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Semiclassical quantization uses classical trajectories and actions to approximate quantum states in a regime where relevant actions are large compared with $\hslash$. Phase consistency on a periodic orbit leads to Bohr-Sommerfeld quantization. Soliton collective coordinates can be quantized this way, while fluctuations supply additional corrections to the leading classical mass.

Beta decay Q values depend on small differences between much larger atomic masses. A smooth semi-empirical mass formula can therefore approximate absolute masses reasonably while predicting the wrong sign for a MeV-scale decay threshold. Nuclear pairing explains offset even-even and odd-odd mass branches, but does not guarantee that every specific even-even neighbor lies below the intervening odd-odd nucleus. Measured masses or additional nuclear shell corrections are needed for quantitative threshold tests near a model's residual error scale.

Sensible heat is transferred heat associated with a temperature change within a single thermodynamic phase. By contrast, latent heat can change the phase fraction while the temperature stays fixed during equilibrium phase coexistence. Neither name denotes an additional stored substance: both describe how transferred heat participates in a process.

For fixed mass $m$ and composition, use the heat capacity appropriate to the heating constraint. Under the quasistatic process assumptions in heat capacity, with only pressure-volume work and unchanged bulk kinetic energy and potential energy, fixed-pressure heating givesHere $c_p$ is specific heat capacity at constant pressure; fixed-volume heating instead uses specific heat capacity at constant volume $c_V$. A cooling step has negative $Q_{\mathrm{sens}}$ with the sign convention in heat. A temperature difference has the same numerical value in units of the kelvin or degree Celsius. The integral follows by adding the small transfers $\delta Q=m{c}_p(T)dT$, rather than assuming a constant heat capacity.

As a stylized ice-to-water example for a fixed-composition closed system at fixed pressure $1\mathrm{atm}$, take $m=0.5\mathrm{kg}$, an initial temperature of $-10{}^{\circ }\mathrm{C}$, melting at $0{}^{\circ }\mathrm{C}$, and a final temperature of $20{}^{\circ }\mathrm{C}$. Assume constant specific heat capacity at constant pressure values $c_{\mathrm{ice}}=2.1\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $c_{\mathrm{water}}=4.18\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and latent heat of fusion $L_f=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. These are stipulated, rounded model values; ignore the container and heat losses, retain all the material, and assume only pressure-volume work under the quasistatic process conditions above. The three steps areOnly the first and third terms are sensible heat; the middle term is latent heat. The plateau begins after $10.5\mathrm{kJ}$ and ends after $177.5\mathrm{kJ}$ of total supplied heat. At $94\mathrm{kJ}$ total input, half the mass is liquid water and the temperature remains $0{}^{\circ }\mathrm{C}$. The equilibrium phase transition temperature depends on pressure; this example stipulates a single fixed pressure. ^[ref]

This split makes heating budgets easy to interpret: melting consumes $167/219.3\approx 76.2\%$ of the supplied heat in this model despite causing no temperature rise. Within either single thermodynamic phase, the heating curve slope is $dT/dQ=1/(m{c}_p)$; a larger heat capacity makes the slope shallower. A flat interval represents changing phase fraction, so it cannot be modeled by applying $Q=m{c}_p\Delta T$ across the phase transition. The complementary constructed example in latent heat tracks that phase fraction explicitly.

Identical linear springs of stiffness $k$ have equivalent stiffness $2k$ in parallel and $k/2$ in series. Parallel springs have equal displacement and add forces; series springs carry equal force and add extensions. A mass $m$ attached to an equivalent massless spring has angular frequency $\sqrt{k_{\mathrm{eff}}/m}$.

In a series RLC circuit, the same electric current $i=q^{\prime }$ flows through the resistor with electrical resistance $R$, the inductor, and the capacitor. Choose $q$ as the signed electric charge on the capacitor plate into which positive $i$ flows. With terminal drops measured in that direction, Kirchhoff's voltage law givesHere $V(t)$ is the applied electric potential difference; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfiesThe angular frequency ${\omega }_0$ is the value for lossless free oscillation; $\alpha$ is the decay parameter. The three possibilities are underdamped RLC response for $\alpha <{\omega }_0$, critically damped RLC response for $\alpha ={\omega }_0$, and overdamped RLC response for $\alpha >{\omega }_0$. ^[ref].

For a fair comparison, start each circuit with the same electric charge $q(0)=q_0$ and zero electric current $q^{\prime }(0)=0$. Set $\tau ={\omega }_{0}t$ and $a=\alpha /{\omega }_0$; then $x=q/q_0$ obeys $d^{2}x/d{\tau }^2+2adx/d\tau +x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and electric charge.

The stored energy is the sum of magnetic energy and capacitor energy:Differentiating and using the circuit equation gives the complete power balance,In the free response, $E^{\prime }=-R{i}^2\le 0$: Joule heating accounts exactly for the lost energy. For the plotted initial conditions, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau {)}^2$. Since every free solution tends to zero for $R>0$,In the lossless limit $R=0$, energy stays constant and shuttles between the capacitor and inductor. Thus an oscillation of electric charge and a decrease of total energy are compatible statements.

The analogy with a damped harmonic oscillator is exact within these ideal models: electric charge corresponds to displacement, electric current to velocity, inductance to mass, $1/C$ to spring constant, and $R$ to the coefficient of a resisting force proportional to velocity. The two stored energy terms play the roles of kinetic energy and elastic potential energy. ^[ref].

Differentiating the driven equation also gives $L{i}^{\prime \prime }+R{i}^{\prime }+i/C=V^{\prime }$. When using this equation to solve for electric current, retain the original constraint $L{i}^{\prime }(0)+Ri(0)+q(0)/C=V(0)$: the initial electric charge fixes the initial rate of change of electric current.

In a finite volume method, orient every face numerical flux to the right. For cells $1,\dots ,N$, write the discrete total as $M^n={\sum }_{j=1}^N{h}_j{U}_j^n$. If $u$ is a mass per unit length, this total approximates mass; for a different conserved field, it measures that field's integral. Multiplying the update by the cell width gives an amount balance, so adjacent cells contributeThe interior face contributes $-k{F}_{j+1/2}^n+k{F}_{j+1/2}^n=0$. Summing all cells givesWith periodic boundary conditions, the two boundary labels denote the same physical face and must use the same numerical flux; then $M^{n+1}=M^n$. On an open interval the total changes by the boundary transfers. Unequal widths require the weighted total $\sum h_j{U}_j$, not the unweighted sum of cell averages.

The diagram uses two unit-width cells, $k=1/4$, cell averages $(2,1)$ and right-oriented numerical fluxes $(1,3,2)$ at their three faces. The updated cell averages are $(3/2,5/4)$: the total changes from $3$ to $11/4$, exactly $k(1-2)=-1/4$. The shared interior transfer $k\cdot 3=3/4$ cancels. If the receiving cell instead used $3+\epsilon$ for that same face, the total would acquire the spurious term $k\epsilon$. Conservation is therefore an algebraic property of sharing the transfer, even when the numerical flux is a poor approximation. In floating-point arithmetic, independently rounded cell updates may still leave small roundoff errors in the computed total.

For a heavy spherical head with a much thinner long rigid tail, let $P$ be its excess weight, $H=6\pi \mu a$ the head translation resistance, and $K=K_{N}L$ the tail normal translation resistance. Neglect tail mass relative to head mass, sphere rotation resistance relative to tail rotation resistance, and $a/L$ in tail lever arms. In shear $\alpha z{\mathbf{e}}_1$, the stable orientation satisfies $\sin \theta =\alpha /(\beta +\sqrt{{\beta }^2+{\alpha }^2})$, $\cos \theta >0$, with $\beta =3P/[L(K+4H)]$. Resolving resistive-force theory force and torque balance gives the displayed signed vertical velocity. Since $0<\gamma <1$, stronger shear increases $U_z$ towards zero and reduces the downward settling speed. The stable head is below the far tail end. Very large shear can make the neglected sphere-rotation contribution significant.

Using the orbital order after 28 supplied in the simple shell model, proton counts 32 and 38 fill $2p_{3/2}$ and $1f_{5/2}$, so the odd 39th proton occupies $2p_{1/2}$. With paired valence neutrons this gives a $1/2^{-}$ yttrium-91 ground-state prediction. The daughter has paired 40 protons and one neutron beyond the 50-neutron closure in $2d_{5/2}$, giving $5/2^{+}$. The inferred ground-state beta transition is therefore first-forbidden unique. This is a stated paired-shell model deduction; an endpoint diagram alone does not determine the spin assignment or an evaluated isotope mass.

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

A uniform elastic beam with endpoint conditions $y=y_{xx}=0$ has normal modes $\sin (n\pi x/L)$ for positive integers $n$. Substitute $y=X{e}^{-i\omega t}$ into $EI{y}_{xxxx}+{\rho }_{\ell }{y}_{tt}=0$ and impose the four endpoint conditions to obtain the displayed spectrum. Here ${\rho }_{\ell }$ is mass per unit length, $I$ is a second moment of area, and $EI$ is bending stiffness. frequencies proportional to $n^2$ distinguish flexural waves from the nondispersive ideal stretched string.

At renormalized coupling $\gamma =8\pi /n$, the kink-antikink transmission poles lie at $\vartheta =i\pi (1-k/n)$. The relativistic bound-state mass from a rapidity pole gives the displayed increasing breather masses. The kink mass is $M=mn/\pi$, and $k=n$ is an excluded threshold state. No breathers occur at $n=1$; scattering involving a physical second breather requires $n\ge 3$.

The static kink ${\varphi }_K(x)=4\arctan e^{x-a}$ joins adjacent scalar-field vacua $0$ and $2\pi$. It obeys ${\varphi }_K^{\prime }=2\mathrm{sech}(x-a)$ and has topological charge $Q=[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )=1$. In the Sine-Gordon theory normalization with physical mass scale $m$ and coupling $\beta$, its classical mass is $8m/{\beta }^2$. A Lorentz boost gives ${\varphi }_K=4\arctan \exp [(x-vt-a)/\sqrt{1-v^2}]$. Spatial reflection gives an antikink.

Consider an original ideal one-dimensional monatomic harmonic oscillator chain with neighbor separation $a$, mass $M$ and nearest-neighbor restoring coefficient $K>0$. Its displacement equation isSubstitute the normal mode $u_r=u_0{e}^{i(qra-\omega t)}$. Since $e^{iqa}+e^{-iqa}-2=-4{\sin }^2(qa/2)$,Choose $a=4$ ångströms and ${\epsilon }_{\ast }=10\mathrm{meV}$. This is one acoustic phonon branch, distinct from the two-atom elastic example. It has $\epsilon (\pi /(2a))=5\sqrt{2}\mathrm{meV}$ and $\epsilon (\pi /a)=10\mathrm{meV}$.

For coherent creation of one phonon in a crystal, the excitation's wave vector $q$ and the measured transfer obey $Q=G+q$, with $G$ a reciprocal lattice vector. To connect the dispersion relation to an illustrative intensity, take zero temperature, longitudinal displacements and the first interval $Q=q\in (0,\pi /a]$, choosing $G=0$. Expand the scattering phase to first order, $e^{iQ(ra+u_r)}\simeq e^{iQra}(1+iQ{u}_r)$. The constant term supplies elastic scattering; the linear term creates one phonon. Its displacement matrix element is proportional to $1/\sqrt{\omega }$, so the squared amplitude has weight $Q^2/\epsilon (Q)$, apart from common factors. Normalize this weight to one at $Q=\pi /a$:The limit $w(0)=0$ follows from $\sin (\pi x/2)\sim \pi x/2$. Set $\sigma =0.4\mathrm{meV}$ and replace the ideal sharp line by a normalized Gaussian function:This chosen broadening illustrates readout width; it is not a measured resolution or a full spectrometer resolution function. Integrating over all $\Delta E$ gives $w(x)$. The small negative-energy tail is a broadening artifact, not a zero-temperature phonon-annihilation signal.

These panels separate three kinds of information: allowed reciprocal lattice positions, cancellation within a crystal basis structure factor, and an excitation's energy as a function of wave vector. The chain has no magnetic degrees of freedom, optical branches, damping or multi-phonon processes. It illustrates how one excitation could be read; it proves nothing about the completeness of a real material's excitation spectrum.

For interaction ${\lambda }_{2k}{\varphi }^{2k}/(2k)!$, one vertex with two labelled external legs has $k-1$ independently paired tadpole diagrams. Counting contractions gives symmetry factor $S=2^{k-1}(k-1)!$. In a normalized Euclidean path integral, its amputated contribution is $-{\lambda }_{2k}{T}_D(m{)}^{k-1}/S$, where $T_D$ is the tadpole integral. It is linear in the coupling but has $k-1$ loops and a highest four-dimensional ultraviolet pole of order $k-1$ for nonzero mass.

First test rest. With no other tangential forces, the required uphill static friction is $f_s=mg\sin \theta$. Its maximum is ${\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction. ThusThe tangent function is increasing on this angle range, and its inverse tangent determines the last permitted resting angle ${\theta }_c$. At equality, the ideal model still permits limiting rest; a larger angle excludes it. Below the threshold, setting $f_s={\mu }_{s}N$ would incorrectly impose the maximum static friction even when less is required. Static friction on an inclined plane and inclined-plane measurement of static friction develop this onset criterion further.

For a block already sliding downhill, uphill kinetic friction has magnitude $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction. Taking downhill as positive, Newton's second law becomesEvery term in the force balance scales with mass; division by $m>0$ removes it. Doubling mass doubles both gravitational weight and the contact forces, while leaving this acceleration unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.

For an original numerical model, stipulate $m=3\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$, ${\mu }_s=0.45$ and ${\mu }_k=0.30$. The critical angle is ${\theta }_c\simeq 24.23^{\circ }$. At $20^{\circ }$, the normal force is $27.627\mathrm{N}$; rest requires $10.055\mathrm{N}$ of uphill static friction, below its $12.432\mathrm{N}$ bound. A resting block can stay there. If it is already sliding downhill, however, the kinetic friction is only $8.288\mathrm{N}$, giving downhill acceleration $a\simeq 0.5891m/{\mathrm{s}}^2$. Specifying the initial state matters.

At $30^{\circ }$, the downhill component of gravitational weight is $14.700\mathrm{N}$, while $N\simeq 25.461\mathrm{N}$ and the static friction bound is only $11.458\mathrm{N}$. Rest is impossible. During downhill sliding,Released from rest on a sufficiently long ramp, this ideal block covers its first $1.00\mathrm{m}$ in $t=\sqrt{2L/a}\simeq 0.9218\mathrm{s}$, reaching speed $v=\sqrt{2aL}\simeq 2.1697m/s$. As an independent energy calculation, the lost gravitational potential energy is $mgL\sin \theta =14.700\mathrm{J}$ and kinetic friction removes $f_{k}L\simeq 7.638\mathrm{J}$ of mechanical energy. The remaining $7.062\mathrm{J}$ equals $m{v}^2/2$ by the work-energy theorem.

At $\theta =\arctan {\mu }_k$, downhill sliding has zero acceleration and therefore constant velocity. Below that angle, a downhill-moving block slows; when it stops, test the static friction bound again. For uphill sliding, kinetic friction points downhill, so the signed downhill acceleration instead is $g(\sin \theta +{\mu }_k\cos \theta )$. The sign follows contact slip, and the downhill formula cannot be used unchanged after a reversal.

For a mass $m$ passing at relative speed $u$ and impact parameter $b$, integration of $Gmb/(b^2+u^2{t}^2{)}^{3/2}$ gives the transverse kick $2Gm/(bu)$. Weak deflection requires $b\gg Gm/u^2$, up to the equal-mass order-one factor. This scale supplies a small-impact cutoff for the weak-scattering diffusion approximation, not a physical exclusion of stronger encounters.

A uniform rigid body disk of mass $m$ and radius $a$ moves in a frictionless plane. Let $N\ge 3$ identical springs attach equally spaced rim points to fixed anchors in their radial directions. The equilibrium spring length is $l$, natural length $l_0<l$, and spring constant $k$. Write the small centre displacement as $\mathbf{q}$ and the rotation as $\theta$. For a spring with radial and tangential unit vectors $\mathbf{n},\mathbf{t}$, its length isThis follows by a second-order Taylor series for the distance between $\mathbf{q}+aR(\theta )\mathbf{n}$ and $(a+l)\mathbf{n}$. In the elastic potential energy $k(s-l_0{)}^2/2$, the equilibrium tension $k\tau$ multiplies the second-order change of length. Thus it contributes to the linearized stiffness even though the displacement is small.

The equally spaced directions satisfy $\sum \mathbf{n}=\sum \mathbf{t}=0$ and $\sum \mathbf{n}{\mathbf{n}}^T=\sum \mathbf{t}{\mathbf{t}}^T=(N/2)I$. Summing the spring energies givesThe kinetic energy is $m|\dot{\mathbf{q}}{|}^2/2+m{a}^2{\dot{\theta }}^2/4$, using the moment of inertia of a uniform disc. The Euler-Lagrange equations give the displayed squared angular frequencies. The translational normal modes are degenerate; rotation is independent. At zero prestretch the rotational frequency vanishes at quadratic order, and finite-angle spring stretching first contributes at higher order.

A sled intercepting snow with zero horizontal velocity in the ground frame is a simple momentum flux example. At arrival rate $\sigma$, immediate removal with outgoing ground-frame horizontal velocity $v_{\mathrm{out}}$ gives $M\dot{v}=-\sigma v_{\mathrm{out}}$ for constant moving mass $M$. Removal sideways in the sled frame has $v_{\mathrm{out}}=v$ and exponential deceleration; removal sideways in the ground frame has $v_{\mathrm{out}}=0$ and constant sled velocity. Retained snow instead gives $(M+\sigma t)v=M{v}_0$. The outgoing reference frame must be specified.

A real triplet with a rotationally invariant common mass term has three conserved internal currents. With the displayed sign convention, the current for $\delta \mathbf{\varphi }=\theta \mathbf{n}\times \mathbf{\varphi }$ is the negative of the conventional canonical current contracted with $\mathbf{n}$. Overall current sign does not affect conservation. Integrated charges require convergent integrals and vanishing net boundary flux; a local continuity equation alone does not guarantee conservation in an open region.

A solute is a component of a chemical solution distinguished from its solvent, typically present in a smaller proportion. There can be several solutes. A concentration must name both what is counted and the denominator: molar concentration uses amount of substance per final chemical solution volume; a solute-to-solvent mass ratio has another denominator.^[ref]

“Added sodium chloride” and “dissolved sodium cations” describe different levels of the same inventory. Dissociation of an ionic compound changes the grouping of the particles; it preserves their chemical elements. A dissolved ion remains an ion, and dissolved sucrose remains molecular unless a separate chemical reaction occurs. Thus the name of the weighed solute need not be the name of every dissolved chemical species.