Mordred

\[{ds^2} = {d\vec{r^2}} +{r^2}[d\theta^2 + {sin^2} d\phi^2]\] 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 ?

truthfully put aside any concern on what the mediator particle is. It really doesn't matter as they can be used offshell being a mediator boson. Reduce the problem to 2 forms of energy. Potential energy and kinetic energy. Study the linear equations first via studying the mathematics of a classical spring. A graph of its momentum over time is dipolar just as the case with the EM field. This will further equate to the harmonic oscillations as well as the wave equations of QM/QFT. Make sure you have a good understanding of the energy/work/power/momentum relations involved. (all included in any decent book on thermodynamics thus far.). study each case separately conduction, convection and radiation. You should note the relationship between the potential energy terms and the kinetic energy terms an objects conductivity can be described via the spring and those energy terms. After that for a field treatment for temperature use a scalar field this is also included in any decent book on the topic. you will need to understand wave equation treatments if you plan on applying QM/QFT.

were using the triangle to help visualize coordinate changes remember ? It is basic geometry and ratios of change in the case of expansion the distance between any ensemble of coordinates will change equally. That is not the case in an explosion scenario. perhaps this will help lets apply actual coordinates. place a point at 0.0 on an x,y graph. Now place a point at 6,0 and 0,6. measure the distance between 0.6 and 6,0. Now change 0.6 and 6,0 to 12,0 and 0,12 and measure the change in distance between 0,12 and 12,0. Does that distance change by 6 which is the ratio of change along the x and y axis ? the length of the Hypotenuse will have a different rate of change as per Pythagorus theorem using the x and y axis allows us to form a right angle triangle . a^2+b^2=c^2

sigh use that same graph and draw a set three points to make a triangle in the explosion case draw a line from the origin to each corner of the triangle. Now move the triangle along the radial lines you just drew keeping the three points on those lines you drew from the center. Can you honestly tell me that the length of each side expands equally and has no angle change ? An even simple example is any two points on the circumference of a circle. if the radius increases so to does the distance between the two points the rate of change isn't linear as the radius increases either. For example take a circle with radius line place two points on the circumference at two different angles. Now just to get a triangle place a point on one of those angles half of the circumference. Will all sides change equally if you shift the triangle outward while preserving the same angle from the origin to each point of the triangle ? the two points on the same radial angle will have no change in distance while the other point will increase in distance between the other two

the argument against an explosion is valid when you look at the angles and not just distance change between multiple points. That is literally the lesson the balloon analogy was originally designed to convey. Though a more accurate analogy for 3d is the raisin bread analogy

see the edit above we cross posted on edit.

I wouldn't say equivalent when you further consider photons scatterings such as with say Thompson scatterings. You can see the issue via this link. https://en.wikipedia.org/wiki/Thomson_scattering however that's not an issue with phonons defined as Phonon (lattice vibration, cause of thermal conduction in electric insulators, and participant in energy conversion involving kinetic energy of solids). as opposed to photons Photon (has the largest range of energy, carriers its energy even through vacuum, and interacts with electric and magnetic entities in energy conversion). Now consider a solid is held together by an EM field.

Do you recall me mentioning phonons under the conduction case ? I do recall providing a gauge invariance requirement with a link detailing where this requirement becomes an issue with a lattice. Now I'm not saying its incorrect to use photons in this case however you must also factor in the conductivity of the material. If you look at the article the same equation for the quanta of phonons are identical to the equation for photons. We use the phonons simply to make the distinction between the conductive case to the radiative case. This also frees up certain key conservation laws with particle to particle scatterings. Those include conservation of charge, color, flavor, energy momentum, spin etc. Using phonons as a quasi particle we can avoid these issues.

The angles would be needed between any 2 points. In an explosion scenario the angles would vary between any two or more points. This would include a metric that is expanding outward from an origin point. (specifically if one wanted to have commoving coordinates radiating outward from an origin for an inhomogeneous and anistropic expansion ) A homogeneous and isotropic expansion no angles change including those of the metric points. That also includes curvature terms once k is set it doesn't change over time. If K=0 in the past it will remain k=o to the present. With k=0 there is no time dilation with its uniform mass distribution.

You do not show a metric for an explosion. An explosion radiates from starting point outward. So it is anisotropic. The metric equations you have are essentially homogeneous and isotropic. You might want to consider the inner product of your vectors in terms of the symmetry relations of the Minkowski metric. \[\mu\cdot\nu=\nu\cdot\mu\] describing an explosion of an ensemble of particles from an origin point would require angles. for example take 3 particles on a sheet of graph paper. Set an origin point a 0,0. Now place a particle at 1.1 another at 3,5 the last at say 5,3. Now expansion the graph lines increase in scale. (using a draftman's scaling ruler would be useful simply set the new scale at 1:2 ratio and redraw your grid lines). None of the angles change between 1.1, 3,5 and 5,3. Those angles would change if those particles were radiating outward from the origin point.

Glad to see you understand where your errors were coming about. GL in your studies on the 3 scenarios.

Those links written by Dean is literally garbage, for his supposed degrees he certainly didn't learn the the first thing about how to write a decent paper. Though quite frankly there is nothing in those links that deals with physics. Seems more appropriate for philosophy but I honestly don't even see anything worth discussing even under philosophy.