Key 3

Up a level : 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?
      The gamma factor is 25, so we get \displaystyle L=\frac{{{{L}_{0}}}}{\gamma }=\frac{{10000}}{{25}}=400\text{ m}
    2. How long time would it take for the muon to reach the ground, as seen from the frame of reference of the muon?\displaystyle t=\frac{s}{v}=\frac{{400}}{{0.9992c}}=1.34\cdot {{10}^{{-6}}}\text{ s}
    3. Will it reach the ground?
      1.34·10-6 s is quite much smaller than the half-life  2.2·10-6 s, so, most likely, yes.
  2. What speed would a spaceship have to have to be seen as half as long as its proper length?
    We have

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

for L=L0/2, so

\displaystyle \frac{{{{L}_{0}}}}{2}=\frac{{{{L}_{0}}}}{\gamma }

or

\displaystyle \frac{1}{2}=\sqrt{{1-{{{\left( {\frac{v}{c}} \right)}}^{2}}}}

This gives us

\displaystyle \frac{1}{4}=1-{{\left( {\frac{v}{c}} \right)}^{2}}

or

\displaystyle \frac{v}{c}=\frac{{\sqrt{3}}}{2}=0.866c=2.60\cdot {{10}^{8}}\text{ m/s}

Up a level : Length ContractionLast modified: Jun 4, 2025 @ 19:38