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For a block sliding right on a stationary horizontal plane, take rightward acceleration $a$ as positive. A gauge reading $F$ for a pull at angle $\alpha$ above horizontal has horizontal component $F\cos \alpha$ and upward component $F\sin \alpha$. With no vertical acceleration or other forces, Newton's second law givesand thereforeThis inference requires continued contact with $N>0$ and rightward contact slip, so its numerator must be nonnegative. An upward pull reduces normal force; a downward pull, represented by $\alpha <0$, increases it. The horizontal constant-velocity formula is the special case $\alpha =0$, $a=0$.

For the original $2\mathrm{kg}$, ${\mu }_k=0.30$ model, an upward pull at $\alpha =30^{\circ }$ maintaining constant velocity needsDividing this gauge reading by $mg$ would give about $0.2953$, rather than $0.30$: the gauge reads the full inclined force, and the pull has changed the normal force.

Even with a horizontal pull, an accelerating measurement needs the $ma$ correction: ${\mu }_k=(F-ma)/(mg)$. In the same model, a horizontal $7.88\mathrm{N}$ pull produces $a=1m/{\mathrm{s}}^2$ while the block slides right; subtracting $ma=2\mathrm{N}$ recovers the $5.88\mathrm{N}$ kinetic friction. Dividing the uncorrected reading by $19.6\mathrm{N}$ instead gives about $0.4020$. Measure acceleration in an inertial frame, along the sliding direction, over the same interval as the gauge reading. These corrections follow from the force balance, independently of whether the resulting coefficient of kinetic friction is constant across runs.

// Target: classical-mechanics.bigb

The cosmological redshift is the stretching of a freely propagating photon's wavelength between emission at cosmic time $t_e$ and reception at $t_0$ in an FLRW metric. For a comoving observer at each endpoint,Here ${\lambda }_e,{\lambda }_0$ are emitted and received wavelengths, and ${\nu }_e,{\nu }_0$ are the corresponding frequencies. A positive cosmological redshift means that the scale factor grew while the photon travelled. The name describes a shift toward longer wavelengths; it applies to light beyond the visible range as well.

For a computed example, let the scale factor double from $a(t_e)=0.5$ to $a(t_0)=1$. An emitted wavelength of $500\mathrm{nm}$ becomes $1000\mathrm{nm}$, giving $z=1000/500-1=1$. The received frequency is half the emitted frequency. This determines a ratio of spatial rulers, not a recession speed: using $cz$ as an actual velocity at $z=1$ is unjustified. The wavelength panel in the illustration under Hubble's law shows this doubling.

A short derivation follows successive wave crests along a radial null geodesic in FLRW spacetime. Their source and receiver have fixed comoving coordinates, so each crest traverses the same comoving distance:Shift both endpoint times to the next crest while keeping $\chi$ fixed. To first order in the very short periods $\delta t_e,\delta t_0$, this gives $\delta t_0/a(t_0)=\delta t_e/a(t_e)$. Locally, wavelength equals the speed of light times the crest period, yielding the displayed cosmological redshift relation. These periods are much shorter than the timescale on which the scale factor changes.

The relation isolates the smooth expansion contribution. An observed redshift can also contain a Doppler effect from source or observer peculiar velocity and a gravitational redshift. See combined kinematic and cosmological redshift for the multiplicative combination of endpoint motion and expansion. Treating every observed redshift as pure cosmological redshift can therefore misassign a nearby source's distance.

The useful insight is that a received wavelength ratio records how much the scale factor changed during the photon's journey. Turning that ratio into a distance requires an expansion history, through the Hubble parameter, and a specified convention such as luminosity distance or proper distance in cosmology; the ratio alone supplies neither.

Consider steady flow of a Newtonian fluid in incompressible flow with constant dynamic viscosity $\mu >0$ between parallel plates at $z=0$ and $z=H$. The bottom plate is stationary and obeys the Navier slip boundary condition with slip length $b$. The top plate moves at velocity $U>0$ and obeys the no-slip boundary condition. There is no streamwise pressure gradient or body force. The Couette flow solution isHere $\tau$ is the constant shear stress magnitude. The lower wall's Navier slip boundary condition holds exactly because $b{u}^{\prime }(0)=bU/(H+b)=u(0)$. At the top, $u(H)=U$ is exactly the prescribed no-slip boundary condition; no finite upper-wall slip length is imposed in this example.

For a numerical example, stipulate $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $U=1\mathrm{mm}{\mathrm{s}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. The bottom velocity is $0.167\mathrm{mm}{\mathrm{s}}^{-1}$ and the shear stress is $0.0833\mathrm{Pa}$, compared with $0.100\mathrm{Pa}$ when both plates obey the no-slip boundary condition. The extrapolated profile reaches zero at $z=-2\mu \mathrm{m}$.

For fixed $H$, $U$ and $\mu$, $b\to 0$ recovers $u=Uz/H$. As $b\to \infty$, $u\to U$ throughout the gap and $\tau \to 0$: the moving upper wall selects the limiting uniform velocity, while the lower wall approaches a stress-free boundary condition. The dynamic viscosity stays positive; the disappearing shear stress comes from the disappearing velocity gradient.

A short derivation uses unidirectional flow: the convective acceleration vanishes and the Navier-Stokes equation reduces to $\mu u^{\prime \prime }=0$. Write $u=Az+B$. The bottom condition gives $B=bA$ and the top condition gives $A(H+b)=U$.

Let $H$ be any antiderivative of the initial velocity, $H^{\prime }=g$. One choice of profiles isOpposite additive constants in $F$ and $G$ cancel, so the displacement is independent of the chosen antiderivative. Even for compact support of $g$, these separate profiles can approach nonzero constants; cancellation in their sum is essential to support widening for the wave equation.

For a short derivation, the characteristic coordinates $\xi =x-ct$, $\eta =x+ct$ give $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }$ by the chain rule. Thus $u=F(\xi )+G(\eta )$, and the Cauchy data imply $F+G=f$ and $-c{F}^{\prime }+c{G}^{\prime }=g$, giving the profiles above.

Critical damping is the repeated-root case of the viscous damping model: its two characteristic roots coincide at $r=-{\omega }_0$. The solution with arbitrary initial conditions isThe extra factor $t$ is essential: one exponential function alone cannot fit arbitrary initial displacement and velocity when the characteristic equation has a double polynomial root. Substitution, or taking the limit of the underdamped formula as its angular frequency tends to zero, gives the displayed expression.

The precise fastest-return statement is this: hold $m$, $k$, $x_0>0$ and $v_0=0$ fixed, and compare all $\zeta \ge 1$. Every response decreases positively toward equilibrium; the critically damped responseis smaller than every overdamped response at each $t>0$. It therefore reaches and stays within any specified band $|x|\le ϵx_0$, $0<ϵ<1$, sooner than those responses. Exact arrival at rest takes infinite time. This is a comparison among nonoscillatory regimes for these particular initial conditions, not a universal minimum of every possible settling-time criterion.

One short comparison argument uses $q=\partial X/\partial \zeta$. Its initial conditions are zero and $q^{\prime \prime }+2\zeta q^{\prime }+q=-2X^{\prime }$. For release from rest, $X^{\prime }<0$ at $\tau >0$; the causal Green function of a damped oscillator is positive at positive time for $\zeta \ge 1$. Its convolution with $-2X^{\prime }$ gives $q>0$: increasing damping beyond the critical value increases the remaining displacement.

There is no repeated oscillation at critical damping, but there can be overshoot once. For $x_0>0$, the factor in square brackets crosses zero precisely when $v_0<-{\omega }_0{x}_0$, atFor example, $v_0=-2{\omega }_0{x}_0$ gives $x=x_0(1-{\omega }_{0}t)e^{-{\omega }_{0}t}$, crossing equilibrium at $t=1/{\omega }_0$ as shown above. An overdamped response can also cross once for suitable initial conditions. Conversely, specially chosen initial conditions can remove its slow exponential decay entirely. Thus the generic slow-root argument must not be applied to a response whose slow-root coefficient is zero.

A critical layer is a region where the leading wave phase speed coincides with the local shear-flow velocity and an outer approximation degenerates. Small viscosity or other corrections must then be retained on a distinguished transverse scale. The width depends on the order of vanishing of $U-c_0$ and on the governing mode equation; simple-shear and stationary-profile layers have different balances.

For $0<k<1$, a neutral mode of the equal-width Hazel model behaves as $w\sim z^{1-k}$ near $z=0$. Its derivative diverges as $z^{-k}$, so it is not a classical continuously differentiable mode across the critical level of an internal gravity wave. The horizontal velocity is proportional to $w^{\prime }/k$; its local kinetic energy is finite only for $k<1/2$, because ${\int }_0^{ϵ}{z}^{-2k}dz$ converges precisely then. One possible neutral-mode convention is the boundary value from $U-c$ with $c_i\downarrow 0$, which fixes the phase of the power on the negative-$z$ side. The Miles–Howard theorem concerns growing modes with nonreal $c$ and does not rule out such singular neutral limits.

Imagine folding a sheet along its horizontal axis by the smooth map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$, $F(x,y)=(x,y^2)$. The inputs $(a,r)$ and $(a,-r)$ have the same output. Along $y=0$, horizontal motion still changes the output, but vertical motion produces no first-order change: $F(a,\epsilon )-F(a,0)=(0,{\epsilon }^2)$. These inputs are critical points of a smooth map even though their differential is not the zero linear map. The missing direction in the target tangent space is what matters.

Precisely, let $f:X^m\to Y^n$ be a smooth map between smooth manifolds without boundary. An input $p\in X$ is a critical point of a smooth map when its differentialis not surjective, equivalently when its rank of a linear map is less than $n$. An input where it is surjective is a regular point; a map with no critical points of a smooth map is a submersion. The threshold is the target dimension, not $\min (m,n)$. In manifold charts, compute the matrix rank of the $n\times m$ Jacobian matrix; changes of charts preserve that matrix rank. ^[ref]

For the fold,Thus its set of critical points of a smooth map is the entire horizontal axis. Away from that axis, its differential can produce every infinitesimal target displacement. At the axis, it can produce only horizontal ones. This diagnoses where the two inverse branches in the picture merge.

For a scalar smooth function $f:X\to \mathbb{R}$, the target dimension is one. A linear map to $\mathbb{R}$ fails to be surjective exactly when it is zero, so here a critical point of a smooth map means $d{f}_p=0$. In Euclidean space this is precisely the familiar critical point condition $\nabla f(p)=0$: every first partial derivative vanishes. On a Riemannian manifold, it is equivalently ${\nabla }_{g}f(p)=0$ for the chosen Riemannian metric; the Riemannian gradient depends on that metric, but whether it vanishes does not. See Loring W. Tu, An Introduction to Manifolds, Proposition 8.23 for the scalar coordinate criterion. The scalar smooth function $f(t)=t^3$ has a critical point at zero without a local maximum or local minimum.

A source of lower dimension gives a different warning. For the smooth map $G:\mathbb{R}\to {\mathbb{R}}^2$, $G(t)=(t,t^2)$, the Jacobian matrix is the column $(1,2t{)}^T$, of matrix rank one at every input. Every input is a critical point of a smooth map, since the target dimension is two, even though this parametric curve has nonzero velocity everywhere and is an immersion. More generally, if $m<n$, no differential $T_{p}X\to T_{f(p)}Y$ can be surjective, so every input is critical. The terminology measures the ability to fill target directions, rather than whether the parametrized curve looks smooth.

A curve parameter labels the points visited by a parametric curve $\gamma (t)$. Choosing time as the parameter determines its velocity; an arc-length parametrization instead measures accumulated arc length. A regular monotone reparametrization changes the labels and speed while preserving the route and its arc length.

For a rotating conducting sphere, a geostrophic azimuthal velocity $U=rV$ couples to the cylinder-integrated magnetic stress $T=\int B_r{B}_{\varphi }dz$. Put $H=\int B_r^{2}dz$ and let $2z_b$ be the cylinder height. With an insulating boundary and no flow through it, cylindrical averaging removes the Coriolis force and gives $U_t=(2z_b{\mu }_0\rho r^2{)}^{-1}{\partial }_r(r^{2}T)$, $T_t=Hr{\partial }_r(U/r)$. The local squared wave speed is $H/(2z_b{\mu }_0\rho )$, the cylinder average of $B_r^2/({\mu }_0\rho )$.

For the wave equation $u_{tt}=c^2{u}_{xx}$ with zero initial displacement and initial velocity $g$, D'Alembert's formula becomesFor $c=1$ this is the original speed-one form $u(t,x)=\frac{1}{2}{\int }_{x-t}^{x+t}f(s)ds$ when the initial velocity is named $f$. The integral records the total initial kick inside the domain of dependence.

For a computed example, set $g(x)=V$ on $[-a,a]$ and zero elsewhere, with $V>0$. This discontinuous datum gives a weak solution:At the centre, $u(0,t)=V\min (t,a/c)$. For $a=1\mathrm{m}$, $c=2\mathrm{m}{\mathrm{s}}^{-1}$ and $V=0.4\mathrm{m}{\mathrm{s}}^{-1}$, this is $0.1\mathrm{m}$ at $t=0.25\mathrm{s}$ and $0.2\mathrm{m}$ at $t=1\mathrm{s}$. Once $ct>a$, a central plateau $u=Va/c$ occupies $|x|\le ct-a$. In contrast, central zero region for odd compactly supported initial velocity follows when the full integral of $g$ vanishes. Finite propagation speed permits an interior tail; it does not imply the Strong Huygens principle.

For constant wave speed $c>0$, the whole-line Cauchy problemhas the unique classical solution, for $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$,The first term evolves displacement released from rest; the second evolves initial velocity. The factor $1/(2c)$ converts an integral of velocity over distance into displacement. On a finite string the formula applies before the backward characteristic curves reach an endpoint, or globally after an extension that enforces the boundary conditions. ^[ref]

For a completely smooth example, choose a bump function with compact support:where $a>0$ sets the half-width and $A$ sets the initial peak. ThenAt $t=0$ the two copies coincide, recovering peak $A$. At $t=a/(2c)$, $u(0,t)=A{e}^{-1/3}\simeq 0.716531A$. At $t=3a/(2c)$ the pulses are separated, centred at $x=\pm 3a/2$, each of width $2a$ and peak $A/2$. Zero initial velocity forces equal left and right contributions; a purely right-moving initial profile instead requires $g=-c{f}^{\prime }$. Thus “released from rest” and “launched to the right” describe different Cauchy data.

The domain of dependence of $(x_{\ast },t_{\ast })$, for $t_{\ast }\ge 0$, is contained in $[x_{\ast }-c{t}_{\ast },x_{\ast }+c{t}_{\ast }]$ on the initial line. More precisely, displacement contributes only through the two endpoints, while initial velocity contributes throughout the interval. With $c=1$, $x_{\ast }=0.5$ and $t_{\ast }=1.5$, that interval is $[-1,2]$. Changing either initial datum outside it cannot change $u(0.5,1.5)$. If both initial data vanish outside $[-a,a]$, the solution vanishes outside $[-a-ct,a+ct]$: this is finite propagation speed. The expanding interval bounds possible influence; it need not be filled by nonzero displacement, as the separated bumps demonstrate.

Historically, d'Alembert studied the vibrating-string problem in “Recherches sur la courbe que forme une corde tendue mise en vibration”, in the Berlin Academy's volume for the year 1747, pages 214–219. The memoir introduces the differential equation and integrates it through two arbitrary one-variable functions. The original geometric problem makes the formula's enduring appeal clear: arbitrary initial shapes can be handled without first decomposing them into normal modes. ^[ref]

For a brief derivation, use characteristic coordinates $\xi =x-ct$ and $\eta =x+ct$. The chain rule gives $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }$, so $u=F(\xi )+G(\eta )$. The initial conditions yield $F+G=f$ and $-c{F}^{\prime }+c{G}^{\prime }=g$. Integrating these relations gives the displayed formula; the arbitrary integration constants cancel in their sum.