- #1

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- 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?

$$x_{AB}=x_{0,AB}+\frac{\cosh{(a\tau)}-1}{a}$$

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?