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.

  1. 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?
  2. What speed would a spaceship have to have to be seen as half as long as its proper length?

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Given the answer to the last question means that the muon might very well reach the ground even if it starts its existence 10 km above the ground.
The situation, as seen from the ground, is that the muon will exist for long enough to reach the ground. As seen from the muon, the distance to the ground is so short that it will reach the ground even if its half-life is very short.

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Next page : Lorentz transformation of spaceLast modified: Jun 10, 2025 @ 13:28