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

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

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

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.

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