Mordred

DM doesn't interact with the electromagnetic spectrum so doesn't interfere with photons That's well known so how would it affect c in regards to any test using photons ? In rhst regards any far field test using stellar objects would have intervening DM and we do not see any evidence of scattering due to DM. So any link I included using any methodology of far field stellar object examination is also testing any intervening DM. Believe me I wish DM would scatter or influence photon path it would be far easier to detect

Well then the thought experiment in essence has already been performed if you look through I provided you would find improved yet equivalent style experiments. Or did you not notice for an example the experiment done in space and quite frankly if you do a Google search you would find related equivalence principle experiment done on the moon. So don't tell me the effort I took is uncalled for or does the simple detail experiments are constantly being performed in dozens of different methodologies elude you.

the interactive parallelogram for determinant's (khan link above we can assign it as a matrix any assignment will do so lets just call it matrix B so that matrix is the box with two rows and two columns. The rows and columns are equal so it is a square matrix. Adding the \(\hat{i},\hat{j}\) coordinates we can place those coordinates in the (subscript= bottom, superscript=top exponent). \(B_{i,j}\) on subscript, \(B^{i,j}\) where " i " is the row vector," j" is the column vector. Now recall those spin states. Each spin has a limited set of allowed values for the S component. for spin 1/2 particles https://en.wikipedia.org/wiki/Doublet_state In quantum mechanics, a doublet is a composite quantum state of a system with an effective spin of 1/2, such that there are two allowed values of the spin component, −1/2 and +1/2. Quantum systems with two possible states are sometimes called two-level systems. Essentially all occurrences of doublets in nature arise from rotational symmetry; spin 1/2 is associated with the fundamental representation of the Lie group SU(2). for Spin 1 bosons. https://en.wikipedia.org/wiki/Triplet_state In quantum mechanics, a triplet state, or spin triplet, is the quantum state of an object such as an electron, atom, or molecule, having a quantum spin S = 1. It has three allowed values of the spin's projection along a given axis mS = −1, 0, or +1, giving the name "triplet". Spin, in the context of quantum mechanics, is not a mechanical rotation but a more abstract concept that characterizes a particle's intrinsic angular momentum hopefully the above helps as a means to see what the math is doing without knowing the math. Going to let you study the above as there is a lot to absorb on the last post https://en.wikipedia.org/wiki/Spin_quantum_number In the last link look at the reflection symmetry for spin up and spin down.

lol you can relax on this one symmetry is in actuality easily understood its learning which mathematical relations are symmetric that tends to get confusion. lets first take a simple everyday example. One of my favorite examples is to use fan blades.. Each blade is identical so they are symmetric to each other. As the fan rotates those blades are unaltered this is called rotational symmetry. if you move the fan around the room the blades remain unaltered. this is called translational symmetry. if you look at an image of the blades in a mirror they look identical (reflection symmetry) if you combine both translation symmetry and reflection symmetry this is called glide symmetry. Now on the mathematics side. if you add a set of numbers or values and the order of operation doesn't change the resultant the equation is commutative https://en.wikipedia.org/wiki/Commutative_property if an equation is commutative then its relations are symmetric. so for example with states this can be expressed as \[|\psi\rangle_{c}=|\psi_a\rangle +|\psi_b\rangle\] the above tells us the order you add the states doesn't matter. So it is a symmetric relation if on the other hand you have \[|\psi\rangle_{c}=|\psi_a\rangle -|\psi_b\rangle\] the order of operations does change the result this is an antisymmetric relation. it is common nomenclature to place the negative sign as the first term on the RHS of the equal sign \[|\psi\rangle_{c}=|-\psi_a\rangle +|\psi_b\rangle\] multiplication is commutative (symmetric) however division is not (antisymmetric) Now consider spin. Boson spin is an integer 1 most common zero in Higgs. You can add or subtract bosons in any sequence and the order of operation doesn't but also other than total energy and number of bosons in the system no other change occurs as a result. This isn't true for fermions with half integer spins ie 1/2 the order of operations does matter . It is also this detail that leads to the Pauli exclusion principle. https://en.wikipedia.org/wiki/Pauli_exclusion_principle In bosonic systems, all wavefunction must be symmetric under particle exchange. for bosons \[\psi(x_1,x_2)=\psi(x_2,x_1)\] the above is a commutative expression (symmetric) for fermions \[\psi(x_1,x_2)=-\psi(x_2,x_1)\] this is an anticommutative expression (antisymmetric) Last two expressions are called Slater determinants https://en.wikipedia.org/wiki/Slater_determinant the matrix in that link can readily be understood here https://www.khanacademy.org/math/multivariable-calculus/thinking-about-multivariable-function/x786f2022:vectors-and-matrices/a/determinants-mvc that link has an interactive graph you can play around with ( Pay close attention to how the parallelogram changes shape as you interact with it this will become important to understand spin later on.) but also a link to 3d Determinants https://www.khanacademy.org/math/linear-algebra/matrix-transformations/inverse-of-matrices/v/linear-algebra-3x3-determinant This should also help better understand that with vector addition the inner product of two vectors \[\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}\] is symmetric (commutes) the order of inner product of two vectors doesn't matter where as the cross product of two vectors anticommute \[\vec{a}\times\vec{b}=-\vec{b}\times \vec{a}\] now some instant recognition rules with matrix/tensors \[\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}\] this matrix is symmetric and orthogonal https://en.wikipedia.org/wiki/Orthogonality the only non vanishing terms are on the diagonal given as 1 in each entry any non vanishing term not on the diagonal above is antisymmetric For QM if all diagonal terms are a real number (set or reals) that matrix is a also Hermitean https://en.wikipedia.org/wiki/Hermitian_matrix as you can see in the above a Hermitean matrix need not be symmetric but all diagonal terms must be a real number. this will help when it comes to spin statistics under QM example a preliminary lesson to understand https://en.wikipedia.org/wiki/Conjugate_transpose but before we deal with the complex conjugate under QM treatments lets familiarize you with what a complex number is. as You learn best from videos https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:complex/x2ec2f6f830c9fb89:complex-num/v/complex-number-intro now recall that interactive video where I asked you to pay attention to the parallelogram. ? replace the two axis as one axis imaginary numbers the other axis Real numbers. Just like the complex number wiki link images and make further note that the parallelogram in the interactive map is identical to the parallelogram in that same link and that we can add two complex numbers using the parallelogram. this will become important later on when we assign operators as those complex numbers as well. (under QM two spinors which are also complex number for what is called a bi-spinor giving us 4 complex numbers in total needed for Dirac matrices later on). This will also apply to linearization of non linear systems. Take your time on this as it will become incredibly useful for a great many different physics treatments including GR/QFT etc. @studiot will also recognize the above applies to classical physics as well as engineering applications. further consideration as your looking at the parallelogram I want you to also consider the following (eigenvectors and eigenvalues) https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors In linear algebra, an eigenvector (/ˈaɪɡən-/ EYE-gən-) or characteristic vector is a vector that has its direction unchanged (or reversed) by a given linear transformation. if you look at the interactive graph on that last wiki link you can see the connection. to the Khan University interative map. the below will also help for those last two terms https://www.mathsisfun.com/algebra/eigenvalue.html note from last link 1 means no change, 2 means doubling in length, −1 means pointing backwards along the eigenvector's direction now look back at the values in top left corner of the first khan interactive map on determinants. The box with the changing numerical values in matrix form (2 by 2 matrix) All the above is called skew symmetry. As you stretch the eugenvector the parellelogram becomes scewed/stressed. (will apply to all energy momentum stress tensors). Further a skew symmetric geometry can be Hermitean (assign real values only on the eugenvector) while the geometry is not orthogonal (its now skewed) (you may also want to notice that the interactive determinant graph is showing rotation and reflection symmetry ie when i axis is in line with j axis. They have reflection symmetry with one another. Vectors where the only change is direction have rotational symmetry.

Lets start with a couple of statements Physicforum doesn't allow any form of speculation or any physics that isn't concordance or found in Peer reviewed literature. However that is irrelevant. DM has little or nothing to do with the isotropy of the speed of light or its constancy. Both isotropy and constancy is a large part of Lorentz invariance tests. The tests you have above are far too behind the times of modern day tests of c constancy or Lorentz invariance. Those precision tests place the error margin at roughly 1 part in 10^15 for any error margin. That is well beyond the tests you have proposed. for example you second proposed test takes what you have into far greater precision. https://arxiv.org/abs/physics/0510169 here is the usage of stellar objects and time of flight https://arxiv.org/pdf/2409.05838 we also use microwave interferometers, laser interferometers, rotating mirrors, to name a few other tests commonly done. here are some Lorentz invariance constraints https://arxiv.org/abs/2406.07140 here is one example using gravity waves https://arxiv.org/abs/1906.05933 here is a small listing of modern methods https://arxiv.org/abs/gr-qc/0502097 https://arxiv.org/pdf/1304.5795 model independent tests https://arxiv.org/abs/1707.06367 tests done in space New Test of Lorentz Invariance Using the MICROSCOPE Space Mission https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.123.231102 All the above clearly show that we are constantly testing c and Lorentz invariance those tests never stop and were constantly seeking higher precision tests. So nothing you have suggested above is new or hasn't been thought of already. In point of detail Modern tests have far far greater precision than what you have above

Lets tackle symmetry before spin as symmetry can be easily shown and I will work up an easy way of understanding symmetry

One of lifes rules for every trade, science, programming etc. All complex problems are composed of simpler problems. lol or whats that expession one bite of the apple at a time.

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.