Length Contraction

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Say we have two points, A and B, along the x-axis, and a spacecraft is moving past the points in the positive direction of the x-axis.

The proper length of the line segment AB is L₀. It is a proper length since the points A and B are fixed in the frame of reference of the x-axis. For our muon, A could be the point where it was created, and B might be the ground. The velocity of the spacecraft (or muon) is now

$\displaystyle v=\frac{{{{L}_{0}}}}{t}$

Here, t is the time according to the frame of reference of the points A and B. From the standpoint of the spacecraft, points A and B are rushing towards it with the velocity v, so we get

$\displaystyle v=\frac{L}{{{{t}_{0}}}}=\frac{{{{L}_{0}}}}{t}$

The travel along the length L, between A and B, will, according to the spacecraft, take the proper time t₀. But we know from the time dilation that t=λt₀, and thus that

$\displaystyle L=\frac{{{{L}_{0}}}}{\gamma }$

This is the formula for the length contraction.

Continuing with the moun problem from the previous page. Say the ground is 10 km down.
1. How far down would that be according to the frame of reference of the muon travelling at 0.9992c?
2. How long time would it take for the muon to reach the ground, as seen from the frame of reference of the muon?
3. Will it reach the ground?
What speed would a spaceship have to have to be seen as half as long as its proper length?

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Last modified: Jun 4, 2025 @ 19:38