If I have a point $(x_0,y_0,z_0,w_0) \in \mathbb{R}^4$ and I rotate it within the $xy$-plane ($0<\alpha<\pi$) and $zw$-plane ($0<\beta<\pi$), how can I determine the length of the arc traced by the rotation of $(x_0,y_0,z_0,w_0)$? I figure it probably involves integrating across a parametrization of the arc, but I'm not sure how to parametrize it in this situation.
Update- I am going to try to calculate a line element for hyperspherical coordinates. Now the struggle is going to be determining what parametrizations coordinate with rotations through which orthogonal planes. The angles in the coordinate system, however, are defined differently than the angles $\alpha$ and $\beta$. Any assistance or feed back would be greatly appreciated.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\vx}{\Vec{x}}\newcommand{\vz}{\Vec{z}}$The orbit of a vector $$ (x_{0}, y_{0}, z_{0}, w_{0}) = (\vx, \vz) $$ under the set of rotations in the $(x, y)$- and $(z, w)$-planes in $\Reals^{4}$ is a (Riemannian) product of circles with respective radii $$ \|\vx\| = \|(x_{0}, y_{0})\| = \sqrt{x_{0}^{2} + y_{0}^{2}},\qquad \|\vz\| = \|(z_{0}, w_{0})\| = \sqrt{z_{0}^{2} + w_{0}^{2}}, $$ namely a flat torus obtained by identifying opposite edges of the rectangle $$ \bigl[0, 2\pi \|\vx\|\bigr] \times \bigl[0, 2\pi \|\vz\|\bigr]. $$
Rotation by $\alpha$ in the $(x, y)$-plane moves $(\vx, \vz)$ a geodesic distance $\alpha \|\vx\|$; rotation by $\beta$ in the $(z, w)$-plane moves $(\vx, \vz)$ a geodesic distance $\beta \|\vz\|$. These displacements are perpendicular in the torus orbit, so their net geodesic distance is given by the Pythagorean theorem: $$ \sqrt{\alpha^{2} \|\vx\|^{2} + \beta^{2} \|\vz\|^{2}} = \sqrt{\alpha^{2}(x_{0}^{2} + y_{0}^{2}) + \beta^{2}(z_{0}^{2} + w_{0}^{2})}. $$
To confirm, note that the one-parameter subgroup $$ \gamma(t) = \left[\begin{array}{@{}c@{}} x_{0}\cos(\alpha t) - y_{0}\sin(\alpha t) \\ y_{0}\cos(\alpha t) + x_{0}\sin(\alpha t) \\ z_{0}\cos(\beta t) - w_{0}\sin(\beta t) \\ w_{0}\cos(\beta t) + z_{0}\sin(\beta t) \end{array} \right],\quad 0 \leq t \leq 1, $$ carries $(\vx, \vz)$ to its image under the combined rotation, and $$ \|\gamma'(t)\| = \sqrt{\alpha^{2}(x_{0}^{2} + y_{0}^{2}) + \beta^{2}(z_{0}^{2} + w_{0}^{2})}. $$