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For a longitudinal displacement field $\psi (z,t)$ and Lagrangian density $\mathcal{L}=A[\rho {\dot{\psi }}^2-K{\psi }^{\prime 2}]/2$, uniform field shifts give Noether current $j^0=A\rho \dot{\psi }$, $j^z=-AK{\psi }^{\prime }$. Its spatial integral is the rod's longitudinal mechanical momentum. A prescribed position-dependent mass density changes $j^0$ to $A\rho (z)\dot{\psi }$ but does not break this internal shift symmetry. Boundary traction can carry momentum into or out of a finite rod, so action symmetry alone does not guarantee conservation of the integrated charge.

The outward flux of the Newtonian gravitational field through a closed surface is $-4\pi G$ times the enclosed mass. A sphere around a point mass establishes the coefficient; superposition extends the integral law to general mass distributions. The minus sign expresses attractive gravity. A symmetric Gaussian surface can turn the flux relation into a local field calculation, but the integral law alone does not imply that the field is uniform on an arbitrary surface.

A circular infinite cylinder of radius $b$ and uniform mass density $\rho$ has linear mass density $\lambda =\pi b^2\rho$. A coaxial Gaussian surface of length $L$ has inward radial gravitational field and no end-cap flux. Gauss's law for gravity gives $2\pi rL{g}_r=-4\pi G\lambda L$ outside, hence the displayed result. Inside, enclosed linear mass density is $\pi r^2\rho$, so $g_r=-2\pi G\rho r$. An exterior circular orbit has speed $\sqrt{2G\lambda }$ independent of radius. The exterior formula presupposes circular axial symmetry; area alone would not specify the near field of a noncircular rod.

Spread mass $M$ uniformly along $-a<z<a$ on an axis, with linear mass density $\lambda =M/(2a)$. In its perpendicular bisector plane, axial field components cancel, and the radial component isThis is an exact line mass model and a thin-rod approximation for a physical cylinder. A centred circular orbit in this symmetry plane has $v^2=GM/\sqrt{R^2+a^2}$. Thus $v\to \sqrt{GM/a}=\sqrt{2G\lambda }$ for $R/a\to 0$, and $v\sim \sqrt{GM/R}$ for $R/a\to \infty$. For a finite-width rod, the first limit is useful only while its transverse size remains much smaller than $R$.

To construct an original example rather than imitate GW150914, set $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, and adopt the rounded solar mass $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$. Compare $10+10$ and $20+20$ solar mass binaries starting at $20\mathrm{Hz}$. Stop each at the stipulated orbital compactness thresholdThis is a conservative pedagogical stopping rule, not a claimed error bound: the relative speed already reaches $\sqrt{0.10}c$. Higher-order corrections accumulate wave phase well before merger, and precision inference requires better models. The two cutoffs are approximately $102.2$ and $51.1\mathrm{Hz}$ respectively; doubling the chirp mass multiplies $\dot{f}$ at fixed $f$ by $2^{5/3}$.

For a nearby stipulated source at distance $D$, neglect redshift and define inclination $\iota$ as the angle between the orbital angular momentum and the line of sight. The leading quadrupole formula yieldsThe numerator of $A$ has units ${\mathrm{m}}^5{\mathrm{s}}^{-4}$, matching $c^{4}D$, so both polarizations are dimensionless. Its $f^{2/3}$ growth and the rising frequency produce the tightening, strengthening oscillations. For face-on viewing the polarization components have equal magnitudes in phase quadrature; for edge-on viewing the cross component vanishes, agreeing with circular-binary quadrupole radiation.

Shifting an arrival time aligns the chirp evolution, but different polarization projections need not overlay after only a sign flip. The last panel divides the ideal projected records by their known envelope factors to expose this phase difference. A physical source model must jointly explain those differences, not erase them by arbitrary rescaling. Independent integration of the separation equation, the radiated power and $d\Phi /dt=2\pi f$ checks the analytic chirp against orbital energy balance over the entire stipulated interval.

The chirp mass is the combination of two component masses controlling the leading Newtonian binary chirp. Inverting its frequency equation givesUse $f$ in hertz and $\dot{f}$ in hertz per second to obtain kilograms. For equal component masses $m$, $\mathcal{M}=m/2^{1/5}$. A stipulated $20+20$ solar mass system therefore has $\mathcal{M}\approx 17.411$ solar masses. In a cosmological observation the time and frequency measured at the detector instead yield $(1+z){\mathcal{M}}_{\mathrm{source}}$, where $z$ is redshift; phase evolution alone does not independently determine the source mass and redshift.

For the quadratic polynomial $f(x)=(x-a{)}^2-b^2$ with $b>0$, consider the Newton root-finding iteration towards the polynomial root $x_{\ast }=a+b$. If $x_k\ne a$, its signed error $e_k=x_k-x_{\ast }$ satisfiesIndeed the iteration for $u_k=x_k-a$ is $u_{k+1}=(u_k+b^2/u_k)/2$, and subtracting $b$ yields $(u_k-b{)}^2/(2u_k)$. This exact identity explains quadratic convergence near the simple polynomial root. If an initial error is asymptotic to $C{\epsilon }^m$ and $x_k-a\to b\ne 0$ as $\epsilon \to 0$, one step changes its leading power to $2m$ and its coefficient to $C^2/(2b)$.

Kamiokande, associated with Masatoshi Koshiba, was an underground water Cherenkov detector viewed by photomultipliers. Improved water purity and reduced background enabled solar neutrino observations. A concentration of reconstructed recoil directions toward the Sun distinguished a source signal above an approximately direction-independent background.^[ref]

For solar neutrinos, electron elastic scattering supplies a directional recoil, with different cross sections for active neutrino flavors. Neither the Homestake experiment deficit nor the SN 1987A neutrino burst alone established neutrino oscillation; later evidence compared complementary channels and propagation effects.^[ref]

For a scalar one-dimensional dispersion $E(k)=\hslash v|k|$ and a potential of period $L$, a reciprocal vector $nG$, $G=2\pi /L$, connects degenerate right- and left-moving states at $k=nG/2$. Their unperturbed energy is $E_n=n\pi \hslash v/L$. Thus consecutive allowed Bragg thresholds have spacing $\pi \hslash v/L$, whereas shifting momentum by a full reciprocal period changes a linear branch's energy by $2\pi \hslash v/L$. The potential's Fourier coefficients and higher-order couplings determine which gaps actually open and their widths, so a geometric period alone does not imply an exactly energy-periodic spectrum. A graphene pseudospin selection rule can further suppress scattering; the scalar-chain model omits that internal structure.

Consider a spin-degenerate two-dimensional free electron model with rectangular periods $a>b$ and two electrons per unit cell. Write $K=\pi /a$, $L=\pi /b$, and $k_F^2=4\pi /(ab)$. If $K<k_F<L$, the circular Fermi contour crosses only the vertical boundaries of the first Brillouin zone. Folding translates its protruding caps back by a reciprocal lattice vector $(2K,0)$. A weak periodic potential mixes the intersecting plane waves, reconnecting the contours into two open first-band sheets and a closed second-band pocket. The pocket is centred on a zone boundary: its halves on opposite sides of a reduced-zone plot are identified periodically and belong to one closed sheet.

For a functional $J[y]=\int F(x,y,y^{\prime },y^{\prime \prime })dx$, two applications of integration by parts leave the boundary termIf the endpoint displacement is fixed but its slope is free, $\delta y=0$ while $\delta y^{\prime }$ is arbitrary. The first variation therefore forces $F_{y^{\prime \prime }}=0$ at that endpoint. For the beam energy density $F=\beta (y^{\prime \prime }{)}^2/2-\alpha (y^{\prime }{)}^2/2$ with $\beta >0$, this gives $y^{\prime \prime }=0$, the zero-bending moment condition. A clamped endpoint instead prescribes the slope and has $\delta y^{\prime }=0$.

Coupled inductors are inductors whose magnetic flux linkages depend on each other's electric currents. A two-circuit linear model uses two self-inductances and a mutual inductance.

An integrated circuit technology using complementary MOSFETs. CMOS names a circuit technology; active-pixel sensor names a detector architecture.

An ideal insulating gate blocks particle exchange between the gate and substrate of a MOSFET. Their electrochemical potentials, represented by their band-diagram Fermi levels, therefore need not coincide. An applied gate voltage fixes their difference; with $e>0$ and voltage defined by an electron reservoir, ${\tilde{\mu }}_g-{\tilde{\mu }}_s=-e{V}_g$. Contact work-function differences also affect band bending, so equal Fermi levels do not imply flat bands. Each conducting region can separately have a spatially constant equilibrium electrochemical potential.

A ketose is a monosaccharide whose open-chain form contains a ketone group. Fructose is a familiar example, with a carbonyl group at carbon 2 in its open-chain structure.

A ketose can form cyclic structures by reaction of its carbonyl group with one of its own hydroxyl groups. The distinction from an aldose concerns carbonyl connectivity, rather than simply the number of carbon atoms.^[ref]

An aldose is a monosaccharide whose open-chain form contains an aldehyde group. Glucose is an example, even though much of the glucose in aqueous chemical solution has a cyclic structure.

Classification uses the open-chain carbonyl position rather than requiring every molecule to remain open-chain. This connects aldose identity with ring formation and with mutarotation.^[ref]

In water, D-glucose mainly occupies cyclic forms. The C5 hydroxy group adds to the C1 aldehyde, forming a cyclic hemiacetal; reversible opening restores the aldehyde. Reclosure can give either anomer. Their changing proportions cause mutarotation, the approach of optical rotation to an equilibrium value. This pathway involves changing chemical bonds, rather than simply rotating a hydroxy group about a single chemical bond.^[ref]

In this conventional Haworth projection, C5's $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ group points up. The alpha anomer's C1 hydroxy group points down, on the opposite face; the beta anomer's points up, on the same face. C2/C3/C4 retain down/up/down hydroxy groups. These are above/below relationships to a reference ring plane, not a claim that the ring is physically flat.^[ref]

The anomeric carbon is C1, the existing carbonyl group carbon that becomes a new stereocenter; no extra carbon chemical atom appears. The two anomers differ at that centre while retaining the other configurations.^[ref]

Trace the ring as C1–C2–C3–C4–C5–O5–C1, where O5 denotes the oxygen originally in the C5 hydroxy group. It contains five carbon chemical atoms and one oxygen; C6 remains attached outside the ring. The C1 carbonyl group oxygen becomes the C1 hydroxy group oxygen. This accounts for every oxygen: O5 in the ring and five hydroxy groups at C1, C2, C3, C4 and C6. Ring closure preserves ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$. A six-membered ring and a six-carbon sugar are thus different counts.

Recoil is the motion acquired by a body when it exchanges momentum with an emitted, absorbed or scattered particle. For isolated emission from a body initially at rest, momentum conservation makes the remaining body acquire the opposite momentum. If its final mass is $M$ and the recoil is nonrelativistic, its kinetic energy is $E_R\simeq p^2/(2M)$. A photon of energy $E_{\gamma }$ has momentum magnitude $E_{\gamma }/c$, so $E_R\simeq E_{\gamma }^2/(2M{c}^2)$. The transition energy is shared between the photon and the recoiling body; nuclear recoil is an important example.

A uniform square of side $s$ has the displayed moment of inertia about the perpendicular axis through its centre. Integrating $x^2+y^2$ over $[-s/2,s/2{]}^2$ gives $s^4/6$; multiplying by the surface mass density $M/s^2$ gives the result. A square prism of any uniform thickness has the same axial formula when expressed in terms of its total mass. If the side is denoted $2a$ instead, the result is $2M{a}^2/3$, so the side-length convention must be specified before substituting numerical data into a normal mode calculation.