Coordinate Conversion $(3$-Sphere to $4D$ Cartesian)

Question

I understand how to convert from a $3$-sphere to Cartesian, but can someone help me convert from $4$D cartesian to $4$D spherical

Thanks

Answer 1

Let $x_1, x_2, x_3, x_4$ be your spatial cartesian coordinates. Let $r,\theta_1,\theta_2, \theta_3$ be the parameters for the spherical coordinates.

Below are the transformations from 4-D spherical to 4-D cartesian. $$x_1=r\cos{\theta_1}$$ $$x_2=r\sin{\theta_1}\cos{\theta_2}$$ $$x_3=r\sin{\theta_1}\sin{\theta_2}\cos{\theta_3}$$ $$x_4=r\sin{\theta_1}\sin{\theta_2}\sin{\theta_3}$$

After some rearranging, we derive the following inverse transformations (4-D cartesian to 4-D spherical):

$$r=\sqrt{{x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2}$$

$$\theta_1=\arccos \frac{x_1}{\sqrt{{x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2}}$$

$$\theta_2=\arccos \frac{x_2}{\sqrt{{x_2}^2+{x_3}^2+{x_4}^2}}$$

$$\theta_3=\begin{cases} \arccos \frac{x_3}{\sqrt{{x_3}^2+{x_4}^2}} & x_4\geq 0 \\ 2 \pi - \arccos \frac{x_3}{\sqrt{{x_3}^2+{x_4}^2}} & x_4 < 0 \end{cases}$$

If you would like a generalization to n-space, see the "Spherical Coordinates" section.