Spacetime Diagrams

Spacetime diagrams (Minkowski diagrams) are the geometric language of special relativity. Each point is an event; each particle traces a worldline; light travels at 45°. The causal structure — what can influence what — is encoded in the light cone.

Key Concepts

Worldline

The path of an object through spacetime: a curve

(t,x(t))

in the

t

x

plane. A particle at rest is a vertical line; one moving at constant velocity is a straight line; light travels at 45° (slope

c

Light Cone

The set of events connected to a given event by light signals. The future light cone contains all events that can be influenced; the past light cone all events that could have influenced. Events outside the light cone are causally disconnected.

Hyperbolic Geometry

Lorentz boosts are hyperbolic rotations in Minkowski space. Lines of constant proper time are hyperbolas; the Lorentz-boosted axes tilt toward the light cone (they always straddle it symmetrically).

Proper Time from Diagram

The proper time along a worldline segment is

\Delta\tau=\sqrt{\Delta t^2-\Delta x^2/c^2}

— the "length" of the worldline in Minkowski metric (using the

+---

convention).

Key Equations

Worldline slope

\text{slope} = c/v = c/\beta c = 1/\beta

In a ct vs x diagram, the slope is c/v; light has slope 1 (at 45°).

Proper time (segment)

\Delta\tau = \sqrt{\Delta t^2 - \Delta x^2/c^2} = \Delta t/\gamma

The "length" of a worldline segment; shorter than coordinate time Δt.

Light cone

ct = \pm x

Light cone boundary; timelike events have |ct| > |x|; spacelike have |ct| < |x|.

Worked Example

Reading Proper Time from a Spacetime Diagram

Problem

A worldline passes through events $A=(t=0,x=0)$ and $B=(t=5\text{ ns}, x=1.2\text{ m})$ ( $c=3\times10^8$ m/s). Find the proper time $\Delta\tau$ between them.

Solution

Convert: $c\Delta t=3\times10^8\times5\times10^{-9}=1.5$ m.

\Delta\tau = \sqrt{\Delta t^2 - \Delta x^2/c^2} = \frac{1}{c}\sqrt{(c\Delta t)^2 - \Delta x^2}

\Delta\tau = \frac{1}{c}\sqrt{1.5^2 - 1.2^2} = \frac{1}{c}\sqrt{2.25-1.44} = \frac{\sqrt{0.81}}{c} = \frac{0.9\text{ m}}{3\times10^8\text{ m/s}} = 3\text{ ns}

Answer

\Delta\tau=3.0

ns.

Practice

Exercises

7 problems

1 of 7

A particle moves at $v=0.60c$ . What is the slope of its worldline (in a $ct$ vs $x$ diagram)?

Your answer

(dimensionless)

2 of 7

Events $A=(ct=0,x=0)$ and $B=(ct=5\text{ m},x=3\text{ m})$ . Find the proper time $\Delta\tau$ (in m/c, or equivalently in units where c=1). $\Delta\tau^2=(5)^2-(3)^2$ .

Your answer

m/c

3 of 7

Is the interval between events $A=(ct=3\text{ m},x=5\text{ m})$ and $B=(ct=0,x=0)$ timelike, lightlike, or spacelike? Find $s^2=(c\Delta t)^2-\Delta x^2$ .

Unlock Exercise 3