4-velocity and 4-acceleration in instantaneous rest frames I am trying to solve this problem:

Consider a rocket moving relative to an inertial frame $\mathcal{F}$ , such that its worldline is given by $$x^{\mu}=c^2/g(\sinh(g\tau/c),\cosh(g\tau/c)-1,0,0).$$ What are the components of four acceleration relative to the instantaneous rest frame of the rocket, $\mathcal{F}'$?

I (think I) understand how to do this using Lorentz transformation: $dt/d\tau=(1/c) \cdot   dx^0/d\tau=\cosh(g\tau/c)$. This is equal to $\gamma$ and it is then straightforward to compute 3-velocity and then use the Lorentz matrix to get $$dx'^\mu(\tau)/d\tau=(c,0,0,0) \quad
; \quad d^2 x'^\mu(\tau)/d\tau^2=(0,g,0,0)$$
However, when I first saw this I immediately thought that by definition the 4-velocity in the instantaneous rest frame of the rocket would be (c,0,0,0) because in $\mathcal{F}'$ the 3-velocity is zero and $\gamma$ would be 1 and I was wondering if this is a valid reasoning. Even if it is, why is the following wrong?
If $v'^\mu=dx'^\mu(\tau)/d\tau=(c,0,0,0)$, then $a'^\mu=dv'^\mu/d\tau=(0,0,0,0)$ since $c$ is a constant. But this contradicts the calculations given by the Lorentz transformation and I don't understand why and given this I have no idea of how to interpret 4-acceleration.
 A: OK I think I see where your confusion lies.
You're talking about the four velocity and acceleration in the instantaneous rest frame $F'$, and as you say in this frame the four velocity is $(1, 0, 0, 0)$. Your mistake is to assume the four velocity is constant in $F'$, because it is not. Remember that after an infinitesimal time $dt$ the rocket is not longer in $F'$ - it is in a new instantaneous rest frame $F''$. The rocket's velocity in the new rest frame $F''$ is still $(1, 0, 0, 0)$, but in old $F'$ frame it has now changed due to the acceleration. Hence $d{\bf u}/dt$ in $F'$ is not zero.
If you're interested chapter 6 of Gravitation by Misner, Thorne and Wheeler derives the equations of motion that you started with.
A: The four-acceleration can be simply defined as a derivative of four-velocity with respect to proper time, in your case four-velocity is
$$v^{\mu} =(c\:\cosh\left(\frac{g}{c}\tau\right),c \:\sinh\left(\frac{g}{c}\tau\right),0,0)$$
where you can check that $v^{0}=c\gamma$ as you guessed right
and therefore four-acceleration is
$$A^{\mu}=(g\:\sinh\left(\frac{g}{c}\tau\right),g \:\cosh\left(\frac{g}{c}\tau\right),0,0)$$
You can also see from here that in this case $\Vert A \Vert=g$.
