# 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.

• Would it not be the volume of an ellipsoid with radius $R/\gamma$ along thei direction of motion, and $R$ along the two perpendicular axes? I.e. $V=4\pi R^3 /3\gamma$ Oct 18, 2022 at 11:08
• @ZadeJohnston How do we determine that volume of ellipsoid?An explanation rather than just the formula will be very helpful. Oct 18, 2022 at 12:29
• Are you aware that a sphere does not show the length contraction visually because of something called the Terrell-Penrose effect? andrewyork.net/Math/TerrellRotation_York.html Oct 18, 2022 at 19:14

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.