SR w/ Acceleration: Distance Measured by A

[Gaussian97]

Homework Statement

Let Alice and Bob be accelerated in Charles RF, with world lines with respect to C given by (1), what is the distance between A and B measured from C? What is the distance of A and B measured from A?

Relevant Equations

$$c = 1$$
$$t_{AB}=\frac{\sinh{(a\tau_{AB})}}{a}, \qquad x_{AB}=x_{0,AB}+\frac{\cosh{(a\tau_{AB})}-1}{a} \qquad (1)$$

Well, using (1) is easy to see that, at a given time in C $t$ both curves are described with the same value of $\tau$, i.e. $\tau_A=\tau_B=\tau$. So the corresponding positions at a given time $t$ are
$$x_{AB}=x_{0,AB}+\frac{\cosh{(a\tau)}-1}{a}$$
and therefore
$$\Delta x \equiv x_B - x_A = x_{0,B}+\frac{\cosh{(a\tau)}-1}{a} - x_{0,A}-\frac{\cosh{(a\tau)}-1}{a} = x_{0,B} - x_{0, A}$$
so A and B are always at the same distance from each other. No big problem here (I think)
My problem comes when I have to compute the distance measured in A.

The solution that I have says that the vector the vectors are invariant (which I agree) and therefore
$$\Delta x \vec{e}_x = \Delta x' \vec{e}'_x +\Delta t' \vec{e}'_t\Longrightarrow \Delta x' = \Delta x \vec{e}_x\cdot \vec{e}'_x = \cosh{(a\tau)} \Delta x$$
But I don't understand why this is the distance measured by A, because the "temporal distance" here is not zero, shouldn't we find a vector with $\Delta t'=0$ and then compute the distance?

 

[TSny]

The solution that I have says that the vector the vectors are invariant (which I agree) and therefore
$$\Delta x \vec{e}_x = \Delta x' \vec{e}'_x +\Delta t' \vec{e}'_t\Longrightarrow \Delta x' = \Delta x \vec{e}_x\cdot \vec{e}'_x = \cosh{(a\tau)} \Delta x$$
But I don't understand why this is the distance measured by A, because the "temporal distance" here is not zero, shouldn't we find a vector with $\Delta t'=0$ and then compute the distance?

I agree with you. I'm not sure what they're doing in the solution that you posted. I take it that the primed frame is an inertial reference frame for which A is instantaneously at rest at the instant that A determines the distance to B. The distance between A and B, as measured by A, would be determined from simultaneous measurements of the positions of A and B in the primed frame (at the instant that A is at rest in this frame). So, as you say, $\Delta t'=0$. The corresponding $\Delta t$ would not be zero.