Question

The density of titanium is 4.5*10^ 3 kg/m^ 3 A cube of titanium with sides of 1.0 m is moving in the direction of one of its sides with a velocity of 0.9c. Determine the relative density of the titanium cube . 

Answer 1

According to the Lorentz transformation, the length in the direction of motion is given as follows:

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

where $l_x = 1m$ is the rest length in that direction, $v = 0.9c$ is the speed of the cube, and $c$ is the speed of light. In the directions perpendicular to the motion the lengths remain the same:

$$l_y' = l_y = 1m\\ l_z' = l_z = 1m$$

Thus, the volume of the cube in laboratory reference frame is:

$$V' = l_x'l_y'l_z' = l_xl_yl_z\sqrt{1-\dfrac{v^2}{c^2}} = V\sqrt{1-\dfrac{v^2}{c^2}}$$

where $V = l_xl_yl_z = 1m^3$ is the rest volume of the cube. Thus, since the mass does not change during the motion, the relative density will be:

$$\rho' = \dfrac{m}{V'} = \dfrac{m}{V\sqrt{1-\dfrac{v^2}{c^2}} } = \dfrac{\rho}{\sqrt{1-\dfrac{v^2}{c^2}}}$$

where $\rho = m/V = 4.5\times 10^3kg/m^3$ is the rest density of the cube. Thus, obtain:

$$\rho' = \dfrac{4.5\times 10^3kg/m^3}{\sqrt{1-\dfrac{(0.9c)^2}{c^2}}} \approx 1.03\times 10^4 kg/m^3$$

Answer. $1.03\times 10^4 kg/m^3$.