From 2c1a42e251e250e798054094588cd41060859d84 Mon Sep 17 00:00:00 2001 From: Aaron Fenyes Date: Wed, 25 Sep 2024 23:07:52 -0700 Subject: [PATCH] doc: Clarify orientation convention --- notes/inversive.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/notes/inversive.md b/notes/inversive.md index 31eb898..6af980b 100644 --- a/notes/inversive.md +++ b/notes/inversive.md @@ -6,12 +6,12 @@ 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. 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 ... | +| 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., $c = r/(\|P\|^2 - r^2)$. | Note that $1/c = \|P\|^2/r - r$, so there is no trouble if $\|P\| = r$; we just get first coordinate to be 0. Using the point representation $I_P$ from below, let's orient the sphere so that its normals point into the "positive side," where $Q(I_P, I_s) > 0$. The vector $I_s$ then represents a sphere with outward normals, while $-I_s$ represents one with inward normals. | +| Plane p with unit normal $(x,y,z)$ through the (Euclidean) point $(sx,sy,sz)$ | $I_p = (-2s, 0, -x, -y, -z)$ | Note that $Q(I_p, I_p)$ is still $−1$. The parameter $s$ is the plane's distance from the origin. We orient planes using the same convention we use for spheres. 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, which have their normals pointing away from and toward the origin, respectively. | +| Point P with Euclidean coordinates (x,y,z) | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$. This gives us the freedom to choose a different normalization. For example, we could scale the representation shown here by $(\|P\|^2+1)^{-1}$, putting it on the sphere where the light cone intersects the plane where the first two coordinates sum to $1$. | | ∞, 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 | $Q(I_P, I) = 0$ | Actually also works if $I$ is the coordinates of a point, in which case "lies on" simply means "coincides with". | -| 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. | +| 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 $−1$. | | Sphere/planes represented by I and J intersect (respectively, don't intersect) | $\|Q(I,J)\| < (\text{resp. }>)\; 1$ | Follows from the angle formula, at least conceptually. | | P is center of sphere represented by I | Well, $Q(I_P, I)$ comes out to be $(\|P\|^2/r - r + \|P\|^2/r)/2 - \|P\|^2/r$ or just $-r/2$ . | Is it if and only if ?   No this probably doesn't work because center is not conformal quantity. | | Distance between P and R is d | $Q(I_P, I_R) = d^2/2$ | |