Volume of a sphere seemed when moving close to the speed of light Let earth be a sphere of radius $R$. A person is moving with a relativistic velocity $v$ along one of the diameters of earth. What will be the volume measured by him?
Here as far as i know,the length contraction will happen only along the diameter of his trajectory. So the shape becomes a bit of ellipsoid. If the contracted radius along that particular direction is $R'$,how can we determine the volume of the sphere?Surely it can't be $\frac{4}{3}\pi R'^3$ since $R'$ is changed in only one direction.
 A: As the sphere is only deformed along one direction due to the length contraction, it will appear as an ellipsoid to the observer. This ellipsoid has a semi-axis of $R/\gamma$ along the direction of motion (where $\gamma$ is the Lorentz factor), and $R$ along the two perpendicular directions.
Using the equation for the volume of an ellipsoid$^\boldsymbol{\star}$, $V=4\pi abc/3$ for the three semi-axes of the ellipsoid, we obtain a volume of
$$
V'= \frac{4\pi}{3}\frac{R^3}{\gamma}=\frac{V_0}{\gamma},
$$
where $V_0$ is the volume of the sphere measured in its rest frame. In general, the volume of a moving object with volume $V_0$ in its rest frame will be $V_0/\gamma$.

$\boldsymbol{\star}$ Deriving Volume of Ellipsoid: The article I linked doesn't show how obtain the formula for the ellipsoid volume, so I thought I'd give a brief sketch.
The ellipsoid is given by the equation
$$
\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 + \left(\frac{z}{c}\right)^2=1,
$$
where $a, b,c$ are the semi-axes. Cutting the ellipsoid in a plane parallel to the $yz$-plane, we get the ellipse
$$
\left(\frac{y}{b\sqrt{1-x^2/a^2}}\right)^2+\left(\frac{z}{c\sqrt{1-x^2/a^2}}\right)^2=1
$$
which has area $A(x)=\pi bc(1-x^2/a^2)$. Integrating this along $x$ from $-a\to a$, we obtain the required volume.
