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If one clock measures a proper time interval $\Delta \tau$ while moving at constant speed $v$ in an inertial frame, that frame measures $\Delta t=\gamma \Delta \tau$, where $\gamma$ is the Lorentz factor. The two spacetime events must lie on that clock's world line.

Suppose a clock moving at $0.6c$ advances by $4\mu \mathrm{s}$ between two ticks. Set $t=x=t^{\prime }=x^{\prime }=0$ at the first tick. The laboratory's synchronized clocks assign $\Delta t=1.25\times 4=5\mu \mathrm{s}$. During that interval the clock travels $\Delta x=0.6c\times 5\mu \mathrm{s}=3c\mu \mathrm{s}$, approximately $899.38\mathrm{m}$. The Minkowski interval checks the result:Both frames therefore recover the same $4\mu \mathrm{s}$ of proper time. This calculation is useful whenever a process is timed on a moving apparatus but described using laboratory clocks.

A light clock makes the mechanism visible. Two mirrors separated by $h$ perpendicular to the motion exchange a pulse; define one tick as its complete return to the lower mirror. In the clock's rest frame the tick takes $\Delta \tau =2h/c$. In the laboratory each half-tick lasts $T$ and the light traverses a diagonal, so the Pythagorean theorem givesFor $h=1\mathrm{m}$ and $v=0.6c$, each leg covers $1.25\mathrm{m}$ while the mirrors move $0.75\mathrm{m}$. The return tick lasts $2.5\mathrm{m}/c$ instead of $2\mathrm{m}/c$: the light has the same speed of light in both accounts. ^[ref].

Reciprocal time dilation is consistent: each inertial frame times the other frame's single moving clock using its own synchronized clock network. These comparisons use different simultaneous spacetime events. Comparing reunited clocks instead requires their complete world lines, as in the twin paradox.

A timelike curve has a nonzero tangent vector with negative Minkowski norm in metric signature $(-,+,+,+)$ at every regular point. A physical massive observer's world line is also future-directed. Parametrizing it by coordinate time gives $-c^2+|\mathbf{v}{|}^2<0$, or $|\mathbf{v}|<c$: the local tangent condition is exactly the subluminal speed condition. It is stronger than checking only that two endpoints have timelike separation.

The twin paradox compares proper time along the different world lines of a travelling observer and a stay-at-home observer who reunite. Reciprocal inertial time dilation produces an apparent contradiction only if the traveller's change of inertial frame is ignored.

The decisive experiment is local: put two clocks together at departure, separate them, then compare their readings together at reunion. Everyone agrees on those two spacetime events and on the final difference. The twin paradox asks why two clocks can accumulate different proper times between the same spacetime events, despite reciprocal time dilation during uniform motion. Its answer is the geometry of their complete world lines, together with the relativity of simultaneity when comparing distant clocks.

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.

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.

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.