Write down the Lorentz transform relating the components of a 4-vector between two inertial frames.

A particle moves along the $x$-axis of an inertial frame. Its position at time $t$ is $x(t)$, its velocity is $u=d x / d t$, and its 4 -position is $X=(c t, x)$, where $c$ is the speed of light. The particle's 4-velocity is given by $U=d X / d \tau$ and its 4 -acceleration is $A=d U / d \tau$, where proper time $\tau$ is defined by $c^{2} d \tau^{2}=c^{2} d t^{2}-d x^{2}$. Show that

$$U=\gamma(c, u) \quad \text { and } \quad A=\gamma^{4} \dot{u}(u / c, 1)$$

where $\gamma=\left(1-u^{2} / c^{2}\right)^{-\frac{1}{2}}$ and $\dot{u}=d u / d t$.

The proper 3-acceleration a of the particle is defined to be the spatial component of its 4-acceleration measured in the particle's instantaneous rest frame. By transforming $A$ to the rest frame, or otherwise, show that

$$a=\gamma^{3} \dot{u}=\frac{d}{d t}(\gamma u)$$

Given that the particle moves with constant proper 3 -acceleration starting from rest at the origin, show that

$$x(t)=\frac{c^{2}}{a}\left(\sqrt{1+\frac{a^{2} t^{2}}{c^{2}}}-1\right)$$

and that, if $a t \ll c$, then $x \approx \frac{1}{2} a t^{2}$.