Length of a Stick Moving at Speed v in an Angle

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SUMMARY

The discussion centers on the relativistic length contraction of a stick with proper length L moving at speed v at an angle theta. The derived formula for the observed length L' in a frame where the stick moves is L' = L √{(1 - v²/c²)/(1 - (v²/c²)sin²{theta})}. Key points include that the stick contracts only in the direction of motion, specifically along the x-direction, while maintaining its full length in the perpendicular y-direction. Participants confirmed the necessity of separating the components of motion to accurately calculate the total length in the moving frame.

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Tony11235
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Prove that a stick of proper length L has a length L' in a frame in which it moves with speed v along a line that makes an angle theta with it's length is given by

[tex]L' = L \sqrt{\frac{1-v^2 / c^2}{1 - (v^2 / c^2)\sin^2{\theta}}}[/tex]

My problem here is the picture I think. So the stick overall is moving with speed v, but the stick is not necessarily parallel to v, but at an angle created by the direction of v and the stick?
 
Last edited:
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So the stick overall is moving with speed v, but the stick is not necessarily parallel to v, but at an angle created by the direction of v and the stick?

Yes.

You need to remember that your stick is only contracted in the direction of motion. So just for definiteness say that [itex]\vec{v}=v\hat{i}[/itex]. Then the stick is contracted in the x-direction but not in the y-direction.

Once you have the two components in the frame in which the stick has velocity [itex]\vec{v}[/itex] you can find its total length.
 
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Got it. I went that direction earlier, but for some reason it didn't look as if the expression I derived was equivalent. BUT after some algebra...
 

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