Mordred

The main advantage is the potential to detect longer wavelengths the arm length of LIGO can only accept a certain range. Longer arm detection range will allow detection of longer GW wavelengths much like that of an antenna for optimal detection is a quarter wavelength. LIGO however uses multiple beams to increase its sensitivity range. Though quite frankly any GW waves generated by Phobos is well out of any practical means of detection.

Just side note I bet I can give you a relation you never seen nor considered with regards to the Hubble parameter. \[H=\frac{1.66\sqrt{g_*}T^2}{M_{pl}}\] could this have anything to do with those equations of state I keep mentioning ? Ie the thermodynamic contributions of all particle species with regards to determining rate of expansion. Maybe this will help Please explain why your other article includes the equations state and thermodynamic relations to pressure Which obviously involves kinetic energy but this article doesn't I just do not get that though from the varying DE term that sounds more in line with quintessence which wouldn't be w=-1

You may find the Christoffels useful at some point in time so here if your interested. If not no worries https://www.scienceforums.net/topic/128332-early-universe-nucleosynthesis/page/3/#findComment-1272671 Even though you believe the SM model method is wrong here is how expansion rates for H is derived as a function of Z. I will let you figure out how your own personal model works from the mainstream physics That onus is yours and not mine. FLRW Metric equations \[d{s^2}=-{c^2}d{t^2}+a({t^2})[d{r^2}+{S,k}{(r)^2}d\Omega^2]\] \[S\kappa(r)= \begin{cases} R sin(r/R &(k=+1)\\ r &(k=0)\\ R sin(r/R) &(k=-1) \end {cases}\] \[\rho_{crit} = \frac{3c^2H^2}{8\pi G}\] \[H^2=(\frac{\dot{a}}{a})^2=\frac{8 \pi G}{3}\rho+\frac{\Lambda}{3}-\frac{k}{a^2}\] setting \[T^{\mu\nu}_\nu=0\] gives the energy stress mometum tensor as \[T^{\mu\nu}=pg^{\mu\nu}+(p=\rho)U^\mu U^\nu)\] \[T^{\mu\nu}_\nu\sim\frac{d}{dt}(\rho a^3)+p(\frac{d}{dt}(a^3)=0\] which describes the conservation of energy of a perfect fluid in commoving coordinates describes by the scale factor a with curvature term K=0. the related GR solution the the above will be the Newton approximation. \[G_{\mu\nu}=\eta_{\mu\nu}+H_{\mu\nu}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}\] Thermodynamics Tds=DU+pDV Adiabatic and isentropic fluid (closed system) equation of state \[w=\frac{\rho}{p}\sim p=\omega\rho\] \[\frac{d}{d}(\rho a^3)=-p\frac{d}{dt}(a^3)=-3H\omega(\rho a^3)\] as radiation equation of state is \[p_R=\rho_R/3\equiv \omega=1/3 \] radiation density in thermal equilibrium is therefore \[\rho_R=\frac{\pi^2}{30}{g_{*S}=\sum_{i=bosons}gi(\frac{T_i}{T})^3+\frac{7}{8}\sum_{i=fermions}gi(\frac{T_i}{T})}^3 \] \[S=\frac{2\pi^2}{45}g_{*s}(at)^3=constant\] temperature scales inversely to the scale factor giving \[T=T_O(1+z)\] with the density evolution of radiation, matter and Lambda given as a function of z \[H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}\] How you choose to get your model working is your problem . I'm simply challenging your model using main stream physics and relevant questions while providing some guidance on the relevant main stream relations. What you do with them is your problem. Particularly since you choose to not include KE or pressure to prevent gravitational collapse. Does your theory have a critical density I have no idea might be relevant Here is a handy aid for the issue of expansion vs gravitational collapse Apply this (its how the critical density formula got derived.) Along with GR of course. https://www.physics.drexel.edu/~steve/Courses/Physics-431/jeans_instability.pdf In particular see equation 28 for \[\frac{3}{5}\frac{GM^2}{R}\]

For the record I did look at your paper where you argued that negative pressure is invalid. My question still remains the value you give in equation 90 is the OLD cosmological problem not the new cosmological problem the old cosmological problem is called the vacuum catastrophe. The new cosmological problem is why is the value measured so low compared to the calculated OLD cosmological problem. The value you have in equation 90 is not the value measured for Lambda. I don't agree with much your other paper either but hey if you think using the vacuum catastrophe value serves you good luck with that the accepted professional value measured is roughly 6.0∗10−10joules/m3 or 10−27kg/m3 but good luck on applying ZPE to the measured value for Lambda. After all its only 120 orders of magnitude off the mark if you want a clear demonstration of the above statement this forum had a recent other related thread on it and Migl posted an excellent video discussing the problem. feel free to watch it here is Sean Caroll's coverage and no I didn't get my previous equations from this article but the article does contain them. They are well known relations that I regularly use and thus took the time to create my own set of notes how each equation of state is derived and how to use QFT to describe each as well as how the Raychaudhuri equations can also derive the first and second Freidmann equations. Some of those notes I have on this forum as a time saver. https://arxiv.org/pdf/astro-ph/0004075 Here are the Raychaudhuri relations I mentioned. https://amslaurea.unibo.it/18755/1/Raychaudhuri.pdf

yep thanks for confirming thread reported you still fail to understand you cannot apply SR over the entire global metric beyond Hubble horizon.

something tells me your a sockpuppet but just in case its already been shown there is relations between cosmological redshift and time dilation however they won't be accurate without corrections beyond Hubble Horizon you need to employ further corrections as the (1+z) relation is only accurate in the near field.

It doesn't take a genius to figure out in order to describe how a uniform mass distribution can expand by one basic equation that exists in the FLRW metric. The equation already exists for that and that equation uses both PE and KE and works with the Newtons Shell theorem. Obviously you don't want to include the necessary KE component. Trying to use just PE is insufficient. That equation is the scalar field equation of state . Of course I'm going to follow examples pioneered by others I know they work I would be foolish not to. Those examples provide key lessons your choosing to ignore. The article I provided clearly demonstrates that it requires both PE and KE terms to arrive at the first FLRW metric equation. If you choose to ignore that detail that's your choice but the method in article works. Yet you choose to ignore it. So I can't really help you as you would rather try to reinvent physics that works in favor of your model that you would not be able to calculate an expansion rate or even answer the question 1) what prevents your universe from collapsing under self gravity 2) what expansion rate will it have So good luck to you. You don't even have a means to derive a pressure term to determine vacuum energy you require KE vs PE for that so your on your own. Quite frankly it's foolish to ignore Kinetic energy =energy due to motion is required for pressure which is required for vacuum energy. That is a very basic classic physics lesson your choosing to ignore. It's the commonly used method by any professional physicist because it works. All you need to do is study the equations for a piston to recognize those pressure equations uses both Potential and kinetic energy. So why you choose to ignore that lesson I have no idea. The equations of state which relate the pressure terms to energy density uses the same hydrodynamic relations from classical physics lessons such as that piston example. So your blooming right I will rely on the work pioneered by 100's of years of collective research. I know those methods work as opposed to trying to determine pressure terms for a vacuum as opposed to your method. Bloody right I will favor the standard method over yours. \[\omega=\frac{P}{\rho}\] pressure to energy density relation above. scalar field equation of state \[\omega=\frac{\frac{1}{2}\dot{\phi}^2-V(\phi)}{\frac{1}{2}\dot{\phi}^2+V(\phi)}\] scalar field equation of state has both pressure and energy density. When the kinetic energy exceeds the pressure you get the negative pressure relation w=-1 for an incompressible fluid and only the w=-1 value is the only value that gives a vacuum pressure that is constant. that is the lesson you choose to ignore There is VERY good reasons why the FLRW metric uses Pressure to energy density relations and those relations involve both Kinetic energy and potential energy. NOT JUST POTENTIAL ENERGY. https://en.wikipedia.org/wiki/Equation_of_state_(cosmology) there is good reasons those equations of state are used by the professional community. Reasons you choose to ignore lol if you really want to understand that last equation you can literally track it back to Bernoulli's Principle in fluid mechanics. It employs the same principles However as the FLRW metric applies GR its better to use the Einstein Field equations. which quite frankly Your method is not compatible with as you are ignoring the Stress energy momentum tensor. \[T^{\mu\nu}=(\rho+P)\mu^\mu\mu^\nu+pg^{\mu\nu}\] this is the equation that those equations of state derived from via Raychaudhuri equations. So Yes I choose the GR method over your any day. I know they work and how they were derived starting from classical physics As an FYI for yourself and other readers one can use the Canonical formalism with the above (QFT) the action for a minimally coupled scalar fields in spacetime is \[S=\int d^4x \sqrt{-g}(\frac{R}{2k^2} +\mathcal{L}_m+\mathcal{L}_\phi\] Where R is the Ricci scalar \(\sqrt{-g}\) is the metric determinant \(\mathcal{L}_m\) is the matter field Langrangian for matter fields. scalar field \[\mathcal{L}_\phi=-\frac{1}{2}g^{\mu\nu}\partial _\mu\phi\partial_\nu\phi-V(\phi)\] where \(V\phi\) is the scalar field potential Terms previous is the kinetic energy terms. gives Klein Gordon equation \[\square\phi-V_\phi=0\] where \(V_phi=\frac{\partial V}{\partial\phi}\) and \(\square\phi=\nabla_\mu\nabla^\nu \phi\) variation with respect to the gravitational metric \(g_{\mu\nu}\) yields \[R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=k^2(T_{\mu\nu}+T^\phi_{\mu\nu})\] leading to the canonical treatment of the energy momentum tensor \[T^\phi_{\mu\nu}=\partial_\mu\phi \partial_\nu \phi-\frac{1}{2}g_{\mu\nu}(\partial\phi)^2-g_{\mu\nu}V(\phi)\] and \(\partial\phi^2=g_{ab}\partial^a\phi\partial^b\phi\) so using the above with k=0 for flat and equation of state \[p=\omega\rho\] with FLRW metric \[ds^2=-dt^2+a^2(t)(dx^2+dy^2+dz^2\] one can derive the Freidmann equations and acceleration equation which is quite lengthy as you must go through the Ricci tensor with all the relevant Christoffels as very members understand Christoffel symbols I will skip that portion. to arrive at \[3H^2=k^2(\rho+\frac{1}{2}\dot{\phi}+V(\phi)\] \[2\dot{H}+3\dot{H}^2==k^2(\omega\rho+\frac{1}{2}\dot{\phi}^2-V(\phi))\] gives energy density as \[\rho_\phi=\frac{1}{2}\dot{\phi}^2+V(\phi)\] \[p-\phi=\frac{1}{2}\dot{\phi}^2-V(\phi)\] which gives the mathematical proof of the scalar field equation for the FLRW metric above. Clearly demonstrates the relevance of kinetic energy and potential energy now doesn't it. You ask why I will use the work of other professional Physicists over yours there is your answer. The GR method as well as QFT method both work with the FLRW metric for all equations of state including the scalar field equation of state.

Sigh your grasping the importance of the KE term. Treat your model above as a toy universe. Then ask two fundamental questions. 1) What prevents your toy universe from collapsing under its own self gravity ? 2) what expansion rate would it have ? Let me know when you want to address those two questions. Also further consider the zero point energy value you have is the vacuum catastrophe value. Do you really want your model to match the worse mistake in QM predictions ? Do you not think it may be better to match the Observed vacuum energy density of the cosmological constant instead of the vacuum catastrophe ? I will let you think about those questions I also don't agree with your gravitational self energy apply Newtons Shell theorem to the Earth case and not just two seperste hemispheres you must look at all contributions of gravitational force not just two points. Specifically sum of all forces from every coordinate in the Earth's shell. If you have a uniform mass distribution then any point can be treated as a center of mass. The sum of all force vectors using Newtons gravitational law will be zero for gravity g=0 in any uniform mass distribution it is non uniform mass distribution that leads to gravity hence under GR the acceleration is handled through the curvature terms. Nice try but Newtons Shell theorem is basic classical physics why didn't you consider that ? On a multi-point analysis ie sum of forces surrounding any arbitrary CoM in a uniform mass distribution ? Of course you could have just asked yourself why does gravitational force vary with the 1/r^2 relation in regards to your proton example as you didn't include a distance between any points you used. How do you have a gravitational self energy that doesn't decrease the greater the distance. That should have been obvious just as obvious that every other point in the Earth's volume would also have influence

No problem we cross posted as I was adding details to my last post. Your referring to the orbitals in the old Bohr model of the atom that's been replaced by the electron shell layout where electrons form probability clouds such as those subshell images on the previous wiki-link. https://en.m.wikipedia.org/wiki/Electron_shell See the dumbbell arrangements etc in the image in that link the old orbit like a solar image isnt accurate the shell system is

A state can be any set of variables or functions that describe a system but must apply only to the system without any previous history such as path taken previous temperature, energy level etc A state will evolve in time but it is set at whatever moment in time it's being examined at. It can be any particle or combination of particles including atoms or any field treatment. Now that automatically has a boundary condition that boundary being time dependent (the state condition depends on how evolves over time) Whatever other boundary conditions depends on what the state system is describing by those variables or math functions. Mathematically these are constraints on the valid ranges the mathematics of the state are accurate. Simple example a state is like a math set but with equations and variables etc. So say the equation only is accurate for a given range (3,4,5,,6) only that is a constrained set hence has a boundary condition. However if no additional constraint range can be infinite (,except time dependency) In thermodynamics and physics it's useful to define closed states and conserved states. A conserved system must be closed. Studiot is more practiced at classical thermodynamic systems and that's one of the better stepping stones to learning states as well as thermodynamics. As it teaches the same requirements for what is needed for a state to be conserved We do however standardized symbology to go with states \(|\vec{a}\rangle\) is initial vector field a (ket) \(\langle \vec{a}|\) is after state vector a (bra) The above is called Dirac Bra-ket notation you have a transpose (an operand or function between the bra and ket \[\lange \vec{a}|transpose|\rangle|] Now that's a quick and dirty on bra-ket notation Common states \(|\phi\rangle\) = scalar field ie magnitude only (\|\psi\rangle\) is often a complex conjugate field such as a spinor field

Electron shell is the physics name https://en.m.wikipedia.org/wiki/Electron_shell

As an assist here is a recent Universe from nothing model based of the wavefunction of the universe Wheeler DeWitt. https://arxiv.org/abs/1404.1207 Might help

This is the equation you have but every term is just the potential energy without the kinetic energy term \[{E_T} = \sum\limits_i {{m_i}{c^2}} + \sum\limits_{i < j} { - \frac{{G{m_i}{m_j}}}{{{r_{ij}}}}} = M{c^2} - \frac{3}{5}\frac{{G{M^2}}}{R}\] Every term above is just potential energy so your subtracting PE from PE which is automatically zero. The article I included shows this relation. the Kinetic energy term \[\frac{3}{10}MH^2R^2\] arises from expansion and the movement of the mass distribution for an expanding volume. and the combination of the two terms for total energy in the Article and not your equation gives us the energy density we see in the FLRW metric \[H^2+\frac{kc^2}{R^2}=\frac{8\pi G}{3}\rho\] which will work even for all three curvature terms by the inclusion of k. Your method doesn't work to give the correct energy density Study the method in this article its why I included it to begin with. https://core.ac.uk/download/pdf/30798226.pdf The author show this is correct by the following \[E_t=E_k+E_p=\frac{3}{10}MH^2R^2-\frac{3}{5}\frac{GM}{R}\] \[=\frac{3M}{10}(H^2R^2-\frac{2GM}{R})\] \[=\frac{3M}{10}(H^2R^2-\frac{2G}{R}\frac{4\pi}{3}R^3\rho\] \[=\frac{3MR^2}{10}(H^2-\frac{8\pi G}{3}\rho)\] requiring E to be constant=conserved mass is already conserved in the above \[H^2-\frac{8\pi G}{3}\rho=\frac{10E}{3M}\frac{1}{R^2}\rightarrow H^2-\frac{10E}{3M}\frac{1}{R^2}=\frac{8\pi G}{3}\rho\] where \[H=\frac{1}{R}\frac{DR}{dt}=\frac{\dot{R}}{R}\] which is the velocity time derivative for expansion. With curvature under GR and k for curvature= constant \[H^2+\frac{kc^2}{R^2}=\frac{8\pi G}{3}\rho\] this is section 1.24 to 1.28 this above correctly includes the KE and PE terms for mass distribution of an expanding volume. I could tell you didn't really read the article as you missed the opening statement. "I will look at ’the total energy of the universe’. This is an interesting issue, because if the energy of the universe turns out not to be conserved, it will be in conflict with our common understanding of energy. So intuitively we expect the energy of the universe to be constant. Furthermore, if this total energy is constant and zero, it means that ’creating’ a universe does not require any energy. Such a universe could then, in principle, just ’pop up’ from nothing. Our universe is dominated by a so-called cosmological constant, or vacuum energy. It has the property that the energy density is constant in volume, so when the universe expands, the total amount of vacuum energy increases. Where does this new energy come from? One might immediately think that it could be energy from other components in the universe that is converted into vacuum energy. But it turns out not to be that simple, since the vacuum energy increases also for universe models which contain vacuum energy only. For flat (Minkowski) spacetimes, a global energy conservation law can be set up without problems. But we know that we need the general theory of relativity to describe the universe more realisticly. General relativity deals with curved spacetimes, and then it is in general not possible to set up a global law for conservation of energy." in essence the paper is an examination of the plausibility and consequences of Universe from Nothing based models She states energy conservation can work in the Newtonian case however will not work in curved spacetimes. This is an identical problem that plagued the zero energy universe model. The model only worked for flat spacetime. It is those curvature terms that will kill your model as well as your simply looking at the Newtonian case without examining curved spacetimes. We do not live in a critically dense universe our universe has a small curvature term that only approximates flat. The paper clearly looks at the issue of Lambda aka the cosmological constant. The very problem of Lambda being constant is something of a large issue with conservation of mass energy. The Newtonian method will not address this as shown in the paper and please don't claim your two papers solves that issue. Your not even close to solving that issue

Why would I think that when I'm referring to the first equation on your article. Where you have total energy but only included potential energy terms That's where the error lies your total energy formula is wrong. If the first formula is wrong any later formulas are likely incorrect as well.

Of course it is its the rest mass\invariant mass The momentum is p \[E=p+m^2\] using normalized units where c=1. The kinetic energy term isn't mc^2 its the momentum given by p If your using just PE your system is static with no particle motion.

Yes I know the first equation is just the potential energy. However under mainstream physics so is the invariant rest mass given by E=mc^2. So my argument still stands your subtracting potential energy from potential energy giving the result of zero. However in your article you claim that it is total energy and not potential energy and are ignoring kinetic energy terms for an expanding volume. Just because your model isn't mainstream doesn't mean it shouldn't match mainstream physics when it comes to the rudimentary physics under Newton treatment. (For the record stating someone's conjecture isn't mainstream is one of the most common excuses used on this forum) it's a very common and lame argument. Potential and kinetic energy relations has been mainstream since Newton if it can stand that test of time applicability then obviously it's a very robust and accurate methodology. You most certainly will not solve the cosmological constant nor any equation of state without including kinetic energy terms it's literally impossible as those terms involve kinetic energy for its momentum equivalence. If your theory has any motion or expansion of a matter field you need PE and KE. Hopefully you dropped negative mass as that has numerous ramifications you haven't considered.

We already run very complex simulations example testing our mathematics involved for Cosmology. https://www.illustris-project.org/ Mathematica is also a very useful tool commonly used by physicists such a the Feycalc extension Even for the Saha equations one requires coded simulations thankfully this one doesn't have huge computing power requirements. None of these require AI.

Don't worry I fully expect you to search through internet looking for every possible instance of negative mass treatments they always boil down to not understanding how that negative mass is applied to a global vs local field treatment. What you are describing is a global treatment not local to a non zero global baseline. Example a solid state lattice network with negative and positive holes. Global energy value with negative local energy compared to the global energy value. If you had a non- zero baseline that would run counter to a Universe from nothing.... I will state this once again Energy is the ability to perform work a property... it does not exist on its own. Mass is resistance to inertia change hence the momentum terms.... As we are describing a Universe any theory must work as a field treatment such as GR. (Good luck on this point). For Newtonian case try it as a momentum vector field not simply scalar lol it would get trickier when you apply contravariant terms and anti- commutations... Lol after all I should be able to throw any SM particle into this universe and its subsequent equations of motion will work under multi particle field treatment. Hence all the above statements

I should have added "Where is this negative mass you claim exists ? No standard model particle has negative mass and even DM doesn't match the characteristics of negative mass. DE matches the characteristics of negative pressure but not a matter field. Just a side note \(E^2=(pc)^2+(m_o c^2)^2\) is the energy momentum relation if you place a minus sign in front of the RHS your not stating negative mass your stating negative momentum. Not even antiparticles has negative mass nor momentum. So the questions still remains name any particle with negative mass or momentum ? Good luck on that question I certainly know I can't think of any particle with negative coupling strength can you ? You may want to rethink your other article as well as this one. That's just a strong suggestion though Particularly since in your E- and E+ momentum that would be the equivalent to two seperate fields one collapsing while the other is expanding. If you don't believe that statement try using the stress energy momentum tensor with negative momentum with the Einstein field equations. Would not the result be two opposite curvature terms? Or did you even consider that problem ? When you apply thermodynamics to that, one field would be cooling down while the other is heating up. Not what we see is it ? What is obvious to me is that you never considered your proposals beyond Rudimentary Newtonian treatment nor considered the ramifications under a field treatment such as under GR. Let alone the absolute lack of any standard model particle to meet your criteria. If only negative mass was that easy we would already have an Alcubierre warp drive

That's clearly not shown in your mathematics The article I linked shows precisely how \[\frac{3}{5}\frac{GM}{R}\] is derived using an expanding volume. So your expression gives a balance of precisely zero at all times. If that is not your intent then that expression is wrong for your purposes. The relation above is specifically a derivative of the mass density evolution for an expanding volume clearly shown in the article I linked. specifically \[E_t=MC^2-\frac{3}{5}\frac{GM}{R}=0\] Now your telling us that your applying negative mass which under GR itself is considered impossible. If you ever applied a force to a negative mass and Newtons laws to negative mass you have inverse relations Specifically the more force applied the less displacement you get. All you need to get expansion is to have a kinetic energy term to exceed the positive energy term as per the scale field equation of state used in Cosmology. In essence pressure the energy density relations. That is how the FLRW metric works Is this paper not suppose to show how a universe from NOTHING is possible ? Is that not the claim of the paper ? the term mass energy if garbage by the way. mass is resistance to inertia change if an object has mass it has resistance to inertia change or acceleration. its already been mentioned energy DOES NOT exist on its own. It is a property just as mass is a property Your attempt to treat mass energy as something seperate from mass is simply wrong by any definition of mass. The definition for mass should make the inverse relations to force obvious had you considered the definition of mass with regards to negative mass. And please don't tell me your going to disagree with the very definition of mass or energy (ability to perform work) that has been around since the 16th century and how they apply to Newtons laws of inertia If you do then you better get busy rewriting the entirety of every physics theory since then.

Here is how total energy can be calculated it took me a bit to find a methodology that matches your \[\frac{3}{5}\frac{GM^2}{R}\] relation The author also has that same relation but for the potential energy here is a dissertation paper where the author has the identical term for potential energy. The kinetic energy term is derived under Newtonian treatment "Total Energy of the Freidmann-Robertson_Walker Universes " by Maria Mouland https://core.ac.uk/download/pdf/30798226.pdf specifically arriving at \[\frac{3}{10}MH^2R^2\] The inclusion of both terms is what you need for total energy She shows further down how the two terms arrive at equation 1.28 \[H^2+\frac{kc^2}{R^2}=\frac{8\pi G}{3}\rho\] which is a well known relation used in the critical density formula. (proof of methodology ) it is a matter only treatment which is fine w=0 as the critical density formula is also a matter only treatment. If I recall correctly Peebles also had a similar treatment in one of his Introductory to cosmology textbooks. I do know he had the 3/5 relation in his density contrast term of his Large scale structure of the Universe article. Anyways hopefully you find the above article useful

A mediator is an offshell boson or field of bosons for EM field its virtual as opposed to real photons. For strong force it's gluons, for Higgs its Higgs boson for the weak force is W+ W- and Z bosons. https://en.m.wikipedia.org/wiki/Force_carrier

Don't be concerned with Hubble tension, I've been following the Research but the gist of the problem isn't a need for new physics. Its a problem more in terms of fine tuning of luminosity to distance relations. Holicow used near field measurements while Planck is far field ie CMB. Both require fine tuning but more so with the Holicow dataset for using standard candles. We will get into the FLRW metric later on (its a good stepping stone to cosmology and GR) This is one of the more recent papers on the Tension https://arxiv.org/abs/2408.06153 On Leavitt Law calibrations https://arxiv.org/pdf/2205.06280 this stuff is too advanced for you but suffice it to say its being researched

found a fundamental flaw with your total energy equation You have only calculated the Potential energy term and didn't include any term for the kinetic energy. The primary mistake results from just using e=mc^2 instead of the full energy momentum relation. You should be using this as your basis equation for the field \[E=\sqrt{(pc)^2+(m_o c^2)^2}\] the \(\frac{3}{5}\frac{GM^2}{R}\) is just potential energy it doesn't include any momentum its just from the mass term. This equation is accurate for the potential energy but not as the total energy your subtracting two equivalent terms by the operation \[E_t=mc^2-\frac{3}{5}\frac{GM^2}{R}=0\] would be the result it can only give zero the second term is a derivative of the first term. So the result can ONLY be zero. Under sphere treatment as the sphere expands via Hubble constant you do not have the equations of motion just the mass term Is that your intent ? if so what gives rise to any later energy content ? If not and if you like I can provide the kinetic energy term as a Newtonian treatment which you seemingly prefer for an expanding Universe. however if it is your intent then it would the equivalent of zero particles for any kinetic energy term as the result would always be zero nor would you have a mass to begin with and no remaining kinetic energy term to drive expansion which does not make any sense whatsoever. A total energy balance of zero is an empty universe with no component to drive expansion. Doesn't work You need to incorporate the kinetic energy side of the gravitational field as a result of expansion as well for total energy (hint the kinetic term can be derived using Hubble constant) if you want I will post how upon request. I know a Newtonian treatment for that. The essence however is that as the universe expands driven by the equations of state the kinetic term becomes involved as required to maintain a homogeneous and isotropic mass distribution. It is that term that is missing.

have you ever thought to look into zero energy universe ? https://arxiv.org/abs/gr-qc/0605063 you may find this helpful still looking if I still have a copy of one of the earlier Universe from nothing papers I once had a copy of