Mordred

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.

sigh honestly do you seriously not know the difference between something that expands with no inherent direction of any particles that volume contains such as Brownian motion as opposed to a inherent direction of gas escaping a previously pressurized container such as a balloon ? Come on man this is seriously getting futile ask yourself what causes the universe to expand and tell me why you would ignore the equations of state that cause expansion in your analysis ? The very purpose is to predict how objects such as stars and galaxies move the way they do in regards to how they separate. The very goal is to explain how the distance of stellar objects increase from one another. I 100% guarantee you will not match observational evidence via an exploding universe beginning....

\[{ds^2} = {d\vec{r^2}} +{r^2}[d\theta^2 + {sin^2} d\phi^2]\] there an exploding spacetime simply use spherical coordinates and designate the radius with a vector. Now all you need is to add your time coordinate for 4d and a scale factor

\[\frac{1}{R}(\frac{{{d^2}R}}{{d{t^2}}}) = - \frac{{4\pi G}}{3}({\rho _m} + {\rho _r} + {\rho _\Lambda } + \frac{{3({P_m} + {P_r} + {P_\Lambda })}}{{{c^2}}})\] that's the math you posted that didn't latex properly as you forgot \ in your opening [math] statement. Massless particles are set up as ultra relativistic in terms of the equations of state. https://en.wikipedia.org/wiki/Equation_of_state_(cosmology) \[\rho=-4\] for ultra relativistic you can use the \(\omega\) for the ratio between pressure and energy density for ultra relativistic particles. That book also doesn't show the cosmological constant portion. At least not what you posted. You don't have the author to see if I have the same textbook .

Your right you and I do have a different notion of explosion. Particularly in so far as what a geometry would look like if the explosion resulted in what we perceive as spacetime expansion with regards to what the resulting metric would look like. Particularly in what the resulting distribution of mass and its behavior would entail. The primary distinction you keep avoiding us an explosion always results in a preferred direction continously posting math that has no preferred direction will not change my mind. Of course there is a reason why any good cosmology textbook will specify expansion as opposed to explosion. You might want to consider that detail in so far that there is reasons that is the case. I don't know perhaps/perhaps not you might notice the difference if you take a multiparticle system and model each case in vector space would help you see the difference. Expansion after all involves thermodynamics.... Perhaps you can ask yourself how something like the critical density formula even work in the explosion case. Perhaps you can describe the difference between an expanding gas due to temperature change as opposed to one being forced in a particular direction such as popping a balloon might help to understand the difference

Ordinarily I avoid posting whiteboard lesson notes however this one has some valuable images to better help understand cross sections than an article that simply has the relevant math. https://www.precision.hep.phy.cam.ac.uk/wp-content/people/mitov/lectures/ParticlePhysics-2018/H01_Introduction.pdf

the cross section is the waveform you get when you have multiple particles interact. It gives us valuable information such as determining say the mass of the Higgs Boson. The cross sections of how that boson interacts with neutrinos, leptons etc. Allows us to determine its mass value as one example. Cross sections for the standard model for all particles are also applied in the CKMS and PMNS mass mixing matrix using the Weinberg angles (requires the cross section). Here is some cross sections I had in my own thread dealing with Early universe processes as an example you can see every particle interaction generates its own cross section. Those cross sections vary depending on the properties such as spin, charge, mass, flavor etc. Early Universe Cross section list Breit Wigner cross section \[\sigma(E)=\frac{2J+1}{2s_1+1)(2S_2+1)}\frac{4\pi}{k^2}[\frac{\Gamma^2/4}{(E-E_0)^2+\Gamma/4)}]B_{in}B_{out}\] E=c.m energy, J is spin of resonance, (2S_1+1)(2s_2+1) is the #of polarization states of the two incident particles, the c.m., initial momentum k E_0 is the energy c.m. at resonance, \Gamma is full width at half max amplitude, B_[in} B_{out] are the initial and final state for narrow resonance the [] can be replaced by \[\pi\Gamma\delta(E-E_0)^2/2\] The production of point-like, spin-1/2 fermions in e+e− annihilation through a virtual photon at c.m. \[e^+,e^-\longrightarrow\gamma^\ast\longrightarrow f\bar{f}\] \[\frac{d\sigma}{d\Omega}=N_c{\alpha^2}{4S}\beta[1+\cos^2\theta+(1-\beta^2)\sin^2\theta]Q^2_f\] where \[\beta=v/c\] c/m frame scattering angle \[\theta\] fermion charge \[Q_f\] if factor [N_c=1=charged leptons if N_c=3 for quarks. if v=c then (ultrarelativistic particles) \[\sigma=N_cQ^2_f\frac{4\pi\alpha^2}{3s}=N_cQ^2_f\frac{86.8 nb}{s (GeV^2)}\] 2 pair quark to 2 pair quark \[\frac{d\sigma}{d\Omega}(q\bar{q}\rightarrow \acute{q}\acute{\bar{q}})=\frac{\alpha^2_s}{9s}\frac{t^2+u^2}{s^2}\] cross pair symmetry gives \[\frac{d\sigma}{d\Omega}(q\bar{q}\rightarrow \acute{q}\acute{\bar{q}})=\frac{\alpha^2_s}{9s}\frac{t^2+u^2}{t^2}\] here is a sample of phonon cross sections. https://e-learning.pan-training.eu/wiki/index.php/The_scattering_cross_section_for_phonons One detail you should note is that the cross sections and subsequent mean lifetime for accoustic phonons will differ from optical phonons. however using phonons vs photons for conduction has far more involved than merely some personal choice of which to use. Which one is chosen is determined via direct experimentation and not some personal or group choice. There is in fact distinctive and measurable differences between phonons (acoustic and optical) and photons or any other gauge boson virtual or otherwise. That last link for example mentions the effects of a particles polarizations on cross sections. Longitudinal polarizations have different effects than transverse polarizations.

nor is there a galactic time per se

here is one paper that applies the phonon cross sections in its research via resonant scatterings https://arxiv.org/pdf/2109.12117.pdf

Like Swansont I too have done related experiments. For example you mentioned your feelings on virtual particles. However when you are conducting experiments that involve resonance in scatterings you will find that many of the exchanges involve the photon mediator. Great you have no issue with photons. However those photons are offshell. They do not have the same energy levels as real photons. The offshell mediator boson is typically offshell and as such falls under the virtual counterpart. Now this actually becomes unavoidable when you start applying Fermi-golden rule in resonance scatterings. A prime example is when electrons absorb a photon. In the lab you can detect this interaction however the energy level of mediator is far less than a real photon energy level. Yet you cannot deny that an intermediate particle isn't involved. Repeatable experimentation clearly shows the intermediate interaction. here is Fermi-Golden rule \[\Gamma=\frac{2\pi}{\hbar}|V_{fi}|^2\frac{dN}{DE_f}\] single phonon operator term \[E_n=\hbar\omega_q(n_q+\frac{1}{2})\] suffice it to say I've had a lot of dealings with detectable resonant scatterings that can one can only equate to virtual intermediate interactions. Good example is photons mediating charge of an EM field. Even though the photon itself has no charge it still mediates charge A large part of the reason I mention this is that the resonance of a photon virtual ensemble or otherwise will differ from that of phonons they each have different resonant signatures. the cross sections of photons/phonons vary depending on the interaction they are involved in. How they vary show distinctions as the two particles do have different properties. Now the point of this is even though you may or may not get the math to work out using photons for conductance. You may very well find the cross sections of the scatterings involved will match phonons and not photons. I didn't bother posting details on the photon as they are readily available. Its more tricky to find the needed details for phonons. There is also no point posting cross section formulas as each scattering will have its own. They all apply the Breit Wigner cross section formula \[\sigma(E)=\frac{2J+1}{2s_1+1)(2S_2+1)}\frac{4\pi}{k^2}[\frac{\Gamma^2/4}{(E-E_0)^2+\Gamma/4)}]B_{in}B_{out}\]

Well the more in depth you get into particle physics you will find the distinction between Real , quasi, and virtual particles becomes less and less important. Phonons however are not virtual particles they are quasi particles. They measurable properties of which they represent. Both real and quasi-particles are field excitations in so far as they have a minimal of a quanta of action. Individual virtual particles however do not hence they are oft treated as permutations in field theory as opposed to an excitation. You can never measure individual virtual particles its impossible. Now here is where the distinction becomes important. In Feymann path integrals virtual particles are represented by the internal lines as a type of field propagator. Quasi and real particles due to the same requirements of field excitation (localization of a wavefunction) fall on the external lines on path integrals. In QFT this doesn't really matter as all particles are simply states where all particles are field excitations represented by that state. QFT is operator driven so it uses the creation/annihilation operator for thermodynamic treatments as well as for particle production. Its variation of all these classical thermodynamic laws for conduction, convention and radiation apply those operators via the Euler Langrangian equations for its path integrals for example Bose Einstein in classical radiation treatment which is used to calculate the number density of bosons at a given blackbody temperature looks like this \[n_i = \frac {g_i} {e^{(\varepsilon_i-\mu)/kT} - 1}\] PS at some point you will need this equation as well as Fermi-Dirac and Maxwell Boltzmann. under QFT looks like this Bose Einstein QFT format. \[|\vec{k_1}\vec{k_2}\rangle\hat{a}^\dagger(\vec{k_1})\hat{a}^\dagger(\vec{k_2})|0\rangle\] \[\Rightarrow |\vec{k_1}\vec{k_2}\rangle= |\vec{k_2}\vec{k_1}\rangle\] note both versions use phase/momentum space. Practice your vector algebra lol you will need it. Anyways enough about which mediator to use. It isn't particularly important at this stage except to be aware of the distinction of its usage in anything you read on conduction vs radiation treatments. note those distinctions may or may not be compatible with photons with regards to phonons how they are used may mathematically differ. for example an electron hole can behave the same as an electron but it isn't a particle but can be represented by a quasi particle

Lets try another thought experiment using space expanding outward from the explosion. Lets set rate of flow outward from origin. Lets have an increase of space set at 1 parsec per second why not. Now place an observer that flows outward from from the center and set the observer 1 sec after the BB. Simply so we have space ahead of that observer. As space is created outward from the origin the space between the observer and the origin increases but if the space 1 second previous to that observer is flowing outward. Then no new space is being formed ahead of that same observer. That isn't uniform expansion where the distance between every coordinate expands uniformly. (ignore time dilation we are just dealing with how space increases in each case) In the FLRW metric I'm using the commoving observer or equivalent.

The mathematics you posted isn't an explosion but a homogeneous isotropic space is precisely why I posted the proof for a metric space. It is not a space resulting from an explosion. We aren't even considering the time dimension there is no time dilation involved in a uniform mass distribution so you don't require it. Stick with 3 dimensional space. If you expand a Euclidean space uniformly that's expansion however in an explosion the space further away from the point of origin will expand faster than the space near the origin point. You can easily test that by using a cone and slicing the cone at various distances from the origin and measuring its area at the end of the cone furthest from the origin point. The cone is your explosion geometry not Euclidean geometry. Though its correctly a sphere expanding outward from the origin however all particle and vectors will have a preferred direction (outward from origin). That isn't the case in our universe. In our universe there is no preferred direction no vectors has a preferred direction due to any previous force applied regardless if those vectors are coordinates of geometry change or particles. A kinetic style explosion certainly has a preferred direction (outward) from its origin. A good way to model a direction preferred spacetime would be to apply spherical coordinates or even cylindrical coordinates such as the rotating Godel universe whose expansion was due to the outward force of rotation. There is a large list of spacetimes (mostly dealing with BH/WH scenarios). A particularly good example is this particular anistropic model Is the Universe anisotropic right now? Comparing the real Universe with the Kasner's space-time https://arxiv.org/pdf/2305.02726.pdf

the expression for the relation above is as follows without Dirac notation. Using metric space \(\mathcal{M}\) d is a distance function A metric space will satisfy three conditions positivity ( simply sets the x and y axis positive norm so were ignoring any - axis.) \(d(x,y\ge 0\) with iff \( "=" x=y\) a Homogeneous and isotropic will be \[d(x,y)=d(y,x)\] triangle inequality as \[d(x,z)\le d(x,y)+d(y,z)\] there's triangle inequality in terms of distance. If your familiar with Pythagorus theorem and triangle inequality I really don't understand how you cannot see the difference between an expansion of a homogeneous and isotropic metric space and a metric space as an anistropic and inhomogeneous explosion.

Let me ask are you familiar enough with Dirac notation and triangle inequality in vector space ? There is a few math expressions that beautifully express the distinction between triangle inequality and homogeneous and isotropic using vector spaces under Dirac notation is most commonly used. the fact is if you change length A and length B by an equal amount for simplicity length C does not change by an equal amount without an change in angles. Pythagorus theorem clearly shows this with right angle triangles. A homogeneous and isotropic change all points must change equally in ratio in all cases on all sides of the triangle

Are you going tell me you could not see that all sides did not change equally in the hypotenuse on a right angle triangle? In the last example I gave above ?