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Suppose I have a coordinate system, call it $S$, in which an observer $O$ is not moving, and $O'$ is moving with constant velocity and another coordinate system $S'$ where $O'$ is not moving and $O$ is moving in a constant velocity, they both have a stop watch and measure the time of the movement of $O'$, what time does each measure?

Since in $S$, $O$ is not moving then he measures the time of $O$ by using the Minkowsky metric, when I change to $S'$ then $O'$ is not moving and he measures time by using a different metric, that is the metric I get after I change coordinates using the Lorrentz transformation.

My question is this idea correct, have I misunderstood something? If yes can you explain what is my mistake?

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  • $\begingroup$ I'm not sure if I understand what you mean by "the observer measures time using a metric". Why wouldn't you simply use a clock? $\endgroup$
    – Jonas
    Jan 22 at 13:41
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    $\begingroup$ A Lorentz transformation, by definition, preserves the metric. $\endgroup$
    – WillO
    Jan 22 at 15:58
  • $\begingroup$ By saying "measure time using a metric", I mean he actually uses the metric to compute the equations of motion. $\endgroup$
    – Kt hamil
    Jan 22 at 17:48
  • $\begingroup$ @WillO the distances are preserved, but the metric looks different in a different coordinate system, although lengths remain unchanged. I am not talking about proper time. $\endgroup$
    – Kt hamil
    Jan 22 at 17:54
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Should be correct, except in special relativity people use the term "frame" rather than "metric", which is used more often in general relativity. You can say $u$ boosted from frame $S$ to frame $S'$ with a Lorentz boost.

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