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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.

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    $\begingroup$ 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$ $\endgroup$ Oct 18, 2022 at 11:08
  • $\begingroup$ @ZadeJohnston How do we determine that volume of ellipsoid?An explanation rather than just the formula will be very helpful. $\endgroup$
    – madness
    Oct 18, 2022 at 12:29
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    $\begingroup$ 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 $\endgroup$
    – D. Halsey
    Oct 18, 2022 at 19:14

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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.

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