref] That denominator must be retained when comparing this result with a total branching fraction.

The evidence was kinematic. Later regenerated-beam quantum interference established that the short- and long-lived states reached the same two-pion final state, excluding alternatives such as distinct pion-like particles.^[ref] The 1964 observation established evidence for CP violation; it did not determine every parameter of the CKM matrix. Inferring time-reversal symmetry violation from CP violation invokes CPT symmetry, rather than a direct comparison of time-reversed transitions in that experiment.

Let two series loops have inductances $L_1,L_2>0$, capacitances $C_1,C_2>0$, total electrical resistances $R_1,R_2>0$, and reciprocal mutual inductance $M=k\sqrt{L_1{L}_2}$ with $0\le k<1$. For a steady sinusoid of angular frequency $\omega >0$, choose complex peak amplitudes with time convention $e^{\mathrm{i}\omega t}$. An ideal source of peak electric potential difference $V$ drives loop 1; loop 2 has no independent source. ThenEliminating $I_1$ proves the response formula. Reversing one coil changes the sign of $M$ and the relative wave phase, but leaves this amplitude response unchanged. The measured sense ratio is $|R_s{I}_2/V|$, with $R_s$ included in $R_2$, not added a second time.

For identical loops at their uncoupled resonant frequency ${\omega }_0=1/\sqrt{LC}$,Writing $x={\omega }_{0}M/R$ makes this $x/[R(1+x^2)]$. Its derivative has the sign of $1-x^2$, so response at this fixed frequency is largest when $x=1$, equivalently $k=1/Q$. Too little coupling transfers little excitation; excessive coupling reduces the response at the original centre frequency. This is a fixed-frequency optimum, not a universal optimum over every possible receiver and drive frequency.

For the chosen values, $k=1/Q=0.063246$ gives $M=0.31623\mathrm{mH}$. At source peak $V=0.100\mathrm{V}$, the secondary peak electric current is $0.500\mathrm{mA}$ and the sense peak electric potential difference is $25.0\mathrm{mV}$. The primary peak electric current is also $0.500\mathrm{mA}$, so source mean power is $25.0\mu \mathrm{W}$: each loop dissipates $12.5\mu \mathrm{W}$, checking conservation of energy. Here $V$ is the equivalent ideal source behind $R_1$; a generator's terminal reading must be converted using its measured Thévenin resistance.

In the lossless identical-loop limit, the normal modes have equal or opposite electric currents, making their effective inductances $L+M$ or $L-M$. ThusAt $k=0.20$ these are $45.944$ and $56.270\mathrm{kHz}$. Finite electrical resistance broadens and shifts the driven peaks, so the exact response formula, rather than these lossless values, generates the plot. Changing a capacitor moves the uncoupled tuning through $f_0\propto C^{-1/2}$: increasing it by $21\%$ lowers $f_0$ by the factor $1/1.1$. The model explains why tuning and coupling must be adjusted together; it supplies no transatlantic propagation or detector-threshold prediction.

Interaction described by the Coulomb potential and its associated Coulomb force.

The Coulomb force is the electrostatic force between electric charges. For point charges in a uniform isotropic medium of permittivity $ϵ$, the force on charge $q_2$ from $q_1$ is ${\mathbf{F}}_{2\leftarrow 1}=q_1{q}_2{\hat{\mathbf{r}}}_{12}/(4\pi ϵr^2)$, where ${\hat{\mathbf{r}}}_{12}$ points from the first charge to the second.

Equal charge signs give repulsion and opposite signs give attraction. The inverse-square dependence allows superposition to construct fields of extended charge distributions. Nearby conductors or spatially varying dielectrics can redistribute charge and alter the field, so the simple uniform-medium pair formula does not automatically solve those boundary problems.

In the straight-trajectory dipole model, retain the Coulomb-coupling convention of $V(t)=Z{e}^2(bx+vty)/(b^2+v^2{t}^2{)}^{3/2}$. In SI units the coupling denoted $e^2$ here is $e^2/(4\pi {ϵ}_0)$. First-order time-dependent perturbation theory expresses a transition through the Fourier transform of this pulse at the transition angular frequency.

For positive $\omega$ put $\beta =\omega b/v$. The two Cartesian pulse transforms arewhere $K_{\nu }$ is a Modified Bessel function of the second kind. One may derive these from the Fourier transform of $(1+s^2{)}^{-1/2}$, differentiating in its scale and integrating its spatial derivative by parts. For $\beta \ll 1$, $\beta K_1(\beta )\to 1$ while $\beta K_0(\beta )\to 0$, so the transverse impulse gives the probability in the title. If the $x$ matrix element vanishes, the smaller longitudinal term can be the leading nonzero contribution.

The conditions include $b$ large compared with the atomic extent for the electric-dipole approximation, a small total transition probability for first-order theory, and a collision time $b/v$ short compared with $1/|\omega |$. Large $\beta$ instead gives exponentially suppressed excitation in this model through the large-argument behavior of $K_{\nu }$.

Taylor-Couette flow is the flow between coaxial rotating cylinders. Its simplest steady homogeneous purely azimuthal viscous state has steady circular Couette profile $v_{\varphi }=Ar+B/r$. Stability of that base state depends on rotation rates, geometry and viscosity; the inviscid Rayleigh centrifugal stability criterion addresses axisymmetric centrifugal disturbances, not every viscous instability.