Mordred

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.

You don't particularly have to worry about it you can simply use the field treatments without specifying the mediator by using operator states. All mediator bosons are massless all you need from there is sufficient energy/momentum for mediation. However you don't need to learn the relations via the quantum world most statistical mechanics textbooks barely touch on the quantum regime. However at the quantum level then even at short range you will need a mediator. Let's put it this way you can mathematically describe everything you need using nothing more than classical mechanics. When you get to the quantum regime the number density of your mediator can then be determined by whatever the energy level of field or state your measuring by simply applying the Bose-Einstein statistics for bosons.So it's something you can worry about later provided your not breaking any laws such as the speed limit.

One of the first lessons in physics is that energy is a property of some state, object etc. Energy doesn't exist on its own, hence the use of mediator particles. Energy as a property of an objects/state etc ability to perform work. Mass is another property that doesn't exist on its own. It too is a property. Mass being the objects/state etc resistance to inertia change or shortly described acceleration. Those definitions apply regardless of physics theory. Including string theory as properties they must have some state etc to be applied to.

This is just an interesting tidbit as radiation equation of state is \[p_R=\rho_R/3\equiv \omega=1/3 \] when you apply Maxwell-Boltzmann statistics which is a mixed combination of the Bose-Einstein and Fermi-Dirac statistics using the effective degrees of freedom of all bosons and fermions one can determine the temperature evolution. The result turns out to be the inverse of the scale factor. \[\rho_R=\frac{\pi^2}{30}{g_{*S}=\sum_{i=bosons}gi(\frac{T_i}{T})^3+\frac{7}{8}\sum_{i=fermions}gi(\frac{T_i}{T})}^3 \] \[S=\frac{2\pi^2}{45}g_{*s}(at)^3=constant\] temperature scales inversely to the scale factor giving \[T=T_O(1+z)\] as a function of redshift so now you can also follow the temperature evolution knowing this relation

There is another set of ratios that may be of interest resulting from the above equations that may prove useful in better visualizing the amount of influence each contributor has. First for all reader the critical density is the density that without the cosmological constant the universe is critically flat. one can use the above to get a ratio of \(\Omega_i=\rho_i/\rho_c\) \[\Omega_M=\frac{8\pi G}{3H^2_0}\rho_M\] \[\Omega_R=\frac{8\pi G}{3H^2_0}\rho_R\] \[\Omega_{\Lambda}=\frac{\Lambda}{3H^2_0}\] with curvature term \[K=\frac{K}{3a_0^2H^2_0}\] those are the ratios of each component \[H^2(a)=H_o^2(\Omega_R\frac{a_0^4}{a^4}+\Omega_M\frac{a_0^3}{a^3}+\Omega_\Lambda+\Omega_K\frac{a_0^2}{a^2})\] now using the same ratios of the density evolutions of matter, radiation and lambda setting k=0 and knowing that the Hubble parameter changes over time. One can produce the following equation of how the universe evolves over time. This equation correlates the Hubble rate compared to the rate today at each redshift. \[H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}\] as well as the density of each at a particular redshift

In any explosion I know of you will always have vectors involved. The vectors of the blast radius as well its fragments. However a simple and very common approach is simply attach to every spacetime coordinate a test particle. Its a very common treatment in GR. The vector field regardless of what's used originating from any central point radiates outward. T The above is a perfect example of a preferred location. ( the origin point) and a preferred direction (the net sum of vectors) a uniform expansion has net no flow of vectors. Nor does it have a center hence the cosmological principle. There is no inherent direction to how the universe expands. nor is there in stellar objects we measure there is no bulk direction flow.

ok first off what those relations are designed to show you is the critical density relations used to determine the geometry. As well as expansion rates of the geometry. in essence of the following statement \[\Omega_{total}=\Omega_m+\Omega_r+\Omega_\Lambda\] so in your equation they are showing the pressure relations on the portion of the LHS of the equal sign where I have the total density. You can normalize the LHS. The RHS should be the percentages of each except they have the wrong values in each category. Matter for example has 0 pressure, Its energy density has zero influence on pressure. Lambda gets funky in that currently its a negative pressure relation w=-1. You would need to use the scalar field equation of state in the link I gave for the kinetic vs potential energy terms. the statement would be properly presented as this \[\frac{\ddot{a}}{a}=-\frac{4\pi G}{c^2}(\rho+3P)+\frac{\Lambda}{3}\] that details the following. \[\frac{d}{dt^2}(\rho a^3)+P\frac{d}{dt}a^3=0\] (notice your 1 to 3 ratio ) where \(\rho_m=0, \ P_r=\rho/3, \) matter exerts no pressure so the term in the bracket is the radiation energy density to pressure relation. I'm not sure why the relations look wrong in what you have if that's from the book you used. However hopefully the format I used is far more clear. it may help to know that the critical density formula uses matter only with no pressure term p=0 \[\rho_{crit} = \frac{3c^2H^2}{8\pi G}\] that may help to understand the ratio between pressure and density

Mathematica should be able to do it but if wolfram alpha runs on Mathematica experience has taught me you usually end up doing a large parcel of the work to force it. Wolpram alpha does have tensor packages to do a ot of the tensor operations so its useful in that regard. I don't see why Mathematica doesn't do the same. Inversing the matrix is very useful good skill to have. its one of the common requirements to test specific gauge conditions. I've never tested either though for Christoffels not sure how you would go about it except simple geometry.

You need to be able to do it for any spacetime curvature term not just the orthogonal case. That's the easiest one. metrics of previous post are added you cross posted even though I stated was still in edit no problem just read my previous post. Please keep in mind I did not include some bulk mass flow in that above. That would require detailing the stress energy momentum tensor which in turned affects the metric. the Bulk flow would be your explosion of energy/mass spewing out of that singularity

Not really determining a metrics Christoffel and subsequently its killing vectors is a rather tricky process that even many in the field can stumble over. Over the years I've seen even professionally peer reviewed articles overturned due to incorrectly determining either. ok you have the spherical coordinate metric for the FLRW great \[(ds)^2 = c^2 (dt)^2 - a^2(t) ((dr)^2 + r^2 ((d\theta)^2 + sin^2 \theta (d\phi)^2))\] OK the equation you have satisfies the Minkowskii metric in so far as Modern cossmology states for a homogeneous and isotropic expansion. You and I both agree on this. We both agree that SR/GR holds. Where we disagree is what happens if you have some preferred direction. A preferred direction can mean many things that direction could be expanding faster or slower than another direction. It could have some variation of flow, it could have some difference in the amount of force or its pressure. it is some feature that requires it to mathematically make it different than any other direction. Same applies for a preferred location. So what an easy example. Well as were also using SR lets simply look at a velocity boost of the metric itself in a given direction. Here is your starting metric [latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex] [latex]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex] Here is how different boosts occur in different directions on that metric. In spherical coordinates as per this scenario. Lorentz group Lorentz transformations list spherical coordinates (rotation along the z axis through an angle ) \[\theta\] \[(x^0,x^1,x^2,x^3)=(ct,r,\theta\\phi)\] \[(x_0,x_1,x_2,x_3)=(-ct,r,r^2,\theta,[r^2\sin^2\theta]\phi)\] \[\acute{x}=x\cos\theta+y\sin\theta,,,\acute{y}=-x\sin\theta+y \cos\theta\] \[\Lambda^\mu_\nu=\begin{pmatrix}1&0&0&0\\0&\cos\theta&\sin\theta&0\\0&\sin\theta&\cos\theta&0\\0&0&0&1\end{pmatrix}\] generator along z axis \[k_z=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}\] generator of boost along x axis:: \[k_x=\frac{1\partial\phi}{i\partial\phi}|_{\phi=0}=-i\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0 \end{pmatrix}\] boost along y axis\ \[k_y=-i\begin{pmatrix}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0 \end{pmatrix}\] generator of boost along z direction \[k_z=-i\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0 \end{pmatrix}\] the above is the generator of boosts below is the generator of rotations. \[J_z=\frac{1\partial\Lambda}{i\partial\theta}|_{\theta=0}\] \[J_x=-i\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&-1&0 \end{pmatrix}\] \[J_y=-i\begin{pmatrix}0&0&0&0\\0&0&0&-1\\0&0&1&0\\0&0&0&0 \end{pmatrix}\] \[J_z=-i\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&0 \end{pmatrix}\] there is the boosts and rotations we will need and they obey commutations \[[A,B]=AB-BA\] You can see you lose the orthogonal condition of the Minkowskii metric regardless of which direction the boost occurs in. Now if you recall an observer moving at relativistic velocity experiences two factors. length contraction and time dilation. Both factors change the metric. Those transforms above restore the changed metric back to the original. That is the very function of the Lorentz transforms. IT IS TO RESTORE orthogonality..... so lets set a preferred direction cause who cares....in the x direction. [latex]\eta=\begin{pmatrix}-c^2&1&0&0\\1&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex] now if I had an inertial observer also moving in the x direction those points are already filled. So I must account for them differently

prove it and in that prove you a preferred direction any mathematics that does not have a preferred direction will be useless to prove your point. Such as the math you have posted so far. While your at it prove you can have a kinetic based exploding volume that matches observational evidence of a cosmological event horizon that exceeds the Hubble Horizon and get recessive velocities in excess of c with your equations above (this occurs beyond the Hubble horizon Don't think using the Homogeneous and isotropic metric of Minkowskii will prove your point. That metric does not describe an inhomogeneous and anisotropic spacetime. By any equivalence principle. For the record its been attempted numerous times by Professors in other cosmology alternative theories. This includes the rotating Godel universe. We were able to even confirm our universe is not rotating. There have been attempts with models such as a universe birthed by a WH or a previous BH. None of these models could properly match observational evidence. so GL I know you likely don't have the skill set for Christoffel connections and Killing vectors so I won't ask you to produce them.