Time Dilation — Moving Clocks Run Slow
Time Dilation
One of the most astonishing consequences of special relativity is that time itself is not absolute — it depends on relative motion. A clock moving relative to an observer ticks more slowly than a clock at rest. This is not an illusion, not a mechanical defect, and not a trick of perception. Moving clocks genuinely, physically run slower. This effect is called time dilation.
The mathematics of time dilation follows directly from the Lorentz factor. If an event takes a time interval delta-t-zero in a reference frame where the clock is at rest — called the proper time — then an observer who sees that clock moving at speed v will measure a longer time interval delta-t equals gamma times delta-t-zero. Since gamma is always greater than or equal to one, the moving clock always appears to tick more slowly.
At everyday speeds — cars, airplanes, even rockets — gamma is extraordinarily close to one, so time dilation is imperceptibly small. A passenger on a jet flying at 900 kilometers per hour experiences time dilation of roughly one part in a trillion. But at speeds approaching the speed of light, the effect becomes enormous. At 99 percent of c, gamma is about 7.09, meaning that one year of ship time corresponds to more than seven years for a stationary observer.
Time dilation is not merely theoretical. It has been confirmed by numerous experiments. In 1971, Hafele and Keating flew atomic clocks around the world on commercial airliners, and the clocks returned showing exactly the time difference predicted by relativity. Every day, the Global Positioning System must correct for time dilation — without these corrections, GPS positions would drift by about 11 kilometers per day. Muons created in the upper atmosphere by cosmic rays travel at 0.998c and have a half-life of only 1.56 microseconds at rest, yet they routinely reach the ground because time dilation stretches their apparent lifetime by a factor of about 16.
The mathematics behind all these phenomena is nothing more than multiplication by the Lorentz factor — the same algebra and square roots you have already learned. Special relativity is a beautiful demonstration that fundamental algebra can describe the most exotic behaviors of the physical universe.
If $\Delta t_0$ is the proper time (measured in the rest frame of the clock), then the dilated time measured by a moving observer is:
$$\Delta t = \gamma \, \Delta t_0 = \frac{\Delta t_0}{\sqrt{1 - \dfrac{v^2}{c^2}}}$$
Since $\gamma \ge 1$, we always have $\Delta t \ge \Delta t_0$: the moving clock runs slower.
The proper time $\Delta t_0$ is the time interval measured by a clock that is at rest relative to the event. It is always the shortest time interval any observer can measure for that event.
A spaceship travels at $v = 0.8c$. If the ship's clock measures 5 years of travel, how much time passes on Earth?
Step 1: Find $\gamma$ at $v = 0.8c$.
$$\gamma = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = \frac{5}{3} \approx 1.667$$
Step 2: Apply the time dilation formula.
$$\Delta t = \gamma \cdot \Delta t_0 = \frac{5}{3} \times 5 = \frac{25}{3} \approx 8.33 \text{ years}$$
Interpretation: While 5 years pass on the ship, about 8.33 years pass on Earth. The astronauts age less than people on Earth.
Muons produced by cosmic rays travel at $v = 0.998c$ and have a rest-frame half-life of $\Delta t_0 = 1.56$ μs. What is their observed half-life?
Step 1: $\gamma = 1/\sqrt{1 - 0.998^2} = 1/\sqrt{1 - 0.996004} = 1/\sqrt{0.003996} \approx 15.82$
Step 2: $\Delta t = 15.82 \times 1.56 = 24.7$ μs
The muon's half-life is stretched from 1.56 μs to about 24.7 μs, allowing it to traverse the atmosphere and reach the ground.
An astronaut on a mission experiences 1 year of proper time. If 10 years pass on Earth, how fast was the ship traveling?
Step 1: $\Delta t = \gamma \cdot \Delta t_0$, so $\gamma = \Delta t / \Delta t_0 = 10/1 = 10$.
Step 2: Solve for $v$:
$$\gamma = 10 \implies 1 - v^2/c^2 = 1/100$$
$$v^2/c^2 = 1 - 0.01 = 0.99$$
$$v = c\sqrt{0.99} \approx 0.995c$$
Practice Problems
Show Answer Key
1. $\gamma = 1/\sqrt{1-0.36} = 1.25$. Earth time $= 1.25 \times 10 = 12.5$ years.
2. $\gamma = 1/\sqrt{1-0.81} \approx 2.294$. Observed half-life $= 2.294 \times 2 \approx 4.59$ μs.
3. $\gamma = 20/5 = 4$.
4. $v/c = \sqrt{1 - 1/16} = \sqrt{15/16} \approx 0.968c$.
5. $v/c = 3.9/300000 = 1.3 \times 10^{-5}$. $\gamma \approx 1 + \frac{1}{2}(1.3 \times 10^{-5})^2 \approx 1.0000000001$.
6. $\gamma = 1/\sqrt{1-0.9025} \approx 3.203$. Traveling twin ages 4 years (to age 34). Earth twin ages $3.203 \times 4 \approx 12.8$ years (to age 42.8).
7. $\gamma = 1/\sqrt{1-0.990025} \approx 10.01$. Lab lifetime $\approx 10.01 \times 2.2 \approx 22.0$ μs.
8. $\gamma = 2 \implies v = c\sqrt{3/4} = c\sqrt{3}/2 \approx 0.866c$.
9. $\gamma \approx 1 + v^2/(2c^2) = 1 + 3.47 \times 10^{-13}$. In 86400 s, the jet clock loses $86400 \times 3.47 \times 10^{-13} \approx 30$ ns.
10. Because $\gamma \ge 1$, every other observer's time is $\gamma$ times the proper time, which is always $\ge$ the proper time.