Derivation: Time Dilation By Using a Light Clock

The time dilation from special relativity describes how time ticks in different reference frames. The time dilation can be derived simply with the help of a light clock. A light clock basically consists of two mirrors at a fixed distance \( L \) from each other. A photon (light particle) is reflected back and forth between the two mirrors. This photon always moves with constant speed of light \( c \).

To get the time dilation, you have to look at the light clock from two different inertial frames (unaccelerated reference frames). The one inertial frame represents the rest observer \(\text{B}\), which is motionless relative to the light clock. The other inertial frame is the \(\text{B}'\), which moves relative to the light clock with the \(-v\) to the left (in the negative x-direction). From the point of view of \(\text{B}'\), the clock moves to the right (in the positive x-direction) with velocity \(v\). Let us now describe the motion of the photon by putting ourselves first in the observer \(\text{B}\) and then in the observer \(\text{B}'\).

From the view of \(\text{B}\) you see the light clock at rest. There the photon oscillates vertically upwards and then downwards (let's say: along the y-axis). From one to the other mirror the photon needs the following time from the point of view of the rest observer \(\text{B}\):

For a period \(T\) (i.e. once back and forth) it needs twice the time: \( T ~=~ 2\Delta t \).

Now you switch to a moving inertial system \(\text{B}'\), which moves to the left with the velocity \(-v\). Now you observe something completely different! While the photon flies from one mirror to the other, the light clock moves in the positive x-direction (to the right) with the velocity \( v \).

From the point of view of \(\text{B}'\), the photon does not fly straight up (as in the case of resting light clock), but makes a zigzag motion (see illustration 1). The distance \( L' \), which the photon travels from one mirror to the other, is from this point of view, LONGER.

The velocity of the photon is also equal to \(c\) from the point of view of \(\text{B}'\). This is one of the postulates on which the special theory of relativity is based: The speed of light is the same in all inertial frames. However, the photon in the reference frame \(\text{B}'\) must travel a longer distance \(L'\) at the speed of light:

Since \( L' \) is greater than \( L \), \( \Delta t' \) must be greater than \( \Delta t \), because \(c\) is constant in both reference frames:

The photon arrives at the opposite mirror from the point of view of \(\text{B}'\) after the time \( \Delta t' \), while from the point of view of \(\text{B}\) it arrives after the shorter time \( \Delta t \).

The question now is: How are \(\Delta t \) and \(\Delta t'\) related? If you combine the observation where the light clock is resting (\( c\, \Delta t \)) and the observation where the light clock is moving \( c \, \Delta t' \) and \( v \, \Delta t' \), you get a right triangle to which you can apply Pythagoras' theorem \( c^2 = a^2 + b^2 \):

The equation 4 can then be rearranged for example for \( \Delta t' \), i.e. for the time the photon needs to reach the other mirror in the reference frame \(B'\). For this, divide the equation by \(c^2\):

Bring the second summand containing \(\Delta t'\) to the left side:

Now you can factor out \( \Delta t'^2 \):

Bring the factor before the \(\Delta t'^2\) to the right side and take the square root on both sides:

The times \(\Delta t\) and \(\Delta t' \) in two different reference frames are linked by the so-called Lorentz factor (lorentz term) \( \gamma \):

The Lorentz factor \(\gamma\) is always greater than 1 and becomes greater the faster the light clock moves relative to the observer (i.e. with increasing \( v \)). As a result, the time \( \Delta t' \) becomes even greater. The time dilation has a larger effect.

With the help of time dilation the second effect of special relativity can be derived, namely the length contraction.