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In an inertial frame, a body with zero net external force remains at rest or moves with constant velocity. For a point mass of constant positive mass, write the net external force as the vector sum of all external forces on that body:Here ${\mathbf{r}}_0$ and ${\mathbf{v}}_0$ are the position and velocity at time $t_0$. The zero net external force must hold throughout the interval, not merely at one instant. Both the magnitude and direction of the velocity stay fixed: constant speed along a curved trajectory still involves acceleration. Rest is the case ${\mathbf{v}}_0=0$. For an extended closed system, the analogous statement concerns its center of mass; zero net external force need not prevent its parts from moving relative to one another.

The short mathematical argument is to use Newton's second law with $m>0$, then take the integral of $d\mathbf{v}/dt=0$ and of $d\mathbf{r}/dt={\mathbf{v}}_0$. This gives the two constants fixed by the initial position and velocity. The physical content also includes the existence of inertial frames in which these relations apply; an arbitrary accelerating reference frame need not show constant velocity for a body with zero actual net external force.

The useful question is therefore which forces change the motion, rather than which force keeps it going. A free-body diagram may show several nonzero forces whose vector sum is zero. Equal opposing forces on the same body can balance; a third-law interaction pair acts on different bodies and cannot be cancelled within that one body's diagram.

In 1687, Isaac Newton stated Law I in Philosophiae Naturalis Principia Mathematica: a body's rest or uniform straight motion persists unless impressed forces compel a change. The following paragraph identifies air resistance and gravity as influences on the trajectory of a thrown body. The vector equations above are a modern expression of the law, not a quotation of his notation.^[ref]

In an inertial frame, a point mass of constant mass $m>0$ obeysThe net force is the sum of all forces acting on that body; momentum, velocity and acceleration refer to the same reference frame. A force changes velocity, rather than sustaining it: zero net force implies constant velocity, including a nonzero one. Doubling the net force at fixed mass doubles acceleration; doubling mass at fixed net force halves it. ^[ref]

For example, an otherwise balanced block of mass $0.5\mathrm{kg}$ receives a constant horizontal net force $2\mathrm{N}$ for $0.5\mathrm{s}$. Its acceleration is $4m/{\mathrm{s}}^2$, so its velocity changes by $2m/s$. Equivalently, the impulse is $F\Delta t=1\mathrm{N}\mathrm{s}$, and $\Delta v=\Delta p/m$. Starting from rest, its displacement during the pulse is $a(\Delta t{)}^2/2=0.5\mathrm{m}$. After the pulse, zero net force leaves it moving at constant velocity.

Applying the law therefore requires both a specification of the net force and initial conditions. In the harmonic oscillator example, the force is $-kx$, so Newton's second law becomes a differential equation for position. In an accelerating reference frame, the measured relative acceleration includes a contribution from the frame's own motion; the equation of motion in a rotating frame illustrates the extra terms required.

Constant mass is essential to the displayed $m\mathbf{a}$ form. A variable-mass system exchanging matter also exchanges momentum; its balance includes the entering or leaving matter's velocity. Simply differentiating $m\mathbf{v}$ and equating it to external force on that open body can omit this flux. The existing variable-mass system equation specifies the needed correction.

In the 1687 Principia, Newton's second law was stated in terms of change of motion proportional to impressed force, directed along that force. Its original statement is verbal; the derivative equation above is modern notation. ^[ref]

Use a translating accelerated reference frame; its axes do not rotate. If $\mathbf{R}=(\xi ,\eta ,\zeta )$ is the block's position relative to the table, then its inertial position is $\mathbf{r}=\mathbf{R}+\mathbf{A}\sin \omega t$. Its relative velocity is $\dot{\mathbf{R}}$ and its inertial acceleration is $\ddot{\mathbf{R}}-{\omega }^2\mathbf{A}\sin \omega t$. Let $N$ be the upward normal force. During contact $\zeta =\dot{\zeta }=\ddot{\zeta }=0$, and the linear tangential resistance givesThese are Newton's second law transformed to the table's reference frame; the extra terms arise from the table's acceleration. The last equation gives $N=m(g-A_z{\omega }^2\sin \omega t)$ while contact persists. If contact is lost, both the contact resistance and $N$ vanish, and free flight must be described by the corresponding gravity-only equations.

For $A_z=0$, assume $\lambda >0$ and put $\gamma =\lambda /m$. Seek a periodic particular solution ${\xi }_p=b\sin \omega t+c\cos \omega t$. Substitution into the first differential equation givesSolving yieldsThe homogeneous solutions are a constant and a multiple of $e^{-\gamma t}$. Choose the phase $0<\theta <\pi /2$ with $\tan \theta =\omega /\gamma =m\omega /\lambda$. Then $b=-A_x{\sin }^2\theta$ and $c=-A_x\sin \theta \cos \theta$, soAlso $\eta (t)={\eta }_{\infty }+C_{\eta }{e}^{-\lambda t/m}$. The decaying transients leave the stated steady orbit. Positive damping is essential to this late-time assertion; for $\lambda =0$ an additional constant drift may persist.

No attractive contact force is available, so $N$ cannot be negative. Over a whole oscillation its minimum is $m(g-{\omega }^2|A_z|)$. Thus strictly positive contact loading throughout requires $\boxed{{\omega }^2|A_z|<g}$, or the printed form with $A_z$ taken as a nonnegative vertical amplitude. If ${\omega }^2|A_z|>g$, contact cannot persist: it would require $N<0$ during part of the cycle. At equality, $N$ vanishes only at isolated instants and the ideal constrained motion can still remain in contact. Hence the strict inequality characterizes positive loading; mere nonnegative contact permits the marginal equality case.

A tidal force measures the difference of gravitational acceleration across an extended object. A freely falling reference frame removes a spatially uniform acceleration, but not its gradient. For a distant point mass $M$ at distance $R$ and direction $\hat{\mathbf{n}}$ from the reference centre, a small internal displacement $\mathbf{x}$ givesThe tidal tensor has eigenvalues $2GM/R^3$ radially and $-GM/R^3$ in each transverse direction: radial stretching and transverse squeezing. Its trace vanishes in the source-free region, consistent with the gravitational Gauss law. An associated relative potential is ${\Phi }_{\mathrm{tide}}=-GM[3(\hat{\mathbf{n}}\cdot \mathbf{x}{)}^2-x^2]/(2R^3)$.

The scale is $GMl/R^3$ across size $l$, not the centre's absolute gravity $GM/R^2$. This distinction explains how a more distant massive object can exert a large force but a modest tide. Opposite-side stretching supports the elementary two-bulge picture; real ocean tides also involve rotation, basin geometry and fluid dynamics, so the static gradient alone does not predict coastal heights or timing. The same gradient stresses satellites and can disrupt a self-gravitating body when it competes with the body's own gravitational restoring acceleration.

A fictitious force accounts for acceleration of the chosen reference frame rather than an additional interaction. For axes rotating at constant $\mathbf{\Omega }$, differentiating a vector uses $(d/dt{)}_{\mathrm{lab}}=(d/dt{)}_{\mathrm{rot}}+\mathbf{\Omega }\times$. Twice applying this givesso moving the last two terms to the force side produces the Coriolis and centrifugal forces. Here $\mathbf{v}$ is measured in rotating axes.

The rotating saddle trap has potential force $(-m{\omega }^{2}x,+m{\omega }^{2}y)$. ThereforeThe trial solution givesBoth modes are bounded for arbitrary sufficiently small initial displacements and velocities precisely when $|\Omega |>\omega$. At equality there is a zero-frequency secular mode; the exactly stationary solution does not establish stability.

For rapidly rotating saddle confinement with $\Omega \gg \omega >0$, real mode ellipses can be writtenThey are nearly circular clockwise motions at rates ${ϵ}_{\pm }\simeq \Omega \pm {\omega }^2/(2\Omega )$. Rotating these axes back to the laboratory cancels the large angular rate: the leading laboratory motion is a slow approximately harmonic ellipse with frequency ${\omega }^2/(2\Omega )$, plus small rapid micromotion. It results from the changing saddle force, not a static potential minimum. At next order the ellipse precesses prograde at ${\omega }^4/(8{\Omega }^3)$. The rotating-potential model and this guiding-centre distinction are analysed in the primary work cited here.^[ref]

One application is the slow steady precession of a rapidly spinning gyroscope supported at one point. Let its centre of mass be a distance $\ell$ from the pivot along its spin axis, at angle $\alpha$ to the upward vertical. Write $J\simeq I_1\omega$ for its dominant spin angular momentum. In a reference frame precessing steadily about the vertical with angular velocity ${\mathbf{\Omega }}_p$, the spin angular momentum is approximately stationary, soThe gravitational torque has magnitude $mg\ell \sin \alpha$. The cross product on the right has magnitude ${\Omega }_{p}J\sin \alpha$, giving, for a nonvertical axis,Its direction follows from the torque equation. This is the slow precession approximation: the additional angular momentum associated with precession is small compared with the spin contribution.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

The spacecraft clock measures proper time $\tau$, while a clock at rest with Earth measures the launch-to-emission interval $t$. Neglect Earth's subsequent motion and gravity, as in the intended special relativity model. The Lorentz factor and time dilation giveIn Earth's inertial frame, the spacecraft's displacement at emission is $d=vt$. ThereforeHere $\tau =43200\mathrm{s}$ and the speed of light is $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$. The distance is evaluated at the emission event, using simultaneous Earth-frame endpoints, as required by relativity of simultaneity; the return signal has not yet reached Earth. A distance assigned on the spacecraft's simultaneous time slice is different: Earth has moved away at $c/3$ for $12\mathrm{h}$ in that frame, giving $c\tau /3=4$ light-hours. Specifying the reference frame is essential to the numerical answer.

At the south pole $\mathbf{r}=-R\hat{z}$. Choose plate C as the instantaneous reference frame. Since $R=4\times 10^9$ mm, the two components of relative angular speed give 20 and 48 mm/year. The plate velocities in the fixed tangent-plane axes areThe plate velocity triangle therefore has sidesThe physical boundary rays point along $+x$ for AB, $+y$ for AC and $(12,-5)$ for BC. Plate A occupies the quadrant between the first two rays, B the narrow wedge between AB and BC, and C the remainder. The stated overriding relations make AB and AC subduction trenches, while BC remains a transform fault: this is a trench-trench-transform triple junction. Put the subduction teeth on B's side of AB and A's side of AC, identifying B above A and A above C.

For kinematic stability there must be a single junction velocity $\mathbf{U}$ compatible with all three boundary-normal motions. Use the standard idealization that each trench's normal velocity is that of its overriding plate, with no independent rollback. AB is parallel to $x$, so $U_y=v_{B,y}=20$. AC is parallel to $y$, so $U_x=v_{A,x}=-48$. The transform requires $\mathbf{U}-{\mathbf{v}}_C$ to be parallel to $(12,-5)$, which is true for this intersection. HenceThe southern triple junction is instantaneously kinematically stable. It is stationary relative to B, migrates along AB relative to A, and migrates along BC relative to C. This compatibility result does not promise long-term stability if plate rotations or trench rollback change. Notice that the three relative velocities close even though the overriding order is not a statement that all three plates move toward one common stationary point.

The three plate velocities plotted in a common reference frame form a triangle whose directed edges are the pairwise relative velocities. Adding the same reference velocity translates the triangle without changing its edges.

Proper motion describes the change in a celestial object's direction on the sky due to its motion relative to the observer. Its angular velocity differs from the annual stellar parallax signal caused by the observer's orbit.

Thus proper motion is an angular velocity, rather than a speed in kilometres per second. It describes the relative transverse motion of source and observer in a specified reference frame. A star moving towards us can have little proper motion, while a nearer star can show larger proper motion than a faster but more distant one. In the example above, the observed coordinate actually decreases over the first half-year, although $\mu$ is positive: the annual stellar parallax excursion temporarily outweighs the steady drift. A short observed displacement alone therefore need not estimate proper motion well.

Real catalogues give two proper motion components. For right ascension $\alpha$ and declination $\delta$, the angular velocities in perpendicular directions on the local sky areThe cosine factor accounts for the shorter east-west arc associated with a given right ascension increment away from the celestial equator. ESA's Gaia DR3 fields pmra and pmdec already contain these two components; pm is their total proper motion. Do not multiply pmra by another cosine. Components of $60$ and $80\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$, for example, give total proper motion $100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$. ^[ref].

For nearby stars in the ordinary nonrelativistic approximation, Euclidean distance $d$ converts total proper motion into transverse speed $v_{\perp }$:The small-angle approximation gives transverse speed as Euclidean distance times angular velocity in radians per unit time. With arcseconds and parsecs, the parsec convention turns their product into astronomical units per year. One astronomical unit per Julian year of $365.25$ days is approximately $4.74047\mathrm{km}{\mathrm{s}}^{-1}$. The coefficient is a unit conversion, rather than a property of a particular star.

Taking the model's $100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}=0.100\mathrm{arcsec}\mathrm{y}{\mathrm{r}}^{-1}$ as the total proper motion, with the other component zero, gives $v_{\perp }=4.74047\times 0.100\times 20\simeq 9.48\mathrm{km}{\mathrm{s}}^{-1}$. At the same total proper motion, Euclidean distances of $10$, $20$ and $100$ parsecs instead give transverse speeds of $4.74$, $9.48$ and $47.40\mathrm{km}{\mathrm{s}}^{-1}$, respectively. Their stellar parallax values are $100$, $50$ and $10\mathrm{mas}$. The same annual angular displacement can therefore describe quite different journeys through space.

The transverse speed still omits the line-of-sight velocity, which can be measured through the Doppler effect in spectral lines. Combining these measurements permits a three-dimensional velocity estimate in a consistent reference frame. ^[ref].

Historically, Edmond Halley's 1718 paper compared contemporary and ancient stellar directions and argued that some supposedly fixed stars had moved. Its title, "Considerations on the change of the latitudes of some of the principal fixt stars", captures the conceptual surprise: long-separated measurements could reveal motion invisible in an ordinary night's watching. Modern astrometry retains that insight, using repeated precise measurements to turn the apparently static sky into a record of changing positions. ^[ref]. ^[ref].

Rapidity is a dimensionless parameter for a signed collinear velocity $u$ with $|u|<c$, defined using the inverse hyperbolic functions byHere $c$ is the speed of light, and $\tanh$ is the hyperbolic tangent. Zero velocity has zero rapidity; negative velocity has negative rapidity. Every finite real rapidity gives $|u|<c$, while $u\to \pm c$ requires $\theta \to \pm \infty$. Thus a light ray has no finite rapidity in this parametrization. The header's formula uses natural units $c=1$: its $v$ is the numerical velocity $u/c$.

For the inertial frames in the velocity-addition formula, $S^{\prime }$ moves at signed velocity $V$ along the $x$-axis of $S$. Set $\varphi =\mathrm{artanh}(V/c)$. Re-expressing the same particle in the moving inertial frame givesThe subtraction is a passive change of reference frame; the inverse adds the particle's rapidity in $S^{\prime }$ to the frame's rapidity relative to $S$. For example, $V=0.6c$ and $u^{\prime }=0.6c$ give $\varphi ={\theta }^{\prime }=\ln 2$, so $\theta =\ln 4$ and $u=c\tanh (\ln 4)=15c/17$, exactly as in the velocity-addition formula.

For a massive particle of rest mass $m>0$ moving in one spatial dimension, its energy and signed momentum areThe hyperbolic cosine is the particle's Lorentz factor, and the hyperbolic sine is $\gamma u/c$. The hyperbolic Pythagorean identity immediately gives the relativistic energy-momentum relation $E^2-c^2{p}^2=m^2{c}^4$. In the older $c=1$ notation, writing the rest mass as $M$ recovers $v=\tanh \theta$, $E=M\cosh \theta$, and $p=M\sinh \theta$. These expressions use natural units in which the factors of $c$ are numerically one; the preceding formulas restore those factors explicitly. ^[ref].

The useful simplification follows from the hyperbolic-function identity $\tanh (a+b)=(\tanh a+\tanh b)/(1+\tanh a\tanh b)$: the nonlinear velocity-addition formula becomes ordinary addition of rapidities. See iterated collinear boosts with equal rapidity for a plotted example. A common collinear Lorentz boost shifts every rapidity by the same constant, so a rapidity difference is invariant. This makes rapidity useful for relativistic factorized scattering in one spatial dimension, where two-body S-matrices depend on the difference of the incoming rapidities, with particle species fixed.

The Lorentz transformation changes the photon four-momentum and therefore its frequency, producing the displayed formula for a boost along the first axis. Here the primed ray angle is measured in the emitting frame and $\beta =v/c$. The angle and frequency transforms must be used together: aberration of light changes the direction, and a photon-number angular fraction need not equal the corresponding radiant-energy fraction.

Here $c$ is the vacuum speed of light, $\gamma$ is the Lorentz factor, and the two frequencies are measured in different inertial reference frames. Unlike the classical Doppler effect for sound, there is no material rest reference frame for light in vacuum. For small radial recession speed $v_r\ll c$, the wavelength shift satisfies $\Delta \lambda /\lambda \simeq v_r/c$. This small-speed limit underlies Doppler spectroscopy; it is not an exact sound formula.

The Technische Universität Darmstadt demonstration uses a steel ball launched outward from the centre of a rotating disc. A fixed lecture-room camera records a straight path, while a camera carried above the disc records a curved path. The instructions specify launching toward the fixed camera so that the laboratory motion is clearly visible. The rotating iPod transmits its view wirelessly through Apple TV; two vertical support rods of length $1\mathrm{m}$ are specified for the slow-motion arrangement. The paired video views are the readout: one physical motion is represented in two reference frames.^[ref]

A useful preparation control appears in the Universität Freiburg instructions: with the disc stationary, check that the ball travels radially toward its edge. A camera is mounted on supports attached to the disc; turn the disc by hand as the ball leaves its centre and observe the camera view.^[ref]

The inference is a dependence of the observed trajectory on the rotating axes; a curved camera trace alone does not isolate the Coriolis force. To compare with the constant-angular velocity free particle in rotating coordinates, measure the disc angular velocity, calibrate lengths in the camera image, and compare launches at rest and with opposite rotation directions. These are checks of the ideal model, not reported measurements from the manuals. A rolling ball can exchange forces with the surface through friction, and hand rotation need not have constant angular velocity; an Euler acceleration then enters too. The ideal free-ball drawing deliberately omits those effects. Neither manual supplies a numerical uncertainty or a precision measurement of the Coriolis acceleration.

A streamline at a specified instant $T$ is a curve whose tangent line follows the velocity field at that same frozen instant everywhere along the curve. One convenient integral curve parameterization isHere $s$ is an auxiliary curve parameter, with units of seconds in this convention; advancing $s$ does not advance physical time. Any regular reparameterization preserves the tangent geometry. Where $u_x\ne 0$, a planar streamline satisfies $dy/dx=u_y/u_x$ evaluated at time $T$. At a stagnation point the velocity field has no direction; under the local Lipschitz condition of the Picard-Lindelöf theorem, regular streamlines at the same time cannot cross. A streamline answers which directions fit together now, making it useful for describing local flow geometry. ^[ref]

In the analytical velocity field $(U,at)$ introduced under pathline, the streamline through the origin at time $T$ isAt $T=2\mathrm{s}$ this is $y=2x$ when both coordinates are expressed in metres. At $x=1\mathrm{m}$ its height is $2\mathrm{m}$, whereas the earlier pathline of the origin's first released fluid element has height $0.5\mathrm{m}$. Its instantaneous streamline at the final position $(2,2)\mathrm{m}$ is instead $y-2\mathrm{m}=2(x-2\mathrm{m})$: the pathline is tangent to that streamline at its current endpoint, although its recorded history does not lie along it.

The green streamline uses the velocity field at $T=2\mathrm{s}$ throughout; the orange pathline uses times $0\le t\le 2\mathrm{s}$; the purple streakline uses different release times at the single observation time $T=2\mathrm{s}$. The grey arrows all show the instantaneous velocity direction at $T$. These are exact analytical curves with chosen units, not measured dye shapes.

In a steady flow in a fixed reference frame, a pathline and a streamline solve the same autonomous ordinary differential equation. Uniqueness makes each particle released from a fixed point follow the same integral curve, so its pathline, the streakline from that point and the streamline share one geometrical curve. Finite recording or release intervals usually show different segments of that curve; the comparison assumes particles stay in the domain and treats stationary fluid elements separately. Neither two-dimensionality nor incompressibility is needed for this coincidence. Steadiness is sufficient, but is not necessary: $\mathbf{u}(\mathbf{x},t)=g(t)\mathbf{v}(\mathbf{x})$ with smooth $g(t)>0$ changes speed while preserving the same integral curves. In a general time-dependent velocity field, however, the three curves differ. Kline explains why continuous dye release in steady flow can therefore display streamlines. ^[ref]

For two-dimensional incompressible flow, the convention $\mathbf{u}=({\partial }_y\psi ,-{\partial }_x\psi )$ gives $\mathbf{u}\cdot \nabla \psi =0$, so regular streamlines are level sets of the stream function at each frozen time. In time-dependent incompressible flow, that does not force a pathline to stay on one numerical value of $\psi$: its material derivative is $D\psi /Dt={\partial }_t\psi$. The distinction is the difference between a geometric snapshot and a parcel's history.

If the origin of a nonrotating reference frame has inertial position $\mathbf{a}(t)$ and a particle has relative position $\mathbf{R}$, then $\mathbf{r}=\mathbf{R}+\mathbf{a}$. Newton's second law becomes $m\ddot{\mathbf{R}}=\mathbf{F}-m\ddot{\mathbf{a}}$. The effective inertial force is therefore $-m\ddot{\mathbf{a}}$. No rotation-dependent terms arise because the coordinate axes remain parallel to the inertial axes.

Velocity is the vector giving the rate of change of position with time, measured in a specified reference frame. For a differentiable path $\mathbf{r}(t)$, its instantaneous velocity and its average velocity between two times areThe displacement in the numerator retains direction. In Cartesian coordinates, each vector component is a derivative, for example $v_x=dx/dt$; its sign tells whether the corresponding coordinate is increasing or decreasing. The SI unit of velocity is $m/s$. ^[ref]

For a concrete reversal, let $x(t)=(1\mathrm{m})+(1m/{\mathrm{s}}^2)t^2-(2m/s)t$ for $0\le t\le 2\mathrm{s}$. Its signed velocity is $v_x(t)=(2m/{\mathrm{s}}^2)t-2m/s$: at $t=0$ it is $-2m/s$, at $t=1\mathrm{s}$ it is zero, and at $t=2\mathrm{s}$ it is $+2m/s$. The opposite endpoint velocities describe opposite directions, although the endpoint speeds agree. The position moves from $1\mathrm{m}$ to $0$ and back to $1\mathrm{m}$.

Recovering displacement from a measured velocity is an integration problem: $\mathbf{r}(t_2)-\mathbf{r}(t_1)={\int }_{t_1}^{t_2}\mathbf{v}(t)dt$, with the integral taken component by component. In the reversal example, ${\int }_0^{2\mathrm{s}}{v}_x(t)dt=0$: the negative and positive contributions cancel. This distinction is useful in navigation, where the final position depends on direction as well as speed.

The velocity-addition formula translates a particle's velocity between two specified inertial frames. Let $S^{\prime }$ move at signed velocity $V$ along the $x$-axis of $S$, with parallel axes and $|V|<c$, where $c$ is the speed of light. Define $u_x=dx/dt$ and $u_x^{\prime }=d{x}^{\prime }/d{t}^{\prime }$ for the same particle's world line, using position and time measured within the corresponding inertial frame. With Lorentz factor ${\gamma }_V=(1-V^2/c^2{)}^{-1/2}$, the forward Lorentz boost gives

For collinear motion, write $u=u_x$ and $u^{\prime }=u_x^{\prime }$. The inverse velocity-addition formula isThis is the composition of the particle's velocity in $S^{\prime }$ and that frame's velocity relative to $S$. The velocities must belong to this same chain of reference frames, with directions specified. Two arbitrary quoted speeds cannot be inserted blindly. Each velocity measurement uses distances and clocks in a single inertial frame, with Einstein synchronisation for separated clocks.

For example, let $V=0.6c$ and let the particle travel at $u^{\prime }=0.6c$ in $S^{\prime }$. Its velocity in $S$ isThe Galilean transformation would instead give $u=u^{\prime }+V=1.2c$, which fails here. For light moving along the positive axis, $u^{\prime }=c$ gives $u=c$; $u^{\prime }=-c$ similarly gives $u=-c$. Negative directions also work for massive particles: at $V=0.6c$, $u^{\prime }=-0.6c$ gives $u=0$, while $u^{\prime }=-0.8c$ gives $u=-5c/13$. Thus “addition” means composition of signed velocities, rather than addition of their magnitudes.

A short derivation differentiates the Lorentz transformation: $d{x}^{\prime }={\gamma }_V(dx-Vdt)$ and $d{t}^{\prime }={\gamma }_V(dt-Vdx/c^2)$. Dividing these differentials gives the longitudinal velocity-addition formula. The transverse formulas use $d{y}^{\prime }=dy$ and $d{z}^{\prime }=dz$, but the same transformed time denominator; omitting it would ignore the relativity of simultaneity. Replacing $V$ by $-V$ gives the inverse. When both relative and particle velocities are small compared with the speed of light, the denominator is close to one and the nonrelativistic limit recovers the Galilean transformation.

The formula lets us compare particle motion in a moving apparatus and a laboratory without losing the invariant speed of light. Einstein's 1905 paper derives the coordinate change in §3 and the composition of velocities in §5, including the collinear formula, its light-speed case, and composition of parallel transformations. ^[ref]. The rapidity parametrization makes the collinear composition particularly simple.