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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 $t\to \infty$.

There is no nonzero, literally constant four-force that is a pure force throughout this motion. Initially $U=(c,0,0,0)$, so purity forces $F^0=0$. If the full four-vector were constant, $U(\tau )=U(0)+F\tau /M_0$; its square would be $c^2+F^2{\tau }^2/M_0^2$, which cannot remain $c^2$ for a nonzero initially spatial $F$. This is a genuine ambiguity in the source wording.

The usual consistent interpretation is constant laboratory relativistic force $f=d{p}_x/dt$, equivalently constant one-dimensional proper acceleration $f/M_0$. Take $f>0$ and let $s=ft/(M_{0}c)$. Integrating momentum conservation with the applied impulse gives $p_x=ft$, soThe corresponding four-force has components $F^{\mu }=\gamma f(\beta ,1,0,0)$, which are not constant, although its invariant magnitude is $F^2=-f^2$. Its temporal component is indeed $F^0=(\gamma /c)dE/dt$. The speed rises from zero and approaches $c$ asymptotically; $\gamma$ starts at one, initially rises quadratically and becomes linear in $t$ at late times.

If instead only the spatial four-force component $F^x=f_x$ is held constant, the distinct consistent answer is most simply parametrized by $u=f_x\tau /(M_{0}c)$:This follows by integrating $d{U}^x/d\tau =f_x/M_0$ and $dt/d\tau =U^0/c$. The temporal component must still vary. These two interpretations must not be conflated with a constant four-vector.

With constant rest mass, $\mathbf{p}=m\gamma \mathbf{v}$ and the chain rule gives $\dot{\gamma }={\gamma }^3(\mathbf{v}\cdot \mathbf{a})/c^2$. Hence the laboratory relativistic force isUse the vector triple product identity $\mathbf{v}\times (\mathbf{v}\times \mathbf{a})=\mathbf{v}(\mathbf{v}\cdot \mathbf{a})-v^2\mathbf{a}$, and ${\gamma }^3(1-v^2/c^2)=\gamma$, to obtainThe relativistic energy is $E=m\gamma c^2$, whose time derivative is $m{\gamma }^3\mathbf{v}\cdot \mathbf{a}$. Taking the dot product of the first expression for $\mathbf{F}$ with $\mathbf{v}$ gives the same quantity, because $\gamma +{\gamma }^3{v}^2/c^2={\gamma }^3$. Thus $dE/dt=\mathbf{F}\cdot \mathbf{v}$. Since $d/d\tau =\gamma d/dt$, the four-force isIts tensor contraction with the four-velocity vanishes: $F^{\alpha }{U}_{\alpha }=-{\gamma }^2\mathbf{F}\cdot \mathbf{v}+{\gamma }^2\mathbf{F}\cdot \mathbf{v}=0$. This orthogonality also follows by differentiating $p^{\alpha }{p}_{\alpha }=-m^2{c}^2$. Resolving the laboratory acceleration relative to $\mathbf{v}$ gives ${\mathbf{F}}_{\parallel }=m{\gamma }^3{\mathbf{a}}_{\parallel }$ and ${\mathbf{F}}_{\perp }=m\gamma {\mathbf{a}}_{\perp }$.