 Solution

With Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$, the relativistic momentum and four-momentum areThe Newtonian formula $a=F/m$ fails because $p$ is not $mv$: both the magnitude and direction of $v$ affect $dp/dt$. Sincethe relativistic force isTaking the dot product with $v$ gives $F\cdot v=m{\gamma }^{3}v\cdot a$, and substitution yields the inverse relationThis is the requested sum of a vector parallel to $F$ and one parallel to $v$.

In the constant electric field, $dp/dt=qE$ and the particle starts with $p(0)=0$, so $p=qEt$. The relativistic energy-momentum relation impliesThus the speed increases monotonically but remains subluminal, and $v$ tends to $c$ in the direction of $qE$ as $t\to \infty$.