Mordred

Lol Though I never push personal favorite theories most my studies revolve around sterile neutrinos being a viable possibility. I do see a ton of lousy examinations of that possibility though. Part of attraction in that regards is that anti-neutrinos have never been detected though predicted by the models. That aside I find all the push of seeking "new physics" support articles rather annoying. I feel in those cases one should only look for "new physics " once all possibility of mainstream or current physics is clearly evident. I recall all the different Universe geometry papers prior to WMAP and Planck datasets the variations of possible geometries were staggering. So I tend to think of the current craze topic articles rather normal, as they've been around for pretty much any physics topic particularly when describing dynamics etc poorly understood or has a good body of clear evidence. Something that is oft overlooked in physics though is the mathematical developments finding ways to reduce computations of complex systems or finding means to organize all the possibilities such as through the use of gauge groups is oft never a big media blitz so you rarely hear of them. Yet they directly relate to development of fundamental physics . Too often it's the big media hits that get recognized rather than all the seemingly little discoveries or improved methodologies yet the little improvements could very well have a far greater range of applicability across other theories where they can be deployed.

While I agree with the comments above a couple other considerations in the past few decades several milestone advances have been reached. Examples being discovery of the Higgs boson this has the effect of causing a rather large paradigm shift. Detecting gravity waves Confirming neutrino mass Improved measurements of the CMB. Reduced number of possible geometries of the universe significantly. Those are just some examples in many ways were still catching up to those new available findings listed above.

good example of "beauty the beast" with that bite descriptive lol

@DimaMazin I would like you to consider the following scenario now in the above signals sent by A and signals sent by B will be received at point P simultaneous as point P is the center between A and B. When you have a change in the magnitude component of velocity although you have length contraction this will not change. Treat this as 3 observers/emitters A,B,P However if that train goes around a corner which is also acceleration this is no longer true. If you were to draw a line between A and B Observer P will not fall on that line. It would not receive that signal. Instead P will only receive signals along the line from A to P and B to P but not the signal between A and B. So you are no longer dealing with one geodesic path of light but now three separate geodesic light paths. The geodesic path between A and B of the second case no longer goes through P but follows the hypotenuse connecting A and B If you were to say add another point in the second graph along the line connecting A and B and place another observer there. say Observer P_2. Then the events seen by P and P_2 observing A and B will not be seen simultaneous to each other by Observer P and P_2. This is a simple example where the notion of simultaneity breaks down. You have the same issue with curved spacetime. That issue is handled via rotations of the Lorentz transformations via a rotation matrix. Whereas is the first case change in magnitude of velocity it is a Boost not a rotation however in curved spacetime the light follows the curvature. These graphs ae simply a simple way to show the distinction between Lorentz transforms involving the two types of acceleration. A way to help understand why Proper time lies on a point of a geodesic path. That point is described by placing a tangent line to the surface of the curvature and have a parallel transported vector 90 degrees to that point connecting the Tangent vector to the curvature vector along with the perpendicular to the tangent connecting at the same point. Your convectors used under GR see figure 1.2 https://amslaurea.unibo.it/18755/1/Raychaudhuri.pdf figure 1.2 can be used to describe an Einstein train following a geodesic path not curvature of a physical object but rather the signal via photons. It show parallel transport along a geodesic path and the related mathematics that not only describe the observers but also how one can measure curvature using parallel transport which when you get down to is a form of train Einstein train PS In parallel transport the length connecting the parallel transported covectors is the affine connection in essence each section of the train following the geodesic please study the article and note how it leads ups to Principle of General Covariance: “The laws of physics in a general reference frame are obtained from the laws of Special Relativity by replacing tensor quantities of the Lorentz group with tensor quantities of the space-time manifold.”

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.