Time Dilation

How can you calculate time dilation?

In order to calculate time dilation, first, you need to imagine which variables resemble which quantities. In order to do that, take a look at the example below.

Consider a train passenger with a clock that is timing a light pulse reflected between two horizontal mirrors in the carriage, one directly above the other at a distance L apart, as illustrated in Figure 1. A second spectator is observing the train as it travels along a track parallel to a platform.

A diagram showing the two views, (a) is the view from the train, and (b) is the view from the platform, Tezcan - StudySmarter Originals

Let's consider the paths followed by light as seen by each observer. The observer inside the train sees the light travel straight across and back with a total distance of 2L. The observer on the platform sees the light travel a total distance of 2s since the train is moving at a velocity v to the right relative to the platform.

We know that the light travels at a speed c = 3.00 ⋅ 10⁸ relative to any observer and the time is distance divided by speed. So the time measured by the observer inside the train is

\[\Delta t_0 = \frac{2L}{c}\]

And it is the proper time. The time observed by the observer on the platform will be

\[\Delta t = \frac{2s}{c}\]

Now let's give a variable name to \(\frac{vt}{2}\) as d. To find the relation between Δt0 and Δt you need to consider triangles formed by L and s, the third side of these similar triangles is d. If you apply the Pythagorean theorem, then s will be

\[s = \sqrt{L^2 + d^2}\]

We have found that Δt is equal to \(\frac{2s}{c}\). In order to place it into the equation, we need to multiply the right side of the equation by 2 and divide it by c :

\[\Delta t = \frac{2s}{c} = \frac{2\sqrt{L^2 + d^2}}{c}\]

Squaring both sides and writing d as \(\frac{vt}{2}\)

\[(\Delta t)^2 = \Big( \frac{2\sqrt{L^2 + d^2}}{c} \Big)^2= \frac{4(L^2+d^2)}{c^2} = \frac{4(L^2+(\frac{vt}{2})^2)}{c^2} =\frac{4L^2 + (vt)^2}{c^2} \]

Notice how \(\frac{4L^2}{c^2}\) equals (Δt₀)²

\[(\Delta t)^2 = (\Delta t_0)^2 + \frac{(vt)^2}{c^2}\]

If we solve it for Δt

\[(\Delta t)^2 \cdot \Big (1 - \frac{v^2}{c^2} \Big) = (\Delta t_0)^2\]

this then gives us

\[(\Delta t)^2 = \frac{(\Delta t_0)^2}{\Big( 1 - \frac{v^2}{c^2} \Big)}\]

Now if we take the square root on both sides it will yield a simple relation between elapsed times.

\[\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}= \gamma \Delta t_0\]

Where

\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]

What is an example of time dilation?

Time dilation may seem like it can't be seen in real life since it occurs at relative speeds but in today's technology humans interact with space more than ever before.

A great example of this is the global positioning system, or GPS. It is used in almost every area of life. GPS signals travel at relativistic speeds, which means that the corrections for time dilation need to be done. Otherwise, the GPS system would fail in minutes.

GPS is a real-life example of why time dilation is important.

Another great example of time dilation is the Frisch-Smith experiment. The Frisch-Smith experiment was conducted in 1963 by David H. Frisch and it consisted of measuring a quantity of muons in the atmosphere emitted by the Sun per unit time. The average speed of muons coming from the sun is really high and close to the speed of light.

The experiment measured the quantity of muons in two points with a difference of altitude of 1907 meters, which it should take 6.4 µs for muons to travel. However, the average lifetime of a muon is 2.2 μs, which means that only about 25% of muons would reach the endpoint. However, approximately 73% of muons reached the endpoint per hour because their proper time is smaller than the proper time of an observer on Earth. This confirmed that time was flowing more slowly for them with a time dilation factor of 8.80.8.

Results of the Frisch-Smith experiment, Tezcan - StudySmarter Originals