Problem — Contraction of a rod

Problem

A rod has proper length $L_0 = 10$ m. What length is it measured to have by an observer relative to whom it moves along its own axis at $v = 0{,}8\,c$?

Solution

Lorentz factor. With $v = 0{,}8\,c$, $\dfrac{v^2}{c^2} = 0{,}64$:

$\gamma = \frac{1}{\sqrt{1 - 0{,}64}} = \frac{1}{\sqrt{0{,}36}} = \frac{1}{0{,}6} \approx 1{,}67$

Length contraction. The length measured in the frame in which the rod moves is

$L = \frac{L_0}{\gamma} = L_0\sqrt{1 - v^2/c^2} = 10\;\text{m} \times 0{,}6$

$\ev{L \approx 6{,}0\;\text{m}}$

The rod appears shorter along the direction of motion: the proper length $L_0$ is always the maximum.

Links

Topics: Special relativity Concepts: Length contraction Skills: Symbolic setup