What is the worldline for a particle under a constant force in SR? If $f^{\mu}$ is constant, what is the worldline given the force is 
$$f^{\mu}=m\frac{d^{2}x^{\mu}}{d\tau^{2}}$$
Is it wrong to integrate this to get 
$$x^{\mu}=\frac{1}{2m}f^{\mu}\tau^2+u^{\mu}\left(0\right)\tau$$
with $\tau=0$ set to the origin $x^{\mu}=0$? I know that we need to require that 
$$\eta_{\mu \nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$$
but if you integrate like I did, this gives 
$$\left(\frac{1}{m}f_{\mu}\tau+u_{\mu}\left(0\right)\right)\left(\frac{1}{m}f^{\mu}\tau+u^{\mu}\left(0\right)\right)=-1$$
and hence 
$$f_{\mu}f^{\mu}=u_{\mu}\left(0\right)f^{\mu}=0$$
Otherwise $\tau$ is constant. I know the second equality is true, but why the other $f_{\mu}f^{\mu}=0$? Is this true? How do I rearrange for $x^{\mu}\left(t\right)$? 
 A: I believe I have found the answer. Contrary to the comments, this is not the Rindler motion. If you calculate the four-force on a Rindler type trajectory 
$$x=K\cosh \alpha \tau$$
$$t=K\sinh \alpha \tau$$
$$y=z=0$$
You get 
$$f^0=\alpha^2K\sinh \alpha \tau$$
$$f^1=\alpha^2K\cosh \alpha \tau$$
$$f^2=f^3=0$$
which is of course not a constant force. The Rindler motion is of a constant force in the rest frame of the worldline, not the "lab-frame". 
My problem in fact has no solution. It is overdetermined since there are five equations: four relativistic Newton ODEs; and the four-veocity normalisation. A constant four-force is incompatible with the four veocity normalisation unless $f_{\mu}f^{\mu}=0$, and thus the general case where $f_{\mu}f^{\mu}\neq0$ is unphysical. 
It turns out you can't just write down any four-force since the fourth component is already determined by normalisation of the four velocity once you've written three components down. 
A: Relativistic you schuld take this equations:
\begin{align*}
 &\frac{d}{dt}\left(\gamma\,m\,v(t)\right)=\gamma\,f\,,c=1\,,\quad\gamma=\frac{1}{\sqrt{1-v(t)^2}}\,,\Rightarrow\\
 &\frac{dv}{dt}=\frac{f}{m}(1-v^2)\\\\
 &\text{with:}\quad \frac{dv}{dt'}=\frac{dv}{dt}\,\frac{1}{\gamma}\,\Rightarrow\quad
 \frac{dv}{dt'}=\frac{f}{m}\left(1-v^2(t')\right)^{3/2}\\\\
& v(t')=\frac{t'\,f/m}{\sqrt{1+t'^2\,(f/m)^2}}\,,\quad v(0)=0\\
& x'(t')=\sqrt{\frac{m^2}{f^2}+t'^2}\\
&x'^2-t'^2=\frac{m^2}{f^2}\,,\text{This is Rindler motion}
\end{align*}
A: The relativistic EOM's in Minkowski space:
\begin{align*}
u^\mu&=\frac{dx^\mu}{d\tau}=\frac{dx^\mu}{dt}\frac{dt}{d\tau}=\gamma\begin{bmatrix}
                                                                      c \\
                                                                      u^i \\
                                                                    \end{bmatrix}\,,\quad i=1,2,3\\
&\text{Newton}\,,\quad f^i=\frac{d}{dt}(m\,u^i)\\
&\text{Minkowski}\,,\quad k^\mu=\frac{d}{d\tau}(m\,u^\mu)=\frac{d}{dt}\begin{bmatrix}
                                                                        \gamma\,m\,c \\
                                                                        \gamma\,m\,u^i \\
                                                                      \end{bmatrix}=
\begin{bmatrix}
  k^0  \\
  \gamma\,f^i \\
\end{bmatrix}\,,\quad k^0=\frac{\gamma\,u^i\,f_i}{c}\\&k_\mu\,k^\mu= \text{Lorentz scalar!!}
\end{align*}
The EOM's 
\begin{align*}
  \frac{d}{dt} (\gamma\,m\,c^2) & ={\gamma\,u^i\,f_i} \\
 \frac{d}{dt} (\gamma\,m\,v)&=\gamma \,f_i\,,\quad
 \text{With:}\quad \frac{d}{dt}=\gamma\,\frac{d}{dt'}\\
 &\Rightarrow\\
 \frac{d}{dt'} (\gamma\,m\,c^2) & ={\,u^i\,f_i} & (1)\\
 \frac{d}{dt'} (\gamma\,m\,u^i)&= \,f_i & (2)\\
 \gamma&=\frac{1}{\sqrt{1-\frac{v^2}{c}}}\,,\quad v^2=u_i\,u^i\\\\
 &\text{Equation (2):}\quad c=1\\\\
 \frac{du_x}{dt'}&=\frac{f_x}{m}\left(1-\left(u_x^2+u_y^2+u_z^2\right)\right)^{3/2}\\
 \frac{du_y}{dt'}&=\frac{f_y}{m}\left(1-\left(u_x^2+u_y^2+u_z^2\right)\right)^{3/2}\\
 \frac{du_z}{dt'}&=\frac{f_z}{m}\left(1-\left(u_x^2+u_y^2+u_z^2\right)\right)^{3/2}\\
 &\text{With:}\quad u_y(t')=0\,,\quad u_z(t')=0\,,\Rightarrow\\
 u_x(t')&=\frac{t'\,f_x}{\sqrt{m^2+(f_x\,t')^2}}\,,\quad v_x(0)=0\\
 x(t')&=\int\,{u_x}\,{dt'}=\frac{1}{f_x}\,\sqrt{m^2+(f_x\,t')^2}\\\\
 x^2-t'^2&=\left(\frac{m}{f_x}\right)^2\,,\quad\text{ Rindler motion}
\end{align*}
New coordinates
\begin{align*}
  &t'\mapsto \frac{1}{\alpha}\sinh(\alpha\,\tau)\\
   &x\mapsto \frac{1}{\alpha}\cosh(\alpha\,\tau) \\
   &\alpha=\frac{f_x}{m}\,,\quad \Rightarrow\\
  &x^2-t'^2=\left(\frac{m}{f_x}\right)^2
\end{align*}
