Answer to Question #116808 in Mechanics | Relativity for nicolee heath

Question

A rectangular painting measures 1.0 m tall along the y' axis and 3.0 m wide along the
x' axis. The painting is hung on the side wall of a spaceship which is moving passed
the Earth at a speed of 0.9c. Assume that the spaceship is moving along the (x, x')
direction.
a) What are the dimensions of the picture according to the captain of the
spaceship?
b) What are the dimensions of the picture as seen by an observer on the Earth?

Answer 1

a) According to the captain of the spaceship the dimensions of the picture remains the same: 1.0 m tall along the y' axis and 3.0 m wide along the x' axis.

b)  The dimensions of the picture as seen by an observer on the Earth along the y axis remains the same, 1.0 m tall, for the direction of the y axis is perpendicular to the spaceship movement.

 The dimensions of the picture as seen by an observer on the Earth along the x axis will reduce accordin to the Lorentz contraction:

$L_x = L_{x'}\sqrt{1 - \dfrac{v^2}{c^2 }}$,

where $L_{x'}$ is the dimensions of the picture along the x axis on the spaceship, $v$ is the speed of the spaceship and $c$ is the speed of light in the vacuum.

Thus, obtain:

$L_x =3\sqrt{1 - \dfrac{0.81c^2}{c^2 }} = 1.31 m$

 

Answer. a) 1m tall, 3m wide. b) 1m tall, 1.31m wide.