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.

• Someone edited the questioned to include the homework tag but this is not homework. I am just studying. May 20, 2014 at 17:28
• If you have a spare minute, you should read this in order to understand when the homework tag is appropriate. May 20, 2014 at 17:30
• @student: ignore the homework tag, it just means the question is homework-like (which it is) and doesn't mean it isn't a good question. As it happens I've recently been reading up on this (Gravitation by Misner et al chapter 6) and I'm currently flicking back through it in an attempt to answer! May 20, 2014 at 17:33
• @Student how did you get $dx'^{\mu}/d\tau=(c,0,0,0)$? I tried using $\gamma=\sinh(g\tau/c)$ in the Lorentz matrix but multiplying out with $x^{\mu}$ and differentiating with respect to $\tau$ didn't give me $(c,0,0,0)$ Jun 27, 2016 at 21:15

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.
• Thank you. In that case, how can one make sense of the definition $a^\mu=dv^\mu/d\tau$ for an instantaneous reference frame? Or is it just not applicable ? May 20, 2014 at 18:03
• @John Rennie, should $d\textbf{a}/dt$ be $d\textbf{v}/dt$ ? May 20, 2014 at 18:18
• @Student: $F'$ is not an instantaneous reference frame. It's an inertial frame that the rocket instantaneously occupies. In the inertial frame $F'$ $a^\mu=dv^\mu/d\tau$ applies just as it does in all inertial frames. The rest frame of the rocket is not inertial so it's not surprising to find that $a^\mu=dv^\mu/d\tau$ does not apply in the rest frame of the rocket. May 21, 2014 at 5:40
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 $$\Vert A \Vert=g$$.