Show that the two-dimensional Lorentz transformation relating $\left(c t^{\prime}, x^{\prime}\right)$ in frame $S^{\prime}$ to $(c t, x)$ in frame $S$, where $S^{\prime}$ moves relative to $S$ with speed $v$, can be written in the form

$$\begin{gathered} x^{\prime}=x \cosh \phi-c t \sinh \phi \\ c t^{\prime}=-x \sinh \phi+c t \cosh \phi \end{gathered}$$

where the hyperbolic angle $\phi$ associated with the transformation is given by $\tanh \phi=v / c$. Deduce that

$$\begin{aligned} &x^{\prime}+c t^{\prime}=e^{-\phi}(x+c t) \\ &x^{\prime}-c t^{\prime}=e^{\phi}(x-c t) \end{aligned}$$

Hence show that if the frame $S^{\prime \prime}$ moves with speed $v^{\prime}$ relative to $S^{\prime}$ and $\tanh \phi^{\prime}=v^{\prime} / c$, then the hyperbolic angle associated with the Lorentz transformation connecting $S^{\prime \prime}$ and $S$ is given by

$$\phi^{\prime \prime}=\phi^{\prime}+\phi$$

Hence find an expression for the speed of $S^{\prime \prime}$ as seen from $S$.