Mordred

The decay does not affect the number of microstates any neutron examined will have identical number of microstates otherwise it wouldn't be a neutron. What alters is the transition amplitudes via CKMS mass mixing matrix a simpler method though being Beit Wigner where under that treatment you treat the proton or neutron as a single particle.

No there isn't a misunderstanding on my part Are you familiar with the S matrix for protons and neutrons ? The number of up down quarks for each is merely the mean average color charge relations If you examine the number of microstates contained within either the proton and neutron via the S matrix your earlier statement makes little sense based on stability Protons and neutrons are composite particles plain and simple so they have internal microstates. The number of microstates within the Proton does not change in the interaction you described. An electron is not a microstate contained within the proton to begin with holographic principle won't help what many fail to understand is is that the holographic principle is conformal The laws of physics would be the same if you use the holographic Principle or QM/QFT or even classical physics. That is the basis of its premise.

I would like to see a mathematical proof of the above as I know you do not know the math show a peer review article that the neutron would not satisfy the third law of thermodynamics. Both protons and neutrons are degeneracy systems. That peer review should show the relevant Fermi- energy for each https://en.wikipedia.org/wiki/Fermi_energy

yeah thanks for the catch corrected above.

I was curious as to a fundamental question, we all know the CMB temperature today is 2.73 Kelvin with the radius of the Observable universe being 46.3 Gly. So I asked what would the radius need to be to reach 1 Kelvin. It turns out the universe would need to have a radius of 140 Gly. The universe would be roughly 28.8 Gyrs old far far into the future assuming nothing changes with the cosmological parameters

Agreed it would have been more intelligent to use \[\rho_R=\frac{\pi^2}{30}{g_{*S}=\sum_{i=bosons}gi(\frac{T_i}{T})^3+\frac{7}{8}\sum_{i=fermions}gi(\frac{T_i}{T})}^3\] To determine how many up and down quarks would be available to form protons and neutrons in the first place. Maxwell Boltzmann above takes into account the laws of thermodynamics. https://en.m.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

This matches a policy any good physicist follows. Spend more time trying to prove your theory wrong otherwise it will never become robust enough under cross examination. For those interested this is about the best article I have been able to locate on QCD dual superconductivity. https://arxiv.org/abs/hep-lat/0510112 This link shows how to incorporate to string theory https://arxiv.org/abs/hep-ph/0301032 As I mentioned numerous times type 1 vs type 2 superconductors the distinction has to do with vortex penetration depth http://lampx.tugraz.at/~hadley/ss2/problems/super/s.pdf

The harmonic oscillator has absolutely nothing to do with stability read post above. Neither does the cosmological constant vacuum catastrophe even though its obvious you've chosen to ignore everything I stated. Thankfully there are other readers even those that haven't gotten involved.

I'm still wondering how long it's going to take some people , to recognize there is a HUGE distinction between a ground state and an excited particle state..... I've given up trying to get that across to certain people. The clue that all quantum fields has a ground state should have indicated those people might just be missing a detail... It literally doesn't matter what quantum field is used. They all have a ZPE. The ground state isn't the particles themselves. Lol providing the field strength formula for QCD obviously wasn't a strong enough clue. The conversation is still discussing particles and not the interaction between particles which is where the ground state is applied in terms of superconductivity. here is the simplest mathematical statement describing the above. Maybe just maybe this will work \[\hat{a}(\vec{k})|0\rangle=0\] the \(\vec{k}\) is the wavefunction the \(|0\rangle\) is the ground state the \(\hat{a}\) is the creation annihilation operators that give the creation and annihilation of particles (ladder operators) From the ground state. the creation/annihilation operators are determined via the quantum harmonic oscillator. So is the Hamilton \[\hat{H}=\omega(\hat{a}^\dagger a+\frac{1}{2})\] the number Operator is \[\hat{N}=\hat{a}^\dagger \hat{a}\] gives the Hamilton a nice simple form \[\hat{H}=\omega(\hat{N}+\frac{1}{2})\] I will leave it at that.

Hydrogen is one elements that doesn't require fusion to produce others being Deuterium and lithium. The constituents protons neutrons and electrons at high temperaturesdrp out of thermal equilibrium at different temperatures. Electrons themselves drop out of equilibrium during electroweak symmetry breaking neutrons and protons drop out at much lower temperature for 75 % dropout roughly 5000 Kelvin where hydrogen at 75 percent dropout being 3000 Kelvin ie roughly the surface of last scatterring. However for truly early star formation prior this involves supercooling the during slow roll a reheating. This plus higher densities granted the means for truly early star formation. No DM is not needed in regards to hydrogen formation however it is needed for early large scale structure formation (stars, galaxies, galaxy clusters etc Boltzmann brain is a highly speculative conjecture that is supported by mathematics I don't particularly follow it but do know of some physicists that have studied it such as Sean Carrol. It's not needful to understand physics. In regards to particles knowing. This descriptive doesn't make sense. Recall those conservation laws I previously mentioned ? They show what conditions are required in order for particles to form via interactions. In essence if a particle can be formed it will be formed when the requirements are met without violating those conservation laws. This is also involved in determining mean lifetime of particles. For example electrons are stable as there is no particle an electron can decay into.

Guess we found the QFT ghost field lol

Not too mention all particles regardless of type contribute to CMB blackbody temperature including weakly interactive. So why aren't these SU(3) atoms detected is a very relevant question. I also noticed no one paid any attention to the SU(3) gauge I posted this involves protons and neutrons as well as mesons. Guess they don't want to think about what energy levels would be involved with those composite particles.

@JosephDavid I understand your not familiar with the mathematics so anytime I supply mathematics it doesn't relate to you as your unfamiliar with the equations. Don't worry its very common on any forum. So your in good company. My challenge has always been how do I get explanations across that don't rely on a mere matter of trust of my opinion. One might think well I could merely post articles and quote sections etc however that actually doesn't work very well. for example pertinent to what I'm going to describe below if I were to say look at equation 44 to 48 of this article on QCD superconductors comparing the QED superconductors later is I couldn't expect a large majority of our members to understand it. https://arxiv.org/pdf/0709.4635 I could for example in terms of this thread explicitly show that the Meissner effect cannot fully describe a QCD vacuum state via the mathematics. The Meissner effect specifically involves the electroweak symmetry vacuums. So here is my challenge in a nutshell a very clear distinction is the electroweak couplings as opposed to the QCD couplings for color gauge using Yang Mills and the Gell-Mann matrices. This is the electroweak couplings to Higgs field for the electroweak bosons. \[W^3_\mu=Z_\mu cos\theta_W+A_\mu sin\theta_W\] \[B_\mu= Z_\mu sin\theta_W+A_\mu cos\theta_W\] \[Z_\mu=W^3_\mu cos\theta_W+B_\mu sin\theta_W\] \[A_\mu=-W^3_\mu\sin\theta_W+B_\mu cos\theta_W\] What this shows is that the charge conjugate mediation has different coupling strengths. It also shows that the above fields are Abelian also the longitudinal components of the above is your mass terms via the energy momentum relation. Which is essentially what is also applied for the zero point energy equations longitudinal plane waves. This however is not the case in color charge mediation Which for the Miessner effect is described by BCS theory in the above (the EM/EW fields) in the above you get Cooper pairs with flux tubes providing current flow between the Cooper pairs this is true in both type 1 and type 2 superconductivity. However those flux tubes involve the E and B fields for the EM field> how that works for weak field superconductivity involves charge conjugation. Earlier I posted the charge conjugation formula. \[Q+I^3+\frac{\gamma}{2}\]. Now here is the problem with the above with color gauge. interactions (sorry this does require math) its unavoidable. a quark is described by two wavefunctions a Dirac wavefunction and a color wavefunction for the SU(3) mediation Dirac wavefunction \(\psi\) color wavefunction \(\Psi\) \[(i\gamma^\mu \partial_\mu-m)\Psi=0\] color wavefunction below \[\Psi=\psi(x)\chi_c\] the colors are below \[\chi_R=\begin{pmatrix}1\\0\\0\end{pmatrix}\] \[\chi_G=\begin{pmatrix}0\\1\\0\end{pmatrix}\] \[\chi_B=\begin{pmatrix}1\\0\\0\end{pmatrix}\] to be a gauge theory one requires invariance under Dirac invariance. (includes Lorentz invariance) \[\acute{\Psi}=e^{i g_s}{2}\alpha_j B_j(x)\Psi\] I won't bother with the mathematical proof of the last expression however the inclusion of the color gauge fields to the Dirac wavefunctions gives the Langrangian of \[\mathcal{L}=\bar{\Psi}(i\gamma_\mu D^\mu-m)\Psi-\frac{1}{4}F_{j,\mu\nu}F_j^{\mu\nu}\] with a field strength tensor of \[F^{^{\mu\nu}_j=\partial G^\mu_j-\partial^\nu_j=g_sf_[j,k,l}G_k^\mu G_l^\nu\] now what that above expression tells us is that the coupling strength for each gluon mediator 8 in total is identical. Now I'm going to skip a bit in order to mediate the color gauges between quarks you require three Operators \(I_\pm\), \(U_{\pm}\),\( V_{\pm}\) these are the ladder operators for quark color charge interchange. \[(i\gamma_\mu \partial^\mu-m)\Psi=\frac{g_s}{2}\gamma_\mu G^\mu_j(x)\psi\lambda_j\chi_c\] this describes a state \(\chi_c\) with color Couples with the field strength \(g_s\) and changes to another color charge C mediated by \(\lambda_j G_j^G\mu\) I won't go into the fuller color operator expression for each color exchange however the amplitudes for color are not 1/2e as per EM charge color charges are 2/3e or 1/3e so the formulas used for ZPE for those amplitudes are different in wavefunction equivalence due to distinctive difference of the 1/2 EM charges involved for the relevant Meissner effect charges for Em=\(1/2_\pm\) charges for color \(2/3_\pm, 1/3_\pm\) The above qualitatively shows that the field mediation for color charge is significantly distinctive from those of the EM field. This demonstrates that the Meissner effect as per BCS theory or the Anderson-Higgs field do not describe a superconducting QCD vacuum state. for 3 reasons. 1) abelion fields vs non abelion fields. 2) color vs EM charge 3) different coupling strength relations across the mediator fields (EM =different coupling strengths for each mediator as opposed to one coupling strengths for all mediator bosons under QCD 4) the interaction field (mediation requires 2 wavefunctions thereby making it a complex field=added degrees of freedom= higher field number) 5)as this would require significantly more the math I won't delve into it the VeV calculations would be different from a QED VeV from a QCD VeV. Everyone might also note the the above article is looking at 6 Cooper pairs for for QCD vacuum its one of the possibilities. another being Dual Miessner

P is momentum the equation Swansont posted is called the energy momentum relation. https://en.m.wikipedia.org/wiki/Energy–momentum_relation

And that's the trick, we can only garnish indirect evidence. We cannot measure anything less than a quanta of action See here for other readers how that connects to Planck constant and ZPE. https://en.m.wikipedia.org/wiki/Action_(physics)#:~:text=Planck's quantum of action,-The Planck constant&text=%2C is called the quantum of,and the de Broglie wavelength. Now the planck length is the smallest theoretical measurable wavelength. How many planck lengths in the Observsble universe ? Give anyone and idea of the momentum space trend to infinite energy? Using the formulas above for ZPE ?

Ok were going to have to teach you latex above is difficult to read. Anyways take a particle under QFT all particles are field excitations. In terms of ZPE the more you determine localize the position via Compton/De-Debroglie wavelengths the more uncertain you are on its momentum. Yes however a Hermitean matrix the orthogonal diagonal elements have a real number for entry. We're using complex conjugate typically so it's complex conjugate of position and momentum under QM for example. Author doesn't give the relevant details to address that question.

Now take my last post apply that to the quoted section. Little lesson on how to recognize a hermitean matrix.

\[\rho_\Lambda c^2=\int^\infty_0=\frac{4\pi k^2 dk}{(3\pi\hbar)^3}(\frac{1}{2}\sqrt{k^2c^2+m^2c^4})\] The above is a sum over plane waves it doesn't include transverse waves so it's constrained to plane waves only. It also is just first order terms. The reason being for that is the specific action of a harmonic oscillator. Linear only with no non linear components in the above. Now if the above shows the catastrophe is linear read the wiki link under QCD as a complex system will comprise of non linear components. (The tensor fields under SU(3) as one example ) which includes longitudinal and transverse components where the above is longitudinal. If you study the oscillator equations that gives rise to the 1/2 term just prior to the energy momentum under the square root. (Hint U(1) symmetry only in above) do not confuse that as EM only. EM field also has non linear terms

Hmm no actually there is one such theory already but no evidence support some physicists do work on it. So the idea is already being examined though not quite in the way you imagine. (Yes alot of this discussion is applicable to this theory) Boltzmann brains. http://arxiv.org/pdf/1812.01909

notice the integral from summing over momentum states ranges to infinite but the term includes a constraint to prevent that RHS of the \(\int^\infty_0\) that was the regularization term that led to vacuum catastrophe Posted on page 1 That is what needs solving so how does SU(3) solve that ? LMAO I lost count the number of times I stated ENERGY DENSITY this thread.... so earlier I also showed the math showing there are 16 fields which means 16 fields each with a ZPE for QCD as SU(3) for simplicity. Just an FYI to everyone reading for the harmonic oscillator the Operators are Hermitian so it will be equal to its adjoint as all Hermitian operators are. Just in case anyone familiar with Hamilton forms are reading. May also help with sections in regards to the wiki link above (operators being the Hermitian position and momentum operators). What this means is the harmonic operator raising lowering (ladder operators) aka creation/ annihilation operators are also Hermitian

let me ask a different question which people have looked at what zero point energy entails ? Ie which field does it apply to ? All quantum fields have a ZPE ground state this includes all quantum field at the closest QM/QFT allows to absolute zero. Doesnt matter if they are massless or massive or complex mixtures. Every quantum field has a ground state. So what difference would it make to leave the Complex SU(3) field untouched via symmetry breaking if every other field still contributes to the total energy or rather all quantum particle fields draws from the infinite quantum ground state under QFT treatment( where the position and momentum operators are treated at every coordinate for the oscillator field) specifically any quantum field that has not been normalized via the reduced Hamilton ? You could have every single quantum field at absolute zero or any other temperature and still have a ground state The uncertainty principle states that no object can ever have precise values of position and velocity simultaneously. The total energy of a quantum mechanical object (potential and kinetic) is described by its Hamiltonian which also describes the system as a harmonic oscillator, or wave function, that fluctuates between various energy states (see wave-particle duality). All quantum mechanical systems undergo fluctuations even in their ground state, a consequence of their wave-like nature. The uncertainty principle requires every quantum mechanical system to have a fluctuating zero-point energy greater than the minimum of its classical potential well. This results in motion even at absolute zero. For example, liquid helium does not freeze under atmospheric pressure regardless of temperature due to its zero-point energy. https://en.wikipedia.org/wiki/Zero-point_energy so Just how precisely is SU(3) going to solve the vacuum catastrophe if every other quantum field still contributes ? Symmetry breaking or not..... look at graph showing the ground constant at all temperatures... perhaps that will help you understand precisely why I kept mentioning Bose-Einstein and Fermi-Dirac statistics for particle number density ALL fields contribute and ALL fields draw from it for particle creation/annihilation at all temperatures. All quantum fields have harmonic oscillations regardless of temperature. Yet the author claims to somehow magically solve this by leaving SU(3) unbroken in symmetry Thats why I mentioned these equations back on page 5 please read the article in that link.. it will reinforce everything I just described. After you do that , think back on the video MigL posted. then relook at the following from page 1 "lets detail the cosmological constant problem then you can show me how your paper solves this problem I will keep it simple for other readers by not using the Langrene for the time being and simply give a more algebraic treatment. ( mainly to help our other members). To start under QFT the normal modes of a field is a set of harmonic oscillators I will simply apply this as a bosons for simple representation as energy never exists on its own \[E_b=\sum_i(\frac{1}{2}+n_i)\hbar\omega_i\] where n_i is the individual modes n_i=(1,2,3,4.......) we can identify this with vacuum energy as \[E_\Lambda=\frac{1}{2}\hbar\omega_i\] the energy of a particle k with momentum is \[k=\sqrt{k^2c^2+m^2c^4}\] from this we can calculate the sum by integrating over the momentum states to obtain the vacuum energy density. \[\rho_\Lambda c^2=\int^\infty_0=\frac{4\pi k^2 dk}{(3\pi\hbar)^3}(\frac{1}{2}\sqrt{k^2c^2+m^2c^4})\] where \(4\pi k^2 dk\) is the momentum phase space volume factor. the effective cutoff can be given at the Planck momentum \[k_{PL}=\sqrt{\frac{\hbar c^3}{G_N}}\simeq 10^{19}GeV/c\] gives \[\rho \simeq \frac{K_{PL}}{16 \pi^2\hbar^3 c}\simeq\frac{10^74 Gev^4}{c^2(\hbar c)^3} \simeq 2*10^{91} g/cm^3\] compared to the measured Lambda term via the critical density formula \[2+10^{-29} g/cm^3\] method above given under Relativity, Gravitation and Cosmology by Ta-Pei Cheng page 281 appendix A.14 (Oxford Master series in Particle physics, Astrophysics and Cosmology)

your welcome and yes it was me as its a good collection of lectures

everything in that math is what the Author ignored when he states SU(3) all those equations are for quark mass terms. the Higgs mixing angles are included for symmetry breaking this is what the author expects you guys to piece together. So its VERY relevant to the discussion Nothing there will give mass to photons...

That above is nothing more than a representation You do not do any calculations from it. That takes further details. There is nearly 30 different tensors hidden under that expression. You need to factor out the relevant terms in order to apply them. lets demonstrate all of this is contained under that above expression and that is ONLY A TINY PORTION. SU(2) \[{\small\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline Field & \ell_L& \ell_R &v_L&U_L&d_L&U_R &D_R&\phi^+&\phi^0\\\hline T_3&- \frac{1}{2}&0&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}&0&0&\frac{1}{2}&-\frac{1}{2} \\\hline Y&-\frac{1}{2}&-1&-\frac{1}{2}&\frac{1}{6}&\frac{1}{6}& \frac{2}{3}&-\frac{1}{3}&\frac{1}{2}&\frac{1}{2}\\\hline Q&-1&-1&0&\frac{2}{3}&-\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}&1&0\\\hline\end{array}}\] \(\psi_L\) doublet \[D_\mu\psi_L=[\partial_\mu-i\frac{g}{\sqrt{2}}(\tau^+W_\mu^+\tau^-W_\mu^-)-i\frac{g}{2}\tau^3W^3_\mu+i\acute{g}YB_\mu]\psi_L=\]\[\partial_\mu-i\frac{g}{\sqrt{2}}(\tau^+W_\mu^-)+ieQA_\mu-i\frac{g}{cos\theta_W}(\frac{t_3}{2}-Qsin^2\theta_W)Z_\mu]\psi_L\] \(\psi_R\) singlet \[D_\mu\psi_R=[\partial\mu+i\acute{g}YB_\mu]\psi_R=\partial_\mu+ieQA_\mu+i\frac{g}{cos\theta_W}Qsin^2\theta_WZ_\mu]\psi_W\] with \[\tau\pm=i\frac{\tau_1\pm\tau_2}{2}\] and charge operator defined as \[Q=\begin{pmatrix}\frac{1}{2}+Y&0\\0&-\frac{1}{2}+Y\end{pmatrix}\] \[e=g.sin\theta_W=g.cos\theta_W\] \[W_\mu\pm=\frac{W^1_\mu\pm iW_\mu^2}{\sqrt{2}}\] \[V_{ckm}=V^\dagger_{\mu L} V_{dL}\] The gauge group of electroweak interactions is \[SU(2)_L\otimes U(1)_Y\] where left handed quarks are in doublets of \[ SU(2)_L\] while right handed quarks are in singlets the electroweak interaction is given by the Langrangian \[\mathcal{L}=-\frac{1}{4}W^a_{\mu\nu}W^{\mu\nu}_a-\frac{1}{4}B_{\mu\nu}B^{\mu\nu}+\overline{\Psi}i\gamma_\mu D^\mu \Psi\] where \[W^{1,2,3},B_\mu\] are the four spin 1 boson fields associated to the generators of the gauge transformation \[\Psi\] The 3 generators of the \[SU(2)_L\] transformation are the three isospin operator components \[t^a=\frac{1}{2} \tau^a \] with \[\tau^a \] being the Pauli matrix and the generator of \[U(1)_\gamma\] being the weak hypercharge operator. The weak isospin "I" and hyper charge \[\gamma\] are related to the electric charge Q and given as \[Q+I^3+\frac{\gamma}{2}\] with quarks and lepton fields organized in left-handed doublets and right-handed singlets: the covariant derivative is given as \[D^\mu=\partial_\mu+igW_\mu\frac{\tau}{2}-\frac{i\acute{g}}{2}B_\mu\] \[\begin{pmatrix}V_\ell\\\ell\end{pmatrix}_L,\ell_R,\begin{pmatrix}u\\d\end{pmatrix}_,u_R,d_R\] The mass eugenstates given by the Weinberg angles are \[W\pm_\mu=\sqrt{\frac{1}{2}}(W^1_\mu\mp i W_\mu^2)\] with the photon and Z boson given as \[A_\mu=B\mu cos\theta_W+W^3_\mu sin\theta_W\] \[Z_\mu=B\mu sin\theta_W+W^3_\mu cos\theta_W\] the mass mixings are given by the CKM matrix below \[\begin{pmatrix}\acute{d}\\\acute{s}\\\acute{b}\end{pmatrix}\begin{pmatrix}V_{ud}&V_{us}&V_{ub}\\V_{cd}&V_{cs}&V_{cb}\\V_{td}&V_{ts}&V_{tb}\end{pmatrix}\begin{pmatrix}d\\s\\b\end{pmatrix}\] mass euqenstates given by \(A_\mu\) an \(Z_\mu\) \[W^3_\mu=Z_\mu cos\theta_W+A_\mu sin\theta_W\] \[B_\mu= Z_\mu sin\theta_W+A_\mu cos\theta_W\] \[Z_\mu=W^3_\mu cos\theta_W+B_\mu sin\theta_W\] \[A_\mu=-W^3_\mu\sin\theta_W+B_\mu cos\theta_W\] ghost field given by \[\acute{\psi}=e^{iY\alpha_Y}\psi\] \[\acute{B}_\mu=B_\mu-\frac{1}{\acute{g}}\partial_\mu\alpha Y\] [latex]D_\mu[/latex] minimally coupled gauge covariant derivative. h Higg's bosonic field [latex] \chi[/latex] is the Goldstone boson (not shown above) Goldstone no longer applies after spontaneous symmetry breaking [latex]\overline{\psi}[/latex] is the adjoint spinor [latex]\mathcal{L}_h=|D\mu|^2-\lambda(|h|^2-\frac{v^2}{2})^2[/latex] [latex]D_\mu=\partial_\mu-ie A_\mu[/latex] where [latex] A_\mu[/latex] is the electromagnetic four potential QCD gauge covariant derivative [latex] D_\mu=\partial_\mu \pm ig_s t_a \mathcal{A}^a_\mu[/latex] matrix A represents each scalar gluon field Single Dirac Field [latex]\mathcal{L}=\overline{\psi}I\gamma^\mu\partial_\mu-m)\psi[/latex] under U(1) EM fermion field equates to [latex]\psi\rightarrow\acute{\psi}=e^{I\alpha(x)Q}\psi[/latex] due to invariance requirement of the Langrene above and with the last equation leads to the gauge field [latex]A_\mu[/latex] [latex] \partial_\mu[/latex] is replaced by the covariant derivitave [latex]\partial_\mu\rightarrow D_\mu=\partial_\mu+ieQA_\mu[/latex] where [latex]A_\mu[/latex] transforms as [latex]A_\mu+\frac{1}{e}\partial_\mu\alpha[/latex] Single Gauge field U(1) [latex]\mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}[/latex] [latex]F_{\mu\nu}=\partial_\nu A_\mu-\partial_\mu A_\nu[/latex] add mass which violates local gauge invariance above [latex]\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}m^2A_\mu A^\mu[/latex] guage invariance demands photon be massless to repair gauge invariance add a single complex scalar field [latex]\phi=\frac{1}{\sqrt{2}}(\phi_1+i\phi_2[/latex] Langrene becomes [latex] \mathcal{L}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+|D_\mu \phi|^2-V_\phi[/latex] where [latex]D_\mu=\partial_\mu-ieA_\mu[/latex] [latex]V_\phi=\mu^2|\phi^2|+\lambda(|\phi^2|)^2[/latex] [latex]\overline{\psi}=\psi^\dagger \gamma^0[/latex] where [latex]\psi^\dagger[/latex] is the hermitean adjoint and [latex]\gamma^0 [/latex] is the timelike gamma matrix the four contravariant matrix are as follows [latex]\gamma^0=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}[/latex] [latex]\gamma^1=\begin{pmatrix}0&0&0&1\\0&0&1&0\\0&0&-1&0\\-1&0&0&0\end{pmatrix}[/latex] [latex]\gamma^2=\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&i&0&0\\-i&0&0&0\end{pmatrix}[/latex] [latex]\gamma^3=\begin{pmatrix}0&0&1&0\\0&0&0&-1\\-1&0&0&0\\0&1&0&0\end{pmatrix}[/latex] where [latex] \gamma^0[/latex] is timelike rest are spacelike V denotes the CKM matrix usage [latex]\begin{pmatrix}\acute{d}\\\acute{s}\\\acute{b}\end{pmatrix}\begin{pmatrix}V_{ud}&V_{us}&V_{ub}\\V_{cd}&V_{cs}&V_{cb}\\V_{td}&V_{ts}&V_{tb}\end{pmatrix}\begin{pmatrix}d\\s\\b\end{pmatrix}[/latex] [latex]V_{ckm}=V^\dagger_{\mu L} V_{dL}[/latex] the CKM mixing angles correlates the cross section between the mass eigenstates and the weak interaction eigenstates. Involves CP violations and chirality relations. Dirac 4 component spinor fields [latex]\gamma^5=i\gamma_0,\gamma_1,\gamma_2,\gamma_3[/latex] 4 component Minkowskii with above 4 component Dirac Spinor and 4 component Dirac gamma matrixes are defined as [latex] {\gamma^\mu\gamma^\nu}=2g^{\mu\nu}\mathbb{I}[/latex] where [latex]\mathbb{I}[/latex] is the identity matrix. (required under MSSM electroweak symmetry break} in Chiral basis [latex]\gamma^5[/latex] is diagonal in [latex]2\otimes 2[/latex] the gamma matrixes are [latex]\begin{pmatrix}0&\sigma^\mu_{\alpha\beta}\\\overline{\sigma^{\mu\dot{\alpha}\beta}}&0\end{pmatrix}[/latex] [latex]\gamma^5=i{\gamma_0,\gamma_1,\gamma_2,\gamma_3}=\begin{pmatrix}-\delta_\alpha^\beta&0\\0&\delta^\dot{\alpha}_\dot{\beta}\end{pmatrix}[/latex] [latex]\mathbb{I}=\begin{pmatrix}\delta_\alpha^\beta&0\\0&\delta^\dot{\alpha}_\dot{\beta}\end{pmatrix}[/latex] Lorentz group identifiers in [latex](\frac{1}{2},0)\otimes(0,\frac{1}{2})[/latex] [latex]\sigma\frac{I}{4}=(\gamma^\mu\gamma^\nu)=\begin{pmatrix}\sigma^{\mu\nu\beta}_{\alpha}&0\\0&-\sigma^{\mu\nu\dot{\alpha}}_{\dot{\beta}}\end{pmatrix}[/latex] [latex]\sigma^{\mu\nu}[/latex] duality satisfies [latex]\gamma_5\sigma^{\mu\nu}=\frac{1}{2}I\epsilon^{\mu\nu\rho\tau}\sigma_{\rho\tau}[/latex] a 4 component Spinor Dirac field is made up of two mass degenerate Dirac spinor fields U(1) helicity [latex](\chi_\alpha(x)),(\eta_\beta(x))[/latex] [latex]\psi(x)=\begin{pmatrix}\chi^{\alpha\beta}(x)\\ \eta^{\dagger \dot{\alpha}}(x)\end{pmatrix}[/latex] the [latex](\alpha\beta)=(\frac{1}{2},0)[/latex] while the [latex](\dot{\alpha}\dot{\beta})=(0,\frac{1}{2})[/latex] this section relates the SO(4) double cover of the SU(2) gauge requiring the chiral projection operator next. chiral projections operator [latex]P_L=\frac{1}{2}(\mathbb{I}-\gamma_5=\begin{pmatrix}\delta_\alpha^\beta&0\\0&0\end{pmatrix}[/latex] [latex]P_R=\frac{1}{2}(\mathbb{I}+\gamma_5=\begin{pmatrix}0&0\\ 0&\delta^\dot{\alpha}_\dot{\beta}\end{pmatrix}[/latex] Weyl spinors [latex]\psi_L(x)=P_L\psi(x)=\begin{pmatrix}\chi_\alpha(x)\\0\end{pmatrix}[/latex] [latex]\psi_R(x)=P_R\psi(x)=\begin{pmatrix}0\\ \eta^{\dagger\dot{a}}(x)\end{pmatrix}[/latex] also requires Yukawa couplings...SU(2) matrixes given by [latex]diag(Y_{u1},Y_{u2},Y_{u3})=diag(Y_u,Y_c,Y_t)=diag(L^t_u,\mathbb{Y}_u,R_u)[/latex] [latex]diag(Y_{d1},Y_{d2},Y_{d3})=diag(Y_d,Y_s,Y_b)=diag(L^t_d,\mathbb{Y}_d,R_d[/latex] [latex]diag(Y_{\ell 1},Y_{\ell 2},Y_{\ell3})=diag(Y_e,Y_\mu,Y_\tau)=diag(L^T_\ell,\mathbb{Y}_\ell,R_\ell)[/latex] the fermion masses [latex]Y_{ui}=m_{ui}/V_u[/latex] [latex]Y_{di}=m_{di}/V_d[/latex] [latex]Y_{\ell i}=m_{\ell i}/V_\ell[/latex] Reminder notes: Dirac is massive 1/2 fermions, Weyl the massless. Majorona fermion has its own antiparticle pair while Dirac and Weyl do not. The RH neutrino would be more massive than the LH neutrino, same for the corresponding LH antineutrino and RH Neutrino via seesaw mechanism which is used with the seesaw mechanism under MSM. Under MSSM with different Higgs/higglets can be numerous seesaws. The Majorona method has conservation violations also these fermions must be electric charge neutral. (must be antiparticles of themselves) the CKM and PMNS are different mixing angels in distinction from on another. However they operate much the same way. CKM is more commonly used as its better tested to higher precision levels atm. Quark family is Dirac fermions due to electric charge cannot be its own antiparticle. Same applies to the charged lepton family. Neutrinos are members of the charge neutral lepton family Lorentz group Lorentz transformations list spherical coordinates (rotation along the z axis through an angle ) \[\theta\] \[(x^0,x^1,x^2,x^3)=(ct,r,\theta\phi)\] \[(x_0,x_1,x_2,x_3)=(-ct,r,r^2,\theta,[r^2\sin^2\theta]\phi)\] \[\acute{x}=x\cos\theta+y\sin\theta,,,\acute{y}=-x\sin\theta+y \cos\theta\] \[\Lambda^\mu_\nu=\begin{pmatrix}1&0&0&0\\0&\cos\theta&\sin\theta&0\\0&\sin\theta&\cos\theta&0\\0&0&0&1\end{pmatrix}\] generator along z axis \[k_z=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}\] generator of boost along x axis:: \[k_x=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}=-i\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}\] boost along y axis\ \[k_y=-i\begin{pmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0 \end{pmatrix}\] generator of boost along z direction \[k_z=-i\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0 \end{pmatrix}\] the above is the generator of boosts below is the generator of rotations. \[J_z=\frac{1\partial\Lambda}{i\partial\theta}|_{\theta=0}\] \[J_x=-i\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&-1&0 \end{pmatrix}\] \[J_y=-i\begin{pmatrix}0&0&0&0\\0&0&0&-1\\0&0&1&0\\0&0&0&0 \end{pmatrix}\] \[J_z=-i\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&0 \end{pmatrix}\] there is the boosts and rotations we will need and they obey commutations \[[A,B]=AB-BA\] SO(3) Rotations list set x,y,z rotation as \[\varphi,\Phi\phi\] \[R_x(\varphi)=\begin{pmatrix}1&0&0\\0&\cos\varphi&\sin\varphi\\o&-sin\varphi&cos\varphi \end{pmatrix}\] \[R_y(\phi)=\begin{pmatrix}cos\Phi&0&\sin\Phi\\0&1&0\\-sin\Phi&0&cos\Phi\end{pmatrix}\] \[R_z(\phi)=\begin{pmatrix}cos\theta&sin\theta&0\\-sin\theta&\cos\theta&o\\o&0&1 \end{pmatrix}\] Generators for each non commutative group. \[J_x=-i\frac{dR_x}{d\varphi}|_{\varphi=0}=\begin{pmatrix}0&0&0\\0&0&-i\\o&i&0\end{pmatrix}\] \[J_y=-i\frac{dR_y}{d\Phi}|_{\Phi=0}=\begin{pmatrix}0&0&-i\\0&0&0\\i&i&0\end{pmatrix}\] \[J_z=-i\frac{dR_z}{d\phi}|_{\phi=0}=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\] with angular momentum operator \[{J_i,J_J}=i\epsilon_{ijk}J_k\] with Levi-Civita \[\varepsilon_{123}=\varepsilon_{312}=\varepsilon_{231}=+1\] \[\varepsilon_{123}=\varepsilon_{321}=\varepsilon_{213}=-1\] SU(3) generators Gell Mann matrix's \[\lambda_1=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix}\] \[\lambda_2=\begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix}\] \[\lambda_3=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\] \[\lambda_4=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}\] \[\lambda_5=\begin{pmatrix}0&0&-i\\0&0&0\\i&0&0\end{pmatrix}\] \[\lambda_6=\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix}\] \[\lambda_7=\begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix}\] \[\lambda_8=\frac{1}{\sqrt{3}}\begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}\] commutation relations \[[\lambda_i\lambda_j]=2i\sum^8_{k=1}f_{ijk}\lambda_k\] with algebraic structure \[f_{123}=1,f_{147}=f_{165}=f_{246}=f_{246}=f_{257}=f_{345}=f_{376}=\frac{1}{2},f_{458}=f_{678}=\frac{3}{2}\] with Casimer Operator \[\vec{J}^2=J_x^2+J_y^2+j_z^2\] All of that is nothing more than than the relevant details for determining quark mass terms via the CKMS mass mixing matrix

Great, how does that help when the author doesn't show how he determined his conclusions ? I really don't understand why you don't grasp the author made no calculations. \[\mathcal{L}=\underbrace{\mathbb{R}}_{GR}-\overbrace{\underbrace{\frac{1}{4}F_{\mu\nu}F^{\mu\nu}}_{Yang-Mills}}^{Maxwell}+\underbrace{i\overline{\psi}\gamma^\mu D_\mu \psi}_{Dirac}+\underbrace{|D_\mu h|^2-V(|h|)}_{Higgs}+\underbrace{h\overline{\psi}\psi}_{Yukawa}\] this solves the cosmological constant problem do you believe me ?