The constancy of the speed of light has a measurable effect on clocks. A clock moving relative to you runs slow, as you measure it. This is not a flaw of the clock: every process, from ticking quartz to a beating heart to a decaying particle, keeps the same pace with it.
🔑 Key idea
The time between two events that happen at the same place in some frame, measured there, is the proper time $\Delta\tau$. In a frame where that clock moves at speed $v$, the same interval lasts $$\Delta t = \gamma\,\Delta\tau,\qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \ge 1.$$ The proper time is always the shortest time between the two events.
💡 Why it works
Build a light clock: a pulse bouncing between two mirrors a distance $D$ apart. In the clock's frame each tick is up and back, $\Delta\tau = \frac{2D}{c}$. In a frame where the clock moves sideways at $v$, the pulse follows a slanted path. In half a tick, $\frac{\Delta t}{2}$, it moves $\frac{c\Delta t}{2}$ along the slant while the mirrors move $\frac{v\Delta t}{2}$ sideways. Pythagoras: $$\left(\frac{c\Delta t}{2}\right)^2 = D^2 + \left(\frac{v\Delta t}{2}\right)^2.$$ Solve: $\Delta t = \frac{2D/c}{\sqrt{1 - v^2/c^2}} = \gamma\,\Delta\tau$. Postulate 2 is the key step: the light runs at $c$ along the slant, not faster. By postulate 1, every other kind of clock in the ship must agree with its light clock, or the crew could detect their motion.
Light clocks at rest and moving at $0.8c$. The moving clock's pulse runs a longer slanted path at the same speed $c$, so it ticks $\gamma = 1.67$ times more slowly.
How $\gamma$ grows with speed. Below about $0.1c$ the effect is under half a percent.
$v/c$
0.10
0.50
0.60
0.80
0.95
0.995
$\gamma$
1.005
1.155
1.25
1.667
3.20
10.0
Worked examples
✏️ Worked example: Muons reach the ground
Muons made high in the atmosphere live on average $2.2$ μs in their own frame. They move at $0.995c$. On average, how long do they last in Earth's frame, and how far do they get?
Distance: $(0.995)(3.00 \times 10^8)(22 \times 10^{-6}) = 6.6$ km. Without time dilation they would average only 660 m.
Answer: About 22 μs and 6.6 km, which is why we detect so many at sea level
✏️ Worked example: A trip to a nearby star
A ship travels 4.0 light-years at $0.80c$ to a star at rest relative to Earth. How long does the trip take according to Earth, and according to the ship's clock?
An airliner flies at 250 m/s for 10 hours. By special relativity alone, how much less time passes on its clock than on the ground?
At small speeds, $\gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2}$, so the difference is $\Delta t - \Delta\tau \approx \frac{1}{2}\frac{v^2}{c^2}\Delta t$.
$\frac{1}{2}(6.9 \times 10^{-13})(36{,}000\ \text{s}) = 1.25 \times 10^{-8}$ s.
Answer: About 12.5 ns: tiny, but atomic clocks flown around the world in 1971 measured effects of this size
GPS satellites move at about 3.9 km/s. Special relativity slows their clocks by about 7 μs per day, while general relativity (weaker gravity up there) speeds them up by about 45 μs per day. The system corrects for the net 38 μs per day; without it, positions would drift by kilometers each day.
⚠️ Common mistake
Applying $\gamma$ backward, making the proper time the longer one. The proper time is measured by a single clock present at both events, and it is the shortest. Check: muons live longer in Earth's frame, not shorter; the ship's crew ages less on the trip, not more. Ask “which clock is at both events?” and that clock reads $\Delta\tau$.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.