Mordred

Honestly I tried looking for extremely low order interactions, even the numerous graphs I looked through for any superconductor studies were well out of range. Consider this a single electron has greater energy than the value you just gave. The least massive SM particle with mass being neutrinos but the momentum term is still an issue In all honesty the best chance to get the idea working and not a terrible idea in and of itself is simply get rid of the 10^123 factor. Use the Bose-Einstein methodology would solve a great deal of the problem. It would be a very easy fix that way. Thst would go a long way back to feasible by just dropping that 10^123. Secondly use pions for the SU(3) atoms being the lightest meson and use the Maxwell boltzmann statistics to calculate number density. No confusion now you have a clear defined state to work from. That would also open up a large body of similar studies involving QCD Meissner. Epions actually require higher energy levels under SUSY so don't use a SUSY particle. If it were me those would be the route to address the issues.

Yeah your right good catch must have entered something wrong on calculator and didn't spot it. Too early in am last night. Still incredibly large energy density Well above current cosmological constant in joules/meter^3 for simplicity we can ignore spherical 10^45 ev/m^3 gives 1.60*10^26 joules/m^3

The authors unit is 10^-15 meters not meters^3 for range of force. Ie radius of each volume but just in case I will check them later after my meeting. It was 1 am when I did that set so will recheck

Try again it's simple powers and division if you want to use inches 3.9374 ×10^-14 How many 10^{-15} meters fits 1 meter. Recall the author specified volume as well. Have you never learned exponent rules ? \[m^{-n}=\frac{1}{m^n}\] So take \(10^{-15}=\frac{1}{10^{15}}\)

The biggest problem here is that the universe can never have a state of quantum nothingness. Trying to solve for the quantum harmonic oscillator action can never be solved using any classical Newtonian method. I had already included the relevant equations previously to the quantum harmonic oscillator I was hoping you would have looked into them but apparently not. As this coincides with another thread on the harmonic oscillator vacuum catastrophe I'm going to port a worked solution showing the zero point energy ground state \[E_b=\sum_i(\frac{1}{2}+n_i)\hbar\omega_i\] 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). That's the calculations that led to the famous vacuum catastrophic. A large part of the corrections is renormalization methods for the reduced Hamilton and the phase summation using Pauli-Villars regularization. Renormalization procedures are not trivial unfortunately the mathematics gets quite complex Here is a paper for renormalizing the harmonic oscillator https://arxiv.org/abs/1311.6936 However a common method is dimensional regularization take the expression \[\int\frac{d^4p}{(p^2+m^2)^n}\] dimensional regularization replaces the integer with \[\int\frac{d^dp}{(p^2+m^2)^n}\] now what this allows us to do is eliminate divergences caused by integer vales of d which can be eliminated by non integer values of d (d being the power

That's a huge part of the problem but I clearly show it doesn't matter how he defines the SU(3) atom if it contains any energy even of the lowest orders he still exceeds the energy budget. Anyone is welcome to experiment with different values with he method I just described to see that and confirm for themselves. You literally cannot place a single massless or massive particle in each SU(3) atom and not do so. By whopping factors. So using SU(3) atoms whatever it contains how could it possibly solve the problem with similar error margins to the the catastrophe value ? That error margin due to the 123 power has been obvious to several of our members right from the start. lets try another route lets compare Current cosmological constant value to SU(3) atoms using one ev per 10^{-15} cubic meters which is the volume described in article shall we. Lets get to equivalent units. converting \[6.0*10^{-10} joules/m^3=3.75*10^9 ev/m^3\] now take 1 ev per SU(3) atom in volume \(10^{-15}\) meters supplied for for volume for each SU(3) atoms gives \(1.0*10^{138} ev/m^3=1.602 *10^{119} joules/m^3\) looks strikingly like the old Cosmological constant problem it was suppose to solve That 1 ev value is equivalent to a photon or any massless particle the ev value will depend on its wavelength. For 1 ev it's wavelength is 1239 nm. All I'm doing is basic mathematics and conversions here nothing fancy

What energy level is contained in each SU(3) atom. no matter what value you choose you will violate energy mass conservation when compared to the total energy of the observable universe. When you have 10^{123} SU(3) atoms. its as simple as that it doesn't matter what SU(3) is meant to describe the power 123 is the very issue here. Do you not understand rudimentary math I gave a precise means to see I am correct on that. lets try your terminology the number of microstates SU(3) atoms is too high for the observable universe that no matter what energy is contained in each microstate you surpass the total energy budget of the Observable universe error margin taking 10^93 in kg first case your proton example (yes you could do the same with the energy equivalent.) and dividing by 10^53 for total mass of the universe. WoW 10^40 times the total mass of the universe. Basic math there quite obvious. just as obvious to apply the energy momentum equation for conversions between mass and energy see that error margin yet ? try it at 1 ev per SU(3) atom Massive massless doesn't matter its the energy contained yet you argued about massless lets see \(10^{123}\) ev =\( 16.02*10^{104}\) joules apply the complete most famous equation \[E^2=\sqrt{(pc^2)+(m_oc^2)}\] that equation handles both massive and massless cases. gives \(1.783*10^{87} kg\) still an error margin of 10^34. simple math

Define energy density I want to hear what you think it is. I also want your definition of vacuum in regards to that energy density definition. You might recall I provided the equations for QCD vacuum. Show me the formula to determine the cosmological constant value for equation of state w=-1 and showing how it is derived using that formula. For all your claims you have yet to show any relevant formulas or calculations and I have done so this entire thread. So don't try to preach to me or tell me I don't know what I am talking about. What formula did I use to calculate that value ? Hint I mentioned it in this thread but never latexed it in. Lol though that's the value today what's the value at z=1100. Your relatively new here so don't really know my skills however Cosmology and particle physics I have credentials in both fields. I know the equations I mentioned they are of fundamental importance to everything I mentioned. SuSY is part of my studies. Some of those Langrangians are included by me in this thread (easily missed though if you don't know them). (Pati-Salam)

BS it is the cosmological constant has a value of roughly \[6.0*10^{-10}\] joules per cubic meter. Multiply that and see if 10^123 is equivalent. we're done its obvious you don't know what your talking about and do not wish to learn and is just trolling GL have fun

that number does not work get that through your head even if you use the lowest possible energy/ mass equivalence its wrong. PERIOD

Gluons are not DE either you cannot have an expanding universe where any particle density stays constant. Lambda is constant. Stop trying to apply the authors crap to me please. He is wrong on so many levels from what you are describing.

great What do you thing was used originally to calculate 10^90 photons of the standard model ? That number was calculated back in the 70's and is still the theoretical bounds for the universes energy budget to this very day not 10^123 using Bose-Einstein both qluons and photons have the same spin so the result will be identical

One day you will actually look at the math itself instead of trying to lecture those that have I'm fully aware of glueballs Have you ever bothered to look at the lightest theoretical value for a glueball ? https://en.wikipedia.org/wiki/Glueball try that in the method I described earlier. Multiply that by the number of SU(3) atoms then use the energy momentum formula to calculate its mass equivalence. The sheer problem is the power of 123. There is no getting around that even using photons and number of quanta a unit of quanta is in joules per hertz How many joules in one eV ? Even ignoring wavelength and just using joules if you multiply by 10 ^123 you get 10^{73} that's without the corresponding wavelength. Which we know would only get worse if you apply the wavelength.

no kidding guess I didn't know that (hear the sarcasm in my response ? did you forget \[E^2=\sqrt{(pc^2)+(m_0 c^2)}\] qluons will be whatever number density is required to mediate the color interaction between any number of quark combinations the energy equivalence to mass can be applied to above formula. That is precisely why the Langrangian includes the four momentum Momentum includes mass and velocity. After all the times I mentioned Bose-Einstein statistic's have you ever bothered to look at how it gets applied? How many times now ? for the record I tried doing the lowest order of quark combination to get a total energy for volume using the 10^{123} the lowest value I could find using the lightest meson was 10^{81} kg equivalence which is still too high. (that was using my access to feycalc at the university I do assistance instruction at.) care to tell me which meson I used ? before you mention SUSY here are the epions for SUSY for MeV energy levels https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=b86e7520a33a331471d8b4ba6e76a3d106f5c70a

no Not by any valid method I wont waste time repeating myself on the valid method via the Langrangian equations of motion, the relevant creation annihilation operators and Maxwell Boltzmann. I wasted enough time repeating myself and being ignored on those points. meson formula between two quarks page 5 example formula for quark quark interaction ground state of a bound system. \[E(r)=2m-\frac{\alpha_s}{r^2_o}+br+\frac{p^2}{m}\] where m is the mass p the momentum the radius of the ground state is \[\frac{2}{mr^3_o}=\frac{\alpha_s}{r^2_o}+b\] here is a table for you http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/quark.html apply any quark combination in that table then do the conversion with the eV Then think back to those calculations on total mass. 1 ev times 10^{123} still gives too much mass you wont get an interaction field of gluons less than 1 ev

Why does the particle keep changing ? Sounds to me a bunch of guesswork. How many pages spent on trying to determine what is meant by SU(3) atom. If it's just gluons why didn't the author simply state a gluon field to begin with... Though you would still need quarks for the gluons to mediate. So actually doesn't solve the issue. Formula already provided for the field strength between two or more quarks without quarks field strength is zero.

The million dollar prize is still available for Yang-Mills mass gap lmao

Glad we don't have to cover the distinction between real and virtual particles you have that part correct. +1

It's also a thread I gave up on getting the notion the convection of simultaneity loses any practicality in curved space-time. So it's essentially only applicable to Maximally symmetric spacetimes such as Minkowsii. You can consider the local vs global ramifications on that. At the OP the norm of the length is not the same as the norm of a vector Length alone won't include changes in direction Ie rotations acceleration due to change in direction as opposed to boosts change in magnitude of velocity.

Mediator particles are bosons so have no degeneracy pressure and yes you already learning. The reason I recommend learning valence first is to understand the significance of the outer shell before getting into ionization of hydrogen. Simplest isotope proton has positive charge which attracts negative charge the electron no neutron. H^1 Deuterium case has a neutron.H^2 Tritium case H^3 has 2 neutrons Latter 2 cases are produced through ionization and not fusion.

Degeneracy applies to all fermionic systems via the Pauli-exclusion principle. https://en.m.wikipedia.org/wiki/Electron_degeneracy_pressure Above is a simpler case but yes allies as well to quarks. To answer in the case of hydrogen your better off starting with deuterium which is a simpler case We have to be careful here as hydrogen has different isotopes and I should have specified deuterium above https://en.m.wikipedia.org/wiki/Deuterium It would be easier to understand how deuterium forms by examining valence from chemistry as a starting point. https://en.m.wikipedia.org/wiki/Valence_(chemistry)#:~:text=Hydrogen has only one valence,has an incomplete outer shell.

It's quite similar to the Yang Mills mass gap problem which is off topic but thought I would mention it formally connecting that under geometric terms is quite formal and tricky. Well above the nature of this thread.

For me it's cold when your spit freezes before it hits the ground lol.

Here is another detail to consider. Photon polarization already exists and can be shown through the Maxwell equations as the primary basis. In QM they make use of this. For example parametric downconversion relies on photon polarization for a huge range of experiments involving entsnglement where the polarization filter seperates the polarization waves from monochromatic light. Here is the mathematical basis (step by step) including a useful mathematica program. https://bingweb.binghamton.edu/~suzuki/QuantumMechanicsFiles/2-1_Photon_polarization.pdf These polarization relations are also used in QM/QFT for charge conjugation involving photons/antiphotons where the charge conjugation relationship distinguishes whether or not a photon is a photon or antiphoton based on its polarization helicity. So how will your theory work under the above ?

Yeah whatever you say pick any particle to particle interaction under the SU(3) descriptive the orders magnitude will still be wrong. I did those calculations on systems without interactions just scalar particle fields. You include interactions that 10^{123} will still be wrong. Pick any interacting system with that number value go ahead do the math on it. Go ahead give your system a single eV value for each SU(3) atom. You certainly can't have interactions without particles so good luck with that one.....