If clear crystal are bonded by temperature particles, where the intermolecular (interatomic) bonds that holds the crystal together involve only \(T^{+}\) and \(T^{-}\) particles, then either the nuclei so bonded lose the outer \(p^{+}\) layer or gain a \(g^{+}\) and a \(T^{+}\) layer.
In the former case and in the case of atomic crystal unit cells, the nulcei would have transmuted to an element one left along a horizontal period of the periodic table. A crystal made from \(Cu\), behave like \(Ni\) in the lattice.
In the former but in the case of molecular crystal unit cells, the molecule quasi nucleus as a whole loses one \(p^{+}\) layer. The final "transmuted" molecule is that which appears as the chemical composition associated with the crystal. For example ruby, \(Al_2O_3:Cr\) gives the transmuted repeating unit of the crystal. \(AlSiO_3\) is the simplest likely starting molecule. The \(AlSi\) pair loses a singular unpaired proton layer that is above the three paired oxygen, \(O\) proton orbits. The exposed \(T^{+}\) orbit that previously provided for the weak field that held the escaped proton acquires a \(T^{-}\) particle and pairs up with another unit. The paired \((T^{+}, T^{-})\) bond links up with other paired units and we have a crystal lattice.
\(AlSi\) is above \(O_3\), when the crystal collapses as in the case of magnetite (post "Split, Stir And Charged" dated 17 Aug 2017), \(O_3\rightarrow Cr \), Chromium is formed.
If this is true, to form Ruby we have to first help the lone outer proton in \(Al_2O_3\) to escape and then cools it with \(T^{-}\) particles and lastly encourage the paired \((T^{+}, T^{-})\) bonds re-orientate and link up into a lattice.
Bloody hell, not. \(AlSi\) has an extra \(g^{+}\) particle that cannot be removed without also removing the \(T^{+}\) particle that forms crystal bonds. The crystal unit is with an isotope of \(Al\) and is denser (more mass as the result of one more \(g^{-}\) particles that balance the \(g^{+}\) particle in the nucleus) than natural ruby.
Natural ruby without an isotopes and without the resulting density anomaly is likely to have a quasi nuclei \(Al_2O_3\) that has acquired first an extra \(g^{+}\) layer followed by a \(T^{+}\) layer.
What about diamonds?
