Time dilation — derivation, twin paradox: Visual walkthrough

We will need only two facts, both from the Special Relativity Postulates:

Step 1 — Draw the tool: a light clock

WHAT. We build a clock out of light. Two mirrors face each other, separated by a vertical gap we call $L$ (just the height between them — a length, measured in metres). A single flash of light bounces up, hits the top mirror, and comes back down. We call one full round-trip = one "tick".

WHY this tool. A normal clock (gears, springs) hides why it slows. A light clock ties its ticking directly to $c$ — the very thing Rule 2 pins down. If light must behave, the clock must too.

PICTURE. The flash leaves the bottom mirror, kisses the top, returns. Emission and final catch happen at the same spot — remember this, it matters later.

Figure — Time dilation — derivation, twin paradox

Step 2 — The rider's view: light goes straight up and down

WHAT. Sit on the clock and ride along with it. For you, the clock is not moving. The flash goes straight up a distance $L$ , then straight down a distance $L$ . Total path travelled by light: $2 L$ .

WHY. Speed is distance divided by time. We know the distance ( $2 L$ ) and we know the speed ( $c$ ). So the time for one tick is forced on us:

$rider’s tick-time Δ t_{0} = \frac{2 L light’s path}{light’s speed c}$

Rearranging to keep $L$ handy for later:

$L = \frac{c Δ t _{0}}{2}$

PICTURE. A single vertical line up, a single vertical line down — nothing slanted, because in this frame the mirrors sit still.

Step 3 — Your view: the clock flies past, light goes diagonal

WHAT. Now stand on the ground and watch that same clock zoom past you sideways at speed $v$ . During one tick the clock drifts sideways while the flash bounces. So the flash cannot go straight up for you — by the time it reaches the top mirror, the top mirror has moved along. You see the light travel along a slanted diagonal up, then a slanted diagonal down: a tent " $Λ$ " shape.

WHY. The rider and you watch the identical physical flash, but you disagree about where "straight up" points because the clock slid under it. This is the only honest picture consistent with the clock moving.

PICTURE. Two diagonal chalk strokes meeting at the top mirror — the flash's path as you see it.

Step 4 — Cut the tick in half and build a triangle

WHAT. Look at just the first half of the tick — the flash going up. In that half-tick, three lengths appear, and they form a right triangle:

Vertical leg — the flash still had to cross the gap, so it rises by $L$ .
Horizontal leg — the clock drifted sideways by $v \cdot \frac{Δ t}{2}$ (speed $\times$ half the tick-time).
Hypotenuse — the slanted path the light actually took.

WHY the hypotenuse length is $c \cdot \frac{Δ t}{2}$ and nothing bigger. Here is Rule 2 doing its job. You might expect light to "go faster" on the longer diagonal to keep up — but it cannot. You must also measure its speed as exactly $c$ . So the diagonal length = speed $\times$ time = $c \cdot \frac{Δ t}{2}$ .

PICTURE. A right triangle: chalk-blue vertical $L$ , chalk-pink horizontal $v Δ t /2$ , pale-yellow slanted hypotenuse $c Δ t /2$ , with the right angle marked at the bottom corner.

Step 5 — Feed in $L$ and solve for your time $Δ t$

WHAT. We already found $L = \frac{c Δ t _{0}}{2}$ in Step 2. Substitute it so the equation talks only about times ( $Δ t$ , $Δ t_{0}$ ) and speeds ( $c$ , $v$ ):

$\frac{c ^{2} Δ t ^{2}}{4} = L^{2} rewritten \frac{c ^{2} Δ t _{0}^{2}}{4} + \frac{v ^{2} Δ t ^{2}}{4}$

WHY. We want a relationship between the rider's tick $Δ t_{0}$ and your tick $Δ t$ . Killing $L$ leaves exactly that.

Multiply every term by $4$ to clear the quarters: $c^{2} Δ t^{2} = c^{2} Δ t_{0}^{2} + v^{2} Δ t^{2}$

Gather the $Δ t$ terms on one side (they are the unknown we want alone): $c^{2} Δ t^{2} - v^{2} Δ t^{2} = c^{2} Δ t_{0}^{2} ⟹ Δ t^{2} (c^{2} - v^{2}) = c^{2} Δ t_{0}^{2}$

Divide by $(c^{2} - v^{2})$ , then divide top and bottom by $c^{2}$ : $Δ t^{2} = \frac{c ^{2} Δ t _{0}^{2}}{c ^{2} - v ^{2}} = \frac{Δ t _{0}^{2}}{1 - v ^{2} / c ^{2}}$

Take the square root:

PICTURE. The triangle re-drawn with the substitution shown, arrows pointing from each geometric side to its role in the final boxed formula.

Step 6 — What $γ$ does: read the stretch factor across all speeds

WHAT. $γ$ is just a number that depends on how fast the clock moves. Let us watch it over every possible speed from standstill to nearly light-speed.

WHY cover all cases. A formula you cannot picture at the extremes is a formula you do not trust. So we check the slow end, the everyday end, and the ferocious end.

$v = 0$ (clock at rest): $γ = 1/ 1 - 0 = 1$ , so $Δ t = Δ t_{0}$ . Both agree — no dilation. ✅ Sanity restored.
$v = 0.01 c$ (a fast jet, still "slow"): $γ \approx 1.00005$ — a stretch of five parts in a hundred thousand. Invisible. This is why Newton "worked".
$v = 0.6 c$ : $γ = 1/ 1 - 0.36 = 1/0.8 = 1.25$ . Moving clock ticks $25%$ slow.
$v \to c$ (approaching light-speed): $1 - v^{2} / c^{2} \to 0$ , so $γ \to \infty$ . The moving clock nearly freezes from your view.
$v > c$ ? Then $1 - v^{2} / c^{2} < 0$ and the square root is imaginary — nonsense. This is the maths politely telling you: nothing can go faster than light.

PICTURE. The $γ$ -versus- $v$ curve: flat and near $1$ for small $v$ , then rocketing to infinity as $v$ climbs toward $c$ (a vertical wall at $v = c$ ).

Step 7 — A real clock in the sky: the muon (worked case)

WHAT. A muon lives, in its own frame, only $Δ t_{0} = 2.2 μ s$ before decaying. It races down through the atmosphere at $v = 0.99 c$ . How long does the lab see it live?

WHY. This is the derivation cashing out into an experiment you can actually run.

$γ = \frac{1}{1 - 0.9 9 ^{2}} = \frac{1}{1 - 0.9801} = \frac{1}{0.0199} \approx 7.09$ $Δ t = γ Δ t_{0} = 7.09 \times 2.2 μ s \approx 15.6 μ s$

PICTURE. Two number-lines side by side: the muon's own short lifetime bar vs. the lab's stretched-out bar — long enough to reach the ground.

The one-picture summary

Everything above collapses into a single triangle. The rider's tick is the vertical side; your tick is the hypotenuse; the clock's motion is the base. A longer hypotenuse at the same light-speed $c$ means more time per tick — that is time dilation, drawn.

Recall Feynman retelling — the whole walkthrough in plain words

Build a clock that ticks by bouncing light between two mirrors. If you carry it, the light just goes straight up and down — short trip, quick tick. If I watch you fly past, that same light has to travel a slanted zig-zag, because the mirrors slid sideways while the light was in flight. The zig-zag is longer. But light isn't allowed to speed up — it's stuck at $c$ for me too. Longer road at the same speed means the trip takes more time. So each tick of your clock takes longer for me: your clock, and everything about you, runs in slow motion. Draw the up-trip as a right triangle — straight-up side, sideways-drift side, slanted-light side — and Pythagoras hands you the exact stretch factor $γ$ . At walking speed $γ$ is basically $1$ (no one notices); near light-speed it blows up toward infinity (clock nearly freezes); past light-speed the maths goes imaginary, forbidding it. That single triangle is the whole theory.

Visual walkthrough — Time dilation — derivation, twin paradox