Coordinate transformation for a comoving frame for abitrary trajectory Let's suppose there is a particle moving through space time. It moves along the x-direction with a time dependent velocity. Seen from an inertial frame it has the velocity $v(t)$ where $t$ is the coordinate time of the inertial frame. Let $t \in [0,T]$.
I want to get the coordinate transformation for the comoving frame of the particle. By that I mean a coordinate system in which the particle is at rest for all times. I tried to get this transformation by assuming the particle moves for short time periods $\Delta t$ with constant velocity. For constant velocities we can use a normal Lorentz transformation as a coordinate transformation. If I now split the time interval $[0,T]$ into $n$ intervals of with length $\Delta t = T/N$ it is possible to find a LT for each time interval of length $\Delta t$.
For $t \in [n \Delta t,(n+1) \Delta t]$ we get 
$$t' = \gamma_n (t- v_n x)+ a_n$$
The $a_n$ is unknown but it needs to be connected to the previous time interval $[n \Delta t,(n+1) \Delta t]$ by
$$a_n = a_{n-1} + n \Delta t(\gamma_{n-1}-\gamma_{n-2})+ \left(\sum_{m=0}^{n}v_m \Delta t\right) (\gamma_{n} v_n - \gamma_{n-1}v_{n-1})$$
Now by iteration and converting the sums into integrals I get
$$a_n =: a(t)= \int_0^t \tilde{t} \frac{\mathrm d\gamma}{\mathrm d\tilde{t}} +r(\tilde{t})\frac{\mathrm d(\gamma v)}{\mathrm d\tilde{t}} \mathrm d\tilde{t}$$
where $r(t)$ is the position of the particle seen from the inertial frame with $r(0)=0$.
So in total we get for the transformed time:
$$t' = \gamma(t) (t- v(t) x)+ \int_0^t \tilde{t} \frac{\mathrm d\gamma}{\mathrm d\tilde{t}} +r(\tilde{t})\frac{\mathrm d(\gamma v)}{\mathrm d\tilde{t}} \mathrm d\tilde{t} $$
Something similar can be done for the space coordinate.
Now my question is if this is the right approach? Or did I make a mistake somewhere? In my new coordinate system the particle should be at rest for all $t'$. Are there other ways to find the comoving coordinate system ? And is there a formula without the integral?
 A: 
Seen from an inertial frame it has the velocity v(t)

$\def\D#1#2{{d#1 \over d#2}}$
Define $x(t)$ such that $dx/dt=v$, then $h(\tau)$ so that
$$\eqalign{
   \D x\tau &= \sinh h(\tau) \cr
   \D t\tau &= \cosh h(\tau).\cr}$$
Note that $\tanh h = v$, $\cosh h = \gamma$, $h$ is rapidity ($\tau$ is proper time.)
The transformation you're looking for is
$$\eqalign{
    x &= \int_0^\xi \sinh h(\xi')\,d\xi' + \eta\,\cosh h(\xi) \cr
    t &= \int_0^\xi \cosh h(\xi')\,d\xi' + \eta\,\sinh h(\xi).\cr}$$
You may see that for $\eta=0$
$$\eqalign{
    dx = \sinh h(\xi)\,d\xi \cr
    dt = \cosh h(\xi)\,d\xi \cr}$$
so that motion of a point fixed at $\eta=0$ is just the one assigned
for $x(t)$ with $\xi=\tau$.
In coordinates $(\xi,\eta)$ the metric is
$$[1 + \eta\,h'(\xi)]^2 d\xi^2 - d\eta^2$$
showing that $(\xi,\eta)$ are generalized Rindler coordinates.
Other properties easily follow, especially for motion of points with
$\eta=\mathrm{const.}\ne0$.
A: I agree with the suggestion to look into the MTW. I really like the section on uniformly accelerating objects.
I don't think you need the complicated step that involves the Lorentz transforms, and taking the continuum limit.
In any inertial frame, a world-line of a moving object can be expressed as $x^\mu=\left(ct,\, \mathbf{r}(t)\right)^\mu$, so $dx^\mu=(c, \mathbf{v}(t))^\mu dt$, and $dx^2=(c^2-v^2)dt^2$, but also, by definition, $dx^2=c^2 d\tau^2$ ($\tau$=proper time), so $dt/d\tau=\gamma=\left(1-v^2/c^2\right)^{-1/2}$, even if velocity is time-dependent.
If you know the expression for the time-dependent velocity $v=v(t)$, simply integrate $d\tau=dt\sqrt{1-v(t)^2/c^2}$ to get the proper time ($\tau$) in terms of lab-time time ($t$).
Once you know how proper time relates to lab-time i.e. $\tau=\tau\left(t\right)$ you can write down the four-velocity $u^{\mu}(\tau)=\frac{dx^{\mu}}{d\tau}=\frac{d\tau}{dt}\cdot\frac{dx^{\mu}}{dt}=\frac{d\tau}{dt}\cdot\left(c, \mathbf{v}(t(\tau))\right)^{\mu}$
Four-velocity will give you the direction of the temporal axis for the inertial frame in which your object is at rest at that specific time. You then have an arbitrary choice on how to span the 3 spatial dimensions that are perpendicular to four-velocity.
