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For a small-amplitude conservative wave packet in a slowly varying fluid flow, its intrinsic energy divided by intrinsic frequency is transported along its group velocity, with the basic advection added in the laboratory frame. A stationary background conserves laboratory angular frequency, but a changing mean velocity changes the intrinsic one through the Doppler effect; intrinsic energy alone need not be conserved. Geometrical spreading and dissipation enter the action-flux balance.

The wave equation is a second-order linear partial differential equation describing waves with finite propagation speed. In one spatial dimension its homogeneous, constant-coefficient form isHere $x$ is position, $t$ is time, $u(x,t)$ is the disturbance, and subscripts denote partial derivatives. For a wave equation on a string, $u$ is transverse displacement; in linear acoustics, it can be a pressure perturbation. In several spatial dimensions, $u_{xx}$ becomes the Laplacian $\Delta u$. The same mathematical structure connects vibrating strings and acoustic waves, although their physical assumptions and wave speeds differ. ^[ref]

For the whole-line Cauchy problem, prescribe both displacement $u(x,0)=f(x)$ and velocity $u_t(x,0)=g(x)$. A classical solution has continuous partial derivatives through order two; $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$ suffice for D'Alembert's formula to produce one. A finite string additionally needs boundary conditions. These statements concern the unforced model with constant $c$; damping, spatially varying coefficients and nonlinear terms change the solution formula.

A travelling wave $F(x-ct)$ carries its entire profile to the right at wave speed $c$; $G(x+ct)$ travels left. Their sum is again a solution by the superposition principle. A standing wave instead has a fixed mode shape whose size and sign oscillate: in $\sin (kx)\cos (ckt)$, each node of a standing wave stays stationary. The distinction is especially clear in the pulse and string diagrams below. For this ideal model, the dispersion relation is ${\omega }^2=c^2{k}^2$: each propagation branch is nondispersive, so a localized profile translates without spreading. This makes the equation a useful reference against which to understand wave dispersion and damping.

The appealing combination is local interaction with global predictability: neighbouring displacements determine acceleration through $u_{xx}$, yet D'Alembert's formula explicitly identifies exactly which initial locations can influence a later observation. Finite propagation speed turns that observation into a domain of dependence, rather than requiring knowledge of the entire initial string.

For the homogeneous one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$ with constant $c>0$, wave speed means the speed at which a travelling wave profile translates. In $F(x-ct)$, a feature initially at $x=s$ follows $x=s+ct$, giving signed velocity $+c$; in $G(x+ct)$ it follows $x=s-ct$, giving signed velocity $-c$. The characteristic lines for the one-dimensional wave equation therefore have slopes $dx/dt=\pm c$. The speed of either profile is $c$, even though the displacement at a fixed point can change sign or pass through zero.^[ref]

The same $c$ bounds the causal cone for the constant-speed wave equation. For a wave equation on a string, $c=\sqrt{T/\mu }$ depends on tension and linear mass density; for small acoustic waves in stationary uniform air it is the speed of sound. This interpretation assumes the stated ideal wave equation. In media with wave dispersion, phase velocity, group velocity and the earliest signal arrival have different meanings: a measured wave crest speed need not be the speed of a wave envelope, and neither alone establishes a causal cone for the constant-speed wave equation.

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]

A world line is an object's history as a curve of spacetime events. In an inertial frame, a particle at rest has constant spatial coordinates, so its world line is vertical in a spacetime diagram with $ct$ upward. A particle at constant velocity $v$ along one axis has $x=x_0+vt$, a straight sloping world line. The coordinate representation changes with the inertial frame, while the physical history is the same.

A massive observer follows a future-directed timelike curve. Along a differentiable world line with velocity $\mathbf{v}(t)$, its clock accumulatesThe Minkowski metric therefore measures clock time along the curve, not just the difference between two coordinate readings. In the straight $O$-to-$A$ example, $v=0.75c$ for $4\mathrm{s}$, giving $4\sqrt{1-0.75^2}\mathrm{s}=\sqrt{7}\mathrm{s}$. If the observer changes velocity along the path, use the integral rather than assigning the straight-segment formula to the whole path.

At zero frequency the graded-layer transfer equation integrates to $\exp [L\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)]$, where $L=\frac{1}{2}\log (Z_b/Z_a)$. Eliminating the outgoing wave amplitude at the left endpoint gives the displayed incoming-to-outgoing scattering matrix. It is real orthogonal, so it preserves total incoming and outgoing flux. It coincides with continuity of velocity and traction at a sharp seismic impedance interface; the profile between the endpoints enters at nonzero frequency.

In the same gravity–Marangoni reduction, zero liquid flux gives $\Gamma =1+(H_0^2-H^2)/3$ and $6JX=(1+H_0^2/3)(H^3-H_0^3)/3-(H^5-H_0^5)/15$. The lower part of the liquid flows backwards while the surface and upper third flow forwards. For zero initial height and concentration drop $\delta$, $J={\delta }^{3/2}(1-3\delta /5)/(2\sqrt{3})$ is largest formally at complete depletion. Since surface velocity is $J/\Gamma$, that endpoint is singular and requires physical regularization by diffusion, capillarity or an edge model. The algebraic maximum is not by itself a regular zero-concentration outflow.

In an incompressible shearing sheet with background velocity $-2Ax{\mathbf{e}}_y$, take both the perturbation velocity and the entire magnetic field to be real single Fourier modes with a common real wavevector and transverse amplitudes. Then every quadratic directional derivative vanishes, including terms involving complex conjugates. There is no uniform background magnetic field and consequently no linear magnetic-tension coupling between the amplitudes. Advecting the phase requires ${\dot{k}}_x=2A{k}_y$, ${\dot{k}}_y={\dot{k}}_z=0$. The velocity amplitude obeys $\dot{\tilde{\mathbf{v}}}-2A{\tilde{v}}_x{\mathbf{e}}_y+2\mathbf{\Omega }\times \tilde{\mathbf{v}}=-i\mathbf{k}\tilde{\psi }-\nu k^2\tilde{\mathbf{v}}$; the magnetic amplitude obeys the displayed equation. The modified pressure includes magnetic pressure; the physical pressure can contain a mean and second harmonic even when the modified pressure is a single mode.

When matter leaves a moving body with zero instantaneous relative velocity, it carries the body's current momentum per unit mass but creates no additional recoil term. A short-interval momentum balance including the newly released material gives $q\dot{v}=F_{\mathrm{ext}}$ for the remaining body's mass $q(t)$. The expression $\frac{d}{dt}(qv)=F_{\mathrm{ext}}$ omits the outgoing momentum flux. Nonzero relative ejection velocity instead produces the thrust term of the rocket equation.

At fixed outer velocity and integral scale of turbulence, the mean viscous dissipation per unit mass approaches a positive finite value as the Reynolds number becomes large. The dimensionless coefficient $C_{ϵ}=ϵ\ell /u^3$ is then of order one. This is a phenomenological property of established three-dimensional turbulence, not a consequence of taking kinematic viscosity to zero in an arbitrary smooth flow.

Extend the velocity rigidly through a moving sphere in an incompressible tube flow. The extended velocity is continuous and divergence-free, so its total flux is independent of height. Quiescent far-field conditions make this flux zero. The sphere contributes $V_z{A}_s$ through a cross-section; the rotational contribution integrates to zero over its circular slice. Thus the fluid flux compensates the solid displacement. A descending sphere generates upward fluid flux in every section cutting its interior. This statement assumes a localized flow with no imposed throughflow and negligible tube-end effects.