Jacobian of the Kelvin transform

Question

The Kelvin transform of the circle in $\mathbb{R}^n$ with centre $\textbf{u}$ and radius $r$ is defined by $$\textbf{y} \mapsto \textbf{u} + r^2|\textbf{y} - \textbf{u}|^{-2}(\textbf{y}-\textbf{u}).$$ If $\textbf{u} = \textbf{0}$ then I know that the Jacobian of this transform is $$-\frac{r^{2n}}{|\textbf{y}|^{2n}}$$ but am unsure about what the Jacobian is when $\textbf{u} \neq \textbf{0}$.

Answer 1

According to http://jan.ucc.nau.edu/~ns46/238/jacobi.pdf, theorem 16, the Jacobian of a chain of transformations is the product of Jacobians of transformations. The Kelvin transform $\mathbf{y} \mapsto \mathbf{u} + r^2 | y - u |^{-2} (\mathbf{y} - \mathbf{u})$ is the composition of transformations $\mathbf{y} \mapsto \mathbf{y} - \mathbf{u}$, $\mathbf{x} \mapsto r^2 |x|^{-2} \mathbf{x}$ and $\mathbf{z} \mapsto \mathbf{z} + \mathbf{u}$. You wrote the Jacobian for the middle one, and Jacobians of two others are identity matrices. Thus, the Jacobian for the entire transform is $$-\frac{r^{2n}}{|y - u|^{2n}}.$$