Following Mermin again, we’ll see how the invariance of the speed of light in all inertial frames leads directly to the relativity of simultaneity. Alice rides a train. In one of the carriages it is arranged to have two photons of light emitted from the center of the carriage, one traveling towards the front and the other towards the back. The events \(E_f\) and \(E_r\) are respectively the photon reaching the front and rear of the carriage. In Alice’s frame of reference these events occur simultaneously — we don’t even have to refer to clock’s in Alice’s frame since we know that light travels at the same speed in all directions.

Now consider the situation from the perspective of a track-side observer, Bob. From his perspective Alice’s train is traveling with a velocity \(v\). Three events take place, first, the photons are emitted from the center ¹ of the carriage. Since they still travel at light speed \(c\) in all directions he ‘sees’, that is, the clocks in his latticework record, the event \(E_r\) occurring before the event \(E_f\). Schematically we have,
This is of course as we’d expect, as the left traveling photon heads to the back of the carriage, the back of the carriage is traveling towards it with velocity \(v\) while as the right traveling photon heads towards the front of the carriage, the front is traveling away from it with speed \(v\). Lets say that in Bob’s frame the length of the train carriage is \(L\). If \(T_r\) is the elapsed time in Bob’s frame between the photons being emitted and the left traveling photon reaching the back of the carriage then we have
\begin{equation}
cT_r=\frac{1}{2}L-vT_r
\label{eq:reartime}
\end{equation}
and after a time \(T_f\) the right traveling photon covers \(cT_f\) given by
\begin{equation}
cT_f=\frac{1}{2}L+vT_f.
\label{eq:fronttime}
\end{equation}
These individual times are not what we’re interested in though. We’re interested in the time difference, let’s call it \(\Delta T\), between the events \(E_r\) and \(E_f\) as observed by Bob, for which we obtain,
\begin{equation*}
c\Delta T=v(T_r+T_f).
\end{equation*}
But the total distance traveled by the photons is \(D=cT_r+cT_f\), the spatial separation Bob observes between the two events. So finally we obtain
\begin{equation*}
\Delta T=\frac{Dv}{c^2}.
\end{equation*}

Two events, \(E_r\) and \(E_f\), which are simultaneous in Alice’s inertial frame of reference, are not simultaneous in Bob’s frame, moving with velocity \(v\) in the direction pointing from \(E_f\) to \(E_r\) relative to Alice’s. In Bob’s frame, the event \(E_r\) occurs a time \(Dv/c^2\) before the event \(E_f\), where \(D\) is the spatial separation of the events as seen by Bob.

Alice’s clocks, that is those synchronised in her frame, will record the events \(E_r\) and \(E_f\) occurring at the same time. Moreover they will also record the fact that Bob’s clocks show the event \(E_f\) occurring a time \(Dv/c^2\) after \(E_r\). Alice’s explanation for this fact will be that Bob’s clocks aren’t properly synchronised. Bob on the other hand says the events aren’t simultaneous and says that Alice’s clocks cannot, therefore, be properly synchronised.

The rule about simultaneous events in one frame not being simultaneous in another can be stated in terms of clocks thus:

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 \(Dv/c^2\).

It will be useful in the next post, where we consider some consequences of the relativity of simultaneity, to have a relation between the two lengths \(L\) and \(D\) in Bob’s frame. From \eqref{eq:reartime},
\begin{equation*}
T_r=\frac{L}{2}\frac{1}{c+v},
\end{equation*}
and from \eqref{eq:fronttime},
\begin{equation*}
T_f=\frac{L}{2}\frac{1}{c-v},
\end{equation*}
so that
\begin{align*}
\Delta T&=\frac{L}{2}\frac{2v}{c^2-v^2}\\
&=\frac{L\gamma^2v}{c^2}
\end{align*}
where we have introduced the Lorentz factor, \(\gamma\), defined as
\begin{equation*}
\gamma=\frac{1}{\sqrt{1-(v/c)^2}}.
\end{equation*}
Notice that for \(v{<}c\), \(\gamma{>}1\), and, combining with our previous expression for \(\Delta T\), we conclude that \(L\) and \(D\) are related according to
\begin{equation}
D=\gamma^2L.
\end{equation}