Particle Masses: A solution of the wave-equation.
5.1
5.2
For 
, equation 5.1 simplifies to
5.3
solving for 
gives
5.4
where 
, G is the curvature scalar.
A spherically symmetric solution for 
is
5.5
Using the new De-Broglie relation,
5.6
for 
the integral in equation 5.4 is
which simplifies to
The wavefunction peaks when 
Solving for r gives
Using 5.6, and 
it can be shown that the wavefunction peaks at an Energy E given by
5.7
For n>0, 
, E reduces to
For n=2, E(2)=172.6GeV which is near the mass of the Top Quark.
From the Lagrangian the energy of a boson is
The Curvature Scalar is determined by the potentials and omega.
Where f=3-6, it can be shown that for spherically symmetric potentials, and neglecting ‘magnetic’ potentials, the above curvature scalar is determined by the potentials
Which in terms of the coupling constants are
The energy of a boson for 
simplifies to
where
Using 
, solving for E gives
The coupling constants at Z energy are
And for 
, the rest energy of the Z is 93.4GeV
This compares with a measured value of 91.18GeV. This difference is due to another interaction with coupling strength 
, this gives a rest energy for the Z of 91.18Gev.
The carrier of this new interaction if uncoupled from the electroweak interation has a rest energy of approximately 420.3GeV and is possibly a dark energy boson.