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Start with the four-dimensional Minkowski metric $d{s}^2=-d{t}^2+d{r}^2+r^{2}d{\Omega }^2$, where $r\ge 0$. Choose an arbitrary length $L>0$, and use the retarded and advanced null coordinates $u=t-r$, $v=t+r$. For the Minkowski conformal compactification, setBoth $p,q$ lie between $-\pi /2$ and $\pi /2$. Since $v\ge u$, we have $R\ge 0$; the remaining inequalities are $|T|+R<\pi$. Moreover,Multiply by the square of the conformal factor ${\Omega }_c=2\cos p\cos q/L$. The resulting metric is regular on the appropriate boundary pieces and preserves the null directions. Suppressing the angular two-spheres gives a triangular Penrose diagram with radial null geodesics at $45$ degrees.

The line $R=0$ is the ordinary timelike centre $r=0$. The upper sloping edge $T+R=\pi$ is future null infinity, reached with $v\to +\infty$ and finite $u$; the lower sloping edge $T-R=-\pi$ is past null infinity, reached with $u\to -\infty$ and finite $v$. The vertices $(T,R)=(\pi ,0)$ and $(-\pi ,0)$ are future and past timelike infinity, denoted $i^{+}$ and $i^{-}$. The vertex $(0,\pi )$ is spacelike infinity, $i^0$. These are limiting endpoints in the conformal completion, rather than ordinary physical events. In particular, finite diagram coordinates at null infinity do not imply finite physical affine parameter.

The four-dimensional radial diagram is the triangle $R\ge 0$, $|T|+R\le \pi$. If one instead draws two-dimensional Minkowski spacetime with a signed Cartesian spatial coordinate, the diagram is the full diamond. The centre is a boundary of the radial quotient, not a boundary of the physical four-dimensional Minkowski spacetime.

Spacelike infinity is the ideal endpoint of spatial directions in the Minkowski conformal compactification, represented by $(T,R)=(0,\pi )$. Unlike null infinity, it is a vertex of the radial Penrose diagram.