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Help Me Gain an Intuitive Understanding of Lorentz Contraction

Frames k and k' are inertial frames. Frame k' is moving at a velocity of magnitude v relative to frame k in the direction of the x-axis. there are rigid rods in each frame oriented along the direction of the relative velocity. Someone in frame k measures the rod that is stationary in k as having length L1 and the rod stationary in frame k' having a length L'2. What are the lengths of the rods measured by someone in frame k'?

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marked as duplicate by Qmechanic, Manishearth, Waffle's Crazy Peanut, Sklivvz Dec 29 '12 at 15:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This looks like a homework assignment... –  Sklivvz Dec 29 '12 at 13:52

2 Answers 2

The answer is length contraction, the lengths of the first is $L1\sqrt{1-v^2}$, it gets contracted, while the length of the moving one is $L2/\sqrt{1-v^2}$, so that after contraction it gives the right answer. The details of the geometry is explained in my answer here: What are the mechanics by which Time Dilation and Length Contraction occur? .

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Reading between the lines, are you thinking there may be a contradiction in the length's of the rods measured by the two observers, or are you thinking of something like the train and platform experiment (see http://en.wikipedia.org/wiki/Relativity_of_simultaneity#The_train-and-platform_thought_experiment)?

In SR it's dangerous to just throw around ideas like length contraction and time dilation because if you're not careful it's easy to make a mistake and end up with a paradox. The only safe way to procede is to take the spacetime co-ordinates of the events you're interested in, e.g. the two ends of the rod, and apply the Lorentz transforms to see where those events are in the other frame. If you can phrase your question more precisely I'll have a go at doing this calculation for you.

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