The following formula is a stated structure of a "water" molecule in the category of metamatter. The word "rhema" represents the metaphysical level of metamatter which definition can be found in the text.

2 metaalphas + 3 alphas - 1 ontoalpha + 1/2 neutral pion + 10 electrons

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First, we now look at each structure to determine the atomic mass unit value.
1 alpha = 2 protons + 2 neutrons - BE (muon + (pion)^1/2 ± )                                                                           4
which is equation (A.1) and
where a (pion)^1/2 ± = (273.12658^)1/2 = 16.526,542 emu     and
where the muon is equal to                 +206.768,40 emu
                                                         223.294,942 emu
which we divide by 4 since there are 4 particles and we now have a binding energy BE = 55.823,735 emu so, substituting in the equation (A.1),
we have 1 alpha = 2(1836.108,9 + 1838.640,2) - 55.823,735
which equals 7293.6745.

Now if we compare a Helium atom 1 helium = 4.0026 amu X 1822.7337 emu/amu = 7295.6739

and if we subtract 2 electrons, we have 1 helium = 7293.6739 emu which is almost exactly the same as the

alpha particle in equation (A.1) which is verified.

And second, we now compare a metamatter molecule with "matter" (or physical) water molecule.

2 metaalphas = 2 X 11.229.8 = 22,459.6 emu

+3 alphas = 3 X 7,293.674 = 21,881.022

44,340.622

-1 ontoalpha = -11,645.354 = -11,645.354

32,695.268

+1/2 neutral pion = 1/2(264.11628) = + 132.05814

32,827.326

+10 electrons = 10 x 1 = + 10.00000

Then 1 metawater molecule is equal to 32,837.326 emu and if we divide by 1822.7337 emu/amu

We calculate 1 metawater molecule = 18.015,427 amu which if we compare this to physical water H2O = 18.015,34 amu which is quite close.

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