Comparison of two proper times in Special Relativity using an in-between frame

Question

I would like to compare two proper times of two inertial observers in Special Relativity.

Let's consider three inertial observers A, O and B freely moving on a line in a flat space-time, with B moving with a constant velocity $v$ with respect to A and O moving with a constant velocity $w$ with respect to A.

In the rest frame of O, A will move with the velocity $-w$ (always), and, by choosing $w$ carefully, B can move with the same velocity but in the opposite direction (+$w$). I just created an in-between frame in which the situation is symmetrical. Applying the basic time dilation formula in the rest frame of O for A and B respectively, I find $d\tau_{A} = d\tau_{B}$.

Mathematical proof

I want the velocity of O seen from B to be $-w$. I therefore need to solve the following system based on the velocity addition formula: $$ -w = \frac{w-v}{1-\frac{wv}{c^{2}}}$$

This leads to

$$w = \frac{c^{2}}{v}\left(1-\sqrt{1-\left(\frac{v}{c}\right)^{2}}\right)$$

In the rest frame of O associated with the coordinate time $t_{o}$, we have:

$$d\tau_{A} = \frac{dt_{o}}{\gamma(-w)}$$ $$d\tau_{B} = \frac{dt_{o}}{\gamma(w)}$$

With $\tau_{A}$ and $\tau_{B}$ the proper times of A and B, and $\gamma(w) = 1/\sqrt{1-(w/c)^{2}}$

Leading to $d\tau_{A} = d\tau_{B}$. Ok.

Now let's go back in the rest frame of A. We associate a coordinate time $t$ to it. We therefore have:

$$d\tau_{B} = \frac{dt}{\gamma(v)}$$

Similarly, in the rest frame of B associated with a coordinate time $t'$ :

$$d\tau_{A} = \frac{dt'}{\gamma(-v)}$$

Which leads to $dt=dt'$. Or we know that:

$$dt' = \gamma(v)(dt-\frac{v}{c^{2}}dx)$$

It seems like there is an inconsistency. Please let me know what is wrong, thank you!

Answer 1

Please let me know what is wrong, thank you!

Nothing is wrong. $\gamma$ is an even function, as you have shown

It seems like there is an inconsistency.

The inconsistency in your edit is the pairs of events that you are choosing to compare.

When you use simultaneity in the middle frame to select the events on the clocks then you get that the pair had equal proper time. That pair of events will have equal proper time in all frames, but will only be simultaneous in in the middle frame.

When you use simultaneity in the A frame to select the events on the clocks then you will get that the B clock had less proper time. With that pair of events the B clock will have less proper time in all frames, but will only be simultaneous in the A frame.

Answer 2

$\let\D=\Delta \let\g=\gamma \def\rA{{\mathrm A}} \def\rB{{\mathrm B}}$ Your problem is that you make a too casual use of $d\tau=dt/\g$. You'd keep in mind its meaning.

If there are two (inertial) frames, A and B, what do you mean by "proper times of A and B"? Time is always measured between two events E, F. And for the same events you may measure time in both frames. The difference is whether E and F happen in the same place of A or of B (or possibly neither).

Proper time of A can only mean time measured in A for events E, F having a common position in A. If you measure the time interval between E and F with B's clocks, you'll find $$\D t_{\rB} = \g\,\D t_\rA$$ and only $\D t_\rA$ is proper time. If E, F have same position in B, then the opposite holds: $$\D t_{\rA} = \g\,\D t_\rB$$ with the same $\g=\g(v)=\g(-v)$,