Occult Chemistry - Clairvoyant Observations on the Chemical Elements by C. W. (Charles Webster) Leadbeater;Annie Wood Besant
page 16 of 126 (12%)
page 16 of 126 (12%)
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puzzling body; the difficulties of observation were very much increased by
the extraordinary activity shown by this element and the dazzling brilliancy of some of its constituents. The gaseous atom is an ovoid body, within which a spirally-coiled snake-like body revolves at a high velocity, five brilliant points of light shining on the coils. The snake appears to be a solid rounded body, but on raising the atom to E 4 the snake splits lengthwise into two waved bodies, and it is seen that the appearance of solidity is due to the fact that these spin round a common axis in opposite directions, and so present a continuous surface, as a ring of fire can be made by whirling a lighted stick. The brilliant bodies seen in the atom are on the crests of the waves in the positive snake, and in the hollows in the negative one; the snake itself consists of small bead-like bodies, eleven of which interpose between the larger brilliant spots. On raising these bodies to E 3 the snakes break up, each bright spot carrying with it six beads on one side and five on the other; these twist and writhe about still with the same extraordinary activity, reminding one of fire-flies stimulated to wild gyrations. It can been seen that the larger brilliant bodies each enclose seven ultimate atoms, while the beads each enclose two. (Each bright spot with its eleven beads is enclosed in a wall, accidentally omitted in the diagram.) On the next stage, E 2, the fragments of the snakes break up into their constituent parts; the positive and negative bodies, marked _d_ and _d'_, showing a difference of arrangement of the atoms contained in them. These again finally disintegrate, setting free the ultimate physical atoms, identical with those obtained from hydrogen. The number of ultimate atoms contained in the gaseous atom of oxygen is 290, made up as follows:-- 2 in each bead, of which there are 110: 7 in each bright spot, of which there are 10; 2 x 110 + 70 = 290. |
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