Mordred

That should help answer this question. Hint will also help to identify the longitudinal and transverse relations of group velocity and phase velocity.

So am I when dealing with spectroscopic examinations as group velocity dispersion occurs in interstellar mediums as well as spacetime vacuum. In many ways using DeBroglie waves is far simpler than wave vector methodologies such as via a Fourier transformation which gets incredibly tricky when dispersion occurs. Particularly since the further you look the mean average density increases which has huge ramifications on dispersion as the medium density isn't consistent. Extremely useful though for determining the medium properties. For the OP De-Broglie waves becomes useful as you will see dispersion in matter waves as the waves propogate through any medium including quantum and spacetime vacuums. It's extremely common it's that most times the mathematical treatments have already factored in the De-Broglie relations for example under QM treatments so it tends to get missed as being involved. Here examine this article to get a better understanding of phase velocity and group velocity including dispersion. https://www.mlsu.ac.in/econtents/784_PHASE VELOCITY AND GROUP VELOCITY.pdf Make note of the detail that the phase velocity doesn't transport energy in terms of your opening post where phase velocity can exceed c but as no energy is transported there is no violation. This link provides some additional details https://www.sciencedirect.com/topics/physics-and-astronomy/phase-velocity#:~:text=It can also be greater,not carry energy or information. In essence keep matter velocity, phase velocity and group velocity as seperate as each has its own distinctive relations with regards to De-Broglie waves.

100 Mpc is the correct currently accepted scale though there was roughly 5 years ago some consideration of using 120 Mpc instead. Never happened as it wasn't really necessary. Here is a counter paper to the DESI findings and it raises a couple of valid points in so far as DESI uses the Hubble tension as part of its argument however the Hubble tension is largely resolved in so far as the later papers brought forward. https://arxiv.org/abs/2404.18579 In essence the paper strongly suggests caution as the evidence isn't strong enough yet.

Thankfully just using energy momentum relation greatly simplifies things hence we rarely use De-Broglie

If your looking into group velocity you need a superposition of waves for example if the two waves have different momentum terms the wavepacket will continually change as it's travelling.

One of the first confirmation tests was the Davisson-Germer experiment. https://en.m.wikipedia.org/wiki/Davisson–Germer_experiment DeBroglie also ties into photoelectric effect ie quantum Tunneling such as done by Einstein for which he received a Nobel prize. (Matter wave duality is needed) Edit the first test is a common classroom test used today so you should find plenty of example and related articles.

I wouldn't worry too much about this formula but to answer you questions \[g_n=\sum^{n-1}_{L=0}(2s+1)(2L+1)=2n^2\] \(g_n\) is the maximum number of electrons states for the shell where n is just a shell identifier ie n=1 for shell 1. If its shell two then n=2. \(\sum\) is a summation the lower value is the minimal value in this case its stating no angular momentum L the n-1 is the upper bound and this gets rather complex as the n-1 indicates its a telescoping series under Calculus. Which is term that is best left for much later. However wiki shows an example https://en.wikipedia.org/wiki/Summation

Agreed with the above its one of the reasons it's often easier to learn more from the classical treatment angles than it is from the physics side such as via Shrodinger and Hartree-Fock or via Spectrography. Lol molecular ionization spectography is incredibly tricky it's one side of it that I am extremely out of practice on lmao. It's also something I rarely if ever deal with. I can't recall the last time I ever dealt with molecules in chemistry or physics lol. That's where you or others would be far suited than myself to describe. +1 for that reminder though lol

On this I would prefer either Studiot or Swansont reply my knowledge on atoms from a chemistry point of view is too rusty. Typically when I deal with atoms it's more in dealing with its ionization and the related mathematics in so far as spectographic readings and related QM/QFT equations which your not ready for Not that I'm ignoring you I just think you would served by others more familiar particularly on the more classical examinations. However that being said here is a couple of details. The number of protons of the nucleus determines the number of electrons for a neutral atom. However each shell can only contain a maximum number of electrons Following your \(2n^2\) rule. I will do this in a format I'm more familiar with in terms of how a spectroscopist would examine the atom. Each n level is called a shell, and historically, observational spectroscopists named the n = 1 level the K shell, the n = 2 level the L shell, the n = 3 level the M shell, the n = 4 level the N shell, etc. n is the principle quantum number l is the angular momentum m is the magnetic quantum number For a given shell, n, each angular momentum state, l, is called a subshell. The subshells are also described by spectroscopic notation, “s” for l = 0, “p” for l = 1, ”d” for l = 2, and “f” for l = 3. These respectively stand for “sharp”, “principle”, “diffuse”, and “fundamental”. A given subshell, nl, for example n = 1 with l = 0, is denoted “1s”. For n = 2 and l = 1, the state is denoted “2p”. The magnetic quantum number is not included in the spectroscopic notation. As an example, there is only a single 1s state (m = 0) and there are three 2p states (m = −1, 0, +1). the applies as well for the Schrodinger model of the atom. So the shells themselves are denoted by the principle quantum number each shell can have subshells which include the angular momentum quantum number "l" the outer shell can have fewer than the maximum by the 2n^2 rule if the total number of electrons equals the number of protons. When ionization occurs there is a transition of the electron to change shells or lose an electron leaving fewer electrons than the number of protons. That atom is now ionized. The spin of a particle is intrinsic and do not change due to its location or atomic configuration. electrons always have spin + or -1/2. bosons typically always have spin 1 with the exception of Higgs boson spin 0 or theoretical graviton spin 2 ( never confirmed or observed| quarks will have a spin value of some multiple of 1/3 ie 2/3 etc. Those do not change they are intrinsic to those particles.. but DO not think of quantum spin as a spinning ball that is wrong Spin is a complex conjugate of the principle quantum numbers so for example spin 1/2 requires a 720 degree rotation to return to its original state. Where as a ball only requires 360 degrees for example.. I don't know if your ready for this yet but under the Schrodinger model the spin orbit couplings which gives a new quantum number J. where J=(S+L). has the form where s=1/2 so J will be multiples of 1/2 ie 3/2 for example BUT DO NOT CONFUSE J with spin. Different quantum number \[g_n=\sum^{n-1}_{L=0}(2s+1)(2L+1)=2n^2\] the 2n^2 relation is from the old Bohr or Rutherford model of the atom. The above equation applies for the Schrodinger atom model. Now this goes a bit more detailed under what is called the Dirac model. in The Dirac model we have an expansion of quantum numbers. n=principle L= angular momentum J= total angular momentum (also called the Orbital azimuthal angular momentum ie the Z axis) \(m_L\) is the orbital magnetic \(m_J\) is the total magnetic \(m_s\) is the spin. Hope that helps. For other readers all the above will relate to an energy diagram called a Grotian diagram https://en.wikipedia.org/wiki/Grotrian_diagram this takes all the above numbers of the Dirac model and equates the relevant energy terms into a diagram. This diagram also shows the transitions and transition levels. If you study Grotian diagrams you will hit a new term called doublet and singlets. Doublet Lyman Doublets=\(\begin{cases}nP_{1/2}-1S_{1/2}\\nP_{3/2}-1S_{1/2}\end{cases}\) this is mainly for the Grotian diagrams so don't worry about it for now. Most people that don't do Spectrograph's won't know this detail. However in regards to Grotian diagrams they are directly involved in the selection rules https://en.wikipedia.org/wiki/Selection_rule as you can see from that link that takes a considerable amount of study to fully understand. example the above applies to Raman series https://en.wikipedia.org/wiki/Raman_spectroscopy and Balmer series https://en.wikipedia.org/wiki/Balmer_series and Lyman series https://en.wikipedia.org/wiki/Lyman_series to name a FEW there a lot of different series depending on the atoms involved It is the above that allows us to use Spectrograph's to determine the composition of stellar objects such as Stars, plasma, atoms, molecules etc as well as their electron configuration. Preliminary to the above being Moseley's Law https://en.wikipedia.org/wiki/Moseley's_law which required corrections from the Rutherford model to the Schrodinger model. It is one of the major pieces of evidence that the Bohr/Rutherford model was wrong. Moseley's Law uses the Bragg equation from Chemistry but historically Mosely was also involved in the layout of the periodic table as a good portion of his work led to the modern day periodic table layout. Previous To Moseley's Periodic table atoms were arranged according to atomic mass (Mendeleev's Periodic table) side note for the x ray scatterings you also need Braggs law LOL welcome to a crash course on how we know atomic structure, via spectrographic studies as well as how the Modern day periodic table got developed

Recessive Velocity corrections past Hubble Horizon approx z=1.46 \[E_Z=[\Omega_R(1+z)^4+\Omega_m(1+z)^3+\Omega_k(1+z)^2+\Omega_\Lambda]^{1/2}\] \[v_{r}=\frac{\dot{a}}{a_0}D\] \[\frac{\dot{a}(t_0)}{a_o}=\frac{H(z_0)}{1+z_o)}=\frac{H_0E(z_o)}{1+z_O}\] \[v_r=\frac{cE(z_o)}{1+z_o}\int^{z^{obs}}_0\frac{dz}{1+z_o}\frac{D_c(Z_o,Z_s)}{D_H}\] \(Z_{os}\) is the reduced redshift \[1+z_{os}=\frac{1+z_s}{1+z_o}\] for observerd source redshift z_s present epoch Observer \(z_0=0 ,E(Z_o)/1+(z_o)=1\) \[v_r=(o,z)=c\int^z_o\frac{dz}{E(z)}=c\frac{D_c(z)}{D_H}\] gives redshift as a multiple of speed of light

Thermodynamic equilibrium \[\rho=\frac{g}{(2\pi)^3}\int d^3 p E f(\vec{p})=\begin{cases}\frac{\pi^2}{30}geffT^4&T\ge m\\m,n&T\le m \end{cases}\] \[g_{eff}=\sum_{i=b} gi (\frac{T_i}{T})^4+\frac{7}{8}\sum_{i=f}gi(\frac{t_i}{T}^4\] \[P=\frac{g}{2\pi^2}^3\int d^3p\frac{p^2}{3E} f(\vec{p}\] further details equation 130 to 138 https://mypage.science.carleton.ca/~yuezhang/Cosmology note.pdf number density \[n=\int_0^\infty n(p)dp\] energy density \[U=\int^\infty_0 n(p)\epsilon_p dp\] pressure \[P=\frac{1}{3}\int^\infty_0 n{p} v_pp dp\] \(v_p=p/m\) and \(\epsilon p^2/m\)= non relativistic, relativistic \(v_p=c, \epsilon_p=pc\) hence \[P_{nr}=1/3\int^\infty_0 2\epsilon_p n(p) dp\rightarrow P=2/3 U\] \[P_{ER}=1/3\int^\infty_0 \epsilon_p n(p) dp\rightarrow P=1/3 U\] where \[n(p)n dp=n(\epsilon()\frac{g}{h^3}4\pi p^2 dp\] g is statistical weight \[n(\epsilon)=\begin{cases}\frac{1}{e^{\epsilon-\mu}/KT+0}& Maxwell\\\frac{1}{e^{\epsilon-\mu}/KT+1}& fermions\\\frac{1}{e^{\epsilon-\mu}/KT-1}&bosons\end{cases}\] Bose_Einstein \[n_i = \frac {g_i} {e^{(\varepsilon_i-\mu)/kT} - 1}\] Fermi_Dirac \[n_i = \frac{g_i}{e^{(\epsilon_i-\mu) / k T} + 1}\] Maxwell \[\frac{N_i}{N} = \frac {g_i} {e^{(\epsilon_i-\mu)/kT}} = \frac{g_i e^{-\epsilon_i/kT}}{Z}\] Saha \[\frac{n_i+n_e}{n_i}=\frac{2}{\omega^3}\frac{g_i+1}{g_i}exp[-\frac{(\epsilon_i+1-\epsilon_i)}{k_BT}\] \[\rho_R=\frac{\pi^2}{30}{g_{*S}=\sum_{i=bosons}gi(\frac{T_i}{T})^3+\frac{7}{8}\sum_{i=fermions}gi(\frac{T_i}{T})}^3\] \[S=\frac{2\pi^2}{45}g_{*s}(at)^3=constant\] Physics of the Interstellar medium The Physics of the Intergalactic Medium.pdf

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.