Events, Intervals, and Minkowski Geometry

All inertial observers agree on the Minkowski interval between two events, its timelike/null/spacelike causal class, and—after choosing the same time orientation—the order of causally related events. They do not generally agree on coordinate differences, simultaneity, or the temporal order of spacelike-separated events.

Required background. Calculus and Linear Algebra provides quadratic forms, linear maps, and basis changes; Variational Mechanics and Symmetries supplies the distinction between a physical trajectory and its coordinate description.

Events before coordinates

An event is something localized in both place and time: for example, a flash at a detector or the meeting of two clocks. A worldline is the sequence of events along one object’s history. Two observers who meet agree that the meeting happened even when they assign it different coordinates.

Minkowski spacetime is affine: subtracting two events $A$ and $B$ produces a displacement vector $\Delta x^\mu=x^\mu_B-x^\mu_A$, but no event is intrinsically the origin. An inertial Cartesian chart assigns four numbers $x^\mu=(t,x,y,z)$ to each event. Physically, its time labels can be supplied by a lattice of clocks at rest relative to one another and synchronized by light signals. A different inertial lattice changes the labels without changing the events.

The invariant quadratic form is

$$s^2=\eta_{\mu\nu}\Delta x^\mu\Delta x^\nu =-(\Delta t)^2+|\Delta\mathbf x|^2.$$

Restoring units gives $s^2=-c^2(\Delta t)^2+|\Delta\mathbf x|^2$. A Lorentz transformation satisfies

$$\Lambda^\mathsf{T}\eta\Lambda=\eta,$$

so $\Delta x'=\Lambda\Delta x$ has the same $s^2$. A translation changes both event coordinates by the same constant and cancels from their difference. Tong derives the invariant interval and Lorentz-group condition in Tong 2013, §§7.3.1–7.3.2, printed pp. 125–130 (PDF).

For two distinct events, the sign classifies their separation:

• $s^2\lt0$ is timelike; a slower-than-light worldline can join the events.
• $s^2=0$ is null; a light signal can join them.
• $s^2\gt0$ is spacelike; no causal signal limited by the light cone can join them.

The cone also has future and past halves. Proper orthochronous Lorentz transformations preserve those halves, so if $B$ lies in the future causal cone of $A$, every such inertial observer agrees that $A$ precedes $B$; see Tong 2013, §7.2.2, printed pp. 117–118 (PDF).

Here proper excludes a spatial reflection and orthochronous excludes reversal of the chosen future direction. The latter is what matters for causal ordering. Coincident events have zero displacement and $s^2=0$, but the zero vector is not a light-ray direction. Also, a timelike separation says that a causal influence is possible; it does not establish that one event actually caused the other.

The causal-cone diagram makes this partition visible. Inspect the null boundary first, then the timelike interiors and spacelike exterior. Its second panel also isolates a low-dimensional topological fact: in $1+1$ dimensions the spacelike exterior has separate left and right components, although the same sign test still classifies every displacement.

An index-one tangent metric has two timelike components bounded by the null cone. The 1+1 cross-section makes the low-dimensional exception explicit: the spacelike set has separate left and right components, so the cone complement has four pieces. Schematic and not to scale.

Canonical frames for each class

For a displacement in $3+1$ dimensions with $\Delta\mathbf x\ne0$, first rotate the spatial axes so that

$$\Delta\mathbf x=(\Delta x,0,0), \qquad \Delta x=|\Delta\mathbf x|\gt0.$$

If $\Delta\mathbf x=0$, a timelike displacement is already in canonical rest-frame form. Otherwise the remaining calculation lies in the $(t,x)$ plane. Take a boost with $|v|\lt1$:

$$\Delta t'=\gamma(\Delta t-v\Delta x), \qquad \Delta x'=\gamma(\Delta x-v\Delta t), \qquad \gamma=(1-v^2)^{-1/2}.$$

If the separation is timelike, $|\Delta x/\Delta t|\lt1$. Choosing $v=\Delta x/\Delta t$ gives $\Delta x'=0$: there is an inertial frame in which both events occur at the same place, and

$$|\Delta t'|=\sqrt{-s^2}.$$

If the separation is spacelike, put $r=\Delta t/\Delta x$, so $|r|\lt1$. Choosing $v=r$ gives $\Delta t'=0$: there is a frame in which the events are simultaneous, separated by $\sqrt{s^2}$. More precisely,

$$\Delta t'=\gamma\Delta x(r-v),$$

so allowed boosts with $v\lt r$ and $v\gt r$ assign opposite signs to $\Delta t'$. Null separation admits neither rest nor simultaneity frame; the cone remains the cone. The boost and interval ingredients appear in Tong 2013, §7.1, printed pp. 108–114, and §7.3.1, printed pp. 125–127 (PDF); Hughes 2024, Lecture 4, §4.3, printed pp. 31–33 (PDF) applies them to the canonical frames.

Coordinates and measurements

A coordinate chart is a labeling convention implemented, in practice, by a synchronized network of clocks and rulers. A local observer is instead characterized by a timelike direction and measures quantities by projections at an event. Coordinates are indispensable for calculation, but a coordinate component alone is not an observer-independent observable.

Transfer check: classify three displacements

In units with $c=1$, let event $A=(0,0)$ and consider $B=(5,3)$, $C=(2,4)$, and $D=(3,3)$ in $(t,x)$ coordinates. Classify each separation and find a useful canonical frame.

Solution

For $A\to B$,

$$s_B^2=-5^2+3^2=-16,$$

so the separation is timelike. The boost $v=3/5$ is allowed and gives $\Delta x'_B=0$. Since $\gamma=5/4$,

$$\Delta t'_B=\frac54\left(5-\frac35\,3\right)=4=\sqrt{-s_B^2}.$$

For $A\to C$,

$$s_C^2=-2^2+4^2=12,$$

so it is spacelike. Choosing $v=2/4=1/2$ makes $\Delta t'_C=0$. Choosing instead $v=3/4$ gives $\Delta t'_C=\gamma(2-3)\lt0$, explicitly reversing the coordinate time order without threatening causality.

Finally, $s_D^2=-3^2+3^2=0$, so $D$ is null-separated from $A$. Substitution into the boost shows $\Delta x'_D=\Delta t'_D$ for every finite $|v|\lt1$; its null class is unchanged.

The sign test is a useful diagnostic: with $(-+++)$, a stationary clock has $ds^2=-dt^2\lt0$. If a calculation labels that displacement spacelike, its signature has been mixed. The canonical-frame construction assumes global inertial charts and flat spacetime. In curved spacetime the causal class of each tangent vector is still exact, but separated events generally do not define one affine displacement measured by a single constant matrix $\eta_{\mu\nu}$.

Lorentz Transformations, Rapidity, and Velocity Addition now derives the boost used above and explains exactly when its rapidity parameter adds.

References

• Hughes, Scott A. Introduction to Relativity and Spacetime Physics. MIT 8.033 Lecture 4, especially §4.3, printed pp. 31–33 (PDF), Fall 2024.
• Tong, David. Dynamics and Relativity. §7.1, printed pp. 108–114; §7.2.2, printed pp. 117–118; §§7.3.1–7.3.2, printed pp. 125–130 (PDF), University of Cambridge, Lent Term 2013. Signature translated from $(+---)$ to $(-+++)$.