Relativity That ClicksEpisode 3 / The spacetime interval — the one invariant that no observer can split apart

Time stretches by γ, length shrinks by 1/γ — so is there nothing at all that is the same no matter who measures it?

The spacetime interval — the one invariant that no observer can split apart Time and distance are all over the place, different for each observer (stage machinery). But joining the two with "−", s² = (ct)² − x² alone is
the same no matter who measures it. This subtraction is the true geometry of spacetime — the backbone of relativity.

Tools you'll need: Pythagoras, γ from Episode 2, one subtraction This episode's invariant: s² = (ct)² − x²

In Episode 2 we saw that when you move, time stretches by $\gamma$ and length shrinks by $1/\gamma$. Time and distance are all over the place for each observer — truly "a quantity with units is stage machinery." At which point you might grow anxious. Is there no longer anything solid, anything the same no matter who measures it? There is. Exactly one thing. The quantity that joins time and space with "−" instead of Pythagoras's "+" — the spacetime interval $s^2=(ct)^2-x^2$ — alone is identical down to the last digit for every observer. The time-stretching and length-shrinking that looked all over the place were, in fact, settling accounts with each other so as to preserve this one invariant. In this episode, the geometric backbone of relativity comes into view.

01Recall Euclid's "distance"

Ordinary plane geometry. Rotate a rod and its endpoint coordinates $(x,y)$ change. Lay it flat and $x$ grows while $y$ shrinks. But the rod's length doesn't change — length² $=x^2+y^2$ is invariant under rotation. The coordinates $x,y$ are stage machinery that change with "how you placed it"; the length is the physics. $x^2+y^2$, preserved under rotation, is the invariant of Euclidean geometry.

Relativity has a structure just like this. In exactly one spot — a sign differs.

02The "distance" of spacetime is joined with "−"

On the spacetime diagram (Episode 1), the vertical axis is time $ct$ and the horizontal is space $x$. Change the observer (switch to a person moving at a different speed) and, for the same event, $ct$ and $x$ change all over the place following the $\gamma$ of Episode 2. And yet —

This episode's invariant — the spacetime interval

$$s^2=(ct)^2-x^2\qquad(\text{the same for every observer})$$

It differs from Euclid's $x^2+y^2$ by just one sign — the time term is "+", the space term is "−". This single minus is the source of all the strangeness of relativity. What corresponds to Euclid's "rotation" is the switching of observers (the Lorentz transformation), and it is the "rotation" of spacetime that preserves this $s^2$.

Let's check it. Episode 2's light clock, to the clock itself, merely lets time $\Delta t_0$ pass at the same place ($x=0$), so $s^2=(c\Delta t_0)^2$. Seen from the ground, time stretches to $\gamma\Delta t_0$ and the location moves by $v\gamma\Delta t_0$. The ground's $s^2$ is —

Try it — the stretch and the shrink preserve s² $$s^2=(c\gamma\Delta t_0)^2-(v\gamma\Delta t_0)^2=\gamma^2(c^2-v^2)\Delta t_0^2$$

Substitute $\gamma^2=1/(1-\beta^2)=c^2/(c^2-v^2)$

$$s^2=\frac{c^2}{c^2-v^2}\,(c^2-v^2)\Delta t_0^2=(c\Delta t_0)^2$$

It matches exactly the $s^2=(c\Delta t_0)^2$ measured by the clock itself. The effect of time stretching by $\gamma$ and the effect of the location moving cancel out precisely in the "−" subtraction. Behind the all-over-the-place $ct$ and $x$, $s^2$ does not budge — this is the invariant.

03Let's play with it — change the observer, and the point slides along a hyperbola

On the spacetime diagram below, we place a single event (the green point). The slider is the speed $\beta$ of the observer you ride. Change the observer and that event's coordinates $(ct,\,x)$ move — but not haphazardly. They only slide along the invariant hyperbola $(ct)^2-x^2=s^2$. When time $ct$ increases, space $x$ increases too, and $s^2$ stays strictly constant. The 45° light ray (Episode 1's wall) is the asymptote that this hyperbola approaches.

In the Euclidean case, the point would circle a circle centered on the origin ($x^2+y^2$ constant). In spacetime, because of the "−", the circle turns into a hyperbola. Switching observers is not a circular rotation but a slide along the hyperbola (a hyperbolic rotation). This is what the Lorentz transformation really is. Wherever the point slides to, the "spacetime separation" $s$ from the origin does not change.

Figure: spacetime diagram (vertical ct, horizontal x). View a single event (green point) at different observer speeds β. The point only slides along the invariant hyperbola (ct)²−x²=s² — ct and x both change, yet s² stays constant. The 45° line (light) is the asymptote

observer speed β 0.00

event point (ct, x per observer) invariant hyperbola, s² constant light (45°, asymptote)

04Proper time τ — the invariant becomes a clock

This $s^2$ has a welcome meaning. When an event is timelike separated from the origin ($s^2>0$, inside the light ray), $s/c$ is — the time actually ticked by a clock that travels to that event. This is called the proper time $\tau$.

Proper time — the invariant time read by one's own clock

$$\tau=\frac{s}{c}=\sqrt{t^2-\frac{x^2}{c^2}}$$

This is the "what is truly invariant is each person's proper time τ" foreshadowed in Episode 2. Even though coordinate time $t$ is all over the place for each observer, the $\tau$ read by a clock that actually traveled the path is one on which everyone's calculation agrees — because it is the $s$ along the path.

This also untangles the twin paradox. A twin who stays on Earth and a twin who travels through space and returns. The two worldlines are different paths joining the same two points on the spacetime diagram (departure and reunion). In the Euclidean case, "straight is shortest," but spacetime is a "−" geometry, so it's the reverse — straight (inertial) = the longest proper time, the detour (accelerating and traveling) = a shorter proper time. So the twin who traveled is younger. This is not a relative illusion but the difference in $s$ (the invariant) along the two paths themselves. The side that accelerated and bent its path returns definitely younger.

Three kinds of separation — a bridge to the next episode By the sign of $s^2$, the relation between two events splits into three kinds. Timelike ($s^2>0$, inside the light ray) = can be cause and effect. Lightlike ($s^2=0$, on the light ray) = can be joined by light exactly. Spacelike ($s^2<0$, outside the light ray) = cannot be joined by any signal, no causal relation. This sorting is the groundwork for Episode 5, "Simultaneity depends on the observer, but causality is invariant." The invariant $s^2$ determines the skeleton of causality in spacetime.

◇ ◇ ◇

The honest line — the reach of "−" and the invariant

That the spacetime interval $s^2=(ct)^2-x^2$ is invariant under the Lorentz transformation, that the Lorentz transformation is a hyperbolic rotation preserving $s^2$ with events moving on the invariant hyperbola, that for a timelike interval $\tau=s/c$ is the proper time (the invariant elapsed time read by that clock), and that inertial motion maximizes proper time so that the accelerating side is younger in the twin problem — these are all established physics.

There are two conventions for the sign choice, and this piece adopts time-positive $(+,-,-,-)$ (some books make space positive and write $s^2=x^2-(ct)^2$; the physics is the same). In 3D, $s^2=(ct)^2-x^2-y^2-z^2$. Also, $s^2=(ct)^2-x^2$ is the interval of flat spacetime (Minkowski); with gravity it becomes a curved geometry in which a metric gives the infinitesimal interval $ds^2$ at each point, varying from place to place (Episodes 6 and 7). The "hyperbolic rotation" figure is a 1+1-dimensional schematic; actual velocity composition and 3D rotations ride on top of it.

Practice problems (solvable with this episode's formulas alone)

What quantity does rotation preserve in Euclidean geometry? What does switching observers (the Lorentz transformation) preserve in spacetime? What's the difference?
Show answer
Euclidean: $x^2+y^2$ (length²). Spacetime: $s^2=(ct)^2-x^2$ (the spacetime interval). The difference is that the space term's sign is "−". This one minus turns the circle into a hyperbola.
An observer measures an event at $ct=5,\ x=3$ (same units). What is $s^2$? If another observer measures $x'=0$, what is their $ct'$?
Show answer
$s^2=5^2-3^2=25-9=16$. Since it's invariant, the other observer also has $s^2=16$. If $x'=0$, then $(ct')^2-0=16\Rightarrow ct'=4$. To this observer it's an event where "4" of time passed (= corresponding to proper time τ=4/c×c=4).
In the twin paradox, why is the traveling side younger? Explain using "straight is longest."
Show answer
Spacetime is a "−" geometry, so of the paths joining two points, straight (inertial) = the longest proper time. The traveling twin accelerates and bends its path, so its proper time is shorter and it returns younger. Not a relative illusion but the difference in s along the two paths.
What relation between two events does each of $s^2>0,\ =0,\ <0$ represent?
Show answer
$s^2>0$: timelike (can be causal). $s^2=0$: lightlike (can be joined by light exactly). $s^2<0$: spacelike (cannot be joined by any signal, no causal relation). The sign determines the skeleton of causality.

Episode 3 summarythe invariant is s² = (ct)² − x² — a geometry of one minus

Time and distance are all over the place for each observer (Episode 2). But the spacetime interval $s^2=(ct)^2-x^2$, joining time and space with "−", alone is the same no matter who measures it. It differs by just one sign from Euclid's invariant $x^2+y^2$ (invariant under rotation), and that one minus is the source of relativity's strangeness. Switching observers (the Lorentz transformation) is not a circular rotation but a slide along a hyperbola — the event slides on the invariant hyperbola $s^2=$constant, and $ct$ and $x$ change while $s$ does not budge.

For a timelike interval, $\tau=s/c$ is the proper time — the invariant elapsed time read by that clock itself (collecting on Episode 2's homework). Because spacetime is a "−" geometry, "straight = longest proper time," and in the twin problem the side that accelerated and bent its path returns definitely younger. The sign $s^2\gtrless0$ separates timelike, lightlike, and spacelike, determining the skeleton of causality (on to Episode 5). The unit-bearing $t,x$ are stage machinery; what remains is only the invariant $s$ — the backbone of relativity.

Next time — Episode 4 "E=mc² — mass is rest energy." Just as spacetime had an invariant separation $(ct,x)$, energy and momentum also have an invariant joined by the same "−" — $(mc^2)^2=E^2-(pc)^2$. For a body at rest ($p=0$), $E=mc^2$. Mass was another name for the energy a body has at rest. Set it moving and $E=\gamma mc^2$; made dimensionless, $E/mc^2=\gamma$. And if the mass is zero, all the energy is in the motion $pc$, and it travels at $c$ — here is where we shake hands with Episodes 2 and 3 of "Fields That Click."

This document is Episode 3 of the "Relativity That Clicks" series, a reading piece for physics-loving high-schoolers and undergraduates. The Lorentz invariance of the spacetime interval $s^2=(ct)^2-x^2$ (in 3D, $(ct)^2-x^2-y^2-z^2$), that the Lorentz transformation is a hyperbolic rotation preserving $s^2$, the proper time $\tau=s/c$ for a timelike interval, the maximization of proper time by inertial motion and the consequence of the twin paradox, and the classification into timelike/lightlike/spacelike by the sign of $s^2$ are all established, standard physics. The metric signature adopted is $(+,-,-,-)$ (the physics is identical under the $(-,+,+,+)$ convention). That $s^2=(ct)^2-x^2$ is the interval of flat spacetime (the Minkowski metric) and that under gravity it becomes a curved spacetime in which the line element $ds^2$ is given place-by-place by the metric tensor (Episodes 6 and 7), and that the figure is a 1+1-dimensional schematic on which velocity composition and 3D rotations ride, are stated in the "honest line" in the main text. The hyperbola in the figure is the invariant hyperbola $(ct)^2-x^2=$constant. — To print, use your browser's "Print" and "Save as PDF" (in the print version the slider and answers are static/hidden). Adjacent episodes: Episode 2, γ / Contents.

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen, change the observer's β and the event's point slides along the invariant hyperbola (s² is constant). "Show answer" opens each solution.