Velocity and acceleration in special relativity I would like to compute what the constant acceleration trajectories are in the Minkowski spacetime $(t, x)$ with $d\tau^2 = dt^2 - dx^2$. So given some trajectory $x(t)$ I know the velocity vector is given by
$$U = \left( \frac{dt}{d\tau}, \frac{dx}{d\tau} \right)$$
In the previous version of this post I made some great confusion by differentiating $\tau^2 = t^2 - x^2$ to obtain $d\tau$ and thus $dt/d\tau$ etc. I realize now that there's perhaps some poor notation chosen in the circumstances. Here $d\tau$ really should mean the line element, defined as acting on some vector by $d\tau(v) = \sqrt{|\eta(v, v)|}$. This has nothing to do with the differential of the function $\tau = \sqrt{t^2 - x^2}$. In retrospect that is clear -- the differential (thought of as a 1-form) is linear, while the line element clearly is not. Unfortunately the standard notation for both these things is $d\tau$, which is why I got so terribly confused.
Now I realize that to find the velocity vector of some curve $\alpha$ we must first parametrize $\alpha$ by arc-length and then differentiate as usual. By doing that I managed to convince myself that, for the trajectory $x(t)$, its velocity (covariant) vector is
$$ U = \gamma(1, \dot{x}) $$
where $\gamma = 1/\sqrt{1 - \dot{x}^2}$ (here I am writing $\dot{x}$ instead of $v$ to emphasize that this trajectory is accelerated).
Differentiating again, however, I seem to get
$$ dU/d\tau = \left( \frac{1}{2} \gamma^4 \ddot{x} , \frac{1}{2} \gamma^4 \dot{x} \ddot{x} + \gamma \ddot{x} \right)$$
I believe each of these components must be constant functions, but I cannot seem to solve the resulting equations. Have I done something wrong again? If not, how to proceed?
Edit: actually, after Rounak's answer I realized I made some stupid mistakes and I should actually get
$$ dU/d\tau = (\gamma^4 \dot{x} \ddot{x}, \gamma^4 \dot{x}^2 \ddot{x} + \gamma^2 \ddot{x}) $$
 A: let $\dot{x}=u$ and 'a' is spatial acceleration.
$U= \gamma(1,u)$
Acceleration, $A = dU/d\tau = \gamma dU/dt = \gamma(\dot{\gamma},\dot{\gamma}u+a\gamma)$  --(1)
as $dt/d\tau = \gamma$
$\dot{\gamma} = \sqrt{1-u^2}$ $\implies$ $\dot{\gamma} = \gamma^3ua$
Plug this in (1) and you will get 4-acceleration. If you say that for constant 4-acceleration, each component must be constant, then one can always find some coordinate in which that is not the case. What you should really be trying is acceleration scalar:
$A^2 =  g_{\alpha\beta} a^\alpha a^\beta$, where $a^\mu$ are the components of $A$ and $g_{\alpha\beta}$ is the metric.
A: Rounak's answer is correct, so I accepted; I just wish to add some details here. The formula I wrote for $dU/d\tau$ simplifies, with a little bit of algebra, to
$$dU/d\tau = \gamma^4 \ddot{x}(\dot{x}, 1)$$
Then indeed if you compute $d\tau^2$ of this vector you end up with $\gamma^8 \ddot{x}^2 (\dot{x}^2 - 1)$. You set that equal to some constant and it's not hard to solve the resulting differential equation to indeed get the hyperbolic motion. This is done e.g. in Landau's book on classical field theory, page 24.
