Length contraction in special relativity- from a sphere to an ellipse

Question

I have an object which is of spherical shape, with equation:

$$x_0^2+y_0^2+z_0^2=R^2 \ ,$$

in a frame of reference $S'$ moving with the object. $S'$ is moving with a velocity $v$ with respect to a stationary observer, with a frame of reference $S$ and coordinates ($x,y,z$).

If I want to write the equation specifying the geometry of the object according to $S$ should it be :

$$\left(x_0 \sqrt{1-\left(v^2/c^2\right)}\right)^2 +y_0^2 +z_0^2=R^2 $$

Or

$$ \frac {x^2} {1-(v^2/c^2)} +y^2+z^2=R^2~? $$

Answer 1

You're using coordinates named $(x_0,y_0,z_0)$ for both the primed and unprimed frames in the first two equations. Never do that or you will become hopelessly confused. I'm going to use $(t',x',y',z')$ as coordinates in $S'$ and $(t,x,y,z)$ as coordinates in $S$. The sphere is then given by

$$x'^2+y'^2+z'^2=R^2.$$

Note that this is not a sphere in $\mathbb R^3$, but a worldtube in spacetime, in the shape of a cylinder with a spherical cross-section, since the sphere exists at all times $t'$.

To express the same worldtube in other coordinates, just use the Lorentz transformation:

$$t' = γ(t-vx/c^2),\quad x' = γ(x-vt),\quad y'=y,\quad z'=z$$

which gives you

$$(γ(x-vt))^2+y^2+z^2=R^2.$$

At $t=0$, this is $(γx)^2+y^2+z^2=R^2$. At any other fixed $t$, it's the same shape but with its center at $x=vt$ instead of $x=0$.

There is no need to memorize a separate length contraction formula. You can always just use the Lorentz transformation and do the algebra.

Answer 2

In the frame S, the object is smaller by a factor of $$\gamma=\frac{1}{\sqrt{1-(v/c)^2}}>1$$ In the frame S', $x_0=R,y_0=0,z_0=0$ is a solution. We want the obect to be shorter in the frame S, so $x_0=R/\gamma,y_0=0,z_0=0$ should be a solution. So the equation should be: $$ (\gamma x)^2+y^2+z^2=R^2 $$ Or: $$ \frac{x^2}{1-(v/c)^2}+y^2+z^2=R^2 $$