Artificial Rubies

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?