KAHIBARO
Table of Contents
Why time does not pass the same for everyone
In everyday life, we usually assume that time is universal. We imagine that one second for you is the same as one second for anyone else. Special relativity shows that this is not true when objects move at speeds close to the speed of light.
Time dilation means that a moving clock is observed to run more slowly than a clock at rest with respect to the observer. This effect is not caused by a bad clock or a mechanical problem. It is a real property of spacetime.
If one observer sees another observer moving very fast, the moving observer's time appears stretched. In other words, less time passes on the moving clock between two events than on the clock that remains at rest in the observer's frame.
Proper time and dilated time
The central idea in time dilation is the difference between proper time and the longer time measured in another frame.
Proper time, written as $\Delta \tau$, is the time measured by a clock that is present at both events. It is the shortest time interval between those two events.
If another observer sees that clock moving with speed $v$, that observer measures a larger time interval $\Delta t$. The relationship is
$$
\Delta t = \gamma \Delta \tau
$$
where
$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
and $c$ is the speed of light.
Since $\gamma \ge 1$, the measured time interval $\Delta t$ is always greater than or equal to the proper time $\Delta \tau$.
Important rule:
$$
\Delta t = \gamma \Delta \tau, \qquad \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
$\Delta \tau$ is the proper time, measured in the frame where the clock is at rest.
$\Delta t$ is the longer time interval measured in a frame where the clock is moving.
The light clock picture
A simple way to understand time dilation is to imagine a light clock. A pulse of light bounces between two mirrors. Each round trip marks one tick.
If the clock is at rest relative to you, the light moves straight up and down. If the clock moves sideways relative to you, the light follows a diagonal path. Because light always moves at speed $c$, a longer path means more time per tick. So the moving clock appears to tick more slowly.
Deriving the formula
Suppose the mirrors are separated by a vertical distance $L$. In the frame of the clock, one half tick takes time
$$
\Delta \tau/2 = \frac{L}{c}
$$
Now consider an observer who sees the clock move sideways at speed $v$. During half a tick, the light travels diagonally. The diagonal path forms a right triangle.
The vertical side is $L$, the horizontal side is $v \Delta t /2$, and the hypotenuse is $c \Delta t /2$.
Using the Pythagorean theorem,
$$
\left(c \frac{\Delta t}{2}\right)^2 = L^2 + \left(v \frac{\Delta t}{2}\right)^2
$$
Substitute $L = c \Delta \tau /2$:
$$
\left(c \frac{\Delta t}{2}\right)^2 = \left(c \frac{\Delta \tau}{2}\right)^2 + \left(v \frac{\Delta t}{2}\right)^2
$$
Multiply through by $4$:
$$
c^2 \Delta t^2 = c^2 \Delta \tau^2 + v^2 \Delta t^2
$$
Rearrange:
$$
(c^2 - v^2)\Delta t^2 = c^2 \Delta \tau^2
$$
So
$$
\Delta t^2 = \frac{c^2}{c^2 - v^2}\Delta \tau^2
= \frac{1}{1 - \frac{v^2}{c^2}}\Delta \tau^2
$$
Taking the square root gives
$$
\Delta t = \frac{\Delta \tau}{\sqrt{1 - \frac{v^2}{c^2}}}
= \gamma \Delta \tau
$$
This is the time dilation formula.
How large is the effect
At ordinary speeds, $v$ is much smaller than $c$, so $\gamma$ is extremely close to $1$. That is why time dilation is not noticeable in daily life.
At high speeds, the effect becomes important.
| Speed | $v/c$ | $\gamma$ | Meaning |
|---|---|---|---|
| $0.1c$ | 0.1 | about 1.005 | very small effect |
| $0.5c$ | 0.5 | about 1.155 | noticeable |
| $0.8c$ | 0.8 | about 1.667 | strong effect |
| $0.9c$ | 0.9 | about 2.294 | moving clock much slower |
| $0.99c$ | 0.99 | about 7.089 | very strong effect |
For example, if a spaceship moves at $0.8c$ and 1 hour passes on a clock inside the ship, an observer who sees the ship moving measures
$$
\Delta t = \gamma \Delta \tau = 1.667 \times 1 \text{ hour} = 1.667 \text{ hours}
$$
So the moving clock advances by only 1 hour while the stationary observer says 1.667 hours have passed.
A worked example
Suppose unstable particles are created in a laboratory and move at speed $0.98c$. In their own rest frame, their lifetime is $2.2 \times 10^{-6}\,\text{s}$. How long do they live in the laboratory frame?
First calculate $\gamma$:
$$
\gamma = \frac{1}{\sqrt{1 - (0.98)^2}}
= \frac{1}{\sqrt{1 - 0.9604}}
= \frac{1}{\sqrt{0.0396}}
\approx 5.03
$$
Now use time dilation:
$$
\Delta t = \gamma \Delta \tau = 5.03 \times 2.2 \times 10^{-6}\,\text{s}
$$
$$
\Delta t \approx 1.11 \times 10^{-5}\,\text{s}
$$
So in the laboratory frame, the particles live about five times longer.
Physical meaning
Time dilation does not mean that a person feels their own time slowing down. In each person's own frame, their own watch ticks normally, their heartbeat feels normal, and their processes proceed as usual.
The difference appears when comparing clocks in relative motion. Each inertial observer finds that clocks moving relative to them run slow.
Key interpretation:
A clock always measures normal time in its own rest frame.
Time dilation appears when one frame observes a clock moving relative to it.
Experimental evidence
Time dilation has been confirmed many times. Fast moving unstable particles survive longer than they would if time were absolute. Atomic clocks carried on aircraft and satellites show measurable differences compared with clocks on Earth. Particle accelerators also provide clear evidence, because particles moving near the speed of light live longer in the laboratory frame.
These results are not small corrections to old physics. They are direct confirmations that time depends on relative motion.
Limits and special cases
If $v = 0$, then
$$
\gamma = 1
$$
so
$$
\Delta t = \Delta \tau
$$
There is no time dilation when there is no relative motion.
As $v$ approaches $c$, the denominator in $\gamma$ becomes very small, so $\gamma$ becomes very large. This means time dilation becomes enormous at speeds very close to the speed of light.
If $v = c$, the formula breaks down because the denominator becomes zero. Objects with mass cannot reach the speed of light in special relativity.
Important limits:
If $v = 0$, then $\gamma = 1$ and there is no time dilation.
If $v$ gets close to $c$, then $\gamma$ becomes very large.
For material objects, $v < c$.
Common misunderstanding
A common mistake is to think that time dilation means only an illusion caused by signal delay. It is true that observers must account for light travel time when making measurements, but even after correcting for that, time dilation remains. It is a real difference in measured time intervals between frames.
Another common mistake is to use the formula backward. The proper time is always the time measured in the frame where the clock is at rest. The dilated time is the larger interval measured in a frame where the clock is moving.
Summary
Time dilation is one of the central results of special relativity. A moving clock is observed to tick more slowly than a clock at rest relative to the observer. The relation is
$$
\Delta t = \gamma \Delta \tau, \qquad \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
The proper time $\Delta \tau$ is measured in the frame where the clock is at rest. The time $\Delta t$ measured in another frame is larger. The effect is tiny at low speeds and dramatic near the speed of light. Experimental evidence strongly confirms this behavior of time.