SR length contraction question

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SUMMARY

The discussion centers on the application of the length contraction formula in special relativity, specifically how to account for the velocity of a rod moving along the x-axis. The length contraction formula is stated as L = L' * sqrt(1 - v²/c²), where L is the proper length and L' is the contracted length. The key insight provided is that the effective speed of the rod relative to another inertial frame S' is given by the relativistic velocity addition formula: (u + v) / (1 + uv/c²). This effective speed should replace v in the length contraction formula to accurately calculate L'.

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A rod moves along x-axis with speed u and has length L relative to Inertial Frame S. What is its length L' relative to the other inertial S' that moves with speed v with respect to S.

I know the famous length contraction formula L=L' * sqrt(1-square(v/c))
but I confuse how to include velocity of the rod (u) in this formula...:confused:
thanks
 
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The speed of the rod relative to inertial system S' would be (u+ v)/(1+ uv/c^2)[/math]. Put that into the contraction formula rather than v.
 
HallsofIvy said:
The speed of the rod relative to inertial system S' would be (u+ v)/(1+ uv/c^2)[/math]. Put that into the contraction formula rather than v.
<br /> <br /> Thanks a lot for making it clear to me
 

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