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A stationary mode shape does not imply absence of energy. Write $A=2a$ and $y=A\sin (kx)\cos (\omega t)$. For the ideal wave equation on a string, the kinetic energy and elastic potential energy per unit reference length add toHere $P$ is signed power crossing a position on the string, positive to the right; $e$ is an energy density per length, not per volume. Differentiation and $\mu y_{tt}=\tau y_{xx}$ give the local conservation of energy identity ${\partial }_{t}e+{\partial }_{x}P=0$. Substitution into $P$ givesIt can be nonzero away from a node of a standing wave or an antinode, but its time average over an oscillation period is zero. Energy flows locally back and forth even though this pure standing wave transports no net energy across a section over a complete cycle. At a node of a standing wave, $y_t=0$, so kinetic energy density is zero, but $y_x$ need not vanish: elastic potential energy density can be positive there.

For a fixed-end normal mode, integration over $0\le x\le L$ givesThe last equality uses $\tau k^2=\mu {\omega }^2$. At a straight-string snapshot, $U=0$ while $K=E$; at an extreme snapshot, $K=0$ while $U=E$. This is normal-mode energy exchange. A driven real string with damping requires continuing input power to replace lost energy, so a visually stationary pattern does not establish zero average supplied power.

A state function assigns a value to each equilibrium thermodynamic state of a specified thermodynamic system. Its change depends only on the initial and final states. Internal energy $U$, entropy $S$ and enthalpy $H=U+pV$ are state functions; here $p$ is pressure and $V$ is volume. Transferred heat $Q$ and work $W$ belong to a thermodynamic process, rather than to a state. A system stores internal energy; it does not store separate inventories of “heat received” and “work received”.^[ref]

Use the explicit convention $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work done by the thermodynamic system on its surroundings. For a closed system with unchanged endpoint bulk kinetic energy and external potential energy, the first law of thermodynamics readsThe notation $dU$ denotes an exact differential of a state function; $\delta Q$ and $\delta W_{\mathrm{by}}$ denote incremental transfers. Along a reversible thermodynamic process of a simple compressible system,where $T$ is absolute temperature. These are rules for particular thermodynamic processes, not definitions of stored heat or stored work. In particular, a quasistatic process with friction can remain near thermodynamic equilibrium while producing entropy.^[ref]

For any state function $F$ and two equilibrium states $A,B$, an integral of its exact differential givesThe proof is telescoping: add the changes $F(x_{i+1})-F(x_i)$ along a subdivided path, and all intermediate values cancel. The smooth limit gives the integral. Conversely, if an integrable differential form has zero integral around every accessible closed path in a connected region, define $F(B)$ by integrating from a reference state. Two routes differ by a closed path, so the definition is unambiguous. This argument needs every accessible loop, rather than a local test that might miss a hole in the domain; see path independence.

The same cancellation does not hold separately for heat and work. A thermodynamic cycle has $\Delta U=0$ but can have nonzero net transfers, with $\oint \delta Q=\oint \delta W_{\mathrm{by}}$. Nor does the fact that entropy is a state function make $Q/T$ its general change: the formula $dS=\delta Q_{\mathrm{rev}}/T$ uses a reversible thermodynamic process and its varying temperature.

This distinction makes endpoint calculations powerful. We can compute an internal energy, entropy or enthalpy difference along a convenient reversible reference path even when the actual thermodynamic process is irreversible. Its actual heat and work must still be obtained from its own boundary interactions. The appealing cancellation is visible in two reversible ideal-gas paths between the same states: a different mechanical output requires a matching difference in thermal input, while stored internal energy arrives at the same value.

Let $\kappa >0$ be an inverse length. In three spatial dimensions, seek the Green's function that decays at infinity and satisfieswhere ${\nabla }^2$ is the Laplacian and ${\delta }^{(3)}$ the three-dimensional Dirac delta function. The point source and the differential operator have spherical symmetry. For $r>0$, the radial Laplacian givesWriting $u(r)=r{G}_{\kappa }(r)$ reduces this to $u^{\prime \prime }-{\kappa }^{2}u=0$. Decay at infinity removes the growing exponential function, leaving $G_{\kappa }(r)=C{e}^{-\kappa r}/r$.

The Dirac delta function fixes the coefficient. Integrating over a ball $B_{ϵ}$ and using the divergence theorem givesThe first term tends to $4\pi C$, while the second is $O({ϵ}^2)$. Hence $C=1/(4\pi )$ andThis solves the source equation in the sense of distributions, including at the origin. Merely verifying the equation for $r>0$ would leave the normalization undetermined. The limit $\kappa \to 0$ at fixed $r>0$ is $1/(4\pi r)$, the Green function of the Laplacian for the convention $-{\nabla }^2$.

For an attractive scalar toy interaction, absorb constants into $A>0$ and setHere $A$ has units of energy times length, and negative $F_r$ points inward. At equal $A$, comparison with the unscreened model $V_0=-A/r$ givesThus the potential energy and force have the same exponential scale but different suppression factors. At $x=1$, these ratios are approximately $0.368$ and $0.736$; at $x=3$, they are $0.0498$ and $0.199$. Finite range means rapid decay, not an abrupt cutoff.

For steady flow through a fixed control volume with one inlet and one outlet, the first law of thermodynamics and conservation of mass giveHere $\dot{m}$ is the mass flow rate, $\dot{Q}$ incoming heat per unit time, ${\dot{W}}_s$ outgoing non-flow work per unit time, $w$ the flow speed, $z$ elevation and $g$ gravitational acceleration. The specific enthalpy $h=u+p/\rho$ includes both the transported internal energy per unit mass $u$ and the pressure work needed to push volume through the boundary; $\rho$ is mass density. The inlet supplies that pressure work, while the outlet consumes it to displace the downstream fluid.^[ref]

For an insulated stationary throttle, require negligible heat transfer, no shaft or other non-flow work, and negligible changes in bulk kinetic energy and gravitational potential energy. Then $h_2=h_1$. With unchanged composition this is also equality of molar enthalpy. The upstream and downstream states must admit equilibrium pressure, temperature and property values, although the constriction itself is irreversible and can be far from thermodynamic equilibrium. Start-up storage of energy in the walls and fluid, heat leakage, or a large outlet speed invalidates the reduced balance. A nozzle can convert specific enthalpy into bulk kinetic energy even when it is insulated.

The same accounting for a fixed transported amount is $U_2-U_1=p_1{V}_1-p_2{V}_2$, hence $U_1+p_1{V}_1=U_2+p_2{V}_2$. This explains why dropping the two pressure work terms would give the wrong conserved quantity.^[ref]

For constant mass $m$, a finite jump from potential energy zero at $x<0$ to $V_0>0$ at $x>0$ requires continuity of the wavefunction and its first derivative. With incidence only from the left and $E>V_0$, put $k=\sqrt{2mE}/\hslash$ and $q=\sqrt{2m(E-V_0)}/\hslash$. Matching $A+B=C$ and $k(A-B)=qC$ gives reflected probability amplitude $B/A=(k-q)/(k+q)$ and transmission amplitude $C/A=2k/(k+q)$. The transmission probability is $(q/k)|C/A{|}^2$, conserving probability current.

For $0<E<V_0$, replace $q$ by $i\kappa$, where $\kappa =\sqrt{2m(V_0-E)}/\hslash$ is the tunnelling decay constant. The right-hand wavefunction is $C{e}^{-\kappa x}$, with zero probability current. Its probability density decays on length $1/(2\kappa )$, while its amplitude decays on length $1/\kappa$. The semi-infinite raised region gives total reflection, unlike a finite potential barrier allowing quantum tunnelling into a second allowed region.

With boundary fluxes suppressed, let $B$ transfer kinetic energy to potential energy, $M$ transfer available potential energy irreversibly to background potential energy, and $W$ transfer internal energy to potential energy. Then $\dot{P}=B+W$, ${\dot{P}}_b=M+W$, $\dot{A}=B-M$, and $\dot{K}=F-B-ϵ$. The internal reservoir has $\dot{I}=ϵ-W$, conserving total energy apart from forcing $F$. The distinction between reversible buoyancy exchange and irreversible mixing is essential outside a quasi-steady negligible-$A$ regime.

Real atoms have multiple internal quantum states. Optical pumping and spatially varying electromagnetic polarization can remove kinetic energy by additional mechanisms, producing sub-Doppler cooling below the ideal Doppler cooling limit. In polarization-gradient cooling, state-dependent energy shifts from the Stark effect and delayed optical pumping make atoms repeatedly climb potential energy hills. This mechanism needs the internal-state structure omitted by the independent-beam two-level system above. ^[ref].

William D. Phillips's team tested unexpectedly cold sodium molasses with time-of-flight atom detection. A probe above the cloud initially saw no atoms; deliberately heating them verified that the detector worked. Moving the probe below the cloud produced arrival signals consistent with temperatures below the Doppler cooling limit. Independent release/recapture checks used sharply apertured beams and included gravity; uncertain cloud and probe geometry still limited the inferred temperature. These controls tested the thermometer as well as the cooling theory. ^[ref].

A linear deep-water gravity wave of surface-displacement amplitude $a$ has equal mean kinetic energy and potential energy, and total mean energy per horizontal area $\mathcal{E}={\rho }_{w}g{a}^2/2$. Its energy transport speed is the group velocity. This amplitude is half the crest-to-trough height.

Consider a symmetric top with fixed vertex $Q$, axial unit vector $\mathbf{e}$, principal moments of inertia $A,A,C$, and centre of mass at $h\mathbf{e}$. Let a force $\kappa \ell (\mathbf{k}-\mathbf{e})$ act at $\ell \mathbf{e}$, where $\mathbf{k}$ is the upward vertical unit vector. Together with gravity, it produces torque $(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ and potential energy $mgh\mathbf{k}\cdot \mathbf{e}+\kappa {\ell }^2(1-\mathbf{k}\cdot \mathbf{e})$.

The axial angular velocity of a symmetric top $n$, vertical angular momentum $\mathbf{k}\cdot \mathbf{L}$ and total mechanical energy are conserved. For steady precession, $\dot{\mathbf{e}}=\Omega \mathbf{k}\times \mathbf{e}$ and $\mathbf{k}\cdot \mathbf{e}=\cos \alpha$. Substituting $\mathbf{e}\times \ddot{\mathbf{e}}={\Omega }^2\cos \alpha \mathbf{e}\times \mathbf{k}$ into $A\mathbf{e}\times \ddot{\mathbf{e}}+Cn\dot{\mathbf{e}}=(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ gives the condition in the title for nonvertical inclination. If the total angular velocity itself is $\Omega \mathbf{k}$, then $n=\Omega \cos \alpha$, and the condition becomes $\kappa {\ell }^2=mgh+(A-C){\Omega }^2\cos \alpha$.

A bead in a liquid keeps undergoing Brownian motion even when an optical trap holds its average position fixed. For a stationary harmonic optical trap in thermal equilibrium at absolute temperature $T$, the equipartition theorem assigns mean potential energy $k_{B}T/2$ to this one quadratic coordinate. With $k_B$ the Boltzmann constant and $x=0$ at equilibrium,At the same temperature, quadrupling the spring constant divides the position variance by four and the standard deviation by two. The Boltzmann factor gives the complete equilibrium probability density function, a Gaussian distribution:The prefactor makes ${\int }_{-\infty }^{\infty }p(x)dx=1$; its second moment is ${\int }_{-\infty }^{\infty }{x}^{2}p(x)dx=k_{B}T/k$. The probability density function has units of inverse length, and its area gives probability. This density is exact for the ideal quadratic model. In a finite optical trap it approximates the occupied well when thermal excursions remain within the harmonic region and escape is negligible during observation. Driven motion, appreciable nonconservative forces or laser heating need not give this equilibrium density.

Here $\ell =\sqrt{k_{B}T/k_0}$, $q=x/\ell$ and $\kappa =k/k_0$. The plotted density per unit $q$ is $p_q(q)=\ell p(\ell q)=\sqrt{\kappa /(2\pi )}{e}^{-\kappa q^2/2}$. Each curve has total area one; their dimensionless standard deviations are $1$, $1/\sqrt{2}$ and $1/2$. The curves are exact model densities, not noisy experimental data.

Brownian motion becomes a measuring tool: a calibrated position measurement supplies ${\sigma }^2=⟨(x-⟨x⟩{)}^2⟩$, and the known temperature supplies $k=k_{B}T/{\sigma }^2$. Once $k$ is known, a slowly applied constant external force $f$ shifts the mean by $⟨x⟩=f/k$, so $f=k⟨x⟩$ in the same harmonic range. Fluctuations therefore calibrate the conversion from displacement to force. Svoboda and Block used this equipartition theorem method when calibrating their bead traps. ^[ref].

For the example stiffness $0.01pN/nm$ and $T=300\mathrm{K}$, $k_{B}T\approx 4.142\mathrm{pN}\mathrm{nm}$ and $\sigma \approx 20.35\mathrm{nm}$. Using a variance measured in uncalibrated detector units would not determine that physical stiffness. Measurement noise, finite exposure time, drift and insufficient sampling of the correlated Brownian motion can bias the estimate; record positions over many optical trap relaxation times and account for the detector response. The deterministic relaxation curve above describes relaxation without thermal forcing; an individual thermal trajectory continues to fluctuate.

For a bond free of applied tension, use a local potential energy $V(a+x)=V(a)+\kappa x^2/2+\gamma x^3/6+\cdots$ with $\kappa >0$. Classical equipartition gives $⟨x^2{⟩}_0=k_{B}T/\kappa$; vanishing mean force then yields $⟨x⟩=-\gamma k_{B}T/(2{\kappa }^2)$ to first order in anharmonicity. Thus the linear thermal expansion coefficient is $\alpha =-\gamma k_B/(2a{\kappa }^2)$. For a Morse potential, $\gamma =-3\kappa /\xi$, giving positive expansion $3k_B/(2a\kappa \xi )$. This is a local bound-state approximation; the isolated cubic truncation is not a globally stable equilibrium potential.

At an equilibrium ${\mathbf{r}}_{\ast }$, the total force vanishes. A small transverse displacement must produce a transverse restoring force; a small axial displacement must also restore the bead. Downstream of a tight focus, the backward optical gradient force can oppose the forward optical scattering force. Equality at one point is insufficient: the total axial force must decrease as the bead moves downstream through that point. The equilibrium can therefore lie downstream of the intensity maximum.

For uncoupled overdamped coordinates, ${\zeta }_i\dot{\delta r_i}=-k_i\delta r_i$ with ${\zeta }_i>0$ decays only when each $k_i>0$. More generally, linearizing gives $\dot{\delta \mathbf{r}}=MJ\delta \mathbf{r}$, where $M$ is the hydrodynamic mobility matrix and $J_{ij}=\partial F_i/\partial r_j$ at equilibrium. All eigenvalues of $MJ$ must have negative real parts for linear asymptotic stability. For a conservative force with symmetric positive-definite mobility, a positive-definite Hessian matrix of the potential energy supplies this condition. A general optical force need not possess a global potential energy.

Increasing numerical aperture admits more strongly converging light rays and a steeper axial variation of the field. But a larger nominal numerical aperture alone does not establish trapping: illumination of the objective pupil, optical aberration, particle properties and the actual in-liquid focus also matter. A harmonic optical trap describes local confinement after this stability question has been answered; it does not predict capture from arbitrary positions or an infinite holding force.

At fixed positive bias ratio $s=I_b/I_c$ with $0<s<1$, the Josephson coupling energy and work of the ideal bias source combine intoThis is the effective potential energy for an unwrapped Gauge-invariant Josephson phase; it includes the bias source and decreases by $2\pi s{E}_J$ on advancing one period. A local minimum and the adjacent downhill barrier areTo verify the barrier, solve $U^{\prime }=E_J(\sin \varphi -s)=0$, select $U^{\prime \prime }({\varphi }_m)>0$ and $U^{\prime \prime }({\varphi }_b)<0$, and subtract their potential energies. The small-oscillation angular frequency is fixed by the curvature:Thus the barrier and oscillation frequency both vanish as $s\to 1^{-}$, but with different powers. For $ϵ=1-s\ll 1$, expanding $\arccos (1-ϵ)$ and $\sqrt{2ϵ-{ϵ}^2}$ givesThe first approximation is for a nearly critical bias. At $s=0$ the exact downhill barrier is $2E_J$; at $s\ge 1$ there is no confining local minimum in this model.

For those stipulated parameters, direct evaluation gives $E_J/h=496.684\mathrm{GHz}$, $\Delta U/h=29.7670\mathrm{GHz}$, and ${\omega }_p/(2\pi )=2.89608\mathrm{GHz}$. Hence $\Delta U/(\hslash {\omega }_p)=10.2784$. The nearly critical barrier approximation gives $29.6165\mathrm{GHz}$, about $0.51\%$ below the exact result. These are original circuit calculations, not parameters assigned to the historical samples.

For a time-independent Hamiltonian operator $H$, an energy eigenstate satisfiesFor a one-dimensional particle of mass $m$ in a real potential energy $U(x)$,Its full time-dependent wavefunction is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$: substitution in the Schrödinger equation checks both the minus sign and the factor $\hslash$. Thus an energy eigenstate has stationary position probability density, even though its quantum phase evolves. The boundary conditions and operator domain are part of this eigenvalue problem, selecting physically allowed solutions; the infinite square well and Radial Schrodinger equation for the hydrogen atom give complementary confined examples.

For a complete discrete orthonormal basis of energy eigenstates, linearity gives $\Psi (x,t)={\sum }_n{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$, with ${\sum }_n|c_n{|}^2=1$. Cross terms in $|\Psi {|}^2$ oscillate at angular frequencies $(E_n-E_m)/\hslash$. A continuous spectrum requires the corresponding spectral integrals; for a free particle on the whole line, individual plane waves are generalized eigenfunctions, and a normalized Gaussian wave packet is their integral quantum superposition. Adding a spatially constant real $C$ to $U$ shifts every energy eigenvalue by $C$ and multiplies the full state by $e^{-iCt/\hslash }$, a global phase; energy differences, position probabilities and probability current are unchanged.

An ideal linear torsion spring exerts a restoring torque proportional to the relative angle between its two ends. Integrating the torque gives its elastic potential energy. A spring connecting two rotating bodies contributes $\kappa ({\theta }_2-{\theta }_1{)}^2/2$ to the potential energy; a spring connected to a fixed frame is obtained by setting one angle to zero. The coefficient $\kappa$ is the torsional spring constant.

Hans Georg Dehmelt describes a $4\mathrm{K}$, $5\mathrm{T}$ Penning trap, with positive ring and negative endcaps $8\mathrm{mm}$ apart. Axial RF emission tracked an injected seven-electron bunch losing particles; reducing the drive retained the last electron. A weak magnetic bottle converted spin changes into axial-frequency changes. At successive probe frequencies, counted spin flips traced a resonance.^[ref]

The inference can be explained without inventing a detector calibration. Let a weak bottle have on-axis field $B_z\simeq B_0+b{z}^2$. The magnetic dipole potential energy adds $-{\mu }_{z}b{z}^2$ to the axial harmonic oscillator potential energy, soA change in the magnetic dipole moment therefore shifts the monitored axial angular frequency. This is a simplified readout model, not a quantitative reconstruction of the historical bottle. It illustrates how a transition in one degree of freedom becomes a detectable change in another.

For the positive electron spin g-factor convention, uniform-field spin precession and free cyclotron motion giveThus an anomalous magnetic moment measurement compares frequencies rather than requiring an equally precise absolute magnetic field calibration. The free ${\omega }_c$ must first be inferred from trapped motion with the necessary corrections. A spin-flip count versus probe frequency supplies evidence for a transition resonance; it does not directly measure a spatial electron radius. Nor is this radio readout an optical fluorescence record: different experiments require different preparations and detectors. In a precision comparison, repeat scans with changed drive strength and monitor magnetic field drift, motional excitation and detector back-action; these are experimental-design checks, not additional historical observations asserted here.

A triangular localized potential energy has forces $+F$ behind and $-F$ ahead of the trap centre. With $v_c=F/\zeta =L{f}_c$, $\alpha =2X_0/L$ and $\beta =f_T/f_c$, its circular response is $f_p/f_c=\alpha \beta /({\beta }^2-1+\alpha )$ for $\beta >1$. A trapped particle has $f_p/f_c=\beta$ for $0\le \beta \le 1$, interpreting the discontinuous force at the cusp by a sticking or rounded-well limit.

Consider one chosen model at a fixed total energy, with two open scattering channels, no absorption and no omitted outgoing modes. Channel 1 leaves the target unchanged; channel 2 excites it. Let $v_1=v>0$ and $v_2=v/2$. For the same projectile mass, conservation of energy then requires a target excitation $\Delta =3m{v}^2/8$. These are hypothetical parameters, not a fit to a particular atom. Each incoming or outgoing mode is normalized to unit probability current; the outgoing modes will have the angular profiles specified below.

Choose the S-matrixThis phenomenological unitary matrix specifies the channel conversion; it is not derived here from an interaction potential energy. For incidence in channel 1, its first column gives the outgoing flux amplitudesTo recover ordinary wave coefficients relative to an incident coefficient one, undo the velocity normalization:Thus $|c_2{|}^2=32/25>1$ is allowed: channel 2 travels more slowly. Its outgoing probability current fraction is $16/25$, not $32/25$. Renormalizing the unweighted squares would give $9/41$ and $32/41$, the wrong probabilities. More generally the ordinary coefficient map is $A=V^{-1/2}S{V}^{1/2}$, with $V=\mathrm{diag}(v,v/2)$; it obeys $A^{†}VA=V$. The conserved quantity is the squared coefficient norm weighted by velocity, which represents probability current.

Now specify an original angular realization of these two outgoing modes, with scattering angle $\theta$ and no azimuth dependence:The first normalization follows from $2\pi {\int }_{-1}^1(1+u^2)du=16\pi /3$, where $u=\cos \theta$. These angular profiles define the modes of this model; they are not asserted to follow from Born's approximation or an atomic potential energy. The joint directional probability densities are

Suppose a detector accepts the cone $D:0\le \theta \le \pi /3$, with full azimuth and unit efficiency. Its angular fractions within the channels areMultiplying the angular fractions by the outgoing flux weights givesFor $1600$ independent incident events, the expected values are therefore $171$ and $256$ accepted detections, compared with $576$ and $1024$ over all directions. The remaining expected $1173$ events fall outside the cone. The chosen incident number makes the expectations integers; actual counts fluctuate. Conditional on a detection inside the cone, the channel probabilities are $171/427$ and $256/427$, rather than $9/25$ and $16/25$. Angular acceptance changes the observed channel proportions even with equal detector efficiencies.

The final atomic states are orthogonal, so ignoring the atomic readout adds the two channel intensities; it does not produce $|b_1{g}_1+b_2{g}_2{|}^2$. Coherent paths that reach the same final state can instead interfere, as the existing probability amplitude and quantum superposition examples demonstrate. The distinction is physical distinguishability of final states, not whether the detector operator chooses to display their labels. Flux normalization and angular integration make the scattering interpretation useful: they turn complex outgoing waves into conserved event weights and then into the statistics of a specified detector.

For motion confined to a plane but retaining potential energy $-\kappa /r$, separation in polar coordinates gives angular wavefunctions $e^{im\varphi }$ with integer $m$. The regular radial factor behaves as $r^{|m|}$ at the origin. Removing this power and the decaying exponential function leaves an Generalized Laguerre polynomial differential equation. polynomial termination requires $N=n_r+|m|+1/2$, yielding the displayed energy. At fixed $q=n_r+|m|$, the spinless quantum degeneracy is $2q+1$. This is not the logarithmic potential of electrostatics formulated entirely in two spatial dimensions.

Two equal uniform rods of length $2l$ and mass $M$, one pivoted to the floor and the other endpoint sliding on it, have arch coordinates $B=(2l\cos \theta ,2l\sin \theta )$ and $C=(4l\cos \theta ,0)$. The kinetic energy decomposition about the center of mass gives $T=\frac{4}{3}M{l}^2(1+3{\sin }^2\theta ){\dot{\theta }}^2$, while gravitational potential energy is $2Mgl\sin \theta$. Conservation of energy from rest at angle $\alpha$ gives the displayed equation and hinge impact speed $\sqrt{6gl\sin \alpha }$, provided contact is maintained. The endpoint's normal reaction becomes negative during this constrained fall if $\sin \alpha >35/54$; a unilateral floor then permits detachment and the constrained formula no longer describes the entire motion.