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For overdamped Brownian motion in potential $\kappa x^2/2$, the relaxation time is $\tau =\zeta /\kappa$ and the equilibrium correlation is $(k_{B}T/\kappa )e^{-|t|/\tau }$. Expanding the squared displacement between two equilibrium samples gives the title expression. Initializing the particle exactly at the centre instead gives variance $(k_{B}T/\kappa )(1-e^{-2t/\tau })$. Both are valid but concern different initial ensembles; their saturation values differ by a factor of two. These formulas neglect inertia and require times long compared with momentum relaxation.

Coherent absorption from one applied beam and stimulated emission into another, transferring an atom between two long-lived states. The energy transfer uses the frequency difference and the momentum transfer the wavevector difference. It is distinct from an absorption-absorption process.

Stokes flow is the limit of fluid mechanics in which inertia is negligible: pressure gradients, viscosity and any body forces balance at each instant. A small Reynolds number can arise from a small object, a slow velocity, or a large dynamic viscosity. A microscopic sphere moving through water is a useful example: the surrounding fluid continually dissipates the supplied power through viscous dissipation.

The picture is a meridional slice of the three-dimensional axisymmetric flow around a sphere, with the sphere held fixed and the distant fluid moving rightward at speed $U$. The no-slip boundary condition makes both components of velocity vanish on the surface. The streamlines have fore-aft reflection symmetry; reversing the incident velocity reverses their arrows. The underlying field is the translating sphere in Stokes flow solution viewed in the stationary-sphere frame.

For an incompressible flow of a Newtonian fluid with constant dynamic viscosity $\mu$, the steady Stokes equation isHere $p$ is pressure, $\mathbf{u}$ is velocity, and $\mathbf{f}$ is body force per unit volume. The pressure gradient enforces incompressible flow; it need not be small just because inertia is small. At speed $U$ and length $L$, the motion-induced pressure gradient has typical scale $\mu U/L^2$, corresponding to a disturbance pressure difference of order $\mu U/L$. See Shreyas Mandre's fluid dynamics notes, sections 9.1.1–9.1.2.

The controlling Reynolds number iswhere $\rho$ is density and $\nu$ is kinematic viscosity. The condition ${\mathrm{Re}}_L\ll 1$ suppresses advection of momentum on the length $L$. Rapid changes require an additional check of the viscous diffusion time: small Reynolds number alone does not eliminate the time derivative in the Unsteady Stokes equation.

Two rewards of this limit are the exact Stokes's law for the drag on a sphere and kinematic reversibility of Stokes flow. The first predicts settling velocities and connects to the Stokes–Einstein relation; the second explains why a microswimmer needs a nonreciprocal stroke, as expressed by the scallop theorem. Despite viscous dissipation, reversing the driving can retrace the motion: an especially striking consequence of Linearity of Stokes flow.

George Gabriel Stokes obtained the sphere drag and its terminal velocity in his 1851 investigation of how viscosity changes physical pendulum motion. He also used the slow fall of small water droplets to discuss clouds. The paper was read on 9 December 1850; 1851 is its publication date. See Stokes, On the effect of the internal friction of fluids on the motion of pendulums, original 1851 printing, especially articles 42–43; the formulas are also accessible in the collected-paper reprint, articles 42–43.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

A circular particle accelerator with bending fields and accelerating fields varied in coordination with the increasing beam momentum.

For a nearly massless incident tau neutrino and stationary target of mass $M_T$, the initial Mandelstam variables give $s=M_T^2+2M_T{E}_{\nu }$. At threshold the final tau lepton and the lightest allowed recoil system have no relative momentum, so $s=(m_{\tau }+M_X{)}^2$. This proves the formula. A free neutron recoil channel gives approximately $3.45\mathrm{GeV}$; an electron target has a different recoil and a threshold near $3.09\mathrm{TeV}$.

In the modern weak charged current interpretation, one allowed assignment isThe electron antineutrino, tau neutrino, muon neutrino and antineutrino of the tau neutrino carry away energy and momentum. There are four undetected particles in this decay assignment. Interchanging which parent decays electronically gives the opposite electric charge assignment. Each parent conserves electric charge and, in this approximation, total lepton number; four-momentum conservation applies separately to production and to each particle decay.

If $B_e$ and $B_{\mu }$ denote the corresponding branching fractions, independent parent decays give an inclusive mixed-flavour fraction $2B_e{B}_{\mu }$. The factor two counts the two assignments to opposite parent charges. For chosen $B_e=B_{\mu }=1/5$, the mixed fraction is $2/25$, before acceptance and detection efficiency. Correlations between the parents' spins can affect angular selection even when the fully inclusive branching fractions factorize.

The neutrinos in that equation are a physical interpretation to test. Neither a deficit in visible energy nor two unusual tracks alone establishes their identities or uniquely counts them. Even a compatible invariant mass and an angular distribution require particle identification, background checks and agreement across further particle decay channels. The direct detection milestone for the tau neutrino was in 2000, separate from the charged tau lepton discovery.

Choose $m_{\tau }=2\mathrm{GeV}$ and $\sqrt{s}=6\mathrm{GeV}$ purely for illustration. Then each parent has $E_{\tau }=3\mathrm{GeV}$, $|{\mathbf{p}}_{\tau }|=\sqrt{5}\mathrm{GeV}$ and $\beta =\sqrt{5}/3$. Treat all six daughters as massless, including the charged electron and antimuon, and generate each three-body particle decay with constant squared decay amplitude. Choose the parent direction isotropically and the two decays independently; this omits the physical production angular law and spin correlations.

The Lorentz-invariant phase space factorizes through $t=q^2$ into a two-body decay to $\ell +q$, followed by a two-body decay of $q$:After angular integration the first factor contributes $1-t/m_{\tau }^2=x$, while the massless second factor is constant. Since $|dt|=m_{\tau }^{2}dx$, the normalized chosen density is $f(x)=2x$. Drawing $x=\sqrt{U}$ with $U$ uniform on $[0,1]$ implements this density. Isotropic directions, followed by a Lorentz boost for the invisible pair and another for the parent, conserve the complete four-momentum of each generated event.

Define transverse acoplanarity by $\alpha =\pi -\Delta \varphi$, where $\Delta \varphi \in [0,\pi ]$ is the angle between the two visible momentum projections perpendicular to the beam. A fully visible two-body event with zero initial transverse momentum has $\alpha =0$ and $M_{\mathrm{miss}}^2=0$. The toy tau lepton events spread away from that point. These observables are complementary: an angular imbalance asks about directions, while the missing invariant mass also uses the visible energies and longitudinal momenta.

A Lorentz-invariant phase space density supplies allowed configurations and their geometric measure. A physical weak interaction decay amplitude reweights them and correlates spins. Consequently the plotted density is a lesson about conservation and available configurations, not a predicted experimental rate or a fit to the discovery sample. The massless charged-daughter approximation is also unsuitable near the muon rest-energy threshold.

For a degenerate isotropic parabolic electron band, electrical conductivity is $\sigma =n{e}^2{\tau }_{\sigma }/m$ and electronic thermal conductivity is ${\kappa }_e={\pi }^{2}n{k}_B^{2}T{\tau }_{\kappa }/(3m)$. Their ratio is ${\kappa }_e/(\sigma T)=L_0{\tau }_{\kappa }/{\tau }_{\sigma }$, where $L_0=({\pi }^2/3)(k_B/e{)}^2$ is the Lorenz number. Nearly elastic scattering that relaxes charge and heat currents similarly gives ${\tau }_{\kappa }\simeq {\tau }_{\sigma }$ and the Wiedemann-Franz law. Inelastic scattering can relax heat current and directed momentum differently, invalidating the simplest identification even in a good metal.

Euclidean finite temperature compactifies time to a circle of radius $R_{\beta }=\beta /(2\pi )$. The string oscillator ground states with zero momentum number and winding $w=\pm 1$ have the displayed mass from the bosonic string mass spectrum. They become massless at ${\beta }_H=4\pi \sqrt{{\alpha }^{\prime }}$ and tachyonic at higher temperature. This thermal winding mode diagnoses breakdown of the perturbative low-temperature string gas at the Hagedorn temperature. The endpoint and order of a transition require interactions and a suitable stable vacuum; the ordinary zero-winding bosonic tachyon is already unstable independently of this thermal threshold.

At fixed gradient and magnet length, the signed thin-lens focal length is proportional to particle momentum. For total energy error $\delta =(E-E_0)/E_0$, differentiating $p^2{c}^2=E^2-m^2{c}^4$ gives $d\ln p/d\ln E=1/{\beta }^2$. The ultrarelativistic result is $f=f_0(1+\delta )$ to first order. If energy means kinetic energy instead, the coefficient is $(K+m{c}^2)/(K+2m{c}^2)$.

Let a strong shock wave of radius $R$ and speed $V$ sweep a uniform mass density ${\rho }_0$ into a shell of mass $4\pi {\rho }_0{R}^3/3$ and speed $2V/(\gamma +1)$. If the cavity pressure is $\alpha$ times the immediate post-shock pressure, shell momentum balance gives $R\dot{V}=3(\alpha -1)V^2$. Constant total energy gives $R^3{V}^2=\mathrm{constant}$ and hence $\alpha =1/2$. Including shell internal and kinetic energy, and treating the cavity volume as $4\pi R^3/3$ to thin-shell accuracy, yields $R=\xi (E/{\rho }_0{)}^{1/5}{t}^{2/5}$ with ${\xi }^5=75(\gamma -1)(\gamma +1{)}^2/[16\pi (5\gamma -3)]$. This coefficient is approximate; resolving the shell's finite volume changes it.

A third-law interaction pair consists of the two forces belonging to the same mutual interaction, with their agent and recipient exchanged. The names action-reaction pair do not imply a delay: in the instantaneous model, the two forces occur together. Test a proposed pair by naming both bodies and the interaction, rather than merely noticing equal magnitudes and opposite directions.

For a book resting on a table, the table's upward normal force and Earth's downward gravitational weight both act on the book. They can balance in the book's free-body diagram, but they are not a third-law interaction pair. The partner of the table's normal force on the book is the book's downward contact force on the table. The partner of Earth's gravitational force on the book is the book's gravitational force on Earth. Their equality does not require equal accelerations: Earth's much greater mass makes its response correspondingly smaller for a given net force.

Similarly, friction on a sliding block from a surface is paired with friction on that surface from the block. For the collection containing only the block, that friction is an external force; including the contacting surface makes both members internal forces. Changing the collection changes the bookkeeping of momentum, not the physical interaction.

The Thomson charge-to-mass experiment infers the magnitude of an electron's charge-to-mass ratio by combining electric and magnetic deflection of cathode rays. The central idea is to determine speed separately from magnetic curvature: a known speed turns the observed gyroradius into a measurement of $e/m$.

In 1897, J. J. Thomson used two grounded slits in an evacuated glass tube, then aluminium plates $5\mathrm{cm}$ long, $2\mathrm{cm}$ wide and $1.5\mathrm{cm}$ apart. A battery supplied their electric potential difference. Deflected cathode rays produced a phosphorescence patch on the glass; an external scale and movable luminous pointer recorded its position in darkness. Greater evacuation reduced gas electrical conductivity that screened the electric field; reversing plate polarity reversed deflection. External coils supplied the magnetic field, calibrated by induction against larger coils. He adjusted equal electric and magnetic deflections. ^[ref]

His comparison included air, hydrogen, carbon dioxide, and aluminium versus platinum cathodes, supporting a common carrier. He explicitly noted uncorrected magnetic bending outside the plates. His tables give $m/e$, the reciprocal of the magnitude used here. The inferred ratio alone could not separate electric charge from mass. ^[ref]

To connect this readout with the Lorentz force, derive a simple nonrelativistic model in modern SI notation. Let the uniform deflecting region have length $\ell$, followed by a field-free drift of length $L$. Assume a narrow speed distribution, small deflection angles, negligible collisions, and matching electric and magnetic field profiles. For an electron with speed $v$, the electric acceleration has magnitude $eE/m$, the transit time is approximately $\ell /v$, and the subsequent drift adds exit transverse velocity times $L/v$. Thus the displacement magnitudes measured separately with only electric field or only magnetic field areThe electric expression follows from constant transverse acceleration; the magnetic expression uses its small-deflection approximation. Equal displacement magnitudes then giveThis is an ideal derivation of the comparison principle, not a numerical reanalysis of the historical tube. Different exterior magnetic field profiles invalidate cancellation of the common geometrical factor. The comparison determines $e/m$ through deflection and the known apparatus dimensions; it does not require a downstream semicircle.

For a second, deliberately different teaching geometry, combine a velocity selector with a separate region containing only a uniform magnetic field $B_a$ perpendicular to the selected velocity. Here $B_s$ belongs to the velocity selector, while $B_a$ belongs to the analyzer: they need not be equal. The magnetic Lorentz force supplies the centripetal acceleration, soIf the analyzer turns the beam through a semicircle before it hits a screen in the entry plane, the separation $D$ between entrance and impact is the diameter: $r=D/2$. For a shorter observed arc, the screen displacement is not generally a radius; fit the appropriate circle geometry. This two-stage velocity selector and curvature apparatus is an original ideal teaching model, not a claim that Thomson's Fig. 2 used a separate selector and semicircular analyzer.

For a worked example, stipulate $E=2.00\times 10^3\mathrm{V}{\mathrm{m}}^{-1}$, $B_s=1.00\times 10^{-3}\mathrm{T}$, $B_a=2.00\times 10^{-4}\mathrm{T}$, and $D=0.1136\mathrm{m}$. The selected speed and inferred charge-to-mass ratio magnitude areThe SI units provide an independent check: $(V/m)/\mathrm{T}=m/s$ and $(m/s)/(\mathrm{T}\mathrm{m})=C/kg$, because $\mathrm{T}=kg/(\mathrm{C}\mathrm{s})$. The charge-to-mass ratio of the electron is negative; this calculation gives its magnitude. The bending direction gives the sign of the electric charge once the magnetic field direction is known. Scaling both $e$ and $m$ by the same positive factor preserves every displayed trajectory, so an independent electric charge measurement is required to determine mass separately.

The curvature formula assumes nonrelativistic momentum $p=mv$. At higher speeds, the relativistic cyclotron orbit gives $p=\gamma mv=e{B}_{a}r$, hence $e/m=\gamma v/(B_{a}r)$, where $\gamma$ is the Lorentz factor; the ideal straight-through selection $v=E/B_s$ still follows from vanishing Lorentz force. The stipulated speed is about $0.0067$ times the speed of light, making $\gamma -1$ about $2.2\times 10^{-5}$. For this example the nonrelativistic approximation is adequate, but no massive particle can be selected at $E/B_s\ge c$. Finite slit widths, collisions with residual gas, uncertain magnetic field calibration, and fields outside the ideal boundaries limit a real inference; repeat at several selected speeds and analyzer strengths to test consistency of $e/m$ rather than trusting one spot position.

For three massive daughters with a smooth nonzero decay amplitude at the decay threshold, the integrated Lorentz-invariant phase space vanishes as $Q^2$, where $Q$ is the parent's mass minus the sum of daughter masses. The spatial Dirac delta function leaves six independent momentum coordinates. Rescaling each by $Q^{1/2}$ contributes $Q^3$, while the conservation of energy Dirac delta function contributes $Q^{-1}$. The factors $1/(2E)$ are smooth nonzero constants at the decay threshold. Thus the decay width also scales as $Q^2$ when the decay amplitude has a nonzero constant term. A vanishing decay amplitude, a massless daughter, or a threshold singularity can change this law. In three spatial dimensions the same argument gives $Q^{(3N-5)/2}$ for $N$ massive daughters.

For $H=p^2/(2m)+V(q)$, time slices of length $\delta$ use $N$ momentum measures $d{p}_j/(2\pi \hslash )$ and $N-1$ intermediate coordinate measures. Each momentum Gaussian integral gives $[m/(2\pi i\hslash \delta ){]}^{1/2}\exp [im(q_j-q_{j-1}{)}^2/(2\hslash \delta )]$. The resulting configuration integral has normalization $[m/(2\pi i\hslash \delta ){]}^{N/2}$. A common endpoint convention for $V$ and the Trotter product formula identify both limits with the same operator transition kernel; a damping or imaginary-time prescription defines the oscillatory integrals.

// Target: perturbative-quantum-field-theory.bigb

A top-hat wavefunction is constant on an interval and zero elsewhere. It has a uniform position distribution and a sinc function momentum amplitude. Its discontinuous edges place it outside the finite-kinetic-energy domain of the free Hamiltonian. Smooth-edged localized packets can approximate its position profile while having different momentum tails.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

A particle detector that samples successive positions along charged-particle trajectories. Together with a known magnetic field, a fitted trajectory measures momentum, charge sign and an interaction or decay vertex.

For ${\varphi }_K(x-X)=c\tanh (c(x-X))$ in the unit kinetic normalization, $\int ({\varphi }_K^{\prime }{)}^{2}dx=M=4c^3/3$. Substitution of $X(t)$ therefore gives the free-particle collective-coordinate effective Lagrangian $-M+M{\dot{X}}^2/2$. Its quantization has momentum $p$ and energy $M+p^2/(2M)$ to this order. The exact uniformly moving classical kink is a Lorentz boost of the static one, with energy $M/\sqrt{1-v^2}$.