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?

a) 1 m tall, 3 m wide

b) 1 m tall, 1.31 m wide

Explanation:

According to the captain of the spaceship, the dimensions of the picture is the same i.e 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 will remain 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 if we are to go by the Lorentz contraction:

L(x) = L(x)' * √[1 - (v²/c²)]

where

L(x)' = the dimensions of the picture along the x axis on the spaceship,

v² = the speed of the spaceship and c² = the speed of light in the vacuum.

On substituting, we have

L(x) = 3 * √[1 - (0.81c²/c²)]

L(x) = 1.31 m