Time dilation - properHoc

Up a level : The Theory of Relativity
Previous page : The Special Theory of Relativity - an introduction
Next page : Time dilation - Interactive

Imagine you are on a spacecraft in a room with height h. Some light source sends a pulse across the room from the floor to the ceiling, where a mirror is placed, as in the figure. The pulse is then reflected to the source. The time for the round-trip is 2t₀, and the distance travelled is 2h.

The speed of the pulse is now

$\displaystyle c=\frac{h}{{{{t}_{0}}}}$

The time is a proper proper since it is measured in the same frame of reference as the light source and the mirror.

Now, let us assume that the spacecraft moves at some speed v to the right, relative to some observer outside the spacecraft. The height h must still be the same as before. We could imagine the spacecraft passing through a hole just wide enough for it. Then it must be just wide enough for all observers – say it leaves some scrape marks on the edge of the hole, and this must happen for all observers.

We get that for the external observer, the light must move in some zig-zag path like in the figure below.

The distance between A and B must be vt, because during the time the light moves from A to M the spacecraft has moved forward the distance vt. Observe that the time t is the time according to the external observer. It might thus not be the same as the proper time of the spaceship – and that is what we will show.

That gives us that the line from A to M will have the length

$\displaystyle L=\sqrt{{{{h}^{2}}+{{{\left( {vt} \right)}}^{2}}}}$

and that

$\displaystyle c=\frac{L}{t}=\frac{{\sqrt{{{{h}^{2}}+{{{\left( {vt} \right)}}^{2}}}}}}{t}$

Here, L is the length according to the outside observer, and t is the time according to the outside observer. Multiplied by t and squared, we get

$\displaystyle {{c}^{2}}{{t}^{2}}={{h}^{2}}+{{v}^{2}}{{t}^{2}}$

We subtract v²t² and factorise to get

$\displaystyle {{t}^{2}}\left( {{{c}^{2}}-{{v}^{2}}} \right)={{h}^{2}}$

From c=h/t₀ , we get h=ct₀ . Substituting this into the above, we get

$\displaystyle {{t}^{2}}\left( {{{c}^{2}}-{{v}^{2}}} \right)={{c}^{2}}t_{0}^{2}$

Dividing through by c² we get

$\displaystyle t_{0}^{2}=\frac{{{{t}^{2}}({{c}^{2}}-{{v}^{2}})}}{{{{c}^{2}}}}={{t}^{2}}\left( {1-{{{\left( {\frac{v}{c}} \right)}}^{2}}} \right)$

The square root of that gives us

$\displaystyle {{t}_{0}}=t\sqrt{{1-{{{\left( {\frac{v}{c}} \right)}}^{2}}}}$

$\displaystyle t=\frac{{{{t}_{0}}}}{{\sqrt{{1-{{{\left( {\frac{v}{c}} \right)}}^{2}}}}}}$

Now, let us set

$\displaystyle \gamma =\frac{1}{{\sqrt{{1-{{{\left( {\frac{v}{c}} \right)}}^{2}}}}}}$

This is called the gamma factor. This will finally get us

$\displaystyle t=\gamma {{t}_{0}}$

as our expression for the time dilation. So, t₀ is how long time it has passed according to the spacecraft, and t is according to the observer who sees itself as stationary. In the figure below, we can see a graph of the gamma factor. As you can see, it says close to one for quite high speeds, and then it increases very rapidly as the speed approaches c.

In these exercises, you may use c=3.00·10⁸ m/s.

Say a sci-fi spaceship flies past the Earth at 0.3c.
1. When one year has passed in the spaceship, how long time has passed on Earth, according to the people on Earth?
2. The people on the spaceship see themself as stationary. How long time has passed on the spaceship when the people on Earth says a year has passed?
3. When one year has passed in the spaceship. How long time has passed on the Earth according to the people in the spaceship?
Say a muon (a type of elementary particle) is created in the upper atmosphere. It has a half-life of 2.2·10^–6 s.
1. It moves close to the speed of light. How far would you expect it to move if you were solving it using classical physics?
2. Say the speed of the muon is 0.9992c. Calculate the gamma factor.
3. How long would the half-life of the muon be as seen from the ground?
4. How far would you expect the muon to travel according to the theory of relativity?
A spaceship moves at 10000 km/h relative to a space station. When one year has passed in the spaceship, how much has passed on the space station (seen as stationary)
Say a car travels at 100 km/h. When one year has passed in the car, how much has passed on the ground (seen as stationary)

Link to answers.

But how about the same situation, as seen from the frame of reference of the muon? It exists for about 2.2·10^–6 s, so how can it have time to reach the ground? For this, we need another concept.

Up a level : The Theory of Relativity
Previous page : The Special Theory of Relativity - an introduction
Next page : Time dilation - Interactive

Last modified: Jun 10, 2025 @ 13:23