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