The ADOMAH Periodic Table

This is a profound and powerful way to visualize the periodic table of the elements –  due to the fact that it mirrors 3-D physical reality:

Geometry of the Periodic Table.

It can be observed that four blocks of ADOMAH Periodic Table, that is built strictly in accordance with the quantum numbers n, l, ml and ms, follow quite curious rule: perimeters of all blocks are equal to 18 units (if elements are placed in rectangular boxes so that any given pair of elements fits in one unit square, see“Description” page): Unit Square or 

s-block is 1 unit high and 8 units long (or wide) (1+8 = 9, is half of the perimeter);

p-block is 3 units high and 6 units long (3+6 = 9);

d-block is 5 units high and 4 units long (5+4 = 9);

f-block is 7 units high and 2 units long (7 +2 = 9);

Therefore, perimeter of each block: P=2×9=18 units.

Those are only four possible rectangles that could have perimeter of 18, if only natural numbers are used. What can it possibly mean? Can it be a coincidence?

Apparently not, this is not a coincidence. In 3D world there is one geometric shape that, if sliced in a certain way, would produce rectangles with the same proportions, orientation, alignment and order as spdf blocks of ADOMAH Periodic Table. This shape is Regular Tetrahedron!

If a regular tetrahedron with edge E is intersected by a plane that is parallel to two opposite edges, cross section will always be a rectangle or a square with perimeter P=2E.

Proportions of fdps blocks of ADOMAH PT, as well as their sequence, alignment and orientation, suggest only one possible conclusion: all four blocks of the Periodic Table are consecutive slices of Regular Tetrahedron, with edges equal to 9 units, produced by the planes spaced 2 units apart, as measured along the edges.

There is another indication that the Periodic System has something to do with regular tetrahedron: Atomic Number of every other alkaline earth element (Be, Ca, Ba, Ubn…) corresponds to every second Tetrahedral Number: 1, 4, 10, 20, 3556, 84120… Remaining alkaline earth elements (Mg, Sr and Ra) have atomic numbers that are equidistant between the Tetrahedral Numbers shown above in bold: 12=(4+20)/2, 38=(20+56)/2, 88=(56+120)/2. See more on this topic below.

Generally, a Periodic System of any size can be described in terms of a regular tetrahedron with edge E:

1) number of the periods NOP=E-1;

2) number of values of the quantum number ‘n’n max=E-1

3) number of blocks representing subshells (number of values of ‘l’): NOS=ceil(0.5(E-1)), where ceil stands for “ceiling function”, that means “rounded to a higher integer if the result has fractional part” (in case of even ‘E’).

4) maximum value of quantum number ‘l’l max=ceil(0.5(E-3))

5) dimensions of the subshell blocks corresponding to l =0,1…, l max:

as measured along the periods a=4l +2 elements; as measured along the groups b=E-(2l +1).

6) number of elements in each block corresponding to quantum number l =0,1…, l max:


7) length of the periods: LP=2(l’ +1)2 ( here, l’ is the maximum value of l for each period).

8) number of the elements (NOE) in the periodic system that corresponds to the tetrahedron with edge ‘E’:

, where lmax=ceil(0.5(E-3)) as defined above.


2 responses

  1. Jess Tauber

    I appear to have been the first to discover this configuration, back in 1979, as a young undergrad chemistry student (and posted to a blog about it back in 2001, the earliest surviving post so far as I can tell). Later versions of the Adomah, though, are more advanced than my model, which was identical to the above. In other illustrations, Valery Tsimmerman uses close packed spheres to build up the tetrahedron, which is a very useful innovation. He also went much more deeply into the mathematics behind the system than I did.

    We’ve been working together since last year. Some months ago I noticed that every other alkaline earth atomic number was identical to every other number in the Pascal Triangle tetrahedral diagonal, and from that he was able to work out that this also made the old system of triads fathomable.

    I’ve been also working on extentions of the Adomah system (I never named it myself) which utilize all spheres/cells in the tetrahedron, rather than doubling them up and leaving half the spaces empty as the old model does. If one takes a stack of layers of spheres flat on a surface (so that each layer contains the triangular diagonal numbers from the Pascal triangle), every pair of successive layers gives a square numbers, starting arbitrarily from zero (0+1=1, 1+3=4, 3+6=9, 6+10=16, and so on). It turns out that periodic table duals (two periods of the same length) also contain square numbers, even ones (BUT using the left-step Janet periodic table, where all periods are shifted rightward two steps and so ending with the alkaline earths) (2+2=4, 8+8=16, 18+18=36, 32+32=64, etc.).

    Unfortunately, simply trying to transpose these numbers onto a layered tetrahedron lying flat on a surface won’t work, as successive layers don’t contain the same numbers of spheres, while dual periods DO. HOWEVER, if one takes dual tetrahedral layers, opening up an angle between layers using an edge as a pivot, you end up with a symmetrical ‘taco shell’ or sleeve that does contain the right number of spheres for dual periods. Four of these taco shells do the trick, the first containing four spheres, then next 16, the next 36, and the last 64. Each of these shells contains two periods each with angled, but concentric rings for each quantum number l. Thus s is in the center, surrounded by p, surrounded again by d, and finally further out by f. When the taco shells are stacked from smallest to largest, the s elements (four mini-tetrahedra) will all form an axis down the center, jacketed by three stacked p ‘rings’, with two d rings and one f ring working outwards.

    This system has a number of advantages- for example, though it breaks up the continuity of successive periods since one side of the ring contains the odd, and the other side the even member of a dual period pairing, the vertical stacking directly connects elements in what is known as ‘secondary periodicity’, a known effect that the traditional table has no way of representing.

    In addition, the tetrahedron is able to directly represent left-justified element connections (where the first few members of each period are known to behave similarly even though they are in different blocks) as well as right-justified element connections (where the last few elements behave similarly, even though in different blocks). The traditional table cannot even begin to handle this.

    More unusual connectivities such as diagonals and knight’s move relations come not from quantum number resemblances, but from relativistic effects and such things as spin-spin coupling, and have no obvious place in either the tetrahedral or traditional renderings. I’m working to try to figure out a way to deal with these geometrically by going beyond the tetrahedron.

    Still, the new all-sphere tetrahedral representation (which I call T3) breaks up the Mendeleev Line into periods, destroying the continuity, as do the vast majority of alternative depictions of the periodic relation.

    However, there is at least one new version, the T4, which actually folds the entire continuous Mendeleev Line into a tetrahedral configuration, maintining each period as a sort of ’tiling’, much like the folding of proteins from domain structures. I’m still exploring this. Unlike the T3, which though 3 dimensional has a rotational axis, T4 is, from Buckminster Fuller’s perspective on tetrahedra, within its own system actually 4 dimensional, and so has unusual properties.

    May 1, 2010 at 8:52 pm

  2. It is all nice, but Paul Dirac has demonstrated that spin quantum number, unlike all other quantum numbers that are integers, comes in halves, that is +1/2 and -1/2. Therefore, each entry in periodic table corresponds to 1/2 and should be placed in rectangular cell instead of square. That is why I believe that ADOMAH approach, where each element corresponds to half of sphere, or to half of square and not to whole one, is the correct solution.

    Valery Tsimmerman

    December 13, 2010 at 7:30 pm

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