The densest possible packing of four spheres in flat 3D space is given by the tetrahedral arrangement, but this arrangement can't be used to fill space because tetrahedrons don't quite "fit together". This is because if five tetrahedrons share a common edge, they take up only 97.96% of the total 2pi arc about that axis, leaving a small gap. However, if the 3D space has positive curvature, the circumference of a circle will be less for a given radius than it would be in flat space. Thus, there must be a closed unbounded 3D space of constant positive curvature in which five tetrahedrons fit together exactly. This is really just another way of describing a 4D Platonic Solid, i.e., a 3-sphere (embedded in 4D space) whose "faces" are regular tetrahedrons. Consider a flat Euclidean 4D space where each point has Cartesian coordinates [w,x,y,z]. A 3-sphere of radius R centered on the origin of this coordinate system is the locus of points such that w^2 + x^2 + y^2 + z^2 = R^2 This locus has the metrical properties of a 3-dimensional space with constant positive curvature. Suppose R is such that we can fit five unit tetrahedrons (i.e., tetrahedrons with unit edge lengths) together without leaving a gap. Looking straight down on their common edge the configuration of vertices would look (roughly) like this 5*--------------*7 / \ / \ / \ / \ / \ / \ / \ / \ / 1/*\2 \ / / | \ \ 3 \ / | \ / 6 \ | / \ | / \ | / \ |/ 4 If we arbitrarily set point 1 at the coordinates [R,0,0,0] then we can easily determine the four vertices of one tetrahedron as follows point 1: [R,0,0,0] point 2: [a,r1,0,0] point 3: [a,b,r2,0] point 4: [a,b,c,r3] where a = R - 1/2R r1 = sqrt(1 - 1/4R^2) b = r1 - 1/2r1 r2 = sqrt(1 - 1/4r1^2) c = r2 - 1/2r2 r3 = sqrt(1 - 1/4r2^2) We can also easily determine point 5, because it is equidistant from points 1,2,3 and opposite point 4. Thus we have point 5: [a,b,c,-r3] This leaves only the two points 6 and 7. We can determine point 6 from the fact that it is equidistant from 1,2,4 and opposite 3. Similarly we can find point 7 equidistant from 1,2,5 and opposite 3. Then, if the tetrahedrons are to fit snugly together, we should be able to determine point 7 again, this time being equidistant from 1,2,6 and opposite 4, and it should coincide with the coordinates for point 7 determined previously. With "edge lengths" of 1, it turns out that the five tetrahedrons fit together snugly only if R = (1+sqrt(5))/2, the so-called "golden proportion". This value defines the radius of the 3-sphere embedded in 4D space, and it also determines the intrinsic curvature of the 3D space. With this value of R the coordinates of the seven vertices are point 1: [r0,0,0,0] point 2: [a,r1,0,0] point 3: [a,b,r2,0] point 4: [a,b,c,r3] point 5: [a,b,c,-r3] point 6: [a,b,d,e] point 7: [a,b,d,-e] where 1 + sqrt(5) sqrt(5+2sqrt(5)) r0 = ----------- r1 = ---------------- 2 1 + sqrt(5) / 15+7sqrt(5) \ / 20+9sqrt(5) \ r2 = sqrt( ------------ ) r4 = sqrt( ------------ ) \ 20+10sqrt(5)/ \ 30+14sqrt(5)/ 2+sqrt(5) sqrt(5+2sqrt(5)) a = --------- b = ---------------- e = 1/2 1+sqrt(5) 5+sqrt(5) 5+2sqrt(5) 1+sqrt(5) c = --------------------- d = - ----------------- sqrt(1230+550sqrt(5)) 2sqrt(10-2sqrt(5)) By the way, the edges of the tetrahedrons are all of length 1 in the 4D embedding space, but the geodesic distances between adjascent vertices within the 3D space is slightly longer. Letting phi denote the golden ratio, the 3D geodesic edges of the tetrahedron are 2phi*invsin(1/2phi). Anyway, this shows that five unit tetrahedrons fit snugly together in a 3D space of constant positive curvature corresponding to the a 3-sphere of radius (1+sqrt(5))/2 embedded in Euclidean 4D space. However, I don't think it necessarily follows that if we continue adding tetrahedrons inside the closed 3D space they will necessarily fill up the space exactly. I suspect they will, and that it will take 120 tetrahedrons to fill up the space, but I don't see how to prove this without actually constructing all of those other tetrhadrons. (I wonder if there is a simpler way of proving global closure. For example, can we argue that twelve sets of 5 tetrahedrons will fit together like the 12 faces of a dodecahedron, giving 60 cells, and then the "interior" of this dodecahedron gives 60 more?) Well, having said all this, I arrive at my question...which actually has nothing to do with spaces of positive curvature! I'm wondering about filling a 3D infinite space of constant NEGATIVE curvature with unit tetrahedrons. In this case we don't have the comfort of a 4D Euclidean space in which our curved 3D space can be embededed. Of course, if we try to fit five tetrahedrons together in a negatively curved 3D space the "gap" will be even bigger than it is in flat space, but I'm thinking if we give the space sufficient negative curvature we will open up the gap far enough that a 6th tetrahedron will fit snugly into place. If so, then this particular infinite 3D space of constant negative curvature can be filled with unit tetrahedrons in a pure hexagonal pattern. For a related discussion, see Packing Universes In Spacetime.

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