A further discussion is necessary before it is permissible to conclude that \Eq{(4.6)}~is the most general possible form for~$ds^{2}$ in terms of ordinary space and time coordinates. If we had reduced~\Eq{(2.1)} to the rather more general form \[ ds^{2} = dx^{2} + dy^{2} + dz^{2} - c^{2}\, dt^{2} - 2c\alpha\, dx\, dt - 2c\beta\, dy\, dt - 2c\gamma\, dz\, dt, \Tag{(4.7)} \] this would have agreed with~\Eq{(4.6)} in the only two cases yet discussed, viz.\ (1)~when $dt = 0$, and (2)~when $dx$,~$dy$, $dz = 0$. To show that this more general form is inadmissible we must examine pairs of events which differ both in time and place. In the preceding pre-relativity definition of~$t$ our clocks had to remain stationary and were therefore of no use for comparing time at different places. What did the pre-relativity physicist mean by the difference of time~$dt$ between two events at different places? I do not think that we can attach any meaning to his hazy conception of what $dt$~signified; but we know one \PageSep{15} or two ways in which he was accustomed to determine it. One method which he used was that of transport of chronometers. Let us examine then what happens when we move a clock from $(x_{1}, 0, 0)$ at the time~$t_{1}$ to another place \index{Clocks, transport of}% \index{Time!convention in reckoning}% $(x_{2}, 0, 0)$ at the time~$t_{2}$. We have seen that the clock, whether at rest or in motion, provided it remains a precisely similar mechanism, records equal \emph{intervals}; hence the difference of the clock-readings at the beginning and end of the journey will be proportional to the integrated interval \[ \int_{1}^{2} ds. \Tag{(4.81)} \] If the transport is made in the direct line ($dy = 0$, $dz = 0$), we shall have according to~\Eq{(4.7)} \begin{align*} -ds^{2} &= c^{2}\, dt^{2} + 2c\alpha\, dx\, dt - dx^{2}\displaybreak[0] \\ &= c^{2}\, dt^{2}\left\{1 + \frac{2\alpha}{c}\, \frac{dx}{dt} - \frac{1}{c^{2}} \left(\frac{dx}{dt}\right)^{2}\right\}. \end{align*} Hence the difference of the clock-readings \Eq{(4.81)} is proportional to \[ \int_{t_{1}}^{t_{2}} dt \left(1 + \frac{2\alpha u}{c} - \frac{u^{2}}{c^{2}}\right)^{\frac{1}{2}}, \Tag{(4.82)} \] where $u = dx/dt$, i.e\Add{.}\ the velocity of the clock. The integral will not in general reduce to $t_{2} - t_{1}$; so that the difference of time at the two places is not given correctly by the reading of the clock. Even when $\alpha = 0$, the moving clock does not record correct time.