Problem — Ruler moving at 0.6c

Problem

A ruler of proper length $1{,}0$ m moves relative to an observer with speed $v = 0{,}60\,c$ along its own direction. How long is it in the observer’s frame?

Solution

Lorentz factor. With $v = 0{,}60\,c$, $\dfrac{v^2}{c^2} = 0{,}36$:

$\gamma = \frac{1}{\sqrt{1 - 0{,}36}} = \frac{1}{\sqrt{0{,}64}} = 1{,}25$

Length contraction. The proper length is $L_0 = 1{,}0$ m; the observer measures

$L = \frac{L_0}{\gamma} = L_0\sqrt{1 - v^2/c^2} = 1{,}0\;\text{m} \times 0{,}80$

$\ev{L = 0{,}80\;\text{m}}$

The ruler, measured in the frame in which it moves, appears shorter by 20% along the direction of motion.

Links

Topics: Relatività ristretta Concepts: Contrazione delle lunghezze Skills: Impostazione simbolica