Observers in different frames disagree about times (dilation) and lengths (contraction). Is there anything about a pair of events they all agree on? In ordinary geometry, turning your axes changes the $x$ and $y$ of a point, but not its distance from the origin. The Lorentz transformation has its own version of distance.
Key idea
For two events separated by $\Delta t$ and $\Delta x$, the spacetime interval $$s^2 = (c\Delta t)^2 - \Delta x^2$$ has the same value in every inertial frame. If $s^2 > 0$ (timelike), one event could cause the other, and $\Delta\tau = s/c$ is the proper time measured by a clock that travels between them at constant velocity. If $s^2 < 0$ (spacelike), no signal can connect them, and their time order depends on the frame. If $s^2 = 0$ (lightlike), only light connects them.
Why it works
Substitute $c\Delta t' = \gamma(c\Delta t - \beta\Delta x)$ and $\Delta x' = \gamma(\Delta x - \beta c\Delta t)$: $$(c\Delta t')^2 - \Delta x'^2 = \gamma^2\left[(c\Delta t)^2 - 2\beta c\Delta t\Delta x + \beta^2\Delta x^2 - \Delta x^2 + 2\beta c\Delta t\Delta x - \beta^2(c\Delta t)^2\right].$$ The cross terms cancel, leaving $\gamma^2(1 - \beta^2)\left[(c\Delta t)^2 - \Delta x^2\right]$, and $\gamma^2(1 - \beta^2) = 1$. In the frame where the two events happen at the same place, $\Delta x = 0$ and $s^2 = (c\Delta\tau)^2$: the interval is the proper time. The minus sign is what makes spacetime differ from space: a clock that moves takes a “shortcut” in time.
Worked examples
Worked example: Proper time between two events
Event B happens 5.0 yr after event A and 3.0 ly away. What is the interval, and what does a clock moving steadily from A to B read for the trip?
- $s^2 = (5.0)^2 - (3.0)^2 = 16$ ly² (with $c = 1$ ly/yr). It is positive: timelike.
- $\Delta\tau = s/c = 4.0$ yr.
- Check with time dilation: the clock moves at $3.0/5.0 = 0.60c$, so $\Delta\tau = 5.0/1.25 = 4.0$ yr.
Answer: $s^2 = 16$ ly², $\Delta\tau = 4.0$ yr
Worked example: Spacelike events
Two supernovae are seen to explode 2.0 yr apart, 3.0 ly from each other. Could one have triggered the other?
- $s^2 = (2.0)^2 - (3.0)^2 = -5.0$ ly²: spacelike.
- Even light would need 3.0 years to cover 3.0 ly, longer than the 2.0 years between them.
Answer: No. And observers moving fast enough would even see them in the opposite order
Worked example: A round trip (the twin paradox)
One twin stays on Earth. The other flies 3.0 ly out and 3.0 ly back, each way taking 5.0 yr of Earth time. How old is each twin's clock at the reunion?
- Earth twin: $10.0$ yr (one clock at rest, at both events).
- Traveler, each leg: $\Delta\tau = \sqrt{5.0^2 - 3.0^2} = 4.0$ yr; two legs: $8.0$ yr.
- The situation is not symmetric: only the traveler changes frames at the turnaround.
Answer: Earth twin 10.0 yr, traveler 8.0 yr
Common mistake
Thinking that since distances and times both change, nothing stays the same between frames. The combination $(c\Delta t)^2 - \Delta x^2$ is the same for everyone, just as a distance is the same however you turn a map. Check with any Lorentz transformation: compute $s^2$ before and after, and it does not change.
Common mistake
Writing $s^2 = (c\Delta t)^2 + \Delta x^2$, as in ordinary geometry. The minus sign is essential; with a plus sign the quantity changes from frame to frame.