\begin{equation} \Delta t_0 = {2 L \over c} . \qquad (1) \end{equation}

Figure 1. (a) In system $S'$ a light pulse is emmitted from a source at $O'$ and is reflected back along the same line, and takes a time $\Delta t_0$ to perform a round trip. (b) Path of the same light pulse, as observed in the system $S$. The speed of the light pulse is the same as in system $S'$, but the path is longer, and hence the moving clock takes a longer amount of time, $\Delta t$, to perform one tick.

The round-trip time measured by the observer seeing the system in motion at speed $v$ is a different interval $\Delta t$. It will be greater than $\Delta t_0$, because the light pulse traces out a longer path, a total round trip distance of $2 D$, and thus, with the same speed of light $c$, measures a longer time: \begin{equation} \Delta t = {2 D \over c} > {2 L \over c} = \Delta t_0 , \end{equation} By Pythagoras' theorem $D$ is given by \begin{equation} D = \sqrt{ L^2 + \left( {v \Delta t \over 2} \right)^2 } . \end{equation} and therefore we have \begin{equation} \Delta t = {2D \over c} = {2 \over c} \sqrt{ L^2 + \left( {v \Delta t \over 2} \right)^2 } . \end{equation} We wish to eliminate $L$ from the equaion, using Eq. (1) we obtain, \begin{equation} \Delta t = {2 \over c} \sqrt{ \left( {c \Delta t_0 \over 2 } \right)^2 + \left( {v \Delta t \over 2} \right)^2 } \end{equation} This becomes \begin{equation} (\Delta t)^2 = \left( \Delta t_0 \right)^2 + \left( {v \over c} \Delta t \right)^2 \end{equation} and solving for $\Delta t$ gives, \begin{equation} \Delta t = {\Delta t_0 \over \sqrt{1 - v^2 / c^2}} . \end{equation} The denominator is always less than unity, and so, as we have already noted, $\Delta t$ is always larger than $\Delta t_0$. Think of a clock in the rest frame, the reading on that clock for the round trip will be $\Delta t_0$. When the round trip is measured by a clock in a frame moving with respect to this first clock, the time interval $\Delta t$ that is recorded will be longer, and this observer thinks the first clock is running more slowly.

3. Length Contraction

The time taken for the light pulse to make the round trip from the source to the mirror and back is

\begin{equation} \Delta t_0 = {2l_0 \over c} \qquad (2) \end{equation}

In the frame in which the ruler is moving, one finds for the time $\Delta t_2$ for the trip from the source to the mirror is $$ c \Delta t_1 = l + v \Delta t_1 , $$

so that

$$ \Delta t_1 = {l \over c-v} . $$ Similarly, one finds for the time $\Delta t_2$ for the return trip from the mirror to the source is $$ \Delta t_2 = {l \over c+v} $$ The total time $\Delta t = \Delta t_1 + \Delta t_2$ for the round trip, as measured by $O$, is \begin{equation} \Delta t = {l \over c-v} + {l \over c+v} = {2l \over c(1-v^2 / c^2)} \end{equation} From the relation between $\Delta t$ and $\Delta t_0$, Eq. (2) becomes $$ \Delta t \sqrt{1 - v^2/c^2} = {2l_0 \over c} . $$ We obtain \begin{equation} l = l_0 \sqrt{1 - v^2 / c^2} . \end{equation}

4. Lorentz Transformations

The question is when an event occurs at a point $(x,y,z)$ at time $t$, as observed in in a frame of reference $S$, what are the coordinates $(x',y',z')$ and time $t'$ of the event as observed in a second frame $S'$ moving relative to $S$ with constant velocity in the $x-$direction. As before, we assume that the origins coincide at $t = t' = 0$. Then in $S$ the distance from $O$ to $O'$ is just $vt$. The distance from $O$ to $P$, as seen in $S$, is

\begin{equation} x = vt + x' \sqrt{1 - v^2 / c^2} . \qquad (3) \end{equation}

Solving for $x'$, we obtain

\begin{equation} x' = {x - vt \over \sqrt{1 - v^2 / c^2}} \qquad (4) \end{equation}

Now we note that the principal of relativity requires that the form of the transformation from $S$ to $S'$ be identical to that from $S'$ to $S$, the only difference is a change in the sign of the relative velocity $v$. Thus from Eq. (3) it must be that

\begin{equation} x' = -vt' + x \sqrt{1 - v^2 / c^2} . \qquad (5) \end{equation}

Equating Eq. (4) and Eq. (5) gives, after some rearangement, an equation between $t'$ and $t$ and $x$,

\begin{equation} t' = {t - vx/c^2 \over \sqrt{1 - v^2 / c^2}} . \qquad (6) \end{equation}

The lengths perpendicular to the direction of relative motion are unaffected, i.e. $y' = y$ and $z' = z$.

Collecting all the transformation equations, we have \begin{eqnarray} x' &=& {x - vt \over \sqrt{1 - v^2 / c^2}} , \nonumber \\ y' &=& y , \nonumber \\ z' &=& z , \nonumber \\ t' &=& {t - vx/c^2 \over \sqrt{1 - v^2 / c^2}} . \end{eqnarray}

4. a Time dilation again

Here we derive the time dilation effect from the Lorentz transformation formula:

\begin{equation} t' = {t - vx/c^2 \over \sqrt{1 - v^2 / c^2}} . \qquad (7) \end{equation}

Refer to Fig. 2. When the clock at rest with respect to the "stationary" system $S$ registers time $t$, the clock at rest with respect to the "moving" system $S'$ is at position $x=vt$. We substitute $x=vt$ into Eq. (7) and obtain \begin{eqnarray} t' &=& {t - v^2 t /c^2 \over \sqrt{1 - v^2 / c^2}} \nonumber \\ &=& t {1 - v^2 /c^2 \over \sqrt{1 - v^2 / c^2}} \nonumber \\ &=& t \sqrt{1 - v^2 / c^2} \end{eqnarray} so that \begin{equation} t' = t \sqrt{1 - v^2 / c^2} . \end{equation}

This says the reading on the clock at rest with respect to the "stationary" system $S$ runs ahead of the reading on the moving clock, and the observer in the "stationary" system concludes that the moving clock is running slow.

Figure 2. ay we have a clock at rest with respect to the system $S$ located at the origin of $S$. We also have a clock at rest with respect to the "moving" system $S'$ located at the origin of $S'$. We assume that the origins coincide at an initial time $t = t' = 0$. Recall this assumption was made when deriving the Lorentz transformation formula.

Draft version

Lorentz Transformation

Contents:

1. Lorentz Transformations

2. Time Dilation

3. Length Contraction

4. Lorentz Transformations

4. a Time dilation again