Hint 3

(a) The 4-velocity of a particle is related to its 3-velocity ${\bf v}$ by

$${\sf v}=\gamma(v)({\bf v},ic).$$

Hence the square of the magnitude of ${\sf v}$ is

$$\vert{\sf v}\vert^2={\sf v}\cdot{\sf v}=-c^2.$$

(b) Since ${\sf v}$ has constant magnitude, its rate of change satisfies

$${\sf v}\cdot{{\rm d}{\sf v}\over {\rm d} t}=0.$$

(Look back at PC117 for more details of the corresponding case of a 3-vector with constant magnitude.)