Spacetime Diagrams

Einstein’s key insight was not just that moving clocks run slow or moving rulers contract. It was that space and time have to be described together if the speed of light is the same for every inertial observer. A spacetime diagram gives us a way to see that geometry directly. Instead of tracking position alone, we track an event by its spatial coordinate xx and its time coordinate tt.

Events, Worldlines, and Frames

An event is a single happening: a flash, a detector click, a spaceship passing a marker. On a spacetime diagram, each event is represented by a single point. A moving object leaves behind a worldline, the path of all its events through spacetime.

Coordinate axes allow us to describe events according to an observer’s frame of reference, i.e. label each event with a coordinate pair (x,t)(x, t). A second observer moving at constant velocity relative to the first uses a different set of axes, and describes the same event with a different coordinate pair (x,t)(x^{\prime}, t^{\prime}). It is convention that the two sets of axes share a common origin, but they could in principle be translated. In special relativity, these coordinate axes correspond to inertial frames, meaning reference frames that are not accelerating or rotating.

The disagreement in spacetime coordinates between two observers should not be surprising. A bit of reasoning would lead you to expect something like:

x=xvtt=t\begin{align*} x' &= x - vt\\ t' &= t \end{align*}

if the primed observer is moving to the right with velocity vv relative to the unprimed observer. Note that one observer is not “wrong” while the other is “right”; it is that they slice spacetime into “space” and “time” differently.

The Light Cone

Worldlines of light that pass through a particular point form a light cone (more of a triangle with a single spatial dimension.) At the origin, these lines are x=ctx = ct and x=ctx = -ct. Different inertial observers may disagree about coordinates, but they agree on the causal structure:

  • events inside the light cone can be connected by slower-than-light motion
  • events on the cone can be connected by light
  • events outside the cone cannot causally influence one another without exceeding cc

Before diving into the particular transformation that describes the coordinate axes of the moving observer, it’s helpful to see how they relate to a light cone. Both sets of axes must agree on the speed of light, which geometrically means the primed axes appear to lean toward part of the light cone. As the relative speed of the two observers increases,

tanθ=vc,\tan \theta = \frac{v}{c},

where θ\theta is the angle between the two sets of axes.

Simultaneity

In ordinary Newtonian intuition, “at the same time” feels universal. In relativity, it is not. For the unprimed observer, events on a horizontal line share the same value of tt, so they are simultaneous in that frame. But the moving observer uses tilted lines of constant tt^{\prime} instead. That means two events can be simultaneous for one observer and not simultaneous for another.

This single geometric fact drives a lot of the rest of the theory:

  • Time dilation emerges when one frame compares elapsed time along a moving clock’s worldline.
  • Length contraction emerges because length must be measured using endpoints recorded at the same time in the measuring frame.

Invariant Interval

Although observers disagree about coordinates, they agree on certain combinations of them. The most important one in this diagram is the spacetime interval

s2=(ct)2x2.s^2 = (ct)^2 - x^2.

That value is invariant under relativistic (Lorentz) transformations. So when you click an event in the simulation and compare its primed and unprimed coordinates, the numbers may change, but the interval does not.

Minkowski Diagrams

Minkowski (spacetime) diagrams place both sets of axes on the same plot, rather than attempting to animate a set of moving axes. It turns out this appears as a rotation/stretching of axes. It’s also common to scale the time coordinates by the speed of light cc. This gives both the xx-axis and ctct-axis units of length and makes sketching the light cone easier.

As you explore the diagram below, try this sequence:

  1. Start with β=0\beta = 0 and notice that the primed and unprimed axes coincide.
  2. Increase β\beta and watch the xx^{\prime} and ctct^{\prime} axes tilt toward the light cone.
  3. Place an event on the graph and compare its coordinates in both frames.
  4. Select the event to see its projections onto both sets of axes.
  5. Turn the light cone, grids, and ticks on and off to isolate the geometric idea you want to study.

Spacetime Diagram

Compare how the spacetime locations of events are measured in two inertial frames.

Problem Solving

Classify an Interval

Problem

Event AA occurs at the origin. Event BB occurs at x=300mx = 300\,\text{m} and t=2.0μst = 2.0\,\mu\text{s}. Find the spacetime interval s2=(ct)2x2s^2 = (ct)^2 - x^2 and classify it.

Given

  • x=300mx = 300\,\text{m}
  • t=2.0μs=2.0×106st = 2.0\,\mu\text{s} = 2.0 \times 10^{-6}\,\text{s}
  • c=3.0×108m/sc = 3.0 \times 10^8\,\text{m/s}

Use

First compute ctct, then evaluate

s2=(ct)2x2.s^2 = (ct)^2 - x^2.

A positive value means the separation is timelike.

Solve

ct=(3.0×108)(2.0×106)=600m.ct = (3.0 \times 10^8)(2.0 \times 10^{-6}) = 600\,\text{m}.

So

s2=60023002=36000090000=270000m2.s^2 = 600^2 - 300^2 = 360000 - 90000 = 270000\,\text{m}^2.

Because s2>0s^2 > 0, the interval is timelike.

Check

Light would need only 1.0μs1.0\,\mu\text{s} to cross 300m300\,\text{m}, so there is enough time for a slower-than-light signal to connect the two events. That matches the timelike result.

Time Dilation for a Moving Clock

Problem

A spacecraft moves at β=0.80\beta = 0.80 relative to Earth. Earth measures Δt=15μs\Delta t = 15\,\mu\text{s} between two ticks of a clock on the ship. How much proper time Δτ\Delta \tau passes on the ship?

Given

  • β=0.80\beta = 0.80
  • Δt=15μs\Delta t = 15\,\mu\text{s}

Use

γ=11β2,Δτ=Δtγ.\gamma = \frac{1}{\sqrt{1-\beta^2}}, \qquad \Delta \tau = \frac{\Delta t}{\gamma}.

Solve

γ=110.802=10.36=53.\gamma = \frac{1}{\sqrt{1-0.80^2}} = \frac{1}{\sqrt{0.36}} = \frac{5}{3}.

Then

Δτ=15μs5/3=9.0μs.\Delta \tau = \frac{15\,\mu\text{s}}{5/3} = 9.0\,\mu\text{s}.

Check

The clock riding with the spacecraft records less elapsed time than the Earth frame does, which is exactly the time-dilation pattern the tilted axes encode on the diagram.

Spacetime Diagram Checkpoint

Question 1 of 3

In the spacetime diagram, why can two events that lie on the same horizontal line $ct = \text{constant}$ fail to be simultaneous in the moving frame?

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