I've just started special relativity and got a question:

Suppose at $t=t'=0$, there is a bar between $x_1=0$ and $x_2=1$ resting in my frame.

Then, to a person moving with velocity $v$ the length will be:

$\Delta x' = \gamma(\Delta x-v\Delta t) = \gamma(1 - v0) = \gamma$

Since $\gamma \gt 1$, the length in moving frame seems to expand, but I'm told that it should contract..

What's wrong in my understanding?


1 Answer 1


You are forgetting the relativity of simultaneity. In the rest frame the time at one end of the bar is t=0 as is the time at the other end of the bar, but in the moving frame the time t=0 at the far end of the bar does not equate to a time t'=0. The length of the bar in the moving frame is the distance between its end points at a fixed time in that frame, which is not what you have calculated.

Indeed, the reason why the bar seems shorter is that in the moving frame, when you are considering the two ends of the bar at the same time, you seem, from the perspective of the rest frame, to be noting the position of the leading end of the bar somewhat earlier than you are noting the position of the trailing end of the bar, which gives the trailing end a chance to move further forward, and thus make the bar seem shorter.


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