Physics Asked on November 19, 2020
In Hobson’s General Relativity book, pg. 125, it says that an instantaneous rest frame $S’$ can be defined for an accelerating observer in frame $S$ such that the observer is momentarily at rest in $S’$.
It also says
since the observer is at rest in $S’$, the timelike basis vector $textbf{e}’_0 $ of this frame must be parallel to the 4-velocity $textbf{u}$ of the observer.
Why is that so?
When looking at the accelerated frame from an inertial frame K, you can always find an inertial frame S' whose velocity matches with the velocity of the accelerated frame at that instant. So in that instant the relative velocity between S' and S is zero. But it's only for that instant because S is accelerating and in the next instant it's velocity would be different.
Edit: I forgot to respond to that quoted part. For a stationary object the four velocity is (c,0,0,0) or if you're working in units with c=1, (1,0,0,0).
Correct answer by Brain Stroke Patient on November 19, 2020
The other answer has noted that the frames are momentarily comoving, since $S$ is accelerating and $S'$ is an inertial frame. At any subsequent moment $S$ will have accelerated slightly and will no longer be comoving with $S'$.
Regarding the part you've quoted, even in the rest frame of an object its 4-velocity is non-zero (ie. it is not the zero vector). In the rest frame of an object its 4-velocity is parallel to the $x_0$ basis vector, $mathbf e_0'$, and so: $$mathbf u=mathbf e'_0.$$
Answered by Charlie on November 19, 2020
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