Suppose a clock, located at point $x'$ in the inertial frame $S'$, registers two events $t_1'$ and $t_2'$. Let $\Delta t'=t_2'-t_1'$. The same two events will be registered by two different (synchronized in their rest frame) clocks, located in two different points of the inertial frame $S$ (which moves with a certain speed relative to $S'$). Let $\Delta t=t_2-t_1$ the elapsed time between these two events as seen from $S$. From the theory of relativity we know that $\Delta t' = \sqrt{1-\beta^2}\Delta t$.

At first glance, the time dilation formula looks asymmetrical. We know that all of the clocks in $S$ are synchronized. Thus, it is tempting to assume that $\Delta t$, which is calculated by two different clocks in $S$, can be calculated by one clock in $S$. But then we have a situation where two perfect and equivalent clocks - one from $S$ and one from $S'$ - are no longer equivalent (one of them is slower than another). But this contradicts the equivalence of all inertial frames of reference.

My textbook says that in order to resolve this "paradox", one must realize that clocks which are synchronized in one inertial frame need not be synchronized in another inertial frame. That is - the clocks "belonging" to the system $S$ aren't showing the same time when seen by an observer in $S'$.

However I don't quite understand this argument. Why do we care what the clocks of $S$ show to an observer in $S'$? As far as I understand, the time interval $\Delta t$ in the formula above is the time elapsed according to the clocks in $S$ as seen by an observer in $S$. In other words - the formula connects what an observer in $S'$ sees to what an observer in $S$ sees = two different observers, not one.

  • $\begingroup$ The situation is symmetrical. For the unprimed observer to measure the elapsed time of the primed observer's moving clock, he has to use two different but synchronized clocks. Vice versa for the primed observer if he wants to measure the unprimed observer's clocks. Each will measure the other's clock ticking more slowly. The key to the apparent paradox is realizing that "same time" measurements made by one observer are not "same time" measurements according to the other observer. I suggest Mermin's "It's about Time: Understanding Einstein's Relativity". $\endgroup$ – Samuel Weir Jan 3 '16 at 21:03
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    $\begingroup$ There's a subtle difference. To the observer in S', the two events occur at the same location. To the observer in S, the two events occur at two different locations. So the same formula used in going from S' to S cannot be applied in going from S to S'. $\endgroup$ – Chet Miller Jan 3 '16 at 21:12
  • $\begingroup$ @ChesterMiller - this does not answer OP's question. $\endgroup$ – cth Jan 4 '16 at 9:13

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