In the Standard Model of Particle Physics [1,2,3,4], electroweak symmetry breaking [5,6,7,8] is responsible for the mass generation of W and Z gauge bosons thus rendering the weak interactions short ranged. The Standard Model scalar potential is: where the Higgs field Φ is a self-interacting ${\mathrm{S}\mathrm{U}\left(2\right)}_{\mathrm{L}}$ complex doublet that has four real degrees of freedom, with weak hypercharge Y =1 and $\mathrm{V}\left(\mathsf{\Phi }\right)$ is the most general renormalizable scalar potential and if the quadratic term is negative the neutral component of the scalar doublet acquires a non-zero vacuum expectation value $v={\left(\sqrt{2}{\mathrm{G}}_{\mathrm{F}}\right)}^{-1/2}$ which is approximately 246,22 GeV and ${\mathrm{G}}_{\mathrm{F}}$ is the Fermi coupling constant. We should also point out that: where ${\mathsf{\Phi }}^0$ and $a^0$ are the CP-even and CP-odd neutral components, and ${\mathsf{\Phi }}^{+}$ is the complex charged component of the Higgs doublet, respectively. The global minimum of the theory defines the ground state, and spontaneous symmetry breaking implies that there is a symmetry of the system that is not respected by the ground state. From the four generators of the ${\mathrm{S}\mathrm{U}\left(2\right)}_L\times \mathrm{U}{\left(1\right)}_Y$ gauge group, three are spontaneously broken, implying that they lead to non-trivial transformations of the ground state and indicate the existence of three massless Goldstone bosons identified with three of the four Higgs field degrees of freedom. The Higgs field couples to the ${\mathrm{W}}_{\mathsf{\mu }}$ and ${\mathrm{B}}_{\mathsf{\mu }}$ gauge fields associated with the ${\mathrm{S}\mathrm{U}\left(2\right)}_L\times \mathrm{U}{\left(1\right)}_Y$ local symmetry through the covariant derivative appearing in the kinetic term of the Higgs Lagrangian: Where the covariant derivative equals:

g and g′ are the $\mathrm{S}\mathrm{U}\left(2\right)$ and $\mathrm{U}\left(1\right)$ gauge couplings, respectively, and ${\mathsf{\sigma }}^a$ where $a=1,2,3$are the typical Pauli matrices. As a result, the neutral and the two charged massless Goldstone degrees of freedom mix with the gauge fields corresponding to the broken generators of ${\mathrm{S}\mathrm{U}\left(2\right)}_L\times \mathrm{U}{\left(1\right)}_Y$ and become the longitudinal components of the Z and W gauge bosons, respectively. The Z and W gauge bosons acquire masses ${\mathrm{M}}_{\mathrm{W}}=\mathrm{g}v/2$and ${\mathrm{M}}_{\mathrm{Z}}=\left({\mathrm{g}}^{\mathrm{\text{'}}}+\mathrm{g}\right)v/2$. The fourth generator remains unbroken since it is the one associated to the conserved $\mathrm{U}{\left(1\right)}_{\mathrm{Q}\mathrm{E}\mathrm{D}}$ gauge symmetry therefore its corresponding gauge field remains massless or in other words, the photon is massless. Similarly the eight color gauge bosons, the gluons, corresponding to the conserved $\mathrm{S}\mathrm{U}{\left(3\right)}_C$ gauge symmetry with eight unbroken generators, also remain massless. Therefore, from the initial four degrees of freedom of the Higgs field, two are absorbed by the ${\mathrm{W}}^{\pm }$ gauge bosons, one by the ${\mathrm{Z}}^0$ gauge boson, and there is one remaining degree of freedom H, that is the physical Higgs boson. The Higgs boson is neutral under the electromagnetic interactions and transforms as a singlet under $\mathrm{S}\mathrm{U}{\left(3\right)}_C$ and hence does not couple at tree level to the massless photons and gluons. The mass of the Higgs boson [9] is given as ${\mathrm{m}}_{\mathrm{h}}=\sqrt{2{\mathsf{\lambda }v}^2}$, where λ is a free coupling parameter and therefore the mass of the Higgs boson is not predicted in the Standard Model. With the Higgs field in the unitary gauge, the ${\mathrm{S}\mathrm{U}\left(2\right)}_L\times \mathrm{U}{\left(1\right)}_Y$ invariant Yukawa Lagrangian For leptons takes the form:

The respective masses of fermions are not predicted since the Yukawa coupling ${\mathsf{\lambda }}_f$ is a free parameter provided in the formula:

in that sense the Higgs mechanism does not predict any of the elementary fermion masses. It is possible to estimate the strength of the fermion-fermion-Higgs interactions: where ${\mathrm{M}}_{\mathrm{e}}$ is the mass of an electron and ${\mathrm{M}}_{\mathrm{u}}$ is the mass of an up quark. A very important consequence of the fermion-fermion-Higgs interaction is its direct dependence on fermion masses. The larger the mass the stronger this interaction becomes. In order to make sure that Yukawa couplings are no longer arbitrary parameters in the SM Higgs mechanism, we have to avoid using the Higgs VEV to calculate the Yukawa couplings.

We will introduce a new equation for Yukawa couplings. This new equation doesn’t depend on the Higgs VEV and it answers why mass ratios of the three generations or families are as such. The equation is: where $\mathsf{\alpha }\left(\mathrm{Q}\right)$ is the running value of the fine structure constant on the scale Q, $\mathrm{N}$ is the generation or family number for the first, second and third generations respectively, ${\mathsf{\theta }}_{\mathrm{W}}\left(\mathrm{Q}\right)$ is the running value of the Weinberg angle and $\left(1+∆{\mathrm{q}}_f\right)={(1+∆\mathrm{q})}^{-1}$ for unstable leptons, where $∆\mathrm{q}$ encapsulates the higher order QED corrections and can be expressed as a power series expansion in the renormalized electromagnetic coupling constant $\mathsf{\alpha }$ where $∆\mathrm{q}={\sum }_{\mathrm{i}=0}^{\mathrm{\infty }}∆{\mathrm{q}}^{\mathrm{j}}$ in which the index $\mathrm{j}$ gives the power of $\mathsf{\alpha }$ that appears in $∆{\mathrm{q}}^{\mathrm{j}}$ but this value can also be experimentally measured by calculating it from the mean lifetimes of unstable leptons or quarks (where we have to include the CKM matrix as well): where ${\mathsf{\tau }}_f$ is the mean lifetime on the unstable fermion. We’re using natural units so the reduced Planck constant has been removed from the equation. All unstable particles have different values of $∆\mathrm{q}$.

Further on $f=l,q$ is the fermion flavor, $l,q$ are lepton and quark flavors, respectively. The quantum number ${\mathrm{n}}_{\mathrm{g}}\equiv 3$ is the number of generations or families, ${\mathrm{n}}_f\equiv 6$ is the number of flavors since both leptons and quarks have six flavors each, $\mathrm{S}=1/2$ is the fermion spin quantum number, ${\mathrm{n}}_{{\mathrm{Y}}_{\mathrm{w}}}$ is the number of particles that interact via the weak hypercharge, where ${\mathrm{Y}}_{\mathrm{w}}=2·({\mathrm{Q}}_{\mathrm{e}}-{\mathrm{T}}_3)$ is the weak hypercharge, ${\mathrm{Q}}_{\mathrm{e}}$ is the (electric) charge quantum number and ${\mathrm{T}}_3$ is the third component of the weak isospin. The quantum number ${\mathrm{n}}_{{\mathrm{Y}}_{\mathrm{w}}}$is defined by the equation: where $\mathrm{B}$ is the baryon quantum number, $\mathrm{L}$ is the lepton quantum number and ${\mathsf{\alpha }}_{\mathrm{G}}\left(\mathrm{Q}\right)$ is the gravitational coupling on the scale $\mathrm{Q}$. Because the gravitational coupling has a tiny value for most particles (around ${10}^{-39}$), we will therefore approximate its value to zero for all quarks and leptons except for the right-handed (or right-chiral) neutrinos, in their case ${\mathsf{\alpha }}_{\mathrm{G}}\left(\mathrm{Q}\right)$ will have a non-zero value.The quantum number $\mathrm{п}={\mathrm{n}}_{\mathrm{u}\mathrm{p}}+{\mathrm{n}}_{\mathrm{u}\overline{\mathrm{p}}}$ is the number of unstable particles ${\mathrm{n}}_{\mathrm{u}\mathrm{p}}$ and unstable anti-particles ${\mathrm{n}}_{\mathrm{u}\overline{\mathrm{p}}}$, whereas $\mathrm{k}$ equals: and $\mathsf{\Psi }=3.3598856\dots$ is the reciprocal Fibonacci constant. The parameters ${\mathrm{c}}_1$, ${\mathrm{c}}_2$, ${\mathrm{c}}_3$ and ${\mathrm{c}}_4$ are defined as: Where ${\mathrm{Q}}_f$ and ${\mathrm{L}}_f$ are quark and lepton flavor quantum numbers, respectively and ${\mathrm{I}}_3$ is the third component of isospin. Then:

then:

then:

After solving the integral we get:

All leptons have a lepton quantum number $\mathrm{L}=1$ and all of their respective anti-particles have a lepton quantum number $\mathrm{L}=-1$. All left-chiral leptons have a weak hypercharge ${\mathrm{Y}}_{\mathrm{w}}=-1$ whereas right-chiral charged leptons have a weak hypercharge ${\mathrm{Y}}_{\mathrm{w}}=-2$ and (if they exist) right-chiral neutrinos don’t have a weak hypercharge, weak isospin or (electric) charge, therefore making them “sterile”.

All quarks have a baryon quantum number $\mathrm{B}=1/3$ and all of their respective anti-particles have a baryon quantum number $\mathrm{B}=-1/3$. All left-chiral quarks have a weak hypercharge ${\mathrm{Y}}_{\mathrm{w}}=1/3$ and their right-handed equivalents have ${\mathrm{Y}}_{\mathrm{w}}=4/3$ for u-type quarks and ${\mathrm{Y}}_{\mathrm{w}}=-2/3$ for d-type quarks.

We managed to explain why there are three generations of particles, both for leptons and quarks respectively. We also explained why the three generations have such mass ratios. We also found a way to measure the Weinberg angle on low energy levels of charged leptons and lighter quarks. We also explained why up and down quarks have the masses that they do, which was previously explained. Our equation doesn’t treat particle masses as constants which was a problem with the Koide formula [18]. We also managed to predict the mass eigenstates of all three left-handed neutrinos and we even made predictions for the upper limits of right-handed sterile neutrinos. There is also a possibility to use this new equation in a ${\mathrm{U}\left(1\right)}_{\mathrm{B}-\mathrm{L}}$ symmetry, or perhaps similar $\mathrm{B}-\mathrm{L}$ GUT symmetries.