Time Dilation - The math - Eric Sciberras

This Article continues from the previous post on Time dilation. Hopefully now we are on the same page to recap from the previous article:

Our current interpretation of time has issues (specifically we can’t find evidence of the Ether that physicists proposed)
The two postulates below are valid:
- The laws of physics are the same in all inertial frames of reference.
- The speed of light in free space has the same value c in all inertial frames of references

Our derivation involves the use of a light clock which is simply a beam of light bouncing between two ‘perfect’ mirrors. We will define a ‘tick’ of the clock as the light bouncing up and then down again.

Source: https://commons.wikimedia.org/wiki/File:Time-dilation-002-mod.svg

Now imagine two light clocks in these two frames:

A stationary frame (your frame, left image) and;
A moving frame (a train moving right past you, right image)

In the stationary frame we observe the light going straight up and down. However in the moving frame as the light bounces between the mirrors the mirrors move some distance so the path taken by the light is at an angle (like a triangle). If you can understand the paths traveled by the light in these two frames then the hard part is over. The rest is simply trigonometry and algebra.

(1) $\begin{equation*} s = \frac{d}{\Delta t} \end{equation*}$

speed equals distance over time.

The proof consists of:

Establishing a relationship between the distance between mirrors and the time it takes for a single ‘tick’ in each frame
Combining the relations
Simplifying and rearranging terms

Stationary Frame

Rearrange equation 1 for distance.

(2) $\begin{equation*} d = s \Delta t \end{equation*}$

Now subbing in values $d = L$ (distance between A and B) and $s = c$ (speed of light) in the stationary frame the light travels a distance of

(3) $\begin{equation*} L = \frac{1}{2} c \Delta t \end{equation*}$

per every ‘tick’

Moving Frame

In the Moving frame we can see that the lights path is at an angle so establishing a relationship with the time elapsed and distance between the mirrors involves using Pythagoras’ theorem

(4) $\begin{equation*} A^2 +B^2 =C^2 \end{equation*}$

From the green triangle in the picture.

(5) $\begin{equation*} A= \frac{1}{2} v \Delta t' \end{equation*}$

This is the distance the mirror moves in half a tick where:

$v$ velocity of the moving frame and;
$\Delta t'$ is the time it takes to complete a tick in the moving frame

Note: $\Delta t$ and $\Delta t'$ are used to distinguish how much time has passed in stationary frame and moving frames respectively.

$B$ is the distance between A and B.

(6) $\begin{equation*} B = L \end{equation*}$

We also know what $C$ is, it is the distance the light moves in half a tick which is:

(7) $\begin{equation*} C = \frac{1}{2} c \Delta t' \end{equation*}$

Where c (lowercase) is the speed of light.

putting everything together:

(8) $\begin{equation*} L^2 + \left( \frac{1}{2} v \Delta t' \right) ^2 = \left( \frac{1}{2} c \Delta t' \right)^2 \end{equation*}$

(9) $\begin{equation*} L^2 + \frac{1}{4} v^2 \Delta t'^2 = \frac{1}{4} c^2 \Delta t'^2 \end{equation*}$

At this stage we will transpose the equation so it is in terms of $\Delta t'$ (this is purely manipulating the equation above)

Rearrange for $L^2$

(10) $\begin{equation*} L^2 = \frac{1}{4} c^2 \Delta t'^2 -\frac{1}{4} v^2 \Delta t'2 \end{equation*}$

Factor out $\frac{1}{4} \Delta t'^2$

(11) $\begin{equation*} L^2 = \frac{1}{4} \Delta t'^2 \left( c^2 - v^2 \right) \end{equation*}$

Rearrange for $t'^2$

(12) $\begin{equation*} \Delta t'^2 = \frac{4L^2}{ c^2 -v^2 } \end{equation*}$

(13) $\begin{equation*} \Delta t' = \sqrt{\frac{4L^2}{ c^2 -v^2 }} \end{equation*}$

(14) $\begin{equation*} \Delta t' = \sqrt{ \frac{4L^2}{ c^2 \left( 1 - \frac{v^2}{c^2} \right) } } \end{equation*}$

we can substitute equation (3), $L = \frac{1}{2} c \Delta t$ (our stationary frame equation) in to get

(15) $\begin{equation*} \Delta t' = \sqrt{ \frac{4(\frac{1}{2} c \Delta t)^2}{ c^2 \left( 1 - \frac{v^2}{c^2} \right) } } \end{equation*}$

(16) $\begin{equation*} \Delta t' = \sqrt{ \frac{ 4 \frac{1}{4} c^2 \Delta t^2}{ c^2 \left( 1 - \frac{v^2}{c^2} \right) } } \end{equation*}$

(17) $\begin{equation*} \Delta t' = \sqrt{ \frac{ \Delta t^2}{ \left( 1 - \frac{v^2}{c^2} \right) } } \end{equation*}$

(18) $\begin{equation*} \Delta t' = \frac{ \Delta t}{ \sqrt{ 1 - \frac{v^2}{c^2} }} \end{equation*}$

Typically physicists define

(19) $\begin{equation*} \gamma= \frac{1}{\sqrt{ 1 - \frac{v^2}{c^2}}} \end{equation*}$

this gives us

(20) $\begin{equation*} \Delta t' = \Delta t \gamma \end{equation*}$

So what does this equation mean? well remember earlier when i said $\Delta t$ and $\Delta t'$ are different, well I assumed that. If I was wrong then the math would simplify to

(21) $\begin{equation*} \Delta t' = \Delta t \end{equation*}$

which would mean time intervals are the same in all frames. This is the old Newtonian representation of time (moving like a arrow) that we discussed earlier.

The new Einstein representation of time is very counter-intuitive, it implies that time behaves differently; instead of being fixed it depends on relative velocities between observers.

Time Dilation – The math

Stationary Frame

Moving Frame

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