Mordred

I decided to do a bit of calculations The observable Universe mass using Critical density is estimated to be \[10^{53} kg.\] So using \(10^{123}\) protons at 936 MeV The corresponding mass is \[1.669 ×10^{96}\] kg Talk about a HUGE mismatch lol thought I would share that. I seriously hope the author isn't using protons or neutrons the theory would automatically be invalid simply on that calculation. There simply put absolutely no way possible to solve the cosmological constant problem with such a large mismatch none whatsoever superconductivity or not. It's literally impossible with 10^123 protons or neutrons Anyone want to try simply multiply 936 MeV times 10^{123} then convert to Kg with e=mc^2

The worse part is I know the mathematics behind every theory that's been mentioned. 35 to 40 years of continous study teaches a lot. So one can only imagine what these condensending tones sound like to me lmao 🤣 😂 😆 😅 Take Maxwell Boltzmann for example SUSY QFT and All apply it. It existed prior to all the above. It's been integrated into all the above. That's the understanding one gains when they sit down and study the mathematics of a given theory.

You can believe what you want about physics here is a little trivia for you it doesn't make any difference whether your describing a system using SUSY or QFT or even classical physics. Every theory must comply with observational evidence. Having \(10^{123}\) protons in our universe exceeds All observational evidence for the mass/energy of the observable universe. Thst detail trumps any theory that states otherwise. Plain and simple. If you ran that mass term through the FLRW matter dominant equations the very universe would collapse. No theory becomes mainstream without rigorous testing via experimental evidence.

It's still amazing that we choose to ignore conservation of mass energy in all the above in favor of a model with no calculations. Sigh I give up if you wish to believe in some paper that on a couple of occasions flat out lies (example mass of photon in OP paper) Feel free I have better things to do. I don't feel like arguing that throwing away decades of active research for mainstream physics that you want to throw away in favor of some paper that doesn't show any qualitative calculations is the wrong approach.

The amplitudes are directly related to the anplitudes inside a proton. Recall All particles are field excitations. Not little balls of matter. Great idea take 936 MeV and multiply it by 10^{123} atoms how much energy does that give ? One doesn't need to be a mathematician to see it will exceed 10^19 GeV which is the total energy density at BB. Exceeding total energy/mass of the universe. (Ps 10^19 GeV is the Planck temp cutoff when you convert to Kelvin) Lol you could for example assume each SU(3) atom has exactly 1 quanta of energy and do the same calculation above just looking at the powers indicate it will exceed also.

I don't particularly have a problem with any chosen particle. I mentioned that numerous times. If you look back though my issue is regardless of any chosen particle or particle field you should still apply Maxwell Boltzmann and not simply use volumes. Secondly all quantum fields has an inherent quantum uncertainty regardless of temperature. I also showed that the calculations for a QCD vacuum is distinctive to a QED vacuum. I also includes peer reviewed links describing dual Meissner for QCD. Not just a single Meissner for QED. This is the details the author didn't include or didn't examine. Let me ask you how many formulas has the author posted showing the numerous amplitudes contained within a proton ? Each field within that proton has inherent uncertainty. So how precisely does that match up to a single vector field calculation for the vacuum catastrophe when not even the electric charges match between quarks and electrons ? The amplitudes mediating the electric charges between protons and electrons don't match each other either. That was part of that examination I did earlier. If the author had applied those missing details I wouldn't have any real problem however he didn't looked deep enough ie into the mathematical proofs of the theories he tries to put together. He doesn't show the first second third and fourth NLO (next leading order integrals involved) In essence he's ignoring a huge set of amplitudes with regards to protons/neutrons etc. Every time you use a Greens Function with regards to any Hamilton has uncertainty and that's every single wavefunction in QFT or QM. You have additional uncertainty adding to a total sum .

What your missing is an essential aspect that I have repeated numerous times. Take a bottle and fill it with neutrons treat each neutron as a microstate. Lets completely ignore internal microstates. With decay over time the number of neutrons will be less over time. This is precisely why I repeatedly mention number density calculations. The number density is also affected by temperature So in terms of entropy you really can't look at stability alone as the stability of neutrons is gained through the formation of deuterium. So if that same bottle is filled with deuterium there is no change in number density. Regardless of any instability of free neutrons. This is something the author never took into consideration he used volume when he should have used number density via Maxwell-Boltzmann which would have given him all the needed requirements to apply for calculating the number of protons or neutrons at any given temperature to apply to entropy. The entropy formulas are a direct result of Maxwell Boltzmann in regards to particle physics and cosmology applications. So why wasn't that method applied to begin with ?

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