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In the parent's rest frame, momentum conservation makes the massive daughter and photon momenta equal and opposite, of magnitude $p$. The relativistic two-body decay formulas giveSince $v=p{c}^2/E_R$ and $r=m/M$,The daughter's Lorentz factor isIf it decays after proper time $\tau$, time dilation gives $t=\gamma \tau$, while uniform motion gives $x=v\gamma \tau$. Therefore the event coordinates in the parent frame are

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 $t\to \infty$.

For collinear motion without external forces, a rocket ejecting material backwards at constant speed $u$ in its instantaneous rest frame obeys $m{\gamma }^{2}dv+udm=0$. Here $m$ is the rocket's rest mass, including stored internal energy, and $\gamma$ is its Lorentz factor in the chosen inertial frame. The equation follows from momentum conservation, conservation of relativistic energy, and the relativistic velocity-addition formula; expelled rest mass is not simply the decrease in rocket rest mass.