Let a ray travelling with velocity~$v$ traverse a distance~$R$ in a direction~$\theta$, so that \[ dt = R/v,\quad dx = R \cos\theta,\quad dy = R \sin \theta. \] \PageSep{20} Let the relative velocity of~$S$ and~$S'$ be small so that $u^{2}/c^{2}$~is neglected. Then by~\Eq{(5.3)} \[ dt' = dt + u\, dx/c^{2},\quad dx' = dx + u\, dt,\quad dy' = dy. \] Writing $\delta R$, $\delta\theta$, $\delta v$ for the change in $R$,~$\theta$,~$v$ when a transformation is made to $S'$'s~system, we obtain \begin{align*} \delta(R/v) &= dt' - dt = uR \cos \theta/c^{2}, \\ \delta(R \cos\theta) &= dx' - dx = uR/v, \\ \delta(R \sin\theta) &= dy' - dy = 0. \intertext{Whence the values of $\delta R$, $\delta\theta$, $\delta (1/v)$ are found as follows:} \delta R &= uR \cos \theta/v, \\ \delta\theta &= -u \sin \theta/v, \\ \delta\left(\frac{1}{v}\right) &= u \cos\theta \left(\frac{1}{c^{2}} - \frac{1}{v^{2}}\right). \end{align*} Here $\delta(1/v)$ refers to a comparison of velocities in the directions~$\theta$ in~$S$'s system and $\theta'$~in $S'$'s~system. Writing $\Delta(1/v)$ for a comparison when the direction is~$\theta$ in both systems \begin{align*} \Delta\left(\frac{1}{v}\right) &= \delta\left(\frac{1}{v}\right) - \frac{\dd}{\dd\theta}\left(\frac{1}{v}\right) · \delta\theta\Add{,} \\ &= \frac{u}{c^{2}} \cos\theta - \frac{u}{v^{2}} \cos\theta + \frac{u\sin\theta}{v}\, \frac{\dd}{\dd\theta}\left(\frac{1}{v}\right) \\ &= \frac{u}{c^{2}} \cos\theta + \tfrac{1}{2} u \sin^{3}\theta\, \frac{\dd}{\dd\theta} \left(\frac{1}{v^{2} \sin^{2}\theta}\right). \end{align*} Hence \[ \Delta\left(\frac{1}{v(\theta)} + \frac{1}{v(\theta + \pi)}\right) = \tfrac{1}{2} u\sin^{3}\theta\, \frac{\dd}{\dd\theta}\left\{\frac{1}{\sin^{2}\theta}\left(\frac{1}{v^{2}(\theta)} - \frac{1}{v^{2}(\theta + \pi)}\right)\right\}. \] By~\Eq{(6.3)} the left-hand side is independent of~$\theta$, and equal to the constant $C' - C$. We obtain on integration \begin{align*} \frac{1}{v^{2}(\theta)} - \frac{1}{v^{2}(\theta + \pi)} &= \frac{C' - C}{u} (\sin^{2}\theta · \log\tan \tfrac{1}{2}\theta - \cos\theta), \\ \intertext{or} \frac{1}{v(\theta)} - \frac{1}{v(\theta + \pi)} &= \frac{C' - C}{C} · \frac{1}{u} (\sin^{2}\theta · \log\tan \tfrac{1}{2}\theta - \cos\theta). \end{align*} It is clearly impossible that the difference of~$1/v$ in opposite directions should be a function of~$\theta$ of this form; because the origin of~$\theta$ is merely the direction of relative motion of $S$ and~$S'$, which may be changed at will in different experiments, and has nothing to do with the propagation of light relative to~$S$. Hence $C' - C = 0$, and $v(\theta) = v(\theta + \pi)$. Accordingly by~\Eq{(6.3)} $v(\theta)$~is independent of~$\theta$; and similarly $v'(\theta)$~is independent of~$\theta$. Thus the velocity of light is uniform in all directions for both observers and is therefore to be identified with the fundamental velocity.