Let $S$ and $S'$ be two frames, moving with velocity $V$ relative to each other. Treat $S$ as a stationary frame, then $S'$ moves away with velocity $V$. Let $\gamma = 7$ so that $\Delta t = 7\Delta t'$. Does this mean that if there are exact clocks in each frame and observers in $S$ establish a video conference call with observers in $S'$, then the clock in $S'$ will show time moving seven times slower?

I don't think that's going to be the case because we can treat things the other way around by taking $S'$ as stationary.

But then, what about time dilation? How should it happen in this case? Will clocks in $S$ and $S'$ differ at all?

• That's basically the classical twin paradox, which is answered here: physics.stackexchange.com/questions/2554/… If both observers continue to move relative to each other with constant velocity, the two observers will never meet and can't compare the clocks. One of the two observers has to accelerate to meet again with the other observer and that acceleration isn't relative. Also note that the signal from the video chat can only travel at the finite velocity $c$ Dec 11, 2016 at 20:54
• See relativity of simultaneity perhaps. Dec 11, 2016 at 20:58
• sequence: "observers in [inertial frame] $S$ [...] observers in [inertial frame] $S'$" -- +1 for your use of "observer" as being associated to any one individual constituent of a reference frame (to any individual "material point"); i.e. consistent with the use by Einstein; especially in "Electrodynamics of Moving Bodies", Ann. Phys. 17, (1905); rather than being associated to all constituents of a reference frame collectively, or to even more remote abstractions. Dec 13, 2016 at 6:38
• Tatsu Takeuchi's diagram on the matter (from his book An Illustrated Guide to Relativity and reproduced in another stack exchange post) gives a nice visual way to understand this problem. Dec 15, 2016 at 1:36

1) Suppose Alice sits on earth while Bob travels away at speed $v$.

2) First do everything in Alice's coordinates: Bob is traveling along the line $x=vt$. Alice, at time $t_0$, sends him a light signal that travels along the line $x=t-t_0$. To see when it arrives, solve $x=vt=t-t_0$ to get $$t={t_0\over 1-v}\qquad x={vt_0\over 1-v}$$

3) Now Lorentz-transform to Bob's coordinates. This gives $$t'={t-xv\over\sqrt{1-v^2}}={t_0(1+v)\over\sqrt{1-v}^2}$$

4) Therefore the speed of the video stream arriving at Bob's ship is slowed down by a factor of $(1+v)/\sqrt{1-v^2}$. This factor is not just your $\gamma$, because that fails to account for the fact that Bob is moving away from Alice as the video streams.

5) By symmetry, exactly the same is true of the speed of the video stream arriving at Alice's screen from Bob's ship.

6) As to your (separate) question about whether Bob's and Alice's clocks differ, that depends, of course, on who you ask. Alice says Bob's runs slower than hers, Bob says Alice's runs slower than his, and Ted, who is moving away from Alice in the same direction as Bob at speed $v/2$, says they don't differ at all.

• I've removed a comment discussion that went in a not-okay direction. Be kind, folks.
– rob
Jul 17, 2021 at 14:10