My name is Alexander FufaeV and here I will explain the following topic:

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 $\text{B}$:

$$ \begin{align} \Delta t ~=~ \frac{L}{c} \end{align} $$

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 $.

Je nach dem, ob sich die Lichtuhr bewegt oder nicht, legt das Photon die Strecke $L$ bzw. eine größere Strecke $L'$ innerhalb einer anderen Zeit zurück.

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:

$$ \begin{align} L' ~=~ c \, \Delta t' \end{align} $$

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

$$ \begin{align} \cancel{c} \, \Delta t' ~&\gt~ \cancel{c} \, \Delta t \\\\
\Delta t' ~&\gt~ \Delta t \end{align} $$

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 $.

Die zurückgelegten Strecken $L'$, $L$ und $v\,\Delta t'$ bilden ein rechtwinkliges Dreieck.

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 $:

$$ \begin{align} (c \, \Delta t')^2 ~=~ (c \, \Delta t)^2 ~+~ (v \, \Delta t')^2 \end{align} $$

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 $c^2$:

$$ \begin{align} \Delta t'^2 ~=~ \Delta t^2 ~+~ \frac{v^2}{c^2} \, \Delta t'^2 \end{align} $$

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

$$ \begin{align} \Delta t'^2 ~-~ \frac{v^2}{c^2} \, \Delta t'^2 ~=~ \Delta t^2 \end{align} $$

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

$$ \begin{align} \Delta t'^2 \, \left( 1 ~-~ \frac{v^2}{c^2} \right) ~=~ \Delta t^2 \end{align} $$

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

$$ \begin{align} \Delta t' ~=~ \frac{1}{ \sqrt{1 ~-~ \frac{\class{blue}{v}^2}{c^2}} } \, \Delta t \end{align} $$

Der Lorentzfaktor (Gamma-Faktor) wird erst bei hohen Relativgeschwindigkeit (nahe der Lichtgeschwindigkeit) sehr groß und damit wird die Zeitverlangsamung umso größer.

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

$$ \begin{align} \gamma ~=~ \frac{1}{\sqrt{1 ~-~ \frac{\class{blue}{v}^2}{c^2}}} \end{align} $$

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.