# Light Cone Tessellations

William R. Olson
bolson@pimeson.com

The tetrahedron is the simplest three-dimensional object. It cannot be used to form a closed lattice in physical space since the angle between two faces is not an integer fraction of a circle. In a light cone the tetrahedron can be used to form a closed lattice since exactly six tetrahedrons fit with one common edge. The vertices of these tetrahedrons are events in space-time that are all contained in a single light cone. The structure of the lattice is the same for all observers since the edges are identical space-like intervals. Connecting a subset of the events in the original lattice can form larger sized lattices. For example the vertices on either side of a triangle have a space-like separation that is sqrt(3) times larger than the separation of events in the original lattice. By connecting all events with this larger separation a new lattice can be formed. The lattice formed in this way is not the same shape as the original lattice since some of the events that would be in the complete larger lattice are not in the original lattice.

In a light cone centered at the origin the coordinates of an events must satisfy the equation X2 + Y2 + Z2 = R2 = T2 where the speed of light is unity. The square of the interval between two events is S2 = (X1 – X2)2 + (Y1 – Y2)2 + (Z1 – Z2)2 – (R1 – R2)2 = 2 * (R1* R2 – X1 * X2 – Y1 * Y2 – Z1 * Z2) = 2 * R1 * R2 * (1 – COS(A)) = 4 * R1 * R2 * SIN2(A/2) where A is the angle between the vectors from the origin to the two events. This shows that the interval between any two events in a light cone is either space-like (S2 > 0) or zero. It is zero only if the angle between the vectors is zero. A lattice of tetrahedrons can be constructed by starting with a tetrahedron consisting of the following four events.

 X11-1-1 Y1-11-1 Z1-1-11 R = Tsqrt(3)sqrt(3)sqrt(3)sqrt(3) event #1event #2event #3event #4

The square of the interval between any two of these events is 2 * (3 – 1+ 1+ 1) = 8.

A new tetrahedron can be formed by replacing one of the events with a new event by simply adding the event coordinates in the three events that do not change and subtracting the event coordinates in the event that changes. As an example a new first event is

-2 -2 -2 2*sqrt(3) new event #1

The square of the interval between this new event and any of the three other old events is 2 * (6 + 2 – 2 – 2) = 8. This forms a new space-like tetrahedron. From this new tetrahedron one can form a new second event with the following results.

-5 -1 -1 3*sqrt(3) new event #2

The square of the interval between this new event and the three old events is also 8. The first event can be changed again producing the following event.

-5 1 1 3*sqrt(3) new event #1

The second event can be change again producing the following event.

-2 2 2 2*sqrt(3) new event #2

The first event can be change again producing the following event, which is the original first event.

1 1 1 sqrt(3) new event #1

The second event can be change again producing the following event, which is the original second event.

1 -1 -1 sqrt(3) new event #2

The result of these six steps is to come back to the original tetrahedron. Hence the lattice formed in this way is closed and by stepping in other directions produces a tetrahedral tessellation of a light cone. The paths through the lattice can be described by assigning numbers to the four vertices of a tetrahedron. A sequence of numbers can be used to represent the vertices that is change at each step. My convention is to write the sequence from right to left. The path 212121 would represent the six steps in the sequence that reproduced the original tetrahedron. The first step is on the right and changed the first event.

The square of the interval between the first event and the first new event is 2 * (6 + 2 + 2 + 2) = 24 = 3 * 8. This interval is sqrt(3) times the original interval. This new interval can be used to form a larger lattice of the same light cone using a subset of the original lattice. A triangle with this interval can be formed by connecting the three #1 events in the path described above. This triangle can be used as the base of two new tetrahedrons.

 1-2-5X 1-21Y 1-21Z sqrt(3)2*sqrt(3)3*sqrt(3)R new event #1, old event #1new event #2, old second event #1new event #3, old third event #1new event #4 to form a new larger tetrahedron

This new event must have the same interval to each of the first three events. There are two solutions.

 -3-3 3-3 -33 3*sqrt(3)3*sqrt(3) one possible event #4second possible event #4

These two new events are along the same vector from the origin as the original events #3 and #4 with coordinates that are three times larger. These events are not in the original tessellation so will not be included in the S2 = 3*8 tessellation. Only events that are in the original tessellation will be included in the larger tessellations. There are many possible larger tessellations and they all have an S2 ratio to the original tessellation of the form N2 + N*M + M2 where N and M are positive integers not both zero. This is the same relation that is true for the square of the cord lengths in the A2 lattice (triangular lattice) of the plane. This is not all that surprising since the collection of all events that form a tetrahedron with any event in the tetrahedral tessellation forms a lattice of triangles with six around each vertex the same as the A2 lattice. The only difference is the triangles in the light cone tessellation are not in a plane.

In order to draw the tessellation the events must be projected onto a plane. One set of coordinates for the starting tetrahedron that can be used to project tessellation onto a plane is as follows.

 X00R*COS(A)R*COS(A+60) Y00R*SIN(A)R*SIN(A+60) Z-R/ZR*ZR*Z-R/ZR*Z-R/Z R = TR/ZR*ZR*Z+R/ZR*Z+R/Z event #1event #2event #3event #4

The square of the interval between any two of these events is S2 = R2. The shape of the projection depends only on the ratios between Z coordinates so the Z axis will be scaled such the R * Z = 1. In the limit as Z approaches infinity the position coordinates approach the following.

 X00R*COS(A)R*COS(A+60) Y00R*SIN(A)R*SIN(A+60) Z0111 event #1event #2event #3event #4

The time coordinate was not included since it is not required for a projection of these events onto a plane. The X and Y coordinates will be divided by the Z coordinate to project them all onto the same plane. This will work for all events except the original event #1 since Z = 0 for this event. The angle A is used to orient the starting triangle on the page. These coordinates transform the same as the original coordinates. Adding three events and subtracting a fourth event forms a new tetrahedron. Steps that change any event except the first event will leave Z = 1 and produce a triangular lattice of the plane. A step that changes the first event will produce an event with Z = 3 in the center of the triangle in which the step was made. In general, the Z coordinate will be proportional to the square of the interval between the new event and the original event #1.

There is a simple way to determine the vertices that are not included in the larger tessellations. For S2 = 3 the missing vertices will all be the same event number. For S2 = 4 the determination is a little more complicated. If the starting tetrahedron contains the missing vertex then that vertex will be assigned a value of zero. The other three vertices will be assigned a value of one. As one walks through the lattice the value of the numbers assigned to the vertices will be propagated the same as the coordinates of the vertices with one exception. The values assigned to the vertices will be computed modulo 4. Any vertex that has a zero value will not be included in the tessellation. For the tessellations with S2 = 7 a similar method is used. In this case two values are assigned to each vertex. The first value is the same as for the S2 = 4 tessellation, zero for the missing vertex and one for the other three vertices. The second value will be zero for the missing vertex and one, three and four for the other three vertices. These values will be propagated modulo 7 and the missing vertices will have zero for both values. By exchanging any two of the second non-zero values in the starting tetrahedron a second S2 = 7 tessellation will be formed. An exchange of a second pair of non-zero values will reproduce the first tessellation. This method is the same for all tessellations with S2 > 4 that are prime numbers. The choice for the second set of starting values depends on the value of S2. For S2 values that are not primes the missing vertices are determined by removing all vertices in the tessellations with S2 = prime factors. For example in the S2 = 84 tessellation the vertices missing in the S2 = 3, S2 = 4 and one of the S2 = 7 tessellations will be missing in this tessellation. There are some special cases that can be determined in a different way. The S2 = 49 tessellation contains two factors of 7. There are three possible combinations. Both of the S2 = 7 tessellations can be the same in which case the S2 = 49 tessellation is the same as the S2 = 7 tessellation. The other case is that the two factors are different S2 = 7 tessellations. In this case a single value can be used similar to the S2 = 4 tessellation. This value is calculated modulo 7. Another special case occurs for tessellations that are the square of prime numbers that do not form tessellations such as S2 = 4 or S2 = 25. There are no S2 = 2 or S2 = 5 tessellations. The missing vertices in these tessellations are calculated the same as the S2 = 4 tessellation except modulo S2. The following is a list of possible starting values for the prime tessellations with S2 < 100.

S2 Second set of values for missing vertex #2

 713193137436167737997 11111111111 00000000000 37122127314852576379 49132528364957646784

The colors for these drawings was chosen by assigning black to the first vertex, red to the second vertex, green to the third vertex and blue to the fourth vertex. The lines connected to a black vertex take on the color of the vertex at the other end of the line. The lines connecting two colored vertices take on the combined colors of yellow, magenta and cyan. The lines are merged by using only the common RGB colors in both lines. This causes the vertices to take on the colors assigned to them.