Mordred

Funny how I get the same results in any cosmology based examination I have done in 35 years regardless of whether or not I use the QM/QFT methodology or that of Relativity. These three methods work quite well in every cosmology study right up until you hit the singularity conditions. For example I can use any of the above methods to calculate the number density of any particle species at a given blackbody temperature (though you do have to calculate each particle in sequence of when they would drop out of thermal equilibrium. (Used for metalicity data for BB nucleosynthesis) I can determine the temperature history of our universe. Give you calculated rates of expansion that matches what we observe etc etc. It doesn't matter what physics problem I need to solve. I can do so with equal success regardless. The only time I can't use GR however SR still applies is the quantum regime. Simply due to the fact that gravity bring the weakest force doesn't have much influence on individual particles. However there still is influence

Has nothing with reality or philosophy of any kind. My question is strictly on the experimental basis. Your trying to state that your brain conjecture deal with specific physics related situations without using any actual physics. I'd like you to prove you have something of interest in that regard that could possibly peak the interest of a physicist like myself.

I believe you missed the point of my last reply. If processing speed of the brain or computer were an issue then how does different tests with different equipment with processors with different processing speeds all arrive at the same constancy of c as well as all the variations of Lorentz invariance tests and tests with regards to time dilation. Each test has its own processing speeds each test has different equipment. Yet they somehow arrive at the same answers.

Just pulling a link to an older thread with some articles and references to use for this thread.

Fine then explain how a previous experiment prior to any computer gave the same value for the constancy of c as that same experiment performed today using computers. You have zero mathematics to show a single error margin due to different processing speeds on any experiment if that experiment was measured with different devices with different processing speeds. You have zero data to show where the processing speed has an impact on the speed of light beyond your declaration of an impact.

Well I skimmed your entire pdf. Not a single formula to validate any claims. Sorry to inform you that is not how physics works. So if your interest is developing something useful for a physicist you might consider studying physics and applying some formulas. Also the brain has its own processing rate just like a PC. You mentioned both even went so far as to detail processing speeds inherent in a CPU etc. Yet for the brain you hint that different observers measure time differently with regards to the speed limit and time dilation but as the brain also has a fixed processing speed that makes no sense. Anyways I come to scienceforums to help others learn physics. I see nothing of interest for me in this particular thread so have fun and good luck

None of this is necessary to understand superposition so it's essentially not of any use. Let's look directly at superposition. The mathematics of QM uses probability equations as per statistical mathematics with its formulas. So those formulas include all possible outcomes. Now if you have some interaction between two particle states to entangle those particles. You have a range of possible outcomes that depend on the numerous conservation laws in particle physics. Ie conservation of charge, spin, flavor, color, momentum,lepton number, isospin etc. Taking those laws, the type of interaction and the detector setup one can determine a correlation function. Now until you measure the resulting entangled states you have the range of possible outcomes. The probability being the correlation function. However once you "observe which is a very confusing term used for measure" an entangled state the probability wavefunction collapses as you have now determined the state.

Time symmetry doesn't bend light nor does time. Time is just a property of a state/object etc that describes a rate of change. Spacetime however is a geometric mathematical process that uses the interval (ct). This gives dimensionality equivalence of length via a vector. It is the interactions of the particle fields via its energy/mass density relations that curves spacetime. For example spacetime without particles to generate a mass term has zero curvature. Gravity under GR is a pseudo-force that results from that spacetime curvature. Hope that helps

I don't know of any either and I keep track of CPT tests on a regular basis.

Relativity doesn't use probabilities in its equations. QM does that's the only difference but everyone believes that's some reality issue when it's nothing more than a different method of mathematical treatment. Nothing more, that includes superposition wavefunction collapse. A wavefunction is a mathematical object that may or may not even involve a physical measurement. Waveforms are a measurable range of values. Wavefunctions are strictly mathematical objects. One of the biggest sources of confusion is where to distinguish when a math statement describes a physical measurement or when it describes a mathematical set of some complex variable.

As one who knows all the major formulas and how they work in Cosmology, GR, QFT, QM and even string theory. After 35 years of examining every major theory. I can attest that for anyone that truly understand why and how thr theories work that are very functional for their intended purposes. Regardless of all the pop media articles and all the articles denying physics and any theory they don't like for mere philosophical reasons. 90 percent of the time it's the ones that don't understand the actual theories that create those articles claiming this or that theory doesn't work . Yes the mathematics of physics is complex. They are used to mathematically describe a huge range of observations. That's the primary reason why statistical mathematics is used by QM. Those mathematics do nothing to determine nor control realism. It's only purpose is to make predictions of cause and effect.

A little warning wiki pages can be written by anyone, sometimes professors but not always. expansion and a BB doesn't rule out the possibility of an infinite universe. Our current observations still hasn't ruled out either possibility. yes but at the same time QM has its Planck unit restrictions which is related to the BB and the resulting singularity condition. What isn't mentioned in most webpages such as wiki is that the singularity is a mathematical singularity. We do have working theories of quantum gravity for everyday observations of our universe. It is only in the extreme range where the issue comes up.

No that's incorrect the speed of light or any information exchange remains c. However keep in mind the our observable universe (shared causality). Was far smaller prior to inflation. However spacetime itself isn't restricted by c. Here is a good article with no or very little math that will greatly help you. You need a many skills or physics skills to understand this article. Entitled "What we have learned from Observational Cosmology" https://arxiv.org/abs/1304.4446

Nice explanation +1 covers the main points in a short and sweet manner

Relativity is used to develop the FLRW metric which is the primary equation describing how our universe expands. The FLRW metric can be applied in full blown GR even QFT. Now as to how the universe expands depends on two primary relations the energy/mass density and the equivalent pressure term those particles generate depending on their momentum term. Matter generates zero pressure. While radiation ie massless particles generate a 1/3 ratio https://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric the first is the FLRW metric, the next link lists the equations of state. https://en.wikipedia.org/wiki/Equation_of_state_(cosmology those two links roughly describe how matter, radiation and the cosmological constant energy/mass and their kinetic energy leads to a thermal expansion much like a gas an unrestrained gas. Now the CMB is an after effect of the BB, inflation, and electroweak symmetry breaking. Due to cosmological redshift the signal strength we receive is in the microwave range of frequencies

For anyone interested @KJW you might this of interest and it does relate to the toy modelling were doing. LOL that and this discussion brings up some fun mental exercises for me in so far as the mathematics that relate. Iin this case its useful to help demonstrate how a metric tensor gets filled from a ds^2 line element and how that affects the Christoffel. here is the Christoffels for the FLRW metric in spherical coordinates. \[ds^2=-c(dt^2)+\frac{a(t)}{1-kr^2}dr^2+a^2(t)r^2 d\theta^2+a^2(t)r^2sin^2d\phi\] \[G_{\mu\nu}=\begin{pmatrix}-1&0&0&0\\0&\frac{a^2}{1-kr^2}&0&0\\0&0&a^2 r^2&0\\0&0&0&a^2r^2sin^2\theta \end{pmatrix}\] \[\Gamma^0_{\mu\nu}=\begin{pmatrix}0&0&0&0\\0&\frac{a}{1-(kr^2)}&0&0\\0&0&a^2r^2&0\\0&0&0&a^2r^2sin^2\theta \end{pmatrix}\] \[\Gamma^1_{\mu\nu}=\begin{pmatrix}0&\frac{\dot{a}}{ca}&0&0\\\frac{\dot{a}}{ca}&\frac{a\dot{a}}{c(1-kr^2)}&0&0\\0&0&\frac{1}{c}a\dot{a}r^2&0\\0&0&0&\frac{1}{c}a\dot{a}sin^2\theta \end{pmatrix}\] \[\Gamma^2_{\mu\nu}=\begin{pmatrix}0&0&\frac{\dot{a}}{ca}&0\\0&0&\frac{1}{r}&0\\\frac{\dot{a}}{ca}&\frac{1}{r}&0&0\\0&0&0&-sin\theta cos\theta \end{pmatrix}\] \[\Gamma^3_{\mu\nu}=\begin{pmatrix}0&0&0&\frac{\dot{a}}{ca}\\0&0&0&\frac{1}{r}\\0&0&0&cot\theta\\\frac{\dot{a}}{c}&\frac{1}{r}&cot\theta&0\end{pmatrix}\] \(\dot{a}\) is the velocity of the scale factor if you see two dots its acceleration in time derivatives. K=curvature term

A common descriptive used by pop media, simplified for layman level readers unfamiliar with the BB that is commonly used unfortunately is explosion of spacetime. However the term explosion typically implies a force vector. However a constant vector is not involved in accordance to a huge bulk of observational evidence. A more accurate description is a "rapid expansion of spacetime" due to reducing average /energy mass densities. This is where the ideal gas laws of thermodynamics steps in. this discussion doesn't make any statements of before the BB. The BB model doesn't describe how the universe began but how it evolved. \(10^{-43}\) after the BB

Just to give an applicable example of what sort of influences a vector field can have in the Einstein field equations. I'm going to provide a couple of links of a related theory that has a constant vector field. In this case its a rotating universe that results in torsion. Einstein-Cartan Theory https://en.wikipedia.org/wiki/Einstein–Cartan_theory notice how the stress tensor is affected in that link. It literally doesn't matter how fast or slow the universe is spinning or how minimal of a vector value one has. The mathematics of that model include the entire range of possible values. This is a good example of how the stress tensor gets affected which in turn affects any metric tensor. now one might think if its spinning too slow for any observer to notice then we can ignore it and believe Minkowskii space would work under Einstein-Cartan. This however isn't true. One of the very important aspects of observer is the world line via the geodesic equations. The full geodesic equation is \[\frac{d^2 x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^{\beta}}{ds}=0\] now the following equation will look somewhat different as the article its from has already factored out the terms it requires that and one can replace any symbol for a tensor with any other identifier and the tensor performs precisely the same way. What one uses to symbolize or name a given tensor is simply convenience. anyways in the article the new geodesic is given by equation 10. \[\frac{d^a x^a}{ds^2}+\Gamma^a_{bc}\frac{dx^b}{ds^2}\frac{dx^c}{ds^2}+2S_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds}=0\] https://api.repository.cam.ac.uk/server/api/core/bitstreams/4d357658-b056-45bf-8d29-919db6fac184/content Now to help those not math savvy. What this essentially shows is an antisymmetric universe that resulted from rotation that generates a torsion term. The effect is that the worldlines are in turn subsequently affected as well as a metrics affine connections, Bianchi identities, Christoffels and killing vectors. Its knowing these details that provides me sufficient reason to doubt any claim that an exploding universe can apply the Minkowskii tensor and get 100 percent of the observational evidence we currently have as a perfect match let alone a near match. Now a consequence of a different geodesic equation is that no Observer will get the same results as any observer in a Minkowskii spacetime (Newtonian limit spacetime) let alone in other spacetimes such as the Schwarzschild metric. Now do we look for anistrophic universes absolutely the research in that never stops, here is a 2016 research paper that places the error margin of 121,000 to 1 in disfavor of an anistropic expansion due to explosion/rotation etc. (the paper studies for any form of directional component. How isotropic is the Universe? https://arxiv.org/abs/1605.07178 I have seen papers with higher numbers in disfavor but as I couldn't put my fingers on them this one will suffice. The question does pertain so I provided my definition with a quick descriptive and the applicable mathematics. I will let @KJW speak for himself on how he thinks an explosion entails and how he is applying his observers. As far as your expansion question what we define as expansion isn't particularly at odds. The debate is what is causing the expansion. In cosmology expansion is literally the results of thermodynamics. Hence the equations of state for cosmology and their incorporation into the FLRW metric acceleration equation. @KJW is attempting an expansion due to an explosion rather than a thermodynamic expansion.

Well for my definition it's quite simple a center of origin with vectors radiating outward at every angle. That's the mathematics I've shown using V(x,y,z)=(x,y,z) for 3d coordinates for simplicity. It doesn't matter what the vectors they are, they are present so cannot be arbitrarily ignored. Having some observer commoving with the vectors doesn't help eliminate the vectors for other observers.

Well whatever you believe if you didn't understand that the Minkowskii tensor which defines the metric would be altered by any vector field. I really can't help you. I even showed you the EFE represention showing that.

No one is arguing that energy density can used to determine gravitational effects. That's an obvious what isn't obvious is amount of gravity one will experience due to that energy/mass will depend on its distribution. You simply have to study and apply Newtons Shell theorem to understand that. Let's put it this way in the Einstein field equations the stress energy momentum term it does not matter if the energy density term is from matter or from radiation. Its simply the energy mass density. How it is distributed in the tensor is what determines how the Stress tensor tells spacetime how to curve to produce gravitational effects. Anyways you can easily convert units such as joules for energy into kg for mass. There are plenty of online calculators for that. It's trivial to take an equal amount of energy/density of matter to an equal amount energy/mass density of photons to see that if doesn't matter if energy mass density is produced by photons or matter but simply the quantity and distribution

Lol the EM field is used as one of the more common Lorentz invariant and constancy of c tests. We've even shown in tests that the EM fields do produce photons via the Aharohm Bohm effect where the particle production in the math was often considered just considered a math sleight of hand

No worries I'm always glad to help anyone trying to better understand physics. As far as experiments for what equates to c the speed limit and there have been a huge number of tests and the error margin resulting from all those tests the constancy of c. This falls under Lorentz invariance so subsequently any tests for Lorentz invariance involves the constancy of c. This lists some of the more modern tests https://arxiv.org/pdf/2111.02029.pdf however its nowhere near a complete listing of test variations. the error margin is incredibly low something on the order of 1 part in 10^(18} for error margin but it would take some digging to find the current error margin

Well unfortunately I still don't agree with that. You may recall I recommended using vector fields ? after all an explosion one common type of vector field that's easily represented. This particular field is a vector field of the first order. a vector field \(V(x,y,z)\) of the first order can be written in the following form of a constant matrix "A" and a constant vector B \[V^T=V(x,y,z))^T=A\cdot [x,y,z]^T+B^T\] the T indicates the transposition of the respective matrix. \[\begin{pmatrix}V_1(x,y,z)\\V_2(x,y,z)\\V_3(x,y,z)\end{pmatrix}=\begin{pmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{pmatrix}\cdot \begin{pmatrix}x\\y\\z\end{pmatrix}+\begin{pmatrix}B_1\\B_2\\B_3\end{pmatrix}\] where B is the constant vector and A the constant matrix. so lets say you have a constant wind in the z direction \(V(x,y,,z),,,B=(0,0,B)\) a rotating vector field would look something like this \(V(x,y,z)=(-y,x,0)\) exploding vector field \(V(x,y,z)=(x,y,z)\) imploding \(V(x,y,z)=(-x,-y,-z)\) Now it doesn't matter the matrix is it could be simply a 3d Euclidean or we can add the 4th dimension to make this 4d or higher dimension. If say we were describing the matter fields or radiation fields we would use the stress energy momentum tensor. However on could have some vector quantity affect geometry as well. Anyways the Minkoskwii metric has no constant vector so there is no B term involved. So claiming it matches any vector space is simply put a mathematically incorrect statement. Recall Minkowskii space is a free fall space with no vectors hence the inner product of two vectors which returns a scalar usage in the Minkowskii tensors. \[\mu \cdot \nu)=(\nu \cdot\mu)\] this statement also tells you its commutative with no preferred direction or reference point. For your observers as per SR observers. and the premises of SR, which a vector field does not satisfy. (the mathematical proof is rather convoluted I'd rather not go into that lmao) but a simplified statement is vector fields fall under SU(N) groups and are not orthogonal where the Minkowskii metric is orthogonal. Under the Poincare group SO(3.1). Now lets skip up to EFE. lets describe our vector as a form of permutations and so we can separate spacetime in any manner of fields including multiple fields. This is regularly done in the EFE its literally one of the most common steps. So lets set our Newtonian limit (MInkowskii) under \(\eta_{\mu\nu}\), lets set our permutations under the premutation tensor \(H_{\mu\nu}\) \[G_{\mu\nu}=\eta_{\mu\nu}+H_{\mu\nu}\] so if you have a constant vector you alter the permutation tensor which gives a new geometry under\(G_{\mu\nu}\) with \( H_{\mu\nu}\) wait a sec that's pretty much the same as what we are doing in the vector spaces. Granted GR loves partial derivatives but an explosion is trivial to describe as divergence of a field using partial derivatives in essence you have some permutation field whether its vectors, scalars etc etc acting upon your baseline metric which forms a new metric statement.

Why do you keep thinking pressure or energy/mass density is gravitational effect ? They do both effect gravity but by different ratios that those equations by themselves do not show if you place the pc^2 portion of radiation it would be the T^00 component of the stress energy momentum tensor. The P in pc^2 is the momentum term not the density or pressure in your equations. So you need further conversions However if both matter and radiation are uniformly distributed gravity at any point would be zero according to Newtons shell theorem. So any gravitational effect depends on its distribution. \[E=pc^2\] where the p is momentum not pressure or density. Another indication of realizing thinking those equations equate to gravity effect is incorrect is that a critically dense universe being flat spacetime has no gravity as there is no curvature term.