From a182b663016b4301b52260d1480899cf44505170 Mon Sep 17 00:00:00 2001 From: Glen Whitney Date: Wed, 18 Sep 2024 20:11:20 -0700 Subject: [PATCH] doc: More elaboration of plane coordinates in inversive notes --- notes/inversive.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/notes/inversive.md b/notes/inversive.md index 7867268..7a5318d 100644 --- a/notes/inversive.md +++ b/notes/inversive.md @@ -7,8 +7,8 @@ These coordinates are of form $I=(c, r, x, y, z)$ where we think of $c$ as the c | Entity or Relationship | Representation | Comments/questions | | ------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | | Sphere s with radius r>0 centered on P = (x,y,z) | $I_s = (1/c, 1/r, x/r, y/r, z/r)$ satisfying $Q(I_s,I_s) = -1$, i.e., $c = r/(\|P\|^2 - r^2)$. | Can also write $I_s = (\|P\|^2/r - r, 1/r, x/r, y/r, z/r)$ -- so there is no trouble if $\|P\| = r$; we just get first coordinate to be 0. | -| Plane p with unit normal $(x,y,z)$ through the (Euclidean) point $(sx,sy,sz)$ | $I_p = (-2s, 0, -x, -y, -z)$ | Note $Q(I_p, I_p)$ is still −1. This plane is at distance $s$ from the origin. | -| Point P with Euclidean coordinates (x,y,z) | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$.  Because of this we might choose  some other scaling of the inversive coordinates, say $(\| | +| Plane p with unit normal $(x,y,z)$ through the (Euclidean) point $(sx,sy,sz)$ | $I_p = (-2s, 0, -x, -y, -z)$ | Note $Q(I_p, I_p)$ is still −1. This plane is at distance $s$ from the origin. And note that these are _oriented_ planes. For example, $(-2, 0, -1/\sqrt3, -1/\sqrt3, -1/\sqrt3)$ and $(2, 0, 1/\sqrt3, 1/\sqrt3, 1/\sqrt3)$ represent planes that coincide in space, just the former has normal pointing away from the origin and the latter pointing toward it. | +| Point P with Euclidean coordinates (x,y,z) | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$. Because of this we might choose some other scaling of the inversive coordinates, say ... | | ∞, the "point at infinity" | $I_\infty = (1,0,0,0,0)$ | The only solution to $Q(I,I) = 0$ not covered by the above case. | | P lies on sphere or plane given by I | $Q(I_P, I) = 0$ | | | Sphere/planes represented by I and J are tangent | If $I$ and $J$ have the same orientation where they touch, $Q(I,J) = -1$. If they have opposing orientations, $Q(I,J) = 1$. | For example, the $xy$ plane with normal $-e_z$, represented by $(0,0,0,0,1)$, is tangent with matching orientation to the unit sphere centered at $(0,0,1)$ with outward normals, represented by $(0,1,0,0,1)$. Accordingly, their $Q$-product is −1. |