For a complete ensemble of downward nuclear de-excitation cascades, let $F_i$ be the direct population of excited level $i$ and $C_{i\to j}$ the total number of transitions from $i$ to $j$. Conservation of population gives the displayed outward-minus-inward formula. A negative $F_i$ rules out a proposed scheme under the assumed complete transition inventory. Gamma ray counts equal transition counts only when unreported internal conversion and other competing paths can be neglected or corrected. Summing over all excited levels cancels internal transitions and leaves the transitions into the ground state.

Gamma decay is electromagnetic nuclear de-excitation by photon emission. It preserves mass number and atomic number, while changing the nuclear state. Angular momentum and parity constrain the allowed radiation multipoles. The gamma ray topic is broader than this nuclear process.

Internal conversion transfers nuclear excitation energy directly to a bound atomic electron, which is emitted, rather than to a real gamma ray. Photon counts alone then underestimate the number of nuclear transitions unless the conversion contribution is included.

For an ultrarelativistic electron scattered elastically by a static, spherical, spin-zero atomic nucleus, a plane-wave Born approximation retains a single electromagnetic interaction. The Fourier transform of its Coulomb potential is proportional to $ZeF(q)/q^2$. Relative to the same point charge, the scattering amplitude is therefore multiplied by $F(q)$ and the differential scattering cross-section by $|F(q){|}^2$:Here $E\gg m_e{c}^2$ is the incident total electron energy, $\theta$ the scattering angle, $d\Omega$ a solid angle element, and $\alpha$ the fine-structure constant. The cosine factor already accounts for the projectile's spin; a spinless projectile would give a different point-charge factor. The formula neglects nuclear recoil, electron rest-mass corrections, radiative corrections and Coulomb distortion in electron scattering. The latter can be substantial at large $Z\alpha$: electron waves then require treatment in the target's Coulomb potential, rather than undistorted plane waves.

Nonzero target spin can introduce magnetic and additional electric contributions, so division by this point-charge expression does not generally isolate one charge form factor for a proton or deuteron. Even for spin zero, finite energy resolution, recoil and Coulomb distortion in electron scattering can change a measured minimum. A zero of the uniform-sphere model is a clear illustration of amplitude cancellation, not a universal prediction of a zero experimental counting rate.

Consider an ideal nuclear charge distribution that is constant inside a ball of radius $R>0$ and zero outside:This is a deliberately sharp-edged model. Substitution into the spherical nuclear charge form factor gives, for $q\ne 0$,At $q=0$ take the continuous limit $F(0)=1$. Direct radial integration gives $⟨r^2{⟩}_{\mathrm{ch}}=3R^2/5$, and the Taylor series provides an independent check:The first positive zero solves $\sin x-x\cos x=0$, equivalently $\tan x=x$, with $x=qR\simeq 4.49340946$. Consequently the first diffraction minimum moves toward smaller momentum transfer as $R$ increases.

For comparison, a smooth Gaussian function charge density with the same nuclear charge radius isThe Fourier transform of a Gaussian stays positive and has no zero. Both models have the same leading low-$q$ curvature, but they disagree at higher momentum transfer: at $qR=2$, the sphere gives $F\simeq 0.653097$ and the Gaussian function gives $F_G\simeq 0.670320$. Measuring only a nuclear charge radius cannot distinguish these two nuclear charge distributions.

As a chosen scale illustration, take $R=3.00\mathrm{fm}$. The predicted first zero is $q=1.49780{\mathrm{fm}}^{-1}$, or $\hslash cq\simeq 295.557\mathrm{MeV}$ using $\hslash c\simeq 197.327\mathrm{MeV}\mathrm{fm}$. With a $200\mathrm{MeV}$ ultrarelativistic electron beam and negligible nuclear recoil, it occurs at $\theta \simeq 95.27^{\circ }$. These are consequences of chosen model parameters, not measured properties of carbon or a claim that all atomic nuclei are uniform balls.

For positive integers $a,b$, the north-east lattice path recurrence isThe boundaries have only straight north-east lattice paths. The origin counts the zero-step north-east lattice path once.^[ref]

Proof outline. A north-east lattice path ending at $(a,b)$ has a final step either from $(a-1,b)$ or from $(a,b-1)$. Deleting that step gives a shorter north-east lattice path; appending it reverses the operation. The two classes are disjoint and exhaust all possibilities, so the rule of sum adds their counts. At $(4,3)$, the arrivals from $(3,3)$ contribute $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$, while those from $(4,2)$ contribute $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$: altogether $20+15=35$.

Set $P(a,b)=0$ if either integer coordinate is negative. Then the same recurrence relation applies on the nonnegative axes away from $(0,0)$: one predecessor contributes one, and the other contributes zero. Initialize $P(0,0)=1$ separately; applying the recurrence relation there would incorrectly give zero.

Computing the grid once in increasing order of $a+b$ is dynamic programming: every entry uses two earlier entries. If a specified point is forbidden, set its count to zero, then continue the same recurrence relation at allowed points. This counts north-east lattice paths avoiding the forbidden points, provided the origin and endpoint are allowed. The last-step argument remains valid because appending a step to an allowed endpoint cannot visit any new intermediate integer lattice point.

Suppose a homogeneous bulk material is unchanged when all spatial coordinates are reversed. Both electric polarization and electric field reverse under this inversion, so a local electric dipole response must obey $P(-E)=-P(E)$. Substituting the nonlinear optical susceptibility expansion givesFor every sufficiently small $E$, all even coefficients vanish. In the full tensor relation, two reversed input electric fields leave their product unchanged while the output electric polarization changes sign, forcing every bulk electric-dipole component of ${\chi }^{(2)}$ to vanish.

The qualification matters: the argument assumes inversion symmetry, a local bulk electric dipole response, and no applied bias that breaks that symmetry. A surface breaks bulk inversion symmetry; spatially varying or magnetic responses also lie outside this local electric-dipole model. Thus detecting second-harmonic generation at a surface or through a weaker magnetic contribution does not contradict the bulk selection rule. Conversely, absence of inversion symmetry permits some quadratic components but does not guarantee a nonzero component in every beam geometry.

If a Lagrangian density depending on first and second derivatives has no undifferentiated field, the scalar-field shift symmetry $\delta \varphi =ϵ$ gives the displayed Noether current. Its divergence is minus the Euler-Lagrange field equation with second derivatives, so it is conserved on shell. The spatial integral of $j^0$ is constant only if the boundary current vanishes. For $\mathcal{L}=\rho {\varphi }_t^2/2+\alpha {\varphi }_x^2/2-\beta {\varphi }_{xx}^2/2$, it gives $j^0=\rho {\varphi }_t$, $j^x=\alpha {\varphi }_x+\beta {\varphi }_{xxx}$. Fixed transverse endpoint positions break the global shift symmetry of the boundary problem; support forces can then change the total transverse momentum even though the bulk continuity equation still holds.

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 $\delta y^{\prime }=0$.