Simultaneity — relativity of simultaneity: Visual walkthrough

Step 1 — What an "event" even is

WHAT: We label two lightning strikes. Strike B at the back of a train, strike F at the front.

WHY: Before we can ask "did these happen at the same time?", we must be able to point at each one with a (where, when) pair. That pair is the only thing physics lets us talk about.

PICTURE: Below, each dot is an event. The horizontal axis is position $x$ ; the vertical axis is time $t$ (later = higher). This kind of chart is a spacetime diagram.

Figure — Simultaneity — relativity of simultaneity

Step 2 — "Simultaneous" drawn as a picture

WHAT: We place Alice's frame — call it $S$ — so that both strikes read the same time. In her world, $t_{1} = t_{2}$ .

WHY: We deliberately start with agreement. If we can show a second observer disagrees about events that Alice calls simultaneous, we've proven simultaneity is not universal. You cannot see a disagreement grow unless you begin from perfect agreement.

PICTURE: The two dots now lie on one flat horizontal line — the "line of now" for Alice. Their only difference is horizontal: the gap in position.

Step 3 — The one tool we're allowed: the Lorentz time equation

Everyday intuition says clocks are universal — one $t$ for everybody. Special relativity forbids that, because of one experimental fact from the postulates: light travels at the same speed $c$ for every observer. To keep that true, moving observers must relabel when things happen. The recipe that does the relabelling is the Lorentz Transformation.

PICTURE: How $γ$ climbs from $1$ as speed grows — gentle at first, then steep near $c$ .

Step 4 — Feed BOTH events through the same machine

WHAT: We apply the Lorentz time equation to Event B and to Event F, one at a time, using the same $γ$ , $v$ , $c$ (they describe the frames, not the events).

WHY: Bob assigns his own time to each event separately. The formula is the same machine; only the event's coordinates $(x, t)$ change as we drop them in. If Bob assigns them different times, simultaneity has broken.

PICTURE: Two events, one machine, two outputs. The front event sits at a larger $x$ , so a bigger chunk $\frac{v x}{c ^{2}}$ gets subtracted from it.

Step 5 — Subtract, and watch the agreement collapse

WHAT: Compute Bob's time gap $Δ t^{'} = t_{2}^{'} - t_{1}^{'}$ by subtracting the two boxed lines.

WHY: A single time doesn't tell us about disagreement — the difference between the two does. If $Δ t^{'} = 0$ , Bob still calls them simultaneous. If not, he doesn't.

$Δ t^{'} = γ (t_{2} - \frac{v x _{2}}{c ^{2}}) - γ (t_{1} - \frac{v x _{1}}{c ^{2}})$

Now use the fact from Step 2 that Alice sees them simultaneous: $t_{1} = t_{2}$ . Those two terms cancel:

$Δ t^{'} = γ (t_{2} - t_{1} - \frac{v x _{2}}{c ^{2}} + \frac{v x _{1}}{c ^{2}}) = - \frac{γ v}{c ^{2}} (x_{2} - x_{1})$

PICTURE: The $t_{1} = t_{2}$ terms crossing out, leaving a clean product of $γ$ , $v$ , and $Δ x$ .

Step 6 — Reading the minus sign: "leading clocks lag"

WHAT: Decide what the sign of $Δ t^{'}$ physically means.

WHY: A formula that says whether order flips is useless if we can't say which way. The sign is the direction indicator.

Take the front at larger $x$ ( $x_{2} > x_{1}$ , so $Δ x > 0$ ) and motion in the $+ x$ direction ( $v > 0$ ). Then $Δ t^{'} = - \frac{γ v Δ x}{c ^{2}} < 0 \Rightarrow t_{2}^{'} < t_{1}^{'} .$

$t_{2}^{'} < t_{1}^{'}$ means the front event (F) has the smaller time in Bob's frame — it happened first for him.
Equivalently, the clock sitting at the front (the leading edge in the direction of motion) reads behind the rear clock. This is the rule "leading clocks lag."

PICTURE: In Bob's tilted "line of now," the front dot dips below the back dot — front is earlier.

Step 7 — Every edge case, drawn

We must never leave the reader in a scenario we didn't show. Here are all the limiting cases.

WHAT & WHY: Push each input to zero or to extremes and check the formula behaves.

The sign flip case: if instead the front is at smaller $x$ (motion toward the back), $Δ x < 0$ and $Δ t^{'} > 0$ — now the back event is first. Same rule, "leading clock lags," just relabeled which edge leads.

PICTURE: Four mini-panels — $Δ x = 0$ , $v = 0$ , high $v$ , and reversed direction — each with its tilt of Bob's "now" line.

Step 8 — Put a number on it (the 300 m train)

WHAT: A $300 m$ train at $v = 0.6 c$ ; ground (Alice) sees front and back struck at once.

WHY: To feel the size of the effect and confirm the sign rule numerically.

The one-picture summary

This single figure compresses the whole walkthrough: Alice's flat "now" line carries two simultaneous events; run each through the same Lorentz machine; Bob's "now" line tilts; the two events split apart in time by exactly $Δ t^{'} = - γ v Δ x / c^{2}$ . The tilt is the relativity of simultaneity.

Recall Feynman: the whole thing in plain words

Picture two firecrackers popping at the two ends of a train — and picture them popping at exactly the same time as far as you, standing on the ground, are concerned. On a chart of space-and-time, you'd draw both pops sitting on one flat line, side by side. Now here's the trick: a friend riding the train doesn't get to keep your flat line. Because light must travel at the same speed for her as for you, the rulebook (the Lorentz equation) forces her to relabel when each pop happened — and the relabelling depends on where the pop was. The front pop sits at a bigger position, so a bigger correction gets subtracted from it. Subtract the two corrected times and everything about your agreement cancels except one leftover piece: minus gamma times speed times the distance between the pops, all divided by $c$ squared. That leftover is her disagreement. If the pops were at the same spot, it's zero — everyone agrees. If she's standing still, it's zero. But the moment there's real distance and real motion, her "flat now" tilts, the front pop slides earlier, and she honestly, correctly says the front went first. Nobody made a mistake. You simply live in slightly different "nows."

Recall Quick self-check

Why doesn't $Δ t^{'}$ depend on $t_{1}$ or $t_{2}$ ? ::: Because we chose $t_{1} = t_{2}$ (simultaneous in $S$ ), so those terms cancel on subtraction, leaving only the position-dependent piece. In $Δ t^{'} = - γ v Δ x / c^{2}$ , whose frame is $Δ x$ measured in? ::: Frame $S$ — the one where the events are simultaneous (Alice's/ground frame). What makes the effect vanish? ::: $Δ x = 0$ (same place) or $v = 0$ (no relative motion). For the 300 m train at $0.6 c$ , what is the gap? ::: $- 0.75 μ s$ (front first in the train frame).