The postulates have a startling first consequence. “At the same time” is not the same for everyone. Two events that happen simultaneously for one observer can happen one after the other for another observer moving relative to the first.
Key idea
Events that are simultaneous but at different places in one frame are not simultaneous in a frame moving along the line joining them. If they are a distance $\Delta x$ apart in frame $S$, then in a frame moving at speed $v$ toward the $+x$ event, that event happens first, earlier by $$\Delta t' = \gamma\,\frac{v\,\Delta x}{c^2},\qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}.$$
Why it works
Put a lamp in the middle of a train car. In the train's frame, light runs at $c$ toward both ends and reaches them at the same moment: two simultaneous events. Now watch from the ground. The light still runs at $c$ both ways (postulate 2), but the rear wall moves toward the light and the front wall runs away from it. So the light reaches the rear wall first. Both observers are right in their own frames; there is no frame-independent “now” for distant places. The formula for $\Delta t'$ comes from the Lorentz transformation in this unit; its $\frac{v\Delta x}{c^2}$ is tiny at everyday speeds, which is why we never notice.
Worked examples
Worked example: Which came first?
Two lightning bolts strike the ends of a long platform at the same time, according to the station clocks. A train speeds along the platform toward the east end. According to clocks on the train, which strike came first?
- In the train's frame, the platform moves west. The train rushes toward the light from the east strike and away from the light from the west strike.
- By the rule, in a frame moving toward one event, that event happens first.
Answer: The east strike (the one the train is heading toward)
Worked example: How much earlier?
Two events 600 m apart happen at the same time in the lab. A ship flies along the line joining them at $0.60c$. How far apart in time are they on the ship's clocks?
- $\gamma = \frac{1}{\sqrt{1 - 0.36}} = 1.25$.
- $\Delta t' = \gamma\frac{v\Delta x}{c^2} = 1.25 \times \frac{(0.60)(3.00 \times 10^8)(600)}{(3.00 \times 10^8)^2}$.
- $= 1.25 \times \frac{0.60 \times 600}{3.00 \times 10^8} = 1.5 \times 10^{-6}$ s.
Answer: $1.5$ μs, the event ahead of the ship first
Worked example: Can cause and effect flip?
A signal can travel at most at $c$. Event A causes event B 3.0 μs later, 600 m away. Could some observer see B before A?
- In 3.0 μs light travels $(3.00 \times 10^8)(3.0 \times 10^{-6}) = 900$ m.
- B is within light's reach of A (600 m < 900 m). For such pairs, $\Delta t' = \gamma(\Delta t - v\Delta x/c^2)$ stays positive for every $v < c$, since $v\Delta x/c^2 < \Delta x/c < \Delta t$.
Answer: No: every observer sees A before B. Only events too far apart for any signal to connect can switch order
Common mistake
Thinking “at the same time” means the same thing to every observer. Simultaneity of distant events depends on the frame. A check from the lamp-in-the-train argument: both observers agree light moves at $c$ and both watch the same walls, yet they disagree about which wall the light reaches first. Only events at the same place that are simultaneous in one frame are simultaneous in all frames.