In the previous post we learnt that if in one frame two clocks are synchronised a distance \(D\) apart, then in another frame in which these clocks are moving along the line joining them with speed \(v\), the clock in front lags the clock behind by a time of \(Dv/c^2\). Let’s now think more about the contrasting perspectives of Alice, riding a train, and Bob, track side, thinking in particular about their respective clock and length readings.

The following picture sums up Alice’s perspective:

Here and below, clocks in either Alice’s or Bob’s frame are denoted by the rounded rectangles with times displayed within. Lengths and times in Alice’s train frame will be denoted by primed symbols, those in Bob’s track frame, unprimed. Above we see that Alice records the two events to occur simultaneously at a time we’ve taken to be 0. We’re free to take the time on Bob’s clock at the rear of the carriage to read 0 at this time in Alice’s frame, but then, since from Alice’s perspective Bob’s clocks are approaching her with speed \(v\), we know that Bob’s clock at the front of the carriage, when Alice’s clock there is showing 0, must be already showing a later time which we denote \(T\). This is unprimed as it’s a time displayed by a clock in Bob’s frame. From our discussion of the relativity of simultaneity we know \(T\) must be given by
\begin{equation}
T=\frac{Dv}{c^2}\label{eq:tracktime}
\end{equation}
where \(D\) is the separation of those two clocks in Bob’s frame as measured in Bob’s frame. Alice measures the length of her carriage to be \(L’\). We call the length of an object measured in its rest frame its proper length so both \(L’\) and \(D\) are proper lengths, whereas \(D’\), the distance between Bob’s clocks as measured by Alice is not proper. Note that, of course, \(L’=D’\).

Now let’s consider Bob’s perspective. We now need two pictures corresponding to two different times in Bob’s frame.

From Bob’s perspective, at time 0 the rear of the carriage is located at his rear clock and the carriage clock there also shows zero. At the later time \(T\), the front of the carriage is located at his front clock and the carriage clock there shows 0. Bob sees Alice’s clocks travel with speed \(v\) towards him so we know that the front clock lag’s the clock to the rear by a time given by
\begin{equation*}
\frac{L’v}{c^2}
\end{equation*}
where \(L’\) is the distance between Alice’s clocks, the length of the carriage, as measured in Alice’s frame. Thus, in the first of the two Bob frame snapshots, the rear carriage clock shows 0 whilst the front carriage clock shows \(-T’\), and in the second snapshot, the rear carriage clock shows \(T’\) while the front carriage clock shows 0 where
\begin{equation}
T’=\frac{L’v}{c^2}.\label{eq:traintime}
\end{equation}

Now, we’re going to be interested in the ratio \(T’/T\), the fraction of track frame time recorded by train frame clocks – a ratio of a moving clock time to a stationary clock time. We see immediately from \eqref{eq:tracktime} and \eqref{eq:traintime} that this is the same as \(L’/D\). But recall that \(L’=D’\) so we have
\begin{equation*}
\frac{T’}{T}=\frac{D’}{D}.
\end{equation*}
\(D’/D\) is the ratio of a measurement of a length moving with speed \(v\), \(D’\), to a measurement of a length at rest, \(D\). This ratio must therefore also be equal to \(L/L’\), the length of the carriage as viewed from Bob’s perspective to the (rest-frame) length of the carriage as measured by Alice. So in fact we have
\begin{equation}
\frac{T’}{T}=\frac{D’}{D}=\frac{L}{L’}.
\end{equation}
Now recall that \(D=\gamma^2L\), where \(\gamma=1/\sqrt{1-(v/c)^2}\), from which it follows that
\begin{equation}
{L’}^2=\gamma^2L^2
\end{equation}
or,
\begin{equation}
L=\frac{1}{\gamma}L’.
\end{equation}
This is length contraction! Recall that \(\gamma{>}1\) so that the length of the carriage as measured by Bob is smaller than the carriage’s proper length measured by Alice. It follows also that
\begin{equation}
T’=\frac{1}{\gamma}T.
\end{equation}
This is time dilation! Whilst stationary clocks record a time \(T\), clocks in motion record a shorter time \(T’\) — moving clocks run slow.