The Euclidean space $\mathbb{R}^{n+1}$ endowed with the pseudo inner-product $$ \langle x, y\rangle=-x_0y_0+x_1y_1+x_2y_3+\ldots+x_n y_n$$ is called Lorenz space and denoted by $\mathbb{L}^{n+1}$.
The well-known hyperboloid model of the hyperbolic space is the $n$-dimensional hypersurface $$ \mathbb{H}^{n}:=\{x \in \mathbb{L}^{n+1}; \,\, \langle x, x \rangle =-1 \,\, \mbox{and}\,\, x_0>0\}.$$
My question is:
What is the most simple isometry between the hyperboloid model and the upper-half space of $\mathbb{R}^{n}$ $$H^{n}:=\{x=(x_1, \ldots, x_{x}) \in \mathbb{R}^{n}; \,\,x_{n}>0\} $$ endowed with the conformal metric $g_{ij}(x)=\dfrac{\delta_{ij}}{x_{n}^{2}},$ where $1 \le i,j \le n$?