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In the ideal kinetic friction model, the loss of combined mechanical energy of the two contacting bodies becomes thermal internal energy at rateThis follows by adding the contact powers $f_{k}v$ and $-f_{k}U$ from Newton's third law. It depends on contact slip, whereas the contact power received by either body separately depends on its velocity in the chosen inertial frame. Thus positive frictional heating is compatible with increasing kinetic energy of one body, as the moving-belt example in kinetic friction shows. Heating need not cause a temperature rise if thermal energy escapes by heat transfer.

Let $R(\alpha )\propto {\alpha }^{-n}$, with $n>0$, and use disturbances $e^{i(kx-\omega t)}$ about uniform velocity $U$. The intrinsic frequency is $\nu =\omega -kU$. Linearized mass conservation and the decrease in friction as area grows give the displayed dispersion relation. For nonzero $k$ and positive flow, one branch grows exactly when $nU>c_0$. At $nU=c_0$, one branch is neutral, $\nu =k{c}_0$, while the other remains damped, $\nu =-k{c}_0-i{R}_0$. The maintained external-pressure gradient supplies the energy; positive friction alone does not imply linear stability of a driven flow.

The front velocity is the propagation speed of the earliest boundary of a newly introduced disturbance's support. It asks when an initially unaffected observer can first respond, whereas group velocity asks how a narrow wave envelope translates. A detector threshold defines a practical arrival time, but can miss arbitrarily small earlier responses; a packet peak is another distinct timing convention. In a causal dielectric model, a high-angular frequency contour argument gives exactly zero field for $t<x/c$, even when phase velocity exceeds $c$. This is a statement about that model's relativistic causality, not a universal identification of group velocity with front velocity.^[ref]

For the stipulated Klein-Gordon equation with compactly supported real Cauchy data, front velocity is bounded by $c$, and $c$ is the model's maximal speed of influence. A short energy method outline usesIntegration outside an expanding interval shows that zero initial energy there remains zero. Thus initial support in $[-a,a]$ cannot influence $x>a$ before $t=(x-a)/c$. The positive energy density includes the restoring term ${\omega }_0^2{u}^2/2$: this adds oscillation behind the front without extending its domain of dependence. A particular disturbance may fail to excite the leading boundary, so the maximal speed is not a promise of a detectable response for every source.

A Gaussian function is nonzero arbitrarily far away, so the Gaussian envelope in a Klein-Gordon medium has no sharp initial support boundary. Its small distant tails are already in the initial data. Likewise, keeping only a narrow interval of the frequency spectrum cannot reproduce an exact abrupt onset. Cutting off a Gaussian tail changes its Fourier transform and the accompanying initial velocity; it is a different initial-value problem. The exponential initial-velocity tail for the Klein-Gordon equation is another concrete example of a pre-existing tail outside the main disturbance.

The water-tank measurement in phase velocity records free surface elevation with calibrated resistance probes and uses a ruler and stopwatch for an independent check. The inspected protocol calls the front or back of a group a timing feature. That visible edge establishes an envelope transit estimate, not the first nonzero disturbance in the sense of front velocity.^[ref] A proposed additional control would repeat the transit fit at several detection thresholds and train durations, using matched probe calibration and the same sampling rate. Dependence on the threshold diagnoses wave envelope reshaping or limited sensitivity; records should be compared before reflections return. Such a test could expose limitations of a group velocity estimate, but would not establish an exact causal front or validate the stipulated Klein-Gordon equation for water waves.

The fundamental theorem of calculus says that accumulation and local change undo one another, under appropriate regularity conditions. If $f$ is a continuous function on $[a,b]$ andthen $A$ is differentiable for $a<x<b$ and $A^{\prime }(x)=f(x)$. Conversely, if $G$ is an antiderivative of $f$ on this interval, continuous at its endpoints, thenThe first statement turns an integral into a function with a known derivative; the second turns an entire accumulation into two endpoint values.

For $f(t)=1+2t$, accumulation from zero gives $A(x)=x+x^2$. At $x=2$, its derivative is $A^{\prime }(2)=5$, exactly the current value $f(2)$. The accumulated value there is $A(2)=6$, and by $x=3$ it has reached $A(3)=12$. The accumulated quantity and its rate are different functions, linked by differentiation.

This connection makes the theorem useful far beyond geometric area. It recovers displacement from velocity, gives a probability from a probability density function, and converts many differential equations into integral equations. The striking economy is that information spread across a whole interval can sometimes be recovered from an antiderivative at its two boundaries.

A short explanation of the first part is thatis the local average of $f$. Continuity makes this average approach $f(x)$. For the second part, $G-A$ has zero derivative and is constant by the mean value theorem. These are the ideas behind the proof; the endpoint and regularity conditions make them valid.

An early geometric form of this connection appears in Isaac Barrow's Lectiones geometricae (1670), Lecture X, Proposition 11. The original volume is catalogued by CESIMA, and Michael Nauenberg's historical research reproduces and examines the geometric argument.^[ref][ref]

For the stipulated Klein-Gordon packet velocities, take the original normalized spectral amplitudewith $k_0>0$ and $0<\sigma \ll k_0$. Here $\sigma$ is the width of the spectral amplitude, not the standard deviation of its squared magnitude. Initially $\psi (x,0)=e^{-{\sigma }^2{x}^2/2}{e}^{i{k}_{0}x}$, a Gaussian function multiplying a carrier wave. Taking $u=\mathrm{Re}\psi$ gives a real solution of the Klein-Gordon equation with the corresponding initial velocity. The tiny negative-$k$ tail is retained in this exact integral; the statement that it is mainly right-going is a narrow-spectrum approximation, not a claim of strictly positive wavenumbers.

The dimensionless figure uses $c={\omega }_0=k_0=1$ and $\sigma =0.18$ at $t=0,4,8$. Blue curves are $\mathrm{Re}\psi$, orange bounds are $\pm |\psi |$, and the marked carrier wave reference is $x=v_{p}t$; it need not be a crest of the full wavefield. The dashed envelope-centre reference $x=v_{g}t$ is a leading approximation. These are calculated curves, not experimental records. The exact dispersion relation makes the packet slightly asymmetric; a quadratic approximation predicts symmetric broadening about $v_{g}t$.

The first-order Taylor expansion translates the initial wave envelope without changing its shape. Retaining curvature instead givesThe distinction between translation and curvature-driven spreading is the same one exposed by the Gaussian pulse calculation in Pulse Propagation by Richard Fitzpatrick (equations 9.8–9.18). Its first-order approximation is not an assertion that a dispersive packet keeps its shape indefinitely.

Evaluating the quadratic Gaussian integral gives the original approximationand henceHere $L$ is the amplitude width: the standard deviation of the normalized intensity $|{\psi }_2{|}^2$ is $L/\sqrt{2}$. The peak falls as the packet widens, while $\int |{\psi }_2{|}^{2}dx$ stays constant. This conserved integral is not the complete physical wave energy of the Klein-Gordon equation.

Rigid translation requires $|\beta |t{\sigma }^2\ll 1$. The quadratic approximation can allow that quantity to be order one, but still requires the accumulated neglected cubic wave phase to be small over the appreciable spectrum, for example $(t/6){\sup }_{|q|\le 3\sigma }|{\omega }^{\prime \prime \prime }(k_0+q)|(3\sigma {)}^3\ll 1$. This sufficient check controls only the stated three-width spectral window; the Gaussian tails must also be negligible at the required accuracy. Higher derivatives or very long times require returning to the exact spectral integral. The width $L(t)$ is therefore a controlled local approximation; it neither predicts front velocity nor establishes a universal identity between group velocity and information velocity.

For an isothermal collisionless tracer without streaming or an escape cutoff, integrate its three-dimensional Maxwell-Boltzmann velocity distribution over the two transverse velocities. The remaining marginal is the displayed normal distribution. Its width and mean do not depend on position. Integrating along the line of sight therefore only multiplies the marginal by the projected surface density; the normalized observed velocity distribution remains the same Gaussian at every projected radius.

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

For an initially resting shallow water layer whose height rises linearly from $-h$ to $h$ across $(-L,L)$, let $R_d=\sqrt{gH}/|f|$ and $\alpha =L/R_d$. Conserved linear shallow-water potential vorticity gives $R_d^2{\eta }_s^{\prime \prime }-{\eta }_s=-{\eta }_0$. Boundedness, oddness, and continuity of height and slope give ${\eta }_s/h=x/L-e^{-\alpha }\sinh (x/R_d)/\alpha$ inside the ramp, and ${\eta }_s/h=\mathrm{sgn}(x)[1-(\sinh \alpha /\alpha )e^{-|x|/R_d}]$ outside. The balanced velocity is $u_s=0$, $v_s=g{\eta }_s^{\prime }/f$. Narrow ramps spread over the Rossby deformation radius; wide ramps retain their initial height except for narrow matching layers.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

At an isolated interface with continuous base velocity, a stable density drop $\Delta \rho /2$ and a jump $[U^{\prime }]$ of base shear, decaying disturbances $w\propto e^{-k|z-z_i|}$ obey the displayed dispersion relation. It follows by inserting their derivative jump $[w^{\prime }]=-2kw$ into the jump conditions for stratified inviscid shear flow. Both the density jump and the vorticity jump contribute to the intrinsic phase velocity $c-U_i$.

The gyroradius is the radius of the transverse circular motion of a charged particle in a uniform magnetic field. In nonrelativistic classical mechanics, with zero electric field and no other forces, it is $m{v}_{\perp }/(|q|B)$, where $v_{\perp }$ is the speed perpendicular to the magnetic field. The parallel component of velocity changes the advance of the helix, rather than its transverse radius.

For fluid velocity $\Omega (-x_3,0,x_1)$ and $B\Omega <1$, the stable deterministic orientation is $\mathbf{p}=(-B\Omega ,0,\sqrt{1-B^2{\Omega }^2})$. Once this orientation has relaxed, swimming at speed $V_s$ shifts the centre of the circular material trajectories to $(-V_s\sqrt{1-B^2{\Omega }^2}/\Omega ,-V_{s}B)$ in the $(x_1,x_3)$ plane. Its distance from the cylinder axis is $V_s/\Omega$. General initial orientations produce an orientation transient before these circular paths.

For $f(E,L)=f_E(E)/L$, with $L=rv\sin \eta$ and $E=\psi -v^2/2$, velocity integration gives $r\rho =2{\pi }^2{\int }_0^{\psi }{f}_E(E)dE$. Thus $f_E(E)=(2{\pi }^2{)}^{-1}d(r\rho )/d\psi$ at $\psi =E$. The factor $1/L$ is locally integrable under the velocity-volume measure. Its angular measure is uniform in $\eta$; averages of ${\cos }^2\eta$ and ${\sin }^2\eta$ are both $1/2$, so the total tangential second moment equals the radial second moment and the velocity-anisotropy parameter is $\beta =1/2$. At the largest allowed radius $\psi (r_E)=E$, the derivative becomes $[\rho +r(d\rho /dr)]/(d\psi /dr)$, not $[\rho +d\rho /d\psi ]/(d\psi /dr)$. Positivity requires $d(r\rho )/d\psi \ge 0$.

The harmonic mean of a nonempty finite list of positive real numbers is the multiplicative inverse of the arithmetic mean of their inverses:For two inputs, $H=2x_1{x}_2/(x_1+x_2)$. Replacing each input by $H$ preserves the sum of inverses, making it useful when a fixed amount divided by a rate determines a cost or duration. This positive-input definition avoids division by zero; negative inputs need not retain the usual averaging or ordering properties. ^[ref].

Travel two legs of equal length $d$ at positive speeds $v_1,v_2$, with no stops. Their travel times are $d/v_1$ and $d/v_2$, so the average speed isFor two $60\mathrm{km}$ legs at $30$ and $60km/h$, the times are two hours and one hour. The average speed is $120/3=40km/h$. The same speeds maintained for one hour each instead cover $90\mathrm{km}$ in two hours, giving the arithmetic mean $45km/h$. A slower equal-distance leg occupies more time, so the data being averaged must include what is held equal. This is total distance traveled divided by elapsed time; a return journey can have zero average velocity while keeping a positive average speed. ^[ref].

The arithmetic-geometric mean inequality applied to the inverses gives $H\le G$; together with $G\le A$ it yields the comparison in geometric mean. For two inputs the arithmetic–harmonic gap has the exact formFor $30$ and $60$ the gap is five. With one positive input held fixed and another approaching zero, the harmonic mean approaches zero too: a very slow fixed-distance leg dominates the travel time. Unequal leg lengths instead require a weighted harmonic mean.

The Hazel model is a smooth stratified shear flow whose base velocity has a hyperbolic-tangent profile and whose squared buoyancy frequency has a hyperbolic-secant-squared profile. The equal-width nondimensional example is $U=\tanh z$ and $N^2=J{\mathrm{sech}}^{2}z$. For $J\ge 0$, its gradient Richardson number is $J{\cosh }^{2}z$, with minimum $J$.

Keep the same bond force constant $k$ and approximate the nuclear masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, and ${}^{35}\mathrm{Cl}$ by $1u$, $2u$, and $35u$, respectively. Here $u$ is the unified atomic mass unit, and $\mathrm{D}$ denotes deuterium. These are chosen mass number approximations, not measured atomic masses or natural-abundance averages. With the same chlorine isotope in both molecules,This is the isotope scaling of diatomic vibration prediction. Doubling the light partner's mass nearly doubles the reduced mass, lowering the frequency by about $28.3\%$. The factor differs from $1/\sqrt{2}$ because the chlorine partner also moves.

For the same small initial stretch $A$ and zero velocity, both models start with the same internal mechanical energy $k{A}^2/2$, but the DCl oscillation period is longer by $\sqrt{72/37}\approx 1.394972$. In the centre of mass frame, $u_{\mathrm{H}}=-35q/36$ and $u_{\mathrm{Cl}}=q/36$ for HCl; $u_{\mathrm{D}}=-35q/37$ and $u_{\mathrm{Cl}}=2q/37$ for DCl. These ratios explain which partner moves and are separate from the change in frequency.

The common-$k$ assumption belongs to the leading Born-Oppenheimer approximation: fixed nuclear positions and electric charges determine the electronic potential energy, while nuclear masses govern motion on that curve. Quantitative comparison requires actual atomic masses, departures from that approximation, and separation of anharmonicity and molecular rotation from the measured band. Even with common $k$, the quantum harmonic oscillator predicts a different zero-point energy for each isotope.^[ref]

In a uniform magnetic field, the Lorentz force leaves the component of velocity parallel to the field constant and rotates the perpendicular component at the nonrelativistic cyclotron frequency $|q|B/m$. The resulting path is a helix around a field line.

The Hernquist model has relative potential $\psi =GM/(r+a)$ and density $\rho =Ma/[2\pi r(r+a{)}^3]$. It has enclosed mass $M{r}^2/(r+a{)}^2$, a central $r^{-1}$ cusp and an outer $r^{-4}$ tail. Since $r\rho =a{\psi }^3/(2\pi G^3{M}^2)$, the half-anisotropic distribution function is $f(E,L)=3a{E}^2/(4{\pi }^3{G}^3{M}^{2}L)$ for $0<E<GM/a$. Its moments are $⟨v_r^2⟩=\psi /4$ and each tangential moment $\psi /8$. Isotropic and other admissible anisotropic distributions can also support this same density; the gravitational potential does not select a unique velocity anisotropy.

In his 1888 reflection experiments, Heinrich Hertz excited a straight conductor with a central spark gap driven by an induction coil. The receiver was a wire circle of $35\mathrm{cm}$ radius with an adjustable small spark gap; tiny sparks were observed in darkness. There was no connecting transmission wire. ^[ref]

He removed hanging metal fixtures, installed a $4\mathrm{m}$ by $2\mathrm{m}$ zinc reflecting sheet, and moved and rotated the receiver. Spark strength alternated with position and gap orientation, supporting specular reflection and wave interference. He checked observations from farther away because the observer's body perturbed the response. Changing oscillator dimensions changed the spatial pattern. ^[ref]

These tests supported propagating electromagnetic waves, rather than mere detection of nearby induction. They did not establish unlimited reception distance. His wire-versus-air wave interference experiment addressed finite velocity; he later judged its quantitative velocity comparison doubtful because of room effects. ^[ref]

The physical inference combines transmission with a spatially structured response to a reflected disturbance. In an ideal monochromatic incident-and-reflected model, standing wave nodes are separated by $\lambda /2$; combining a reliable wavelength with independently known frequency gives phase velocity $v=\nu \lambda$. A spark is a threshold event, however: its presence or length is not a calibrated wave intensity or a direct arrival-time measurement. The actual loop also responds to its orientation and finite size, so the simple standing-wave model is an explanation of the test's logic, rather than a reconstruction of Hertz's measured points. Poynting theorem explains how source energy can reach and drive the receiver's discharge.