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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.

A light-second is a unit of length: the distance vacuum light travels in one second. It is useful in a spacetime diagram because a time of one second corresponds to a $ct$ coordinate of one light-second, making the slopes $x=\pm ct$ of the light cone particularly easy to read. The unit is a distance, rather than an elapsed time.

Choose event $A$, when the trailing end reaches the entrance, as $(ct,x)=(0,0)$ and as the primed origin. Write $T=ct$, $T^{\prime }=c{t}^{\prime }$. Before impact, the shed-frame worldlines are entrance $x=0$, back $x=1.5$, trailing end $x=0.8T$, and leading end $x=0.8T+1.2$. The Lorentz transformation isThus in the javelin frame the ends have $x^{\prime }=0$ and $x^{\prime }=2$, while the entrance and back have $x^{\prime }=-0.8T^{\prime }$ and $x^{\prime }=0.9-0.8T^{\prime }$. These four lines in each frame produce the requested spacetime diagrams. The diagrams show initial motion; dashed portions extrapolate that motion beyond first impact and are not a model of collision dynamics.

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.

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.

The one-dimensional Isentropic Euler equations areDefine the Riemann invariants for one-dimensional isentropic flowCombining the two equations directly givesso $R_{\pm }$ is constant on the corresponding characteristic curve $dx/dt=u\pm c$.

Initially both invariant disturbances are supported in $0\le x\le L$. Once the two characteristic families have separated, the left-moving packet is a simple wave with $R_{+}$ at its equilibrium value, while the right-moving packet is a simple wave with $R_{-}$ at equilibrium. Between and outside them, both invariants retain their equilibrium values. The schematic spacetime diagram below uses the equilibrium slopes $\pm c_0$ and shows separation after $t=L/(2c_0)$.

For $\beta =0.99$, the Lorentz factor is $\gamma =1/\sqrt{1-{\beta }^2}=7.0888$. In the tracks' rest frame the train undergoes length contraction to $100/\gamma =14.1067\mathrm{m}$, which fits in the $90\mathrm{m}$ tunnel. There is a time interval during which it is entirely inside. In the train's rest frame the tunnel instead has length $90/\gamma =12.6961\mathrm{m}$, so it never simultaneously contains the whole $100\mathrm{m}$ train:These refer to different sets of simultaneous events, as required by relativity of simultaneity. Quantitatively let $A$ be the rear entering the near portal, with $(x_A,t_A)=(0,0)$, and $B$ the front reaching the far portal. In the track frame,The train is wholly inside between these events. Transforming their time separation givesThus the front leaves before the rear enters in the train frame. The two events are spacelike separated, so their time order may reverse without a causal contradiction. No doors or forced acceleration need be added to the problem.

During the photon's runner-frame flight, the tunnel entrance moves left by $v\Delta t_{\gamma }^{\prime }=\gamma \beta L$. It began ahead of the runner by $\gamma {\beta }^{2}L$, so at detection it lies behind byDivide this distance into the tunnel by the simultaneously measured tunnel length $L/\gamma$:The runner is stationary in $S^{\prime }$, but their location relative to the moving tunnel still has this fraction. Omitting the initial entrance gap would incorrectly count the runner's progress from an emission event at which they had not yet entered in this frame. These simultaneous positions give the tunnel photon timing in two inertial frames.

Plot position $x$ horizontally and $ct$ vertically, with equal length scales, so vacuum light rays have slopes $\pm 1$. A possible cause of the spacetime event $A$ must lie on or inside its past light cone, with $\Delta t=t_A-t_{\mathrm{cause}}>0$. Equivalently, causality requires $|\Delta x|\le c\Delta t$.

From $B$ to $A$, $\Delta x=6\mathrm{m}$ and $\Delta t=10^{-8}\mathrm{s}$. Using $c\simeq 3.00\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, the available light-travel distance is only $c\Delta t=3.00\mathrm{m}$. In the $(-,+,+,+)$ metric signature, the Minkowski interval isThere is spacelike separation, so a signal would need speed $6\times 10^8\mathrm{m}{\mathrm{s}}^{-1}\simeq 2c$:From $C$ to $A$, $\Delta x=-2\mathrm{m}$ and $\Delta t=1.5\times 10^{-8}\mathrm{s}$. Now $c\Delta t=4.50\mathrm{m}>|\Delta x|$, givingThis is timelike separation, with a possible signal velocity $-2/(1.5\times 10^{-8})\simeq -1.33\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$. Therefore

Frame definitions and relativistic effects. Let $S$ be the inertial frame of Earth and the stationary asteroid, with $+x$ pointing toward the asteroid. Bob's inertial frame $S^{\prime }$ moves at $+v$ relative to $S$. For two events at Bob's fixed position, $\Delta x^{\prime }=0$, the inverse Lorentz transformation givesThe elapsed proper time on Bob's clock is $\Delta \tau =\Delta t^{\prime }$, so time dilation means $\Delta t=\gamma \Delta \tau$: the separation of his clock ticks is larger in Earth coordinates. For a rod at rest in $S^{\prime }$ with proper length $L_0$, a length measurement in $S$ uses simultaneous endpoints, $\Delta t=0$. The direct spatial Lorentz transformation gives $L_0=\Delta x^{\prime }=\gamma \Delta x$, hence length contraction:The simultaneity condition is essential: the endpoint events used to measure length are different from the two clock-tick events used to demonstrate time dilation.

Bob's speed. Food consumption measures Bob's own proper time, ${\tau }_B=4\text{ months}=1/3\mathrm{yr}$. The Earth-frame distance is $D=1/4\mathrm{ly}$ and the Earth travel time is $D/v$. ConsequentlyPutting $\beta =v/c$, ${\beta }^2/(1-{\beta }^2)=9/16$, soThe Earth travel time is $5/12\mathrm{yr}=5$ months. Equivalently Bob measures the contracted Earth-asteroid separation $D/\gamma =0.20\mathrm{ly}$ and sees the asteroid approach at $0.60c$, taking $1/3\mathrm{yr}$ to reach him.

Events A and B. Set both coordinate origins at Event A. Alice reaches the asteroid at Event B, travelling at $u=2c/3$ in $S$, soThe Lorentz transformations into Bob's frame giveHere $c(1\mathrm{yr})=1\mathrm{ly}$ has been used consistently. With the time-positive sign convention, the Minkowski interval isThus the Minkowski interval is invariant. It is timelike, as expected for two events on Alice's subluminal worldline. Using the opposite metric sign convention would change both signs together, without changing invariance.

Relative velocity and spacetime diagram. Divide the differential spatial and temporal Lorentz transformations for the same trajectory:For Alice,The spacetime diagram uses horizontal $x^{\prime }$ and vertical $c{t}^{\prime }$. In these coordinates Bob's worldline is $x^{\prime }=0$, Alice's is $x^{\prime }=c{t}^{\prime }/9$, and the asteroid's isThus Alice has slope $d(c{t}^{\prime })/d{x}^{\prime }=9$, while the asteroid has slope $-5/3$. Their intersection is exactly B, $(x^{\prime },c{t}^{\prime })=(1/32,9/32)\mathrm{ly}$. The asteroid later reaches Bob at $(0,1/3)\mathrm{ly}$. These intercepts and gradients show the relativistic velocity addition geometrically.

A future-directed causal curve from the origin has $|dx/dt|\le c$. Integrating this inequality gives the causal futureThe straight timelike or null segment proves sufficiency. This is the blue region above the two null rays in the upper-left spacetime diagram of part c.

The upper-left panel shows the three causal structure regions, with null boundaries included in the appropriate causal future and causal past. For the charged trajectory, use part b with $a=q{E}_x/m$, choosing the $x$ direction so that $a>0$. The electric field gives the constant three-force $q{E}_x$ and hence the shifted hyperbolaIt starts with a vertical tangent at the origin, stays strictly inside the future light cone, and asymptotes to $x=ct-c^2/a$. The other panels implement the coordinate and topology changes used below; they are original spacetime diagrams, not copies of source artwork.

The same Lorentz transformation maps the past light cone to itself. In the primed axes the causal past isthe orange region in the upper-right spacetime diagram of part c. Future and past are not interchanged by the standard boost.

The causal past follows from the same covering-space argument with reversed endpoints:For $t\le -\pi R/c$, the whole spatial circle can send a causal signal to the origin. The lower-left spacetime diagram in part c is therefore completely orange sufficiently far below the origin.

Let $e>0$ be the magnitude of the electron's electric charge, $m=m_e$, $F=eE>0$, and $\ell =m{c}^2/F$. The Lorentz force is $d{p}_x/dt=qE$, so the electron and Positron acquire opposite momenta:Using the relativistic energy-momentum relation, $u_x=p{c}^2/\mathcal{E}$, and integrating from rest gives the world linesFor $t<0$, the world lines are vertical at $x=0,d$ on a spacetime diagram. They join smoothly to the accelerated branches with zero initial velocity. At small $t$, $x_e\simeq -F{t}^2/(2m)$ and $x_p-d\simeq F{t}^2/(2m)$; at large $t$, their velocities approach $-c,+c$, respectively. Each accelerated world line is a branch of a hyperbola, since $(x_e-\ell {)}^2-c^2{t}^2={\ell }^2$ and $(x_p-d+\ell {)}^2-c^2{t}^2={\ell }^2$. The electric field separates the opposite electric charges in this orientation. The asymptotes are lightlike, while every finite-time tangent is timelike.

A spacetime diagram represents events using spatial and time coordinates, often with horizontal $x$ and vertical $ct$. A particle's world line records successive events on its trajectory; lines describing the propagation of light define the light cone.

The axis convention matters when reading a slope. With $ct$ vertical, a constant-velocity trajectory has $dx/d(ct)=v/c$, so the drawn vertical-to-horizontal slope is $c/v$. Light rays then lie at $45^{\circ }$ when the two axes use equal length scales. A stationary object has a vertical world line.

Such a diagram makes special relativity geometric: changing an inertial frame changes how events are assigned spatial positions and times, while causal relations remain constrained by light propagation. A diagram showing only one spatial coordinate suppresses other motion rather than proving it absent.

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.

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.