Answer to Question #188822 in Physics for Jerick Sapa

Question

A box at rest has the shape of a cube 2.6 m on a side. This box is loaded onto the flat floor of a spaceship and the spaceship then flies past us with a horizontal speed of 0.80c. What is the volume of the box as we observe it?

Answer 1

Let the length of the sides in all directions at rest be $L_x = L_y = L_z = 2.6m$. Assume that the box is moving in x-direction with velocity $v = 0.80c$, where $c$ is the speed of light. Than the length contraction occures in x-direction (according to the Lorentz transformations), and the length of the moving box will be:

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

In the directions perpendicular to x the lenghts do not change:

$$L_y' = L_y,\space \space L_z' = L_z$$

The volume of the box as we observe it is the following:

$$V' = L_x'L_y'L_z' = L_xL_yL_z\sqrt{1 -\dfrac{v^2}{c^2}}\\ V' =(2.6m)^3\cdot \sqrt{1 -\dfrac{(0.8c)^2}{c^2}} \approx 10.5m^3$$

Answer. $10.5m^3$.