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The clock hypothesis says that an ideal clock following a timelike curve measures its proper time: its instantaneous rate depends on velocity through $d\tau =dt\sqrt{1-v^2/c^2}$, with no separate acceleration term. Acceleration still changes the world line and hence the accumulated proper time. A physical clock may suffer disturbances under forces or environmental changes, so experiments must distinguish its mechanism's errors from the geometric prediction.

Hyperbolic motion in special relativity is rectilinear motion with constant proper acceleration. A representative positive-$x$ world line is $x=c{t}_0\cosh (\tau /t_0)$, $t=t_0\sinh (\tau /t_0)$, with proper acceleration $c/t_0$. This describes a hyperbola in spacetime, not a hyperbolic Kepler orbit.

The light cone of a spacetime event $O$ consists of $O$ together with the spacetime events having null separation from it. In the coordinates of an inertial frame centered on $O$, it obeysFor $t>0$ it traces an expanding spherical flash of light, with radius $ct$ at time $t$; for $t<0$ it describes possible incoming light rays. A spacetime diagram with one spatial coordinate shows the two lines $x=\pm ct$. The apparent $45^{\circ }$ angle uses equal scales for $x$ and $ct$ and is not a separate physical angle.

The light cone is the boundary, while its interior has timelike separation from $O$. An outgoing light ray follows the boundary; a future-directed massive observer's world line remains inside it. The exterior has spacelike separation: no signal travelling at or below the speed of light can connect its spacetime events directly to $O$. This separates possible causal influence from merely being assigned a later coordinate time.

In the diagram, $A=(4\mathrm{s},3\mathrm{ls})$ is inside the future boundary, $N=(3\mathrm{s},3\mathrm{ls})$ is on it, and $B=(2\mathrm{s},3\mathrm{ls})$ is outside. Their squared Minkowski intervals are respectively $-7,0,5$ in ${\mathrm{ls}}^2$. Thus $B$ occurs later than $O$ in the displayed inertial frame while still being unable to receive a light-speed-or-slower signal sent from $O$.

The usefulness is immediate for an experiment: a detector's reading can depend on a source event only if a causal influence can reach it. A Lorentz boost changes coordinates but preserves this classification. For a boost with $\beta =0.8$ and Lorentz factor $\gamma =5/3$, applying the Lorentz transformation gives, in light-second length units,The corresponding squared Minkowski intervals remain $-7,0,5$. In particular $B$ is now earlier than $O$, which is allowed for spacelike separation; $A$ remains later.

Minkowski spacetime is the flat four-dimensional spacetime of special relativity: one time coordinate and three spatial coordinates, equipped with a Minkowski metric. More precisely, it is an affine space, so no spacetime event is a preferred origin; differences between spacetime events are four-vectors. Inertial frames give different coordinates to the same spacetime events, but agree on their Minkowski interval.

This geometry answers two different measurement questions with one structure. Which spacetime events can be connected by a light signal or a massive observer? The light cone classifies them. How much time does a moving clock record? Its world line and the Minkowski metric determine its proper time. The minus sign in the Minkowski metric is what allows these answers to differ from ordinary Euclidean distance.

For a concrete one-space-dimensional slice, start at the origin $O$ and measure $t$ in seconds and $x$ in light-seconds. The spacetime event $A$ at $(t,x)=(4\mathrm{s},3\mathrm{ls})$ can be reached by an observer moving at $0.75c$, whereas $B=(2\mathrm{s},3\mathrm{ls})$ would require $1.5c$ to be reached directly from $O$. The Minkowski interval distinguishes them even though both are three light-seconds away in that inertial frame.

Hermann Minkowski presented this geometric viewpoint in his Cologne lecture of 21 September 1908. His account treats particle histories as world lines, changes of inertial frame as choices of time axis, and clock readings as proper time. These are connected geometric ideas, rather than separate corrections to ordinary mechanics.^[ref]

Before its second decay A moves uniformly with $x=-Vt$. Its Lorentz factor isThe proper time along its world line satisfies $d\tau =dt/{\gamma }_A$. Since the first decay occurs at the spacetime origin, integration givesEquivalently these are the components of A's constant four-velocity times $\tau$. The assumption $m>0$ is needed for a daughter rest frame and its proper-time clock; a massless daughter has neither.

Choose the barn entrance as $x=0$, its exit as $x=3$, and the positive $x$ direction as the javelin's motion. Put $t=0$ at the trailing end entering the barn. Write $s=ct$ and $s^{\prime }=c{t}^{\prime }$ in metres. In the javelin frame let the trailing end have $x^{\prime }=0$, the leading end $x^{\prime }=4$, and use the same origin event. The Lorentz transformation isIn the barn frame, the four world lines areIn the javelin frame they areThe two spacetime diagrams below use vertical time coordinates and horizontal spatial coordinates. Event labels refer to the same physical events in both panels; dashed lines show light-ray directions.

Event $A$ is the chosen origin. At $B$, the leading-end world line reaches $x=3$, soThe Lorentz transformation givesFor example $s_B^{\prime }=(13/5)(19/12-36/13)=-37/12$. The squared spatial separation minus squared time separation isequally $16-1369/144$ in the other frame. Thus $A$ and $B$ have spacelike separation.

In the barn frame, for $0<s<19/12$ the trailing end has entered while the leading end has not yet exited, so the whole javelin is inside at each simultaneous time slice. In the javelin frame the barn is shorter than the javelin and there is no simultaneous slice on which it contains the whole javelin. Also $B$ precedes $A$ there. This is consistent because relativity of simultaneity changes which endpoint events are simultaneous; the ordering of events with spacelike separation is not invariant. The physical intersections of the four world lines are the same in both diagrams.

The world lines in the train frame are $x_A=L/2$, $x_B=-L/2$ and $x_C=-vt$. In the station frame, length contraction givesThe emission event is $O=(0,0)$ in both frames. The signal world lines in $S^{\prime }$ are $x^{\prime }=u{t}^{\prime }$ and $x^{\prime }=-u{t}^{\prime }$; their intersections with the train endpoints define the reception events $R_A,R_B$. Their velocities in $S$, from the relativistic velocity-addition formula, are $(u-v)/(1-uv/c^2)$ and $-(u+v)/(1+uv/c^2)$.

The two platform-alignment events occur at $(x^{\prime },t^{\prime })=(\pm L/(2\gamma ),0)$ in $S^{\prime }$. The Lorentz transformation puts them at $(x,t)=(\pm L/2,\mp vL/(2c^2))$ in $S$, on the tilted simultaneity line $t=-vx/c^2$. The spacetime diagrams show all three observers, both signals, their receptions, and these distinct endpoint-alignment events:

Let $P$ be the platform's proper length, measured in its rest frame $S^{\prime }$. At the single station-frame time $t^{\prime }=0$, its ends coincide with train endpoints whose simultaneous station-frame separation is $L/\gamma$. ThusThe platform endpoint world lines are vertical at $x^{\prime }=\pm P/2$ in the station spacetime diagram. In the train frame they are $x=-vt\pm P/(2\gamma )$, so their separation at a common train-frame time is $\boxed{P/\gamma =L/{\gamma }^2}$. The two alignment events lie on the tilted line $t^{\prime }=0$ in the train diagram and are not simultaneous there; their separation $L$ cannot be used as the platform length measured at one train-frame time. This is the distinction enforced by length contraction.

In the station frame, intersecting the signal trajectories with the endpoint world lines givesSince $u>v$, both receptions occur at positive times,The inverse Lorentz transformation then givesSubtracting and simplifying yieldsFor $u<c$, both differences are negative: the rear receives first in both frames, but by different time intervals.

With light signals, $u=c$, the train-frame relativistic velocity-addition formula gives signal velocities $+c,-c$. The endpoints are equally distant from the source in that frame, andThe two reception events are simultaneous in $S$ but not in $S^{\prime }$: in the station frame the rear moves toward its light signal while the front moves away. Directly, $\Delta x=x_B-x_A=-L$ and $\Delta t=0$, so the Lorentz transformation gives $\Delta t^{\prime }=\gamma (\Delta t+v\Delta x/c^2)=-\gamma Lv/c^2$. This is relativity of simultaneity, concerning spatially separated events; it does not contradict the invariant light speed or the causal ordering of events on a single observer's world line.

In the Earth inertial frame, the return journey takes $2d/v$. Bob is at rest in this frame, while Alice's proper time is reduced by the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$. Neglecting the short turnaround interval, their increases in age are

The twin paradox is the apparent contradiction from applying reciprocal time dilation to claim that each twin must be younger than the other. Alice changes inertial frames at the turnaround, so a single comoving inertial frame cannot describe her whole journey; Bob uses one inertial frame throughout. The twins follow different world lines, and integrating proper time along those world lines gives the displayed, consistent ages. The change in Alice's notion of simultaneity at turnaround removes the apparent contradiction. Acceleration distinguishes the paths; its magnitude is not an additional local rule for aging.

The meetings occur where Bob's constant position $x=\alpha c{t}_0$ intersects Alice's positive-$x$ branch of the hyperbolic motion:Bob is at rest in this inertial frame, so his elapsed proper time is

Parametrize Alice's world line bySince $t$ increases with $\eta$, the proper time element isAt the meetings, $\cosh \eta =\alpha$, giving endpoints $\eta =\pm \mathrm{arcosh}\alpha$. Thus Alice's elapsed proper time isHere $\mathrm{arcosh}$ is the nonnegative real inverse hyperbolic cosine. As a check, if $\alpha =\cosh {\eta }_{\ast }$ with ${\eta }_{\ast }>0$, then Bob ages by $2t_0\sinh {\eta }_{\ast }$ and Alice by $2t_0{\eta }_{\ast }$; $\sinh {\eta }_{\ast }>{\eta }_{\ast }$ confirms that Bob again ages more.

The past light cone of a spacetime event $O$ is the past-directed part of its light cone. In Minkowski spacetime, its spacetime events satisfy $|\Delta \mathbf{x}|=-c\Delta t$ with $\Delta t<0$. They are the possible emission events of light rays arriving at $O$. The interior also contains events that can send slower signals to $O$. The retarded potential of a moving point electric charge, given by the Liénard–Wiechert potential, uses the emission event where the source's world line intersects this boundary.

Proper time is the time measured by a clock following a timelike worldline. With metric signature $(-,+,+,+)$ and units $c=1$,along that worldline.

Restoring the speed of light, the same metric signature gives $d\tau =\sqrt{-d{s}^2}/c=dt\sqrt{1-|\mathbf{v}{|}^2/c^2}$ for a future-directed timelike curve. Integrating along a clock's world line gives its elapsed proper time, even when its velocity varies. For uniform motion this becomes the time dilation formula; it is not generally the square root of the interval between endpoints of a curved world line.

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.