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The 1972 paper Around-the-World Atomic Clocks: Observed Relativistic Time Gains by J. C. Hafele and others reports four cesium-beam atomic clocks flown around the world in October 1971, eastward and westward. Their readings were compared with the U.S. Naval Observatory reference time scale before and after travel. An electronic time-interval counter resolved differences to the nearest nanosecond; intercomparisons among the flying clocks continued during travel. The final ensemble results were $-59\pm 10\mathrm{ns}$ eastward and $+273\pm 7\mathrm{ns}$ westward; the quoted spreads are standard deviations. These are corrected clock offsets, not raw readings or aircraft travel times. Four clocks allowed identification of individual rate changes through patterns in their mutual differences. The authors also report magnetic shielding and laboratory orientation checks. Their analysis could not detect a disturbance common to every flying clock, and the small ensemble limits statistical conclusions. This is a concrete reunited-clock experiment with apparatus, readout and drift controls.^[ref]

The companion paper Around-the-World Atomic Clocks: Predicted Relativistic Time Gains by J. C. Hafele and others computes $-40\pm 23\mathrm{ns}$ eastward and $+275\pm 21\mathrm{ns}$ westward from the recorded flight paths. These are predictions, with flight-data uncertainty estimates. Both time dilation from motion and gravitational time dilation matter: aircraft clocks at altitude experience a different gravitational potential, while the rotating Earth means a ground clock is not the inertial home clock of our simple model. Eastward and westward motion change the aircraft's velocity relative to an approximately nonrotating inertial frame, not merely its speed relative to the ground. The measured agreement therefore tests combined special relativity and general relativity; the westward gain does not contradict the maximum-proper time result for fixed endpoints in Minkowski spacetime.^[ref]

A complementary microscopic test is Measurements of relativistic time dilatation for positive and negative muons in a circular orbit by J. Bailey and others. The authors measured muon decay lifetimes in a storage ring at Lorentz factor $\gamma =29.33$: $64.419\pm 0.058\mu \mathrm{s}$ for positive muons and $64.368\pm 0.029\mu \mathrm{s}$ for negative muons. The reported positive-muon lifetime agrees with the independently measured lifetime at rest and special relativity to a fractional error of $2\times 10^{-3}$ at 95% confidence. This is a decay-lifetime readout under continuous centripetal acceleration, rather than a reunion of portable clocks. It supports the clock hypothesis for this physical process and regime; the inspected abstract does not establish universal immunity of every clock mechanism to acceleration. ^[ref]

During either inertial leg, the traveller can correctly say that the moving home clock runs slowly. The missing step is deciding which distant home-clock reading is simultaneous with turnaround. By the Lorentz transformation, the outward inertial frame's constant-time line through $P$ obeys $t-vx/c^2=\text{constant}$. At $x=0$ it meets the home world line atFor the return inertial frame, use $t+vx/c^2=\text{constant}$ through the same spacetime event $P$; its home-clock assignment isThe dashed lines in the spacetime diagram illustrate this relativity of simultaneity. For $v=0.6c$ and $T=10\mathrm{yr}$, they assign $3.2\mathrm{yr}$ and $6.8\mathrm{yr}$ to the home clock. The difference of $3.6\mathrm{yr}$ is a change in the traveller's distant-time assignment, not a sudden physical advance of the home clock.

On each leg, the traveller's $4\mathrm{yr}$ of proper time corresponds to $4/\gamma =3.2\mathrm{yr}$ of home-clock advance according to that leg's inertial frame. Including the twin paradox simultaneity change gives $3.2+3.6+3.2=10\mathrm{yr}$, exactly the reading at reunion. Actual received light signals remain continuous across a physical turnaround; their arrival times are distinct from the distant spacetime events assigned as simultaneous by Einstein synchronisation.

An instantaneous reversal is an idealization. A real traveller changes velocity over a finite time, follows a smooth world line, and has finite proper acceleration. Special relativity still calculates the elapsed proper time using the home inertial frame and the same integral; general relativity is not needed merely because the traveller accelerates in otherwise flat spacetime. The clock hypothesis identifies an ideal clock's reading with this geometric proper time.

One explicit smooth trip between the same $O$ and $R$ starts and ends at rest in the home inertial frame. ChooseIts velocity reverses continuously at $T/2$, and its proper acceleration is bounded because $v_{\max }<c$ and $dv/dt$ is finite. With $v_{\max }=0.6c$ and $T=10\mathrm{yr}$, numerical integration givesThis comparison fixes the reunion spacetime events and maximum speed, not the turnaround distance: the smooth trip reaches $v_{\max }T/\pi \simeq 1.910$ light-years, whereas the ideal trip reaches $3$ light-years. The smooth traveller spends less time at high speed and therefore accumulates more proper time than the ideal traveller, while still accumulating less than the home clock.

For a sufficiently short smooth reversal replacing the corner, the proper time approaches the piecewise inertial result. A reversal interval of home duration $\delta t$ contributes between zero and $\delta t$ to the clock reading; shrinking that interval cannot create a finite extra loss concentrated at the corner. The finite difference survives in the long coasting legs. Keeping the same reversal manoeuvre while lengthening those legs changes the age difference. Conversely, two travellers following mirrored paths with the same speed history accumulate equal proper times, even though both accelerate. Thus acceleration distinguishes paths and enables reunion here, but an amount of acceleration alone does not determine the age difference.

The changing simultaneity assignment also becomes continuous. At a traveller spacetime event $(t,x(t))$ with instantaneous velocity $v(t)$, the momentarily comoving inertial frame assigns home time $t_H=t-v(t)x(t)/c^2$. Differentiating givesDuring the reversal at positive $x$, the last term advances the assigned home time rapidly. It describes the traveller's changing relativity of simultaneity, not a locally measured alteration of the home clock's ticking.

Work in Minkowski spacetime, neglect gravity, and choose the stay-at-home clock's inertial frame. Let departure be $O=(t,x)=(0,0)$ and reunion be $R=(T,0)$. The home clock follows $x_H(t)=0$. A traveller moves outward with velocity $+v$ and returns with velocity $-v$, where $0<v<c$ and $c$ is the speed of light. The ideal world line has a sharp turnaround at $P=(T/2,vT/2)$:Each straight leg lies in an inertial frame, but no single traveller inertial frame covers both legs.

For any differentiable timelike curve in this home inertial frame, its clock accumulates proper time through the integralApply it separately to the traveller's two legs. With the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For $v=0.6c$ and $T=10\mathrm{yr}$, the Lorentz factor is $\gamma =1.25$. The turnaround is $3$ light-years away at home time $5\mathrm{yr}$. The traveller records $4\mathrm{yr}$ outward and $4\mathrm{yr}$ returning: the reunited clocks read $10\mathrm{yr}$ and $8\mathrm{yr}$. These are illustrative model values, not experimental measurements.

The same integral gives a short geometric explanation: $\sqrt{1-v(t{)}^2/c^2}\le 1$, so every such competing timelike curve between these fixed spacetime events has $\Delta \tau \le T$. Equality requires zero velocity throughout, apart from sets of zero duration. Thus the straight home world line maximizes proper time in this Minkowski spacetime problem. This statement concerns fixed endpoints and an inertial home clock; in a gravitational field, compare paths using the appropriate metric tensor.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

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.

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.