Mordred Posted Tuesday at 11:13 PM Share Posted Tuesday at 11:13 PM (edited) 24 minutes ago, JosephDavid said: If you can solve something with a simple relation, why complicate it by adding more language ? As Newton once said, "Truth is ever to be found in simplicity, and not in the multiplicity and confusion of things." It’s like in language: why say "a round object used in games" when you can just say "ball"? The author found a simple, logical relationship that ties together the relevant measurements to address the issue. 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 Edited Tuesday at 11:36 PM by Mordred Link to comment Share on other sites More sharing options...

MJ kihara Posted Wednesday at 12:00 AM Share Posted Wednesday at 12:00 AM Pause a little bit what are you deriving at ? In relation to cosmological constant. Link to comment Share on other sites More sharing options...

Mordred Posted Wednesday at 12:15 AM Share Posted Wednesday at 12:15 AM (edited) 17 minutes ago, MJ kihara said: Pause a little bit what are you deriving at ? In relation to cosmological constant. 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... Edited Wednesday at 12:19 AM by Mordred Link to comment Share on other sites More sharing options...

swansont Posted Wednesday at 12:38 AM Share Posted Wednesday at 12:38 AM 1 hour ago, JosephDavid said: Physics is fundamentally an empirical science, its foundation is built on measurements, observations, and experiments. In the beginning of the thread I asked about the path to experimental confirmation. If you’re going to appeal to this, let’s see the evidence of these SU(3) atoms or the superconducting nature of dark energy. Link to comment Share on other sites More sharing options...

MigL Posted Wednesday at 01:44 AM Share Posted Wednesday at 01:44 AM 4 hours ago, JosephDavid said: we know that protons don’t decay and are associated with SU(3) symmetry, then it makes perfect sense to use the volume of the proton as the benchmark for this remnant state You don't see that this is an assertion with no evidence, physical or mathematical, to back it up ? It just so happens to slot into your pet theory, so you run with it. Why not pick the hydrogen atom as the remnant volume ? Or why not up/down quarks as remnant volume since they are also stable, and involved with SU(3) symmetry ? I suspect it's because it doesn't produce a neat answer with numbers that match the magnitudes you require for your pet theory. Until the foundation is solid, and if you ( the author ) can connect all the dots, it is just a theory based on numerology. 1 Link to comment Share on other sites More sharing options...

JosephDavid Posted Wednesday at 02:04 AM Share Posted Wednesday at 02:04 AM 16 minutes ago, MigL said: You don't see that this is an assertion with no evidence, physical or mathematical, to back it up ? It just so happens to slot into your pet theory, so you run with it. Why not pick the hydrogen atom as the remnant volume ? Or why not up/down quarks as remnant volume since they are also stable, and involved with SU(3) symmetry ? I suspect it's because it doesn't produce a neat answer with numbers that match the magnitudes you require for your pet theory. Until the foundation is solid, and if you ( the author ) can connect all the dots, it is just a theory based on numerology. Good questions. Why not the hydrogen atom? Well, hydrogen is a little more complicated. It’s made of a proton and an electron and that electron’s got U(1) symmetry written all over it, which breaks at low temperatures. So, hydrogen isn't the best candidate because it's tied to electromagnetic interactions, not exactly stable if you're looking at what happens when U(1) breaks. The proton, though? That’s a different story. Now, about quarks. Sure, up and down quarks are part of the SU(3) game, but they’ve got a problem: you never find them alone. They’re always confined, stuck inside protons, neutrons, whatever. Quarks are like that friend who never leaves the party alone, they’re always bound together, thanks to quark confinement. The proton, on the other hand, is the smallest, most stable package of SU(3) stuff we can actually observe. It’s been tested again and again, and it’s never been seen to decay. So why pick the proton as the remnant volume? Because it’s the simplest stable bound state that’s pure SU(3). It’s the one structure that stays unbriken under the third law of thermodynamics, when you get to those really low temperatures, it’s still standing. That’s what makes it perfect for this context. We need something stable, something minimal, and something that stays intact with SU(3) symmetry. That’s the proton. It’s not just a random pick, it’s the only choice that makes sense if you want a fundamental remnant that’s consistent with what we observe in the vacuum. 1 Link to comment Share on other sites More sharing options...

Mordred Posted Wednesday at 03:48 AM Share Posted Wednesday at 03:48 AM (edited) 1 hour ago, JosephDavid said: Good questions. Why not the hydrogen atom? Well, hydrogen is a little more complicated. It’s made of a proton and an electron and that electron’s got U(1) symmetry written all over it, which breaks at low temperatures. So, hydrogen isn't the best candidate because it's tied to electromagnetic interactions, not exactly stable if you're looking at what happens when U(1) breaks. The proton, though? That’s a different story. Now, about quarks. Sure, up and down quarks are part of the SU(3) game, but they’ve got a problem: you never find them alone. They’re always confined, stuck inside protons, neutrons, whatever. Quarks are like that friend who never leaves the party alone, they’re always bound together, thanks to quark confinement. The proton, on the other hand, is the smallest, most stable package of SU(3) stuff we can actually observe. It’s been tested again and again, and it’s never been seen to decay. So why pick the proton as the remnant volume? Because it’s the simplest stable bound state that’s pure SU(3). It’s the one structure that stays unbriken under the third law of thermodynamics, when you get to those really low temperatures, it’s still standing. That’s what makes it perfect for this context. We need something stable, something minimal, and something that stays intact with SU(3) symmetry. That’s the proton. It’s not just a random pick, it’s the only choice that makes sense if you want a fundamental remnant that’s consistent with what we observe in the vacuum. 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 On 10/23/2024 at 9:17 PM, Mordred said: Here is a simpler breakdown using operators ie QM or QFT which would have been far easier to apply symmetry breaking with regards to the paper https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://faculty.pku.edu.cn/_resources/group1/M00/00/0D/cxv0BF5mDfKAOPDbACEBbOQBol4139.pdf&ved=2ahUKEwj-jdiwhaaJAxWvJTQIHZQ3C58QFnoECDYQAQ&usg=AOvVaw0bF6zGAJhSK_UcqzuzVZ4o There is a reason why the vacuum catastrophe is also called the EM field ultraviolet catastrophe the problem is directly related to how it was renormalized... Here is a quick breakdown of the method I would have liked the author to have applied . Bose Einstein QFT format. |k1→k2→〉a^†(k1→)a^†(k2→)|0〉 ⇒|k1→k2→〉=|k2→k1→〉 number operator N^=a^†(k⃗ )a^k⃗ ) Hamilton operator H^=∫d3kωk[N^(k⃗ )+12] momentum of field P^=∫d3kk⃗ [N^(k⃗ )+12] renormlized Hamilton Hr^=∫d3kωka^†(k⃗ )a^(k⃗ ) Now for the full SU(3) Langrangian L=ψ¯fiγμ∂μψf0ψ¯fψf+goψ¯fγμtaψf−14FaμνFμνa where Fμνa=∂νAνa−∂νAνa+goFabcAμbAνc where a=(1,2.....8) for the gluon fields =26 fields=6 flavors+3 colors+8 gauge bosons gives 7 parameters+1 coupling There is SU(3)c notice that this also applies to the weak force with 6 flavors and the 8 gauge bosons for the strong force ? that's the full QCD langrangian the SU(3) langrangian but that still doesn't include the Higgs couplings? So once again I ask what the bugger is an SU(3) atom as the only Langragian the author included was the QED langrangian. 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) Edited Wednesday at 04:05 AM by Mordred Link to comment Share on other sites More sharing options...

Mordred Posted Wednesday at 04:48 AM Share Posted Wednesday at 04:48 AM (edited) 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 Edited Wednesday at 05:49 AM by Mordred Link to comment Share on other sites More sharing options...

MJ kihara Posted Wednesday at 11:17 AM Share Posted Wednesday at 11:17 AM 6 hours ago, Mordred said: includes a constraint to prevent that RHS of the ∫∞0 Which constraint? Link to comment Share on other sites More sharing options...

swansont Posted Wednesday at 11:24 AM Share Posted Wednesday at 11:24 AM 9 hours ago, JosephDavid said: So why pick the proton as the remnant volume? Because it’s the simplest stable bound state that’s pure SU(3). It’s the one structure that stays unbriken under the third law of thermodynamics, when you get to those really low temperatures, it’s still standing. That’s what makes it perfect for this context. We need something stable, something minimal, and something that stays intact with SU(3) symmetry. That’s the proton. It’s not just a random pick, it’s the only choice that makes sense if you want a fundamental remnant that’s consistent with what we observe in the vacuum. What you’ve described is convenience, not something based on either experiment or theory. There’s nothing wrong with picking a convenient reference up until you assert there’s something physically meaningful about it. Link to comment Share on other sites More sharing options...

Mordred Posted Wednesday at 11:48 AM Share Posted Wednesday at 11:48 AM (edited) 1 hour ago, MJ kihara said: Which constraint? \[\rho_\Lambda c^2=\int^\infty_0=\frac{4\pi k^2 dk}{(3\pi\hbar)^3}(\frac{1}{2}\sqrt{k^2c^2+m^2c^4})\] 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 Edited Wednesday at 12:21 PM by Mordred Link to comment Share on other sites More sharing options...

Mordred Posted Wednesday at 12:54 PM Share Posted Wednesday at 12:54 PM (edited) 8 hours ago, Mordred said: 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 Now take my last post apply that to the quoted section. Little lesson on how to recognize a hermitean matrix. Edited Wednesday at 12:56 PM by Mordred Link to comment Share on other sites More sharing options...

MJ kihara Posted Wednesday at 12:56 PM Share Posted Wednesday at 12:56 PM ....In Quantum Field Theory (QFT), as discussed by Steven Weinberg [19], the vacuum energy density arises from summing the zero-point energies of quantum fields, including massless bosons like photons and gravitons. It is calculated as: ρ QFT vac c2 = (2π1ℏ)3 Z 0PPl d3p ℏω p 2 , (17) where PPl = EPl/c is the Planck momentum, and ωp = c|⃗p| for massless particles. Integrating up to the Planck scale yields: ρ QFT vac c2 ≈ E 4 Pl ( ℏc )3 ≈ 1076 GeV4/(ℏc)3 Page 18 of the author article....they are considering all fields,and his cut off is planck scale, meaning that it's taking care of volume smaller than the on considered by SU(3) symmetry unit. Your approach and this one which one is more advantageous? Link to comment Share on other sites More sharing options...

JosephDavid Posted Wednesday at 02:36 PM Share Posted Wednesday at 02:36 PM 3 hours ago, swansont said: What you’ve described is convenience, not something based on either experiment or theory. There’s nothing wrong with picking a convenient reference up until you assert there’s something physically meaningful about it. Well, the choice of the proton isn’t about convenience, it’s about its physical significance. The proton is stable; we’ve never seen it decay, and it stands strong under SU(3) symmetry, even when U(1) breaks down. That’s what gives it physical meaning beyond just being a handy reference. It’s not a random choice, it’s a backed by what we know from experiment and theory. Link to comment Share on other sites More sharing options...

TheVat Posted Wednesday at 03:43 PM Share Posted Wednesday at 03:43 PM (edited) Following only tiny (Planck scale) bits and pieces of the chat here. So these hermitian operators are describing observable quantities for a quantum system, right? So matching actual experimental observations is key? In dumbed-down-for-biology-majors terms, what kind of observations would confirm or disprove the OP hypothesis? Why would a quark flavor symmetry in chromodynamics apply so broadly? 1 hour ago, JosephDavid said: The proton is stable; we’ve never seen it decay, This strikes me as having a Popperian "black swan" issue, in terms of a definitive observation establishing that protons don't decay. Edited Wednesday at 04:00 PM by TheVat pyto Link to comment Share on other sites More sharing options...

Mordred Posted Wednesday at 04:09 PM Share Posted Wednesday at 04:09 PM 3 hours ago, MJ kihara said: ....In Quantum Field Theory (QFT), as discussed by Steven Weinberg [19], the vacuum energy density arises from summing the zero-point energies of quantum fields, including massless bosons like photons and gravitons. It is calculated as: ρ QFT vac c2 = (2π1ℏ)3 Z 0PPl d3p ℏω p 2 , (17) where PPl = EPl/c is the Planck momentum, and ωp = c|⃗p| for massless particles. Integrating up to the Planck scale yields: ρ QFT vac c2 ≈ E 4 Pl ( ℏc )3 ≈ 1076 GeV4/(ℏc)3 Page 18 of the author article....they are considering all fields,and his cut off is planck scale, meaning that it's taking care of volume smaller than the on considered by SU(3) symmetry unit. Your approach and this one which one is more advantageous? 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. 26 minutes ago, TheVat said: Following only tiny (Planck scale) bits and pieces of the chat here. So these hermitian operators are describing observable quantities for a quantum system, right ? 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. 30 minutes ago, TheVat said: In dumbed-down-for-biology-majors terms, what kind of observations would confirm or disprove the OP hypothesis? Why would a quark flavor symmetry in chromodynamics apply so broadly? Author doesn't give the relevant details to address that question. Link to comment Share on other sites More sharing options...

MigL Posted Wednesday at 05:32 PM Share Posted Wednesday at 05:32 PM 15 hours ago, JosephDavid said: Good questions. Why not the hydrogen atom? Sorry my mistake; meant to write Helium atom. Stable, and a superfluid. 1 hour ago, TheVat said: what kind of observations would confirm or disprove the OP Observation of a Casimir effect between separations of less than the 'proton volume' ? I can't think of a method for doing this, but , it would indicate vacuum energy must be considered at much smaller scales. Link to comment Share on other sites More sharing options...

Mordred Posted Wednesday at 06:08 PM Share Posted Wednesday at 06:08 PM (edited) 42 minutes ago, MigL said: Observation of a Casimir effect between separations of less than the 'proton volume' ? I can't think of a method for doing this, but , it would indicate vacuum energy must be considered at much smaller scales. 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 ? Edited Wednesday at 06:16 PM by Mordred Link to comment Share on other sites More sharing options...

swansont Posted Wednesday at 07:06 PM Share Posted Wednesday at 07:06 PM 4 hours ago, JosephDavid said: Well, the choice of the proton isn’t about convenience, it’s about its physical significance. The proton is stable; we’ve never seen it decay, and it stands strong under SU(3) symmetry, even when U(1) breaks down. That’s what gives it physical meaning beyond just being a handy reference. It’s not a random choice, it’s a backed by what we know from experiment and theory. I didn’t say it was random. You have not connected it with anything. It’s not like it’s the volume of the proton that keeps it from decaying. Free neutrons decay. Link to comment Share on other sites More sharing options...

JosephDavid Posted Thursday at 02:06 AM Share Posted Thursday at 02:06 AM 8 hours ago, MigL said: Sorry my mistake; meant to write Helium atom. Stable, and a superfluid. Good question indeed. As you get closer to absolute zero, the third law of thermodynamics is saying, "I want a system that’s as orderly and low-entropy as possible." Now, look at liquid helium. Sure, it’s fascinating stuff—flows without friction, does all sorts of quantum tricks. But even down there near zero, it’s still got a lot of atomic hustle and too many possible states, which means higher entropy. Enter the proton. This thing is rock-solid, It’s stable, and has very few microstates to mess with, exactly what the third law ordered for a low-entropy, high-order setup at the coldest extremes. So, if we are looking for the perfect fit for a minimal-entropy remnant, the proton’s our guy. 8 hours ago, MigL said: Observation of a Casimir effect between separations of less than the 'proton volume' ? I can't think of a method for doing this, but , it would indicate vacuum energy must be considered at much smaller scales. This is an interesting note. Can you give more details? 7 hours ago, swansont said: You have not connected it with anything. It’s not like it’s the volume of the proton that keeps it from decaying. Free neutrons decay. It’s not just about volume, what really matters is remnant volume with minimal entropy. The third law of thermodynamics is looking for something stable and with the least possible disorder. Now, the neutron might look similar in size to the proton, but here’s the catch: free neutrons don’t stick around, they decay, and that means extra entropy. The proton, on the other hand, is rock solid. It’s stable, no decay, and has minimal entropy because it doesn’t have extra states to fall into. So if you’re looking for a low-entropy remnant that satisfies the third law, it’s the proton that wins. Link to comment Share on other sites More sharing options...

Mordred Posted Thursday at 04:09 AM Share Posted Thursday at 04:09 AM (edited) @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 Edited Thursday at 05:42 AM by Mordred Link to comment Share on other sites More sharing options...

MigL Posted Thursday at 12:53 PM Share Posted Thursday at 12:53 PM 10 hours ago, JosephDavid said: free neutrons don’t stick around, they decay, and that means extra entropy. The proton, on the other hand, is rock solid. It’s stable, no decay, and has minimal entropy because it doesn’t have extra states to fall into. Neutrons, being composed of the same building blocks as protons would seem to have as many states as a proton, except a much much longer lifetime.. You are using the fact that we have never seen proton decay, to assume hey never do, and the further assumption that this indicates that the proton volume is equivalent to the SU(3) stability volume because of thermodynamic considerations related to proton stability. But how is this related to the SU(3) minimum volume stability ? You cannot use the same thermodynamic considerations to argue for 'volume' stability. And 'volume' is what you are using to subdivide the unreasonably high vacuum energy, not protons. Link to comment Share on other sites More sharing options...

JosephDavid Posted Thursday at 01:37 PM Share Posted Thursday at 01:37 PM 40 minutes ago, MigL said: Neutrons, being composed of the same building blocks as protons would seem to have as many states as a proton, except a much much longer lifetime.. You are using the fact that we have never seen proton decay, to assume hey never do, and the further assumption that this indicates that the proton volume is equivalent to the SU(3) stability volume because of thermodynamic considerations related to proton stability. But how is this related to the SU(3) minimum volume stability ? You cannot use the same thermodynamic considerations to argue for 'volume' stability. And 'volume' is what you are using to subdivide the unreasonably high vacuum energy, not protons. Your questions made me understand the author arguments more clearly. The proton and neutron might seem similar since they’re both built from quarks held together by the strong force, but they’re not the same when it comes to stability and entropy. Neutrons? They decay, breaking down into a proton, an electron, and a neutrino. That breakdown matters, it means more ways for things to fall apart, more “messiness” or entropy in the system. The proton, though, is different. It doesn’t decay (at least, we’ve never seen it happen), and it doesn’t have any available decay channels in the standard model that wouldn’t violate fundamental conservation laws, like charge conservation. Simply put, it’s stable because the rules of physics don’t give it a way out. So, the proton’s stability is directly tied to these conservation laws. This stability means the proton has minimal entropy, it doesn’t have all those extra states it could move into. And here’s where the third law of thermodynamics kicks in: as a system gets down to close to absolute zero, it demands a setup with as little entropy as possible. This is why the proton fits the bill perfectly. When we’re looking to break down the vacuum energy into stable, low-entropy “units” in the context of SU(3), it’s not just about taking any volume, we need something as stable as they come, with minimal entropy. The proton’s locked-down, conservation-law-protected structure makes it the ideal candidate for this job. So, when we talk about the proton as representing an SU(3) vacuum atom, it’s not just about its size; it’s about its role as the ultimate low-entropy, stable unit in the structure of the vacuum. Link to comment Share on other sites More sharing options...

joigus Posted Thursday at 01:45 PM Share Posted Thursday at 01:45 PM (edited) So these atoms are there. They're not a vacuum. They're a background. Why is it high-precission tests of the standard model haven't found them yet? Hadrons of all kinds, from high-energy beams and jets would certainly scatter off SU(3) bound states, with a very sizable cross section. On top of that, they are 10^{43} times more abundant than ordinary neutrons, so the luminosities would be over the roof. Why haven't we seen the littlest inkling that they're there? Don't think for a moment I haven't noticed you aren't answering any of this. This thread is starting to smell with a really foul stink. Edited Thursday at 01:45 PM by joigus minor correction 1 Link to comment Share on other sites More sharing options...

swansont Posted Thursday at 01:58 PM Share Posted Thursday at 01:58 PM 16 minutes ago, JosephDavid said: So, the proton’s stability is directly tied to these conservation laws. This stability means the proton has minimal entropy, it doesn’t have all those extra states it could move into The neutron doesn’t decay because it has a different entropy. Protons can decay, after all (beta plus). Just not free protons. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now