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A position-dependent force is a conservative force when its work depends only on endpoints. Equivalently, its work around any closed allowed path vanishes, and one can define a potential energy $U$ satisfyingThe potential is defined up to an additive constant. A vanishing curl, $\nabla \times \mathbf{F}=0$, is a local test; it guarantees a global single-valued potential on a simply connected region with suitable regularity. Holes or excluded singularities require checking the global path condition.

The work-energy theorem gives $dK/dt=\mathbf{F}\cdot \mathbf{v}$. For a static potential this implies $d(K+U)/dt=0$. If $U$ depends explicitly on time, then instead $d(K+U)/dt=\partial U/\partial t$ under that force alone. Friction and other nonconservative forces add their own power and can convert mechanical energy into heat.

Examples include the gravitational potential energy $-GMm/r$ outside a spherical mass and the electrostatic energy $U=qV$, where $\mathbf{E}=-\nabla V$. For a point source charge $Q$, $U=qQ/(4\pi {ϵ}_{0}r)$, giving attraction or repulsion through the sign of $qQ$. A spring has $U=k{x}^2/2$ and force $-kx$. An equilibrium has $\nabla U=0$; a strict local minimum is stable against small displacements when the quadratic curvature is positive. Expanding about that minimum explains the harmonic restoring force used in many small-oscillation models.

Magnetic force on a moving point charge does no work, but it is velocity-dependent and is not generally a position-only force $-\nabla U$. Likewise a time-varying magnetic flux induces a circulating electric field with $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$, so that electric field cannot globally be represented solely by an electrostatic scalar potential. These distinctions prevent treating every field force as conservative simply because an energy description exists for the complete system.

In SI units, the microscopic Maxwell's equations for total charge density $\rho$ and current density $\mathbf{J}$ areThe first is Gauss's law: electric charge sources electric flux. The second is Gauss's law for magnetism, expressing the absence of magnetic monopole sources in this theory. Faraday's law relates changing magnetic flux to circulating electric field. The Ampere-Maxwell law relates magnetic circulation to both material current and changing electric flux.

For fixed loops and surfaces, the integral versions areSurface and loop orientations are related by the right-hand rule. These forms follow from the divergence theorem and Stokes' theorem. In material media, equivalent macroscopic equations can separate free and bound sources using $\mathbf{D}$ and $\mathbf{H}$ together with constitutive relations; the displayed forms use total sources.

The displacement-current term is essential for charge conservation. Taking the divergence of the last differential equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot \mathbf{J}=0$. It also makes magnetic circulation consistent for surfaces spanning the same loop through either a charging capacitor's wire or its dielectric gap.

In source-free vacuum the curl equations combine to give ${\nabla }^2\mathbf{E}={\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$, with the same equation for $\mathbf{B}$. ThusA plane wave is transverse, with $\mathbf{E}$, $\mathbf{B}$ and propagation direction mutually perpendicular and $E=cB$. The fields transport energy with density $u={ϵ}_0{E}^2/2+B^2/(2{\mu }_0)$ and Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$. Together with the Lorentz force law, the equations connect charges, currents, forces, induction and electromagnetic radiation.

A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

Use a stationary wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$ solving the time-independent Schrödinger equation. For energy above the positive step potential, the positive wavenumbers areIncidence only from the left permits a right-moving incident plane wave and a left-moving reflected plane wave for $x\le 0$, but only a right-moving transmitted plane wave for $x>0$:The omitted right-incident component would describe a different scattering experiment. Wavefunction matching at a finite potential step requires continuity of both $\psi$ and $d\psi /dx$ for a finite jump and the same mass on either side:Eliminate $B=C-A$ from the second condition:The first ratio is a transmission amplitude; a transmitted probability current also depends on its different propagation velocity.

For $E\gg V_0$, the two wavenumbers nearly coincide:Consequently $B/A\simeq V_0/(4E)\to 0$ and $C/A\to 1$. To leading order the complete wavefunction isThe small fractional change in kinetic energy scarcely changes the propagation wavenumber, so reflection vanishes and the particle behaves approximately as a free right-moving plane wave. A more accurate phase at large positive $x$ retains $k_{2}x$ on the transmitted side; even a small wavenumber difference accumulates over a sufficiently long distance.

For a downward step potential, $V_0=-|V_0|$, the region on the right is classically allowed at every positive incident energy. The same matching calculation applies withAs $E\to 0^{+}$, $k_1\to 0$ and $k_2\to k_{\ast }=\sqrt{2m|V_0|}/\hslash >0$. Henceand the flux coefficients satisfyThis quantum reflection at a downward step occurs even though a classical particle would simply accelerate into the lower-potential region. The abrupt change in wavenumber causes strong reflected-wave interference.

Both this limit and part (d) give total reflection, but their probability amplitudes and spatial densities differ. At the upward-step threshold, $B/A\to +1$ and $C/A\to 2$: an antinode meets the origin and the right-hand density stays finite while its propagation velocity vanishes. Here $B/A\to -1$ and $C/A\to 0$: the incident and reflected waves nearly cancel at the origin, and the transmitted density vanishes while the right-hand propagation velocity remains finite. In both descriptions the incident probability amplitude is held fixed for comparison.

The coefficients above are limiting flux ratios for $E>0$. At exactly $E=0$ the incident probability current vanishes, so directly dividing zero-energy fluxes would not define a scattering probability.

Experimental evidence. Tests of special relativity check linked predictions about light propagation, clocks and four-momentum. Several independent kinds of experiment support those predictions.

The observed partial-drag coefficient agrees with this first-order prediction rather than simple Galilean addition of $u$. It tests a different consequence of the Lorentz transformation from a particle-lifetime measurement.^[ref]

Atmospheric muons and the two frame descriptions. Atmospheric cosmic rays create unstable particles whose decay chains produce muons. A stopped-muon experiment measures the time between a stopping signal and the subsequent decay electron, obtaining the proper mean lifetime ${\tau }_0\simeq 2.2\mu \mathrm{s}$. The associated light-travel distance is only $c{\tau }_0\simeq 660\mathrm{m}$, yet substantial numbers of energetic atmospheric muons reach detectors after travelling several kilometres.

For an ideal selected cohort produced height $h$ above a detector and moving downward at constant speed $v$, the flight time in the terrestrial frame is $h/v$. Time dilation predicts a laboratory mean lifetime $\gamma {\tau }_0$, hence survival fractionWithout time dilation, the predicted exponent would be $-h/(v{\tau }_0)$ and the surviving population much smaller. Altitude comparisons constitute a cosmic-ray muon time-dilation experiment: select comparable momenta or energies and account for detector acceptance, atmospheric energy loss and backgrounds before interpreting the count ratio.

For illustration, not as a claimed measurement, take $h=10\mathrm{km}$ and $v=0.998c$. Then $\gamma =15.82$, the laboratory flight time is $33.4\mu \mathrm{s}$, and the predicted surviving fraction is about $0.38$. Omitting time dilation would give only about $2.5\times 10^{-7}$ on the same idealized constant-speed model.

In the muon rest frame, its decay still follows $e^{-\Delta \tau /{\tau }_0}$. The atmosphere and ground move upward. Their separation, measured simultaneously in that frame, is shortened by length contraction:The same survival fraction follows. The terrestrial description uses a longer-lived moving particle; the muon-frame description uses a shorter atmospheric flight path. They describe the same production and detection events. This consistency illustrates both time dilation and length contraction, while the survival measurement is not two independent experimental tests merely because it can be described in two frames.

Energy-momentum invariant. For a particle with rest mass $m$,Since $E=m{c}^2+K$, substitution and cancellation of $m^2{c}^4$ giveThis relativistic energy-momentum relation reduces to $K\simeq p^2/(2m)$ when $K\ll m{c}^2$.

Laboratory beam energies. If both final muons are at rest in the laboratory, their total momentum is zero and their total energy is $2m_{\mu }{c}^2$. Four-momentum conservation therefore requires equal and opposite incoming momenta. Because the electron and Positron have equal rest masses, they then have equal total energies:Subtract each particle's own rest energy:Using the usual handbook rest energies gives approximately $105.1\mathrm{MeV}$ for each beam. These are the ideal kinematic threshold values. Just above threshold, the pair has small, equal and opposite momenta in this laboratory frame.

Positron momentum. Apply the relativistic energy-momentum relation to $E_{+}=m_{\mu }{c}^2$ and use its stipulated negative-$x$ direction:The electron has the opposite momentum, $p_{e,x}=+p_0$ with $p_0=c\sqrt{m_{\mu }^2-m_e^2}$.

Boost to the electron rest frame. The electron's velocity is $v=p_{e,x}{c}^2/E_{-}=p_0/m_{\mu }>0$. Thus $S^{\prime }$ moves in the positive-$x$ direction, and its Lorentz factor satisfiesThe positron has the same laboratory total energy as the electron, so $v{E}_{+}/c^2=v{m}_{\mu }=p_0$. The supplied Lorentz transformation then givesAs a sign check, the same boost gives $p_{e,x}^{\prime }=\gamma (p_0-v{E}_{-}/c^2)=0$, as required for the electron rest frame.

For injected velocity $\mathbf{v}=v\hat{\mathbf{x}}$, the Lorentz force isFor nonzero electric charge and nonzero operating current, the velocity selector condition isThe requested nonrelativistic kinetic energy is consequentlyBoth electric current and electric charge cancel. Reversing the sign of $q$ reverses both component forces and preserves their cancellation. The dimensions are consistent because $R/{\mu }_0$ has units of speed, while $w/d$ is dimensionless. The assumed nonrelativistic particle regime requires $Rw/({\mu }_{0}d)\ll c$; the problem supplies no values with which to evaluate a numerical energy.

The relativistic kinetic energy is the total energy minus the rest energy:Here $\gamma$ is the Lorentz factor and $c$ the speed of light. The binomial series, with $b=v/c$, givesFor $v\ll c$, the leading term is the classical kinetic energy $M{v}^2/2$. The ratio of the fourth-order term to that leading term is $3v^2/(4c^2)$. ConsequentlyThe condition concerns one term of the binomial series; it is not the condition that the entire relativistic correction be exactly $1\%$.

Write $a=0.020\mathrm{m}$ for the side length. In an equilateral triangle, the distance from centre to a vertex is $a/\sqrt{3}$; it is two thirds of the altitude $\sqrt{3}a/2$. By the superposition principle, the electric potential at the centre due to the three fixed electrons isHere $e>0$ is the elementary charge and ${\epsilon }_0$ the vacuum permittivity. The approaching electron has electric charge $-e$, so its potential energy, measured relative to infinity, is positive:The fixed sources do not move, and radiation is excluded. Hence conservation of mechanical energy givesUsing $e=1.60218\times 10^{-19}\mathrm{C}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ for the electron mass, and $(4\pi {\epsilon }_0{)}^{-1}=8.98755\times 10^9\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}$ givesThis speed is small enough to justify classical kinetic energy. There is an idealization in the word “stops”: along the line normal to the triangle through its centre, this threshold energy makes the electron approach the central equilibrium asymptotically. A smooth, conservative trajectory cannot reach an equilibrium with exactly zero velocity in finite time unless it was already there. The energy threshold, and therefore the requested initial speed, is nevertheless unambiguous.

For an elastic eccentric impact of a thin disc, take upward velocity as positive and let $J$ be the short upward impulse. The disc's moment of inertia about a horizontal diameter is $I=M{a}^2/4$. With impact offset $d=a/2$,The contact point has upward velocity $v_C+d\omega$. Unit restitution against the fixed post requires that velocity to be $+u$, soSince $I=M{d}^2$, $J=Mu$ and the centre's instantaneous speed is zero. The disc rotates with $\omega =2u/a$; its rotational kinetic energy $I{\omega }^2/2=M{u}^2/2$ equals the entire initial energy. Gravity's impulse during the impact is neglected. Using the polar-axis moment of inertia would describe the wrong rotation axis.

The vacuum wave equation gives ${\omega }^2=c^2[k_g^2+(\pi /a{)}^2]$. The cutoff frequency is $f_c=c/(2a)=1.499\mathrm{GHz}$. In a lossless empty guide, energy travels at group velocity,The larger phase velocity $\omega /k_g=c^2/v_E$ is not the speed of energy transport.

Use energy units $m=m_p{c}^2$, $M=m_A{c}^2$, $e=E_{\gamma }$ and $P=pc=\sqrt{E^2-m^2}$. The soft-photon resonant capture threshold follows from four-momentum conservation:For fixed proton energy the left side is largest at head-on incidence. Its threshold is $E+P=2E_0$. Combining that with $(E-P)(E+P)=m^2$ givesEquality is allowed for a head-on photon. Here $e=k_{B}T=2.3482\times 10^{-4}\mathrm{eV}$, soA very soft photon needs an enormous proton Lorentz factor to reach the specified invariant mass. A true thermal photon distribution and a finite-width resonance would smear this monoenergetic threshold.

For $\theta =0$, $E-P=m^2/(E+P)\le m$, whereas $2E_0\gg m$, so production is impossible. For each nonzero angle, sufficiently large energy is possible, and $P\simeq E$ givesTo find the maximum deflection in soft-photon capture, note that the final particle's transverse momentum comes entirely from the photon:Writing $x=\cos \theta$, maximization gives $(1-x)(2x+1)=0$ and the interior maximum $x=-1/2$. Therefore, to the same extraordinarily accurate ultrarelativistic order,The neglected relative corrections are of order $m^2/E_0^2$ and $e/E_0$, both around $10^{-24}$. The final particle is effectively collinear with the proton despite a large photon incidence angle.

In the weak-field Zeeman effect, spin-orbit coupling keeps $L,S,J$ good and a level shifts by ${\mu }_B{g}_{J}B{m}_J$. At fields large compared with fine-structure coupling but still within the linear electronic Zeeman approximation, $L$ and $S$ decouple: the Paschen-Back effect gives shifts ${\mu }_{B}B(m_L+2m_S)$. Electric-dipole transitions obey $\Delta m_L=0,\pm 1$, $\Delta m_S=0$ in that limit, recovering an orbital triplet pattern. Intermediate fields mix the coupled states.

For the weak-field sodium D-line Zeeman pattern, the given Landé g-factor formula givesAllowed photon shifts are $\delta E={\mu }_{B}B(g_u{m}_u-2m_l)$ with $m_u-m_l=0,\pm 1$. ThusThese are the four- and six-component patterns. Both $\pi$ and $\sigma$ components are represented, so an observation strictly along the field, which removes $\pi$ emission, would not show this complete pattern. Hyperfine structure is neglected.

To determine the magnetic field from a Zeeman diagram, the unsplit separation provides its energy scale:The smallest adjacent split spacing $s_E=2{\mu }_B{B}_1/3$ is approximately $0.070$ of that separation in the sketch. ThereforeThis is a graphical estimate; the drawing does not warrant high precision. Its splitting energy is small compared with $\Delta E_D$, consistent with the weak-field approximation.

For Doppler resolution of a Zeeman pattern, thermal velocities give Gaussian Doppler broadening with widthAt $350\mathrm{K}$, the photon-energy FWHM near $589\mathrm{nm}$ is $5.88\times 10^{-6}\mathrm{eV}$. Equating the smallest adjacent Zeeman spacing to this width gives $B\sim 0.15\mathrm{T}$ for loss of individually resolved components. There is no sharp Gaussian resolution threshold; using whole-pattern width instead gives a smaller field of order $0.03$–$0.04\mathrm{T}$ for complete disappearance into a single envelope. Instrumental and collisional broadening would also affect the estimate.

For the quantum harmonic oscillator, using $[\hat{x},\hat{p}]=i\hslash$, direct multiplication givesThe commutators areThus each ladder operation shifts energy by one quantum. The ladder operator norms $\Vert a|n⟩{\Vert }^2=n$ and $\Vert a^{†}|n⟩{\Vert }^2=n+1$, with conventional phases, giveLet $\ell =\sqrt{\hslash /(m\omega )}$. In position space $\hat{p}=-i\hslash {\partial }_x$, so $a{\psi }_0=0$ becomes $(x+{\ell }^2{\partial }_x){\psi }_0=0$. Normalizing and then applying $a^{†}$ givesTheir overlap vanishes because ${\psi }_0^{\ast }{\psi }_1$ is odd on the full real line; the Gaussian second moment verifies $\int |{\psi }_1{|}^{2}dx=1$.

For photons emitted back to back in the measurement frame, their total invariant mass obeys $m^2{c}^4=(E_1+E_2{)}^2-(E_1-E_2{)}^2=4E_1{E}_2$. ThusThe parent's total laboratory energy is consequently about $160.2\mathrm{GeV}$, exceeding its rest energy. The measured-frame opposite-directions condition is essential: opposite directions only in the parent's rest frame would hold automatically and would not determine the second laboratory energy by itself.

In a photoelectric effect experiment, illuminate a clean metal photocathode with monochromatic light and use a retarding voltage to measure the maximum emitted-electron kinetic energy. The relation $e{V}_{\mathrm{stop}}=h\nu -\varphi$ has a frequency threshold and a slope $h$. Increasing intensity chiefly changes the number emitted, while increasing frequency increases the available energy per emission. This supports energy transfer in units $h\nu$, with $\varphi$ the work function. The retarding-potential measurement distinguishes the electron energy from the photocurrent; the two respond differently to changes of the light.

For equally nondegenerate levels, the partition function is $Z=e^{-\beta ϵ}(1+e^{-\beta \Delta })$. The mean energy per particle is $U=ϵ+\Delta /(1+e^{\beta \Delta })$, so the Schottky anomaly has heat capacityAt low temperature, $C\sim k_B(\Delta /k_{B}T{)}^2{e}^{-\Delta /(k_{B}T)}$ because excitation is suppressed. At high temperature, $C\sim {\Delta }^2/(4k_B{T}^2)$ because the two populations approach equality and the energy saturates. The peak is about $0.439k_B$ at $k_{B}T\simeq 0.417\Delta$. The common offset $ϵ$ does not affect heat capacity. Multiply by particle number for the total heat capacity, or divide by the corresponding mass for a mass-specific quantity.

The infinite square well levels for width $L$ are $E_n=n^2{h}^2/(8m_e{L}^2)$. Starting in the ground state, the smallest excitation is $1\to 2$, and its electric-dipole matrix element is nonzero. It therefore sets the longest absorption wavelength:The corresponding energy separation is about $12.4\mathrm{eV}$. Elastic scattering of light is a different process from the stipulated absorption into an excited well state.

The transverse-wave impedance of a string is $Z=\sqrt{T\mu }$ for common tension $T$. The density ratio four gives $Z_2/Z_1=2$. Continuity of displacement and transverse force gives transmitted amplitude ratio $t=2Z_1/(Z_1+Z_2)=2/3$. The power transmission coefficient includes the impedance ratio:The reflected fraction is $1/9$, checking energy conservation. Squaring the displacement amplitude alone would give the wrong transmitted power.