Spacetime Diagrams and the Invariant Interval
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the Lorentz transformation equations and understand why Galilean relativity fails for light. Familiarity with basic vector geometry in Euclidean space is helpful but not strictly required.
Why this matters
Spacetime diagrams turn abstract algebra into visual intuition, revealing the geometric unity of space and time.
In previous lectures, we derived the Lorentz transformation as a mathematical necessity to preserve the speed of light. However, these transformations can feel like a set of disconnected algebraic rules. By moving to a geometric representation, we see that space and time are not separate entities but components of a single four-dimensional manifold. This geometric perspective simplifies the understanding of relativistic effects like time dilation and length contraction, showing them as natural consequences of the geometry rather than mysterious physical distortions. It also introduces the invariant interval, the one quantity that all observers agree upon, which serves as the foundation for the causal structure of the universe.
The Geometry of Spacetime
Imagine plotting the position of an object on a piece of paper. In ordinary Euclidean geometry, the distance between two points is calculated using the Pythagorean theorem: the square of the distance is the sum of the squares of the differences in coordinates. This distance is invariant; it does not change if you rotate your coordinate axes. In special relativity, we extend this idea to include time. We plot time on one axis and space on the other. However, the "distance" in spacetime is not calculated the same way as in Euclidean space. Instead of adding the squares of the time and space differences, we subtract them. This difference in sign is the key to understanding relativity. It means that the geometry of spacetime is not Euclidean but Minkowskian. This geometric structure dictates how events are related to one another and why certain sequences of events are possible while others are not.
Constructing the Diagram
To draw a spacetime diagram, we typically use the \(ct\) axis for time and the \(x\) axis for space. We use \(ct\) rather than \(t\) so that both axes have units of length, allowing them to be plotted on the same scale. The worldline of an object is the path it traces through this diagram. For an object at rest in the \(x\)-direction, the worldline is vertical, parallel to the \(ct\) axis. For an object moving at a constant velocity \(v\), the worldline is a straight line with a slope determined by \(v\). The speed of light, \(c\), corresponds to a slope of 1 (or 45 degrees if the axes are scaled equally). Light rays always travel at 45 degrees, forming the boundaries of the light cone. Events inside the light cone are timelike separated from the origin, while events outside are spacelike separated. This visual representation makes it clear that no massive object can exceed the speed of light, as its worldline would have to tilt beyond the 45-degree line, entering the spacelike region.
The Invariant Interval
The central geometric quantity in special relativity is the spacetime interval, denoted as \(s^2\). For two events separated by \(\Delta t\) in time and \(\Delta x\) in space (assuming motion only in the \(x\)-direction for simplicity), the interval is defined as:
Classifying Intervals
The classification of intervals provides a rigorous way to understand causality. Timelike intervals (\(s^2 < 0\)) imply that the time separation is dominant. In such cases, the order of events is absolute; all observers agree on which event happened first. This is because the time difference is large enough to overcome any spatial separation, ensuring a causal link. Spacelike intervals (\(s^2 > 0\)) imply that the spatial separation is dominant. In these cases, the order of events is relative; different observers may disagree on which event happened first. This is possible because no signal traveling at or below the speed of light can connect the two events, so there is no causal link. Null intervals (\(s^2 = 0\)) represent the path of light. All observers agree that light travels at speed \(c\), so the time and space separations are exactly balanced such that \(c \Delta t = \Delta x\). This classification is not just a mathematical curiosity; it defines the structure of the universe, determining which events can influence which others.
The derivation
Invariance of the Spacetime Interval
For any two events in flat spacetime, the quantity \( s^2 = -c^2\,\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 \) takes the same numerical value in every inertial frame related by a Lorentz transformation (boost, rotation, or any composition thereof). This frame-independent \( s^2 \) is the spacetime interval, and its sign partitions event pairs into timelike (\(s^2<0\)), null (\(s^2=0\)), and spacelike (\(s^2>0\)) classes that fix the causal structure of spacetime.
A common misconception is that the invariant interval \(s^2\) is always positive. In fact, the sign depends on the convention used. Some texts define \(s^2 = c^2 \Delta t^2 - \Delta x^2\), which flips the signs of the classifications. Always check the convention used in your course. The physical meaning remains the same: timelike, spacelike, and null intervals are distinct geometric classes, but their numerical signs may vary. Consistency is key when calculating and interpreting results. C
Worked example
Problem. Two events occur in a laboratory frame. Event A happens at \(t_A = 0\ \mathrm{s}\) and \(x_A = 0\ \mathrm{m}\). Event B happens at \(t_B = 3.0\ \mathrm{s}\) and \(x_B = 2.0 \times 10^8\ \mathrm{m}\). Calculate the spacetime interval \(s^2\) and determine whether the interval is timelike, spacelike, or null. Assume \(c = 3.0 \times 10^8\ \mathrm{m/s}\).
- Calculate the time separation in meters: \(\Delta t = 3.0\ \mathrm{s}\), so \(c \Delta t = (3.0 \times 10^8\ \mathrm{m/s})(3.0\ \mathrm{s}) = 9.0 \times 10^8\ \mathrm{m}\).
- Calculate the square of the time term: \((c \Delta t)^2 = (9.0 \times 10^8\ \mathrm{m})^2 = 8.1 \times 10^{17}\ \mathrm{m^2}\).
- Calculate the square of the space term: \((\Delta x)^2 = (2.0 \times 10^8\ \mathrm{m})^2 = 4.0 \times 10^{16}\ \mathrm{m^2}\).
- Compute the interval using \(s^2 = -c^2 \Delta t^2 + \Delta x^2\): \(s^2 = -8.1 \times 10^{17}\ \mathrm{m^2} + 4.0 \times 10^{16}\ \mathrm{m^2}\).
- Combine terms: \(s^2 = -8.1 \times 10^{17} + 0.4 \times 10^{17} = -7.7 \times 10^{17}\ \mathrm{m^2}\).
- Since \(s^2 < 0\), the interval is timelike.
Units check. The term \(c^2 \Delta t^2\) has units \((\mathrm{m/s})^2 \mathrm{s}^2 = \mathrm{m}^2\). The term \(\Delta x^2\) has units \(\mathrm{m}^2\). The result is in \(\mathrm{m}^2\), which is consistent.
Three quick tests
- If two events have a spacelike interval, can a signal traveling at the speed of light connect them?
Answer
No. A spacelike interval means the spatial separation is greater than the light-travel distance in the given time, so no causal connection is possible. - What is the slope of the worldline of a photon on a spacetime diagram where the vertical axis is \(ct\) and the horizontal axis is \(x\)?
Answer
The slope is 1 (or 45 degrees), because \(x = ct\), so \(ct/x = 1\). - Is the spacetime interval invariant under a Lorentz boost?
Answer
Yes. The interval \(s^2\) is the same in all inertial frames related by Lorentz transformations.
Problem set for PU-105 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.