Time dilation — derivation, twin paradox: Worked examples

Before anything, recall the symbols so nothing is used un-earned:

Definition The two times and the speed ratio (re-anchored)

$c$ = the speed of light in vacuum, about $3.0 \times 1 0^{8}$ metres per second. It is the same number for every observer (postulate 2) — that constancy is the whole reason time dilation exists.
$Δ t_{0}$ = proper time — the time between two events measured by a clock present at both events (they happen at the same place in that clock's frame). Think: the personal time of the thing that is moving.
$Δ t$ = the time the other observer measures for those same two events, watching the clock fly past.
$β$ (Greek "beta") $= \frac{v}{c}$ is the speed as a fraction of light-speed. So $β = 0.6$ means "moving at $60%$ of $c$ ". It just packages $v / c$ into one tidy symbol that shows up everywhere below.
$γ = \frac{1}{1 - v ^{2} / c ^{2}} = \frac{1}{1 - β ^{2}} \geq 1$ is the "stretch factor". Because $γ \geq 1$ , we always have $Δ t \geq Δ t_{0}$ : the moving clock ticks slower than the watcher's clock.

The scenario matrix

Every time-dilation question is really one of these cells. The examples below are labelled with the cell they cover.

Cell	What varies / the trap	Which example
A. Find $Δ t$ from $Δ t_{0}$	you know the moving clock's own time, want the watcher's	Ex 1
B. Find $Δ t_{0}$ from $Δ t$	you know the watcher's time, want the moving clock's	Ex 2
C. Find $v$ (or $γ$ ) from two times	inverse problem — solve for speed	Ex 3
D. Degenerate: $v = 0$	clock at rest → no dilation	Ex 4
E. Limiting: $v \to c$	$γ \to \infty$ , time freezes	Ex 4
F. Low-speed check: $v ≪ c$	$γ \approx 1$ , Newton returns	Ex 4
G. Real-world word problem	identify which frame holds $Δ t_{0}$	Ex 5 (muon), Ex 6 (GPS-style)
H. Round-trip / twin twist	asymmetry, who turns around	Ex 7
I. Exam trap: wrong variable	plugging watcher-time into $Δ t_{0}$	Ex 8

The "same-place" test is easiest to see on a worldline diagram — a picture with position across (space) and time up, where a clock traces a line as it moves. Study this before the examples.

Figure — Time dilation — derivation, twin paradox

Cell A — you know the moving clock's time, find the watcher's

Worked example Example 1 — spaceship pilot's stopwatch (Cell A)

A pilot's onboard stopwatch measures $Δ t_{0} = 4.0$ years for a journey at $v = 0.6 c$ (so $β = 0.6$ ). How many years pass on Earth for the same journey?

Forecast: More years on Earth, or fewer? (Guess: the pilot is the moving clock, so Earth should see more — the moving clock ran slow, so it logged fewer.)

Compute $γ$ : $γ = \frac{1}{1 - β ^{2}} = \frac{1}{1 - 0.36} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25$ . Why this step? $γ$ is the bridge between the two times; we always build it first from $β$ .
Identify $Δ t_{0}$ : the stopwatch is on the ship, present at both departure and arrival events — same place in the ship's frame. So $Δ t_{0} = 4.0$ yr. Why this step? This is the "same place" test — it decides which number is $Δ t_{0}$ .
Multiply: $Δ t = γ Δ t_{0} = 1.25 \times 4.0 = 5.0$ years on Earth. Why this step? We want the watcher's (Earth) time, and $Δ t = γ Δ t_{0}$ enlarges the proper time.

Verify: $Δ t = 5.0 > Δ t_{0} = 4.0$ ✓ (moving clock ran slow, so Earth's clock logged more). Units: years in, years out ✓.

Cell B — you know the watcher's time, find the moving clock's

Worked example Example 2 — Earth clock given, ship clock wanted (Cell B)

Earth measures $Δ t = 10$ years for a probe flying at $v = 0.8 c$ ( $β = 0.8$ ). How much time elapses on the probe's own clock?

Forecast: (Guess: the probe is the moving clock, so it should log fewer than 10 years.)

$γ = \frac{1}{1 - β ^{2}} = \frac{1}{1 - 0.64} = \frac{1}{0.36} = \frac{1}{0.6} = 1.6 \overline{6}$ . Why this step? Same bridge factor as always.
The two events (launch, arrival) happen at the probe → probe's clock is present at both → probe holds $Δ t_{0}$ . Why this step? Same-place test again: it tells us we must solve for $Δ t_{0}$ , not multiply.
Rearrange $Δ t = γ Δ t_{0}$ to get $Δ t_{0} = \frac{Δ t}{γ} = \frac{10}{1.6 6} = 6.0$ years. Why this step? Here the watcher's time is the known quantity, so we divide by $γ$ to shrink back to proper time.

Verify: $6.0 < 10$ ✓ (probe logged fewer). Check consistency: $γ \times 6 = 1.6 \overline{6} \times 6 = 10$ ✓.

Cell C — the inverse problem, solve for speed

Worked example Example 3 — how fast were they going? (Cell C)

A crew logs $Δ t_{0} = 3.0$ years while mission control logs $Δ t = 5.0$ years. Find the speed $v$ .

Forecast: (Guess: the times differ by a lot, so $v$ is a large fraction of $c$ .)

Form the ratio: $γ = \frac{Δ t}{Δ t _{0}} = \frac{5.0}{3.0} = 1.6 \overline{6}$ . Why this step? Both times are known, and their ratio is $γ$ directly — no need for $v$ yet.
Undo the definition of $γ$ . Start from $γ = \frac{1}{1 - β ^{2}}$ , square both sides: $γ^{2} = \frac{1}{1 - β ^{2}}$ . Why this step? The unknown $β$ is trapped inside a square root; squaring frees it.
Solve for $β^{2}$ : $1 - β^{2} = \frac{1}{γ ^{2}} \Rightarrow β^{2} = 1 - \frac{1}{γ ^{2}}$ . So $v = c 1 - 1/ γ^{2}$ . Why this step? Algebraic isolation — we want $v$ alone.
Plug in: $\frac{1}{γ ^{2}} = \frac{1}{( 5/3 ) ^{2}} = \frac{9}{25} = 0.36$ , so $β^{2} = 1 - 0.36 = 0.64$ , giving $β = 0.8$ , i.e. $v = 0.8 c$ . Why this step? Nice numbers confirm we picked a "movie-friendly" speed.

Verify: At $v = 0.8 c$ , $γ = 1/ 1 - 0.64 = 1/0.6 = 1.6 \overline{6}$ , and $1.6 \overline{6} \times 3 = 5$ ✓.

Cells D, E, F — the three limits (one figure, three readings)

The whole behaviour of $γ$ lives in one curve. Read it three ways.

Worked example Example 4 — the three limiting cases (Cells D, E, F)

Evaluate $γ$ (and hence what happens to the moving clock) at three speeds: $v = 0$ , $v \to c$ , and $v = 0.01 c$ .

Forecast: (Guess: at rest nothing happens; near light-speed time nearly stops; at slow speed almost nothing happens.)

Cell D — $v = 0$ (clock at rest):

$γ = \frac{1}{1 - 0} = 1$ . So $Δ t = 1 \cdot Δ t_{0} = Δ t_{0}$ . Why this step? A clock at rest relative to you is your own clock — no stretching. Look at the white dot at the left end of the curve: $γ$ starts flat at $1$ .

Cell E — $v \to c$ (approaching light speed): 2. As $v \to c$ , $β^{2} \to 1$ , so $1 - β^{2} \to 0^{+}$ , and $γ = 1/ tiny \to \infty$ . Then $Δ t = γ Δ t_{0} \to \infty$ : one tick of the moving clock takes forever for you — time appears frozen. Why this step? This is the amber wall of the curve — the vertical asymptote at $v = c$ . It's why nothing with mass reaches $c$ : it would need infinite time-stretch.

Cell F — $v = 0.01 c$ (everyday speed): 3. $γ = 1/ 1 - 1 0^{- 4} \approx 1 + \frac{1}{2} (1 0^{- 4}) = 1.00005$ . Why this step? Use the approximation $\frac{1}{1 - x} \approx 1 + \frac{x}{2}$ for tiny $x$ . This is the near-flat lower shelf of the curve — dilation of $5$ parts per hundred thousand, utterly invisible.

Verify: $γ (0) = 1$ exactly ✓; $γ$ increases without bound as $v \to c$ ✓; $γ (0.01 c) \approx 1.00005$ matches the exact $1/ 0.9999 = 1.0000500 \dots$ ✓.

Recall Why is

$\frac{1}{1 - x} \approx 1 + \frac{x}{2}$ for tiny $x$ ? (used in Ex 4 and Ex 6) Answer ::: It is the binomial approximation $(1 + u)^{n} \approx 1 + n u$ for $∣ u ∣ ≪ 1$ . Here $\frac{1}{1 - x} = (1 - x)^{- 1/2}$ , so $u = - x$ and $n = - \frac{1}{2}$ , giving $1 + (- \frac{1}{2}) (- x) = 1 + \frac{x}{2}$ . The higher-order terms are of size $x^{2}$ , negligible when $x$ is tiny.

Cell G — real-world word problems (find the right frame)

Worked example Example 5 — the muon that shouldn't reach the ground (Cell G)

Muons form high in the atmosphere — take a typical production height of $H \approx 15$ km (about $10$ km up in the stratosphere plus the slant path, the standard textbook value for cosmic-ray muon studies). Their proper lifetime is $Δ t_{0} = 2.2 μ s$ , moving at $v = 0.99 c$ ( $β = 0.99$ ). Newtonianly a muon travels only $c Δ t_{0} \approx (3 \times 1 0^{8}) (2.2 \times 1 0^{- 6}) \approx 660$ m before decaying. How far does one typically travel in the lab (ground) frame — and why do so many nevertheless cross the full $H$ to reach detectors?

Forecast: (Guess: in the lab frame the muon lives much longer, so its typical range grows by a factor of $γ$ .)

$γ = 1/ 1 - 0.9 9^{2} = 1/ 1 - 0.9801 = 1/ 0.0199 \approx 7.089$ . Why this step? Build the bridge factor from the given speed.
Which frame holds $Δ t_{0}$ ? The muon's own clock is present at both "born" and "decay" events → the $2.2 μ s$ is proper time. The lab (ground) is the watcher. Why this step? Same-place test: birth and death happen on the muon.
Lab lifetime: $Δ t = γ Δ t_{0} = 7.089 \times 2.2 μ s \approx 15.6 μ s$ . Why this step? We want how long the muon lives as the ground sees it — Cell A multiply.
Typical lab-frame range: $d = v Δ t = 0.99 (3 \times 1 0^{8}) (15.6 \times 1 0^{- 6}) \approx 4.63 \times 1 0^{3}$ m $\approx 4.6$ km — up from just $0.66$ km. Why this step? Turns the dilated time into the physical range.

A subtlety — how do they still cross the full $H = 15$ km? The $4.6$ km is the mean range for one proper lifetime; it is not a wall. Radioactive decay is exponential, so a fraction of muons survive several lifetimes. Without dilation the survival fraction over the production height $H = 15$ km would be $e^{- H /660 m} = e^{- 15000/660} \approx e^{- 22.7} \approx 1 0^{- 10}$ — essentially none. With dilation the relevant exponent uses the dilated range, $e^{- 15000/4630} \approx e^{- 3.24} \approx 0.039$ — about $4%$ survive, a hundred-million-fold boost that turns "none detected" into "plenty detected."

Verify: $4.6$ km $≫ 0.66$ km ✓; the survival ratio $e^{- 3.24} / e^{- 22.7} = e^{19.5} \approx 3 \times 1 0^{8}$ ✓ — the factor of $γ$ in the exponent is what carries muons through $H = 15$ km. This matches the observed muon flux at ground level. (In the muon's own frame the distance shrinks instead — see Length Contraction — same physics, viewed from the other seat.)

Worked example Example 6 — a fast-orbit clock drifts (Cell G, GPS-style)

A clock moves at $v = 4.0 \times 1 0^{3}$ m/s (roughly orbital speed) for one day of ground time, $Δ t = 86400$ s. How far behind does the moving clock fall due to velocity time dilation alone?

Forecast: (Guess: at a few km/s, $β$ is tiny, so the offset is only microseconds.)

$β = v / c = \frac{4.0 \times 1 0 ^{3}}{3.0 \times 1 0 ^{8}} = 1.333 \times 1 0^{- 5}$ , so $β^{2} = 1.778 \times 1 0^{- 10}$ . Why this step? At small speeds we work with $β^{2}$ directly to keep precision.
Time lost by the moving clock per elapsed ground time: $Δ t - Δ t_{0} = Δ t (1 - \frac{1}{γ}) \approx Δ t \cdot \frac{1}{2} β^{2}$ . Why this step? From the binomial approximation (see the collapsible above Ex 5): $γ = (1 - β^{2})^{- 1/2} \approx 1 + \frac{1}{2} β^{2}$ , so $\frac{1}{γ} \approx 1 - \frac{1}{2} β^{2}$ , giving fractional lag $\frac{1}{2} β^{2}$ . This is the only workable route — the raw difference of two nearly-equal seconds would drown in rounding.
Plug in: $Δ t \cdot \frac{1}{2} β^{2} = 86400 \times \frac{1}{2} \times 1.778 \times 1 0^{- 10} \approx 7.68 \times 1 0^{- 6}$ s $\approx 7.7 μ s$ per day. Why this step? Turns the fractional lag into an actual clock offset.

Verify: Order of magnitude — microseconds per day — is the real velocity contribution to satellite clock corrections ✓. Units: $s \times (dimensionless) = s$ ✓.

Cell H — the round-trip twin twist

Worked example Example 7 — twins with a stopover (Cell H)

Twin B flies to a star $6$ light-years away (Earth frame) at $v = 0.6 c$ ( $β = 0.6$ ), then returns at the same speed. Twin A stays home. Find each twin's aging, and confirm the asymmetry.

Forecast: (Guess: B ages less; the gap is a few years.)

Earth (frame S, one inertial frame throughout) times the round trip: distance out-and-back $= 12$ ly at $0.6 c$ , so $Δ t_{A} = \frac{12}{0.6} = 20$ years. Why this step? Twin A never changes frames, so distance/speed gives A's aging cleanly.
$γ = 1/ 1 - 0.36 = 1/0.8 = 1.25$ . Why this step? The bridge for B's clock.
B's clock is present at departure, turnaround, and return — on each straight leg B holds proper time. So $Δ t_{B} = \frac{Δ t _{A}}{γ} = \frac{20}{1.25} = 16$ years. Why this step? Cell B (divide) applied leg-by-leg; the turnaround itself is instantaneous here.
Asymmetry check: only B turned around (switched inertial frames, felt the rocket's push). A stayed in one frame. So the situation is not symmetric and it is legitimate that B ages less. Why this step? This is the resolution of the paradox — the turning twin is the younger one.

Verify: $Δ t_{B} = 16 < Δ t_{A} = 20$ ✓; B returns $4$ years younger. Consistency: $16 \times 1.25 = 20$ ✓. (A deeper accounting uses the invariant Spacetime Interval, which every frame agrees on.)

The asymmetry is a shape fact: B's worldline is bent, A's is straight. See it drawn:

Cell I — the exam trap

Worked example Example 8 — the deliberately mislabelled question (Cell I)

"A spaceship passes Earth at $v = 0.6 c$ . Earth observers time the trip between two Earth landmarks as $8.0$ s. A student writes $Δ t = γ \times 8.0 = 1.25 \times 8 = 10$ s for the ship. Is the student right?"

Forecast: (Guess: no — they've mislabelled which time is proper.)

Locate the two events: the ship passes landmark 1, then landmark 2. These happen at two different places on Earth, but at the same place on the ship (right at the ship's nose). Why this step? The same-place test decides everything, and here it flips the naive assumption.
Therefore the ship holds the proper time, and the Earth reading $8.0$ s is the dilated $Δ t$ . Why this step? The student assumed Earth held $Δ t_{0}$ — the classic Cell I error (see the parent note's Mistake 2).
Correct computation: $Δ t_{0}^{(ship)} = \frac{Δ t}{γ} = \frac{8.0}{1.25} = 6.4$ s. Why this step? We must divide, not multiply — the ship is the single clock present at both passing-events, so it holds proper time and reads fewer seconds than the two synchronized Earth clocks.

Verify: $6.4 < 8.0$ ✓ (the single moving ship-clock reads less than the two synchronized Earth clocks). The student's $10$ s is wrong precisely because they multiplied the wrong number: they treated Earth's $8.0$ s as proper time when it is actually the dilated time.

Recall Which frame holds

$Δ t_{0}$ in each example? Ex 1 ship ::: ship (events on ship) → multiply for Earth Ex 2 probe ::: probe → divide from Earth's 10 yr Ex 5 muon ::: muon (born & decays on it) → multiply for lab Ex 8 trap ::: ship (both landmark-passings at its nose) → divide, not multiply

Worked examples — Time dilation — derivation, twin paradox

The scenario matrix

Cell A — you know the moving clock's time, find the watcher's

Cell B — you know the watcher's time, find the moving clock's

Cell C — the inverse problem, solve for speed

Cells D, E, F — the three limits (one figure, three readings)

Cell G — real-world word problems (find the right frame)

Cell H — the round-trip twin twist

Cell I — the exam trap

Active recall