TheCosmologist Posted February 19, 2023 Share Posted February 19, 2023 I am the sole author of this work, all rights are reserved to me. Elastic Action: A Wedding of Quantum Field Theory with the General Relativiatic Action This tackles a question on how Sakharov's ground state field for virtual particles enters the gravitational action. In it I conclude that maybe the cosmological constant is in fact a renormalization constant which is only set to zero for flat Euclidean spacetime. The jury us still out, but most respected astrophysicist tend to agree that while spacetime looks quite flat, it probably isn't exactly flat, it's just a very good approximation and his equations on a cosmological scale would predict a small curve like we expect. Let's identify variables [math]A[/math] - action [math]g[/math] - metric [math]x[/math] - variable spatial coordinate [math]c[/math] - speed of light [math]\mathbf{R}[/math] - Ricci curvature scalar [math]\hbar[/math] - Planck constant, reduced [math]G[/math] - Newtons constant [math]k[/math] - wave number * We will use [math]\mathbf{k}[/math] as a constant [math]\frac{8πG}{c^4}[/math] which is the upper value of the gravitational constant In the style of Sakharov, we'd like to write a Langrangian of the ground state fluctuations which has a contribution of geometry by off-shell virtual particles. It's a rare paper to find, but his original ideas can be found here, (a link can be found in references to his paper), https://www.atticusrarebooks.com/pages/books/719/a-d-sakharov-andrei/vacuum-quantum-fluctuations-in-curved-space-and-the-theory-of-gravitation-in-soviet-physics His original ideas can be taken as a precursor to Bogoliubov transformations used to describe how gravity jiggles these off-shell particles at the horizon of supermassive Black holes, owing to their name as Hawking radiation. We then use Sakharov's curvature corrected Langrangian [math]\mathcal{L} = \mathbf{R}\ \hbar c\ k\ \int dk + \mathbf{R}^2\ \hbar c\ \int \frac{dk}{k} + C[/math] [math]= \mathbf{R}\ \mathbf{A}\ k\ \int dk + \mathbf{R}^2\ \mathbf{B}\ \int \frac{dk}{k} + C[/math] [math]= \mathbf{R}\ \mathbf{A}\ k\ \int dk + \mathbf{R}^2\ \mathbf{B}\ \int d\log k + C[/math] Where [math]\int \frac{dk}{k} \approx 137[/math] (the inverse fine structure). C is a renormalization constant set to zero for flat Euclidean spacetime - loosely it could be seen as the cosmological constant. At no point in spacetime is the cosmological constant really zero, so this argues for a microscopic description of curvature. This isn't meant to mean that flat spacetime is devoid of fluctuations however - modern field theory assures us that there is no such thing as empty space as fluctuations are being produced constantly, both on curved backgrounds and in flat spacetime. To meld field theory with the gravitational action, we first identify the action as [math]A = c \int\ d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\mathbf{R} +…][/math] *the presence of the speed of light as a coefficient of the integrand is because of the dimensiones that requires an energy multiplied by a time, which is also the units of action, ie one can use [math]dx = cdt[/math]. In the Landau and Lifshiftz theory of fields, the action becomes [math]A = \frac{c^3}{8\pi G}\ \int\ d^4x\ \sqrt{-g}[\mathbf{R} +…][/math] This is just another way of writing [math]A = c\ \int\ d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\mathbf{R} +…][/math] Let's show quickly why. Using [math]\frac{G}{c^2} = \frac{\ell}{m}[/math] Then we can crunch the dimensions down easily on, [math]A = \frac{c^3}{8\pi G}\ \int\ d^4x\ \sqrt{-g}[\mathbf{R} +…][/math] We invert the length to mass ratio and plug in, which leaves a coefficient of c on the mass term, and the RHS absorbs the inverse lengths to produce the mass density, and we absorb one factor of c from the four dimensional volume element [math]\frac{mc}{\ell} (\ell^4) \cdot \frac{1}{\ell^2} \cdot dxdydz\ cdt = \rho c^2 dt \cdot dxdydz\ dt[/math] Which has units of energy times time, which are the dimensions of action. Plugging in the nodes from the first formula, (which is the background curvature correction of off shell virtual particles discovered by Sakharov), we unify quantum field theory with gravity in the following way, and name it the "elastic action," and we drop the constant of integration for brevity, [math]A = c\ \int\ d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\mathbf{R}\ \hbar c\ k\ \int dk[/math] [math]+ \mathbf{R}^2\ \hbar c\ \int \frac{dk}{k}]\frac{16\pi G}{c^4}[/math] It may look funny that we have had to introduce yet another [math]\frac{16\pi G}{c^4}[/math] but this is very much required as a coefficient that "undoes" what we did mathematically inside of the square parenthesis. I'll show why in the footnotes. This quantum correction is applied to the theory of relativity, melding the two nicely. In short it says the gravitational action involves the ground state zero point energy fields. First Varied Action Using the formulation set above, we now can apply the varied action. Notice the variation does not apply to the constants or as usual, to do this is a straightforward process where I will use Einstein summation. [math]\delta A = c\ \int d^4x\ \frac{\sqrt{-g}}{\mathbf{k}}[\delta g^{\mu \nu} \mathbf{R}_{\mu \nu}\ \mathbf{A}\ k\ \int dk[/math] [math]+ \delta g \mathbf{R}^{\mu \nu}\mathbf{R}_{\mu \nu}\ \mathbf{B}\ \int \frac{dk}{k}]\frac{16\pi G}{c^4}[/math] This is the varied action, we simply split the corrected Reimann curvature R^(μv)R_(μv) like so, as it acts like the fourth inverse power over the wavelengths. Notes: To understand why we had to correct the dimensions for a new coefficient [math]\frac{16\pi G}{c^4}[/math] plays to functional roles. The constants 16 \pi removes the constants found in [math]\mathbf{k}[/math] and [math]\frac{G}{c^4}[/math] is the (only) reasonable gravitational coefficient that removes the dimensions of the Sakharov terms. We just want to focus on how this altered the dimensions and how we must fix those dimensions. Focusing on this, [math]\mathbf{R}\ \hbar c\ k[/math] We know what is, and so it's dimensions are still inverse length squared. The k is called the wave number and has inverse unit of length. All-in-all, we have dimensions of charge squared divided by a length, giving an energy, further with another inverse length cubed, giving the appropriate dimensions of energy density. What we "put in" those brackets, must be undone, and there's a straight-forward way to do it. We don't need to "undo" what we have in since it already features in the action, but we will concentrate on [math]\hbar c\ k\ \int dk = \frac{\hbar c}{\ell^2}[/math] We understand that the following dimensions must hold true: [math]\frac{G}{c^2} \equiv \frac{\ell}{m}[/math] Since dimensionally-speaking [math]\frac{\hbar c}{\ell^2} = \frac{Gm^2}{\ell^2}[/math] Then we can decompose it in the following way: [math]\frac{Gm}{\ell^2} = \frac{Gm}{\ell}\frac{c^2}{G} = \frac{m}{\ell} \cdot c^2 = \frac{c^4}{G}[/math] Interesting isn't it? It seems then the solution has been found. In order to "undo" what we did, it requires a correction coefficient of the upper limit of gravity as [math]\frac{c^4}{G}[/math] (by taking its inverse). Why its inverse? Simply because if [math]\frac{\hbar c}{\ell^2} = \frac{Gm^2}{\ell^2} = \frac{c^4}{G}[/math] Then we must invert to remove these unwanted dimensions. Second Varied Action Starting from the Lagrangian density we derived earlier: [math]\mathcal{L} = c\ \frac{\sqrt{-g}}{\mathbf{k}} \left[\mathbf{R} \hbar c\ k \int dk + \mathbf{R}^2 \hbar c\ \int \frac{dk}{k} \right] \frac{16 \pi G}{c^4}[/math] To compute the variation of the action, we need to vary the metric tensor g_{\mu\nu} and integrate the variation over all spacetime coordinates: [math]\delta A = \int d^4x\ \delta\left(\mathcal{L}\sqrt{-g}\right)[/math] Using the product rule of variation, we have: [math]\delta A = \int d^4x\ \left[\delta\mathcal{L}\sqrt{-g} + \mathcal{L}\delta(\sqrt{-g})\right][/math] We can simplify the second term using the variation of the determinant of the metric tensor: [math]\delta(\sqrt{-g}) = \frac{1}{2}\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}[/math] Substituting this expression and our variation of the Lagrangian density into the equation for the variation of the action, we obtain: [math]\delta A = c \int d^4x\ \frac{\sqrt{-g}}{\mathbf{k}} [\delta g^{\mu \nu}\ \mathbf{R}_{\mu \nu}\ \mathbf{A}\ k \int dk[/math] [math]+ \delta g \mathbf{R}^{\mu \nu} \mathbf{R}_{\mu \nu}\ \mathbf{B}\ \int \frac{dk}{k}]\frac{16 \pi G}{c^4}[/math] where we have defined [math]\mathbf{A} = \hbar c[/math] and [math]\mathbf{B} = \hbar c[/math]. Note that the variation of the determinant of the metric tensor has been expressed in terms of the variation of the metric tensor [math]g_{\mu\nu}[/math], using the inverse metric [math]g^{\mu\nu}[/math] and the Ricci tensor [math]\mathbf{R}_{\mu\nu}[/math]. Substituting our expression for [math]\delta(\sqrt{-g})[/math] gives: [math]\delta A = -\frac{c}{2}\ \int\ d^4x\ \frac{\sqrt{-g}\ g^{\mu\nu}\delta g_{\mu\nu}}{\mathbf{k}}[\mathbf{R}\ \hbar c\ k\ \int dk[/math] [math]+ \mathbf{R}^2\ \hbar c\ \int \frac{dk}{k}]\frac{16\pi G}{c^4}[/math] We can simplify this expression by using the fact that [math]\sqrt{-g}\ g^{\mu\nu} = -2g^{\mu\nu}[/math], which gives: [math]\delta A = c\int\ d^4x\ \frac{g^{\mu\nu}\delta g_{\mu\nu}}{\mathbf{k}}[\mathbf{R}\ \mathbf{A}\ k\ \int dk[/math] [math]+ \mathbf{R}^2\ \mathbf{B}\ \int \frac{dk}{k}]\frac{8\pi G}{c^4}[/math] Higher Powers Sakharov concludes the higher powers are taken like so: [math]\int \frac{dk}{k}(\mathbf{B}\ \mathbf{R}^2 + \mathbf{C}\ \mathbf{R}^{ik}\mathbf{R}_{ik} +\mathbf{D}\ \mathbf{R}^{ikjm}\mathbf{R}_{ikjm} … + higher\ powers)[/math] Where [math]\int \frac{dk}{k} \approx 137[/math] [math]\delta A = c\int\ d^4x\ \frac{g^{\mu\nu}\delta g_{\mu\nu}}{\mathbf{k}}[\mathbf{R}\ \mathbf{A}\ k\ \int dk + \int \frac{dk}{k}(\mathbf{B}\ \mathbf{R}^2 + \mathbf{C}\ \mathbf{R}^{ik}\mathbf{R}_{ik} +\mathbf{D}\ \mathbf{R}^{ikjm}\mathbf{R}_{ikjm} )]\frac{8\pi G}{c^4}[/math] Higher Powers of Fluctuations [math]\mathbf{R}\ \hbar c\ k \int dk[/math], taking higher powers of [math]\hbar c[/math] requires that the dimensions are scaled appropriately… Say the higher powers don't just affect the curvature, but affects higher powers of [math]\hbar c = (A, B,C,D) \approx 1[/math] So taking higher powers of [math]\hbar c[/math] in gives [math]\delta A = c\int\ d^4x\ \frac{g^{\mu\nu}\delta g_{\mu\nu}}{\mathbf{k}}[\mathbf{R}\ \mathbf{A}\ k\ \int dk + \int \frac{dk}{k}(\mathbf{B}^2\frac{1}{Gm^2}\ \mathbf{R}^2[/math] [math]+ \mathbf{C}^3 \frac{1}{G^2m^4}\ \mathbf{R}^{ik}\mathbf{R}_{ik} +\mathbf{D}^4 \frac{1}{G^6m^8}\ \mathbf{R}^{ikjm}\mathbf{R}_{ikjm} )]\frac{8\pi G}{c^4}[/math] Would be the same as writing [math]\delta A = c\int\ d^4x\ \frac{g^{\mu\nu}\delta g_{\mu\nu}}{\mathbf{k}}[\mathbf{R}\ \mathbf{A}\ k\ \int dk + \int d\log k(\mathbf{B}^2 \alpha^{-1}_G\ \mathbf{R}^2[/math] [math]+ \mathbf{C}^3\alpha^{-2}_G \ \mathbf{R}^{ik}\mathbf{R}_{ik} +\mathbf{D}^4 \alpha^{-3}_G\ \mathbf{R}^{ikjm}\mathbf{R}_{ikjm} )]\frac{8\pi G}{c^4}[/math] Where we use the gravitational fine structure to normalise the higher dimensions, [math]\mathbf{R}^2 \hbar^2 c^2(\frac{1}{Gm^2}) k \int dk = \mathbf{R}^2 \hbar c\ \alpha^{-1}_G k \int dk[/math] While the fine structure constant and Newton's gravitational constant are both fundamental constants of nature, they describe completely different physical phenomena. Therefore, it is not accurate to say that higher powers of the fine structure constant are equivalent to correcting gravity at higher powers. However, there are some theories, such as string theory, that suggest a connection between the values of fundamental constants and the properties of space-time, including the strength of gravity. In these theories, it is possible that changes in the fine structure constant could lead to changes in the properties of space-time and therefore, to corrections in the theory of gravity. But this is a highly speculative area of research and is not yet well understood. Taking higher powers of the fine structure constant can reveal higher order corrections to physical phenomena that cannot be accounted for by classical or first-order quantum mechanical calculations. In particular, higher-order quantum corrections are important in understanding the behavior of subatomic particles and the interactions between them. For example, the anomalous magnetic moment of the electron can be calculated with higher and higher accuracy by taking into account higher order QED corrections involving higher powers of the fine structure constant. The electron's anomalous magnetic moment has been measured experimentally to extremely high precision, and the agreement between theory and experiment is a remarkable demonstration of the power of higher-order quantum corrections. Similarly, in quantum chromodynamics (QCD), the theory that describes the strong nuclear force, higher order corrections involving higher powers of the strong coupling constant (the analog of the fine structure constant for the strong force) are important for understanding the properties of hadrons (particles made of quarks), and for calculating the scattering amplitudes of quarks and gluons. In general, higher order corrections involving higher powers of dimensionless coupling constants are important for understanding the behavior of quantum field theories at energies far beyond the scales of current experiments. Such corrections can also provide clues to the existence of new physics beyond the Standard Model of particle physics. References http://ayuba.fr/pdf/sakharov_qvf.pdf Link to comment Share on other sites More sharing options...

joigus Posted February 20, 2023 Share Posted February 20, 2023 Just a few preliminary questions. If it's all about Sakharov's Lagrangian from 1968, the action --which is nothing but the Lagrangian density integrated to all of space--, must lead to exactly Sakharov's action from 1968. So, In what sense is it new? In what sense is it elastic? Why no mention of current work on Sakharov's Lagrangian? --as late as 2018, from what I gather. Why do you not consider Planck's length as a reasonable cutoff, and concentrate only on EM, though make comments on QCD, which has a completely different coupling constant? About notation: What does it mean when you write bold-face symbols? That's usually reserved for vectors or tensors, not for scalars? In what sense are you singling out these scalars? And last but not least, What are your conclusions? Link to comment Share on other sites More sharing options...

TheCosmologist Posted February 20, 2023 Author Share Posted February 20, 2023 It's new because the gravitational action has never been written to encorporate the fluctuations of the ground state fields. It's called the elastic action because Sakharov coined the term metrical elasticity in his paper which is recited. Bold R is the curvature scalar, bold k is simply a constant which is found in the action. All bold notation has been throughly explained in the OP. I have made a mention of Sakharov including the original Langrangian. My conclusions are the following: fluctuations can and do effect the background of space and has strange analogy to the Casimir effect (as mentioned by Sakharov) but more curiously we reach a chicken and egg problem. What came first, curvature or fluctuations? Can curvature of the manifold produce the fluctuations? Moreover when the curvature is extremely significant, these virtual off shell particles can become real on shell particles by virtue of what is called a "gravitational boost." It's not a well-known subject, but gravity can affect the way these particles are created, and is very similar to how Hawking radiation becomes long-lived fluctuations from the curvature decay of black holes. As mentioned, this effect of curvature on fluctuations, and fluctuations on curvature holds vital and important questions about the nature of quantum gravity. Link to comment Share on other sites More sharing options...

joigus Posted February 20, 2023 Share Posted February 20, 2023 1 hour ago, TheCosmologist said: It's new because the gravitational action has never been written to encorporate the fluctuations of the ground state fields. Quote 5. Identify the regulated one-loop effective action as the leading contribution to gravity. The gravitational action induced at one-loop order consists of contributions to vacuum energy, curvature terms up to second order and torsion terms up to fourth order It seems there have been other people working on the very same thing. Why don't you quote other people's work on exactly the same? Sorry I said, 1 hour ago, joigus said: If it's all about Sakharov's Lagrangian from 1968, the action --which is nothing but the Lagrangian density integrated to all of space--, must lead to exactly Sakharov's action from 1968. So, I should have said, "the action --which is nothing but the Lagrangian integrated over time." The Lagrangian density integrated over space gives the Lagrangian. Anyway, you seem to be using a very peculiar renormalisation scheme. You just plug in some constants and do some dimensional analysis from there. What renormalisation technique are you using? Link to comment Share on other sites More sharing options...

joigus Posted February 20, 2023 Share Posted February 20, 2023 Let me clarify further: It's not a new elastic action, is it? It seems to be good-old Sakharov's action in a bizarre notation. So what appears to be 'new' is how cavalierly you deal with divergent integrals by substituting them for a number and then using some numerical analysis by means of a handwaving 'technique.' It's actually your method for regularising the integrals that's to be subject to scrutiny. Experts are likely to ask you about that. Can you justify the only thing that's new, please? Link to comment Share on other sites More sharing options...

TheCosmologist Posted February 20, 2023 Author Share Posted February 20, 2023 39 minutes ago, joigus said: Let me clarify further: It's not a new elastic action, is it? It seems to be good-old Sakharov's action in a bizarre notation. So what appears to be 'new' is how cavalierly you deal with divergent integrals by substituting them for a number and then using some numerical analysis by means of a handwaving 'technique.' It's actually your method for regularising the integrals that's to be subject to scrutiny. Experts are likely to ask you about that. Can you justify the only thing that's new, please? Yes it is new You seem seem a bit confused over this matter. It's very much new as no one has ever wrote the action in terms of field theory like this. That's what constitutes the equation to be "new." Link to comment Share on other sites More sharing options...

joigus Posted February 20, 2023 Share Posted February 20, 2023 16 minutes ago, TheCosmologist said: Yes it is new No, it's not. 19 minutes ago, TheCosmologist said: You seem seem a bit confused over this matter. No, I'm not. This has taken me back to when I was aged seven, and having arguments with my classmates. Not very long after the paper you're trying to get credit for was published. Thanks for the memories. Link to comment Share on other sites More sharing options...

iNow Posted February 20, 2023 Share Posted February 20, 2023 ChatGPT is similarly confident when declaring plainly wrong answers as entirely valid 1 Link to comment Share on other sites More sharing options...

TheCosmologist Posted February 20, 2023 Author Share Posted February 20, 2023 7 minutes ago, iNow said: ChatGPT is similarly confident when declaring plainly wrong answers as entirely valid Curious statement, in what sense of the matter am I relatable to a chatbot? 10 minutes ago, joigus said: No, it's not. No, I'm not. This has taken me back to when I was aged seven, and having arguments with my classmates. Not very long after the paper you're trying to get credit for was published. Thanks for the memories. Sorry what? I am confused now! You do realise the equations in the OP are based on the melding of field theory with the gravitational action, and has not been don3 before. What part am I trying to take credit for except which was my own contribution of the melding of the two? I've been very specific claiming this. Link to comment Share on other sites More sharing options...

joigus Posted February 20, 2023 Share Posted February 20, 2023 2 minutes ago, TheCosmologist said: Curious statement, in what sense of the matter am I relatable to a chatbot? That's exactly what ChatGPT would say. Link to comment Share on other sites More sharing options...

exchemist Posted February 20, 2023 Share Posted February 20, 2023 1 minute ago, joigus said: That's exactly what ChatGPT would say. This all has an air of Reiku/Graviphoton/Gareth Meredith...... and all stations to Cardiff......about it. 1 Link to comment Share on other sites More sharing options...

TheCosmologist Posted February 20, 2023 Author Share Posted February 20, 2023 6 minutes ago, joigus said: That's exactly what ChatGPT would say. you're very strange. I am very much a fleshy human being. Link to comment Share on other sites More sharing options...

joigus Posted February 20, 2023 Share Posted February 20, 2023 5 minutes ago, exchemist said: This all has an air of Reiku/Graviphoton/Gareth Meredith...... and all stations to Cardiff......about it. Pardon? 😅 Link to comment Share on other sites More sharing options...

iNow Posted February 20, 2023 Share Posted February 20, 2023 25 minutes ago, TheCosmologist said: in what sense of the matter am I relatable to a chatbot? In most senses, specifically herein Link to comment Share on other sites More sharing options...

joigus Posted February 20, 2023 Share Posted February 20, 2023 19 minutes ago, TheCosmologist said: you're very strange. I am very much a fleshy human being. Nay, it is thou who discombobulates me with thy maelstrom in a way. 1 Link to comment Share on other sites More sharing options...

TheCosmologist Posted February 21, 2023 Author Share Posted February 21, 2023 (edited) 9 hours ago, joigus said: Nay, it is thou who discombobulates me with thy maelstrom in a way. I thought you'd appreciate the time that I took to rebuke the statements you have made about nothing being "new" here as I finally came to realise what your issue is. So please tell me if anything I've said in the forthcoming post, is incorrect in any way to the way your mindset works on this issue On The Elastic Paper: Part II A Rebuttal To Accusation and Further Insights When I proposed this new theory, a critique, and not a very insightful one, couldn't ascertain why my formulas where new. Here I point out a historical reference to the Dirac equation - Dirac knew about the mass energy equivalence in its relativistic form, equally, from Einstein’s own special theory. He was also aware of how to make energy and momentum operators. When upgrading observables to their operator equivalences, we call this process a second quantization method. Now, like all theoretical scientists today, we “stand on the shoulders of giants” as Einstein was modest to admit. The point of Dirac, was to meld quantum theory in a “wedding” with relativity. We do not say, like I was accused of, of taking credit of already known formula's. Indeed, the formulas I constructed are but an extension not dissimilar to the process Dirac took, except mine concentrated on general relativity. Would we say he “stole the work” and then accuse of nothing being “new” about the Dirac equation…? Of course not. Further Insights to Build a Mental Picture Now I'll provide a deeper insight into one of my last formulae, namely the higher curvature corrections and higher powers of the Planck constant which revealed equally higher powers of the gravitational fine structure. [math]\delta A = c\int\ d^4x\ \frac{g^{\mu\nu}\delta g_{\mu\nu}}{\mathbf{k}}[\mathbf{R}\ \mathbf{A}\ k\ \int dk[/math] [math]+ \int \frac{dk}{k}(\mathbf{B}^2 \alpha^{-1}_G\ \mathbf{R}^2 + \mathbf{C}^3\alpha^{-2}_G \ \mathbf{R}^{ik}\mathbf{R}_{ik}[/math] [math]+\mathbf{D}^4 \alpha^{-3}_G\ \mathbf{R}^{ikjm}\mathbf{R}_{ikjm} )]\frac{8\pi G}{c^4}[/math] The higher powers of the Planck constant could mean we are dealing with more virtual particles given in some differential volume element, which take the form in curved spacetime as [math]dV = \sqrt{g}\ dxdydz[/math] It would be naive to think we are dealing with a single virtual particle in all cases, moreover, further folly to think that the prediction made in the first paper about how curvature can produce fluctuations, to neglect a notion that higher power curvature corrections produce a flurry of energetic virtual particles in a given space to be devoid from the theoretical model presented. Indeed, during the first instant of our universe when it nucleosynthesized into existence, there was present the strongest gravitational forces that have ever existed in our universe. This is what I call the “birthplace of real fluctuations,” as the gravitational field again, was able to boost a reasonable quantity of the off-shell particles into long-lived real on shell matter, the same observable “stuff” that we see in present cosmology which encompasses anything between 1–4% of all spacetime volume. It must have been a very fast process which does owe some merit to inflation - if it has been a slow process during the initial expansion phase, much more observable matter would have been present today and the observed density would be of orders much larger. There's around [math]3 \times 10^{80}[/math] particles roughly speaking in our universe, give or take a power of ten. If the expansion phase had been much slower, the magnitudes could have been tens upon hundreds of powers larger. So rapid initial expansion seems to concur why spacetime appears so dilute in this respect also. The Singularity Sakharov mentioned the initial singularity in his paper - for many reasons, many cosmological models now avoid singular solutions to spacetime. One quantum law which should forbid singularities for example, is that you cannot squeeze a particle into a region of spacetime that is smaller than its own wavelength, meaning the initial phase was not [math]R =0[/math]. * just as a side note, even Hawking and Penrose abandoned their own singularity theorems as some historical reference. An important feature of the aforementioned equation is that it contains the wave number of the fluctuations. As mentioned earlier, it would be naive to think we are dealing with a unit spacetime cell, with one fluctuation, the equation deals with higher powers of curvature closely tied to the element volume - it’s basically telling us that any region of spacetime larger than a discrete cell can have two, and even more fluctuations in a given increasing volume that we integrate over and is the true frequency divided by the speed of the wave and thus equal to the number of waves in a unit distance. This is why melding Sakharovs equation and the action is important as it deals with distances as it is an action principle. Further, Bosons can fall into the same energy states, whilst the Fermions cannot - so the physics can be as exotic as the model allows for given fluctuations in a region of spacetime. In previous investigations I made on the Bekenstein entropy and other entropy equations, I discovered that non-commutation was something seemingly written into the structure of black holes. Don't worry, we won't digress from the topic too much, I just want to introduce quickly why the wave number is interesting for any calculation when done properly. [math]\mathbf{S} \leq k_B\frac{2 \pi mc^2R}{4 \hbar c} = k_B\frac{2 \pi\ c\ mR}{4 \hbar}[/math] This encompasses the deBroglie wavelength as [math]\mathbf{S} \leq k_B\frac{2 \pi mc^2R}{4 \hbar c} = k_B\frac{2 \pi\ c\ R}{4 \lambda}[/math] It just so happens we can replace [math]\frac{2\pi\ R}{\lambda}[/math] Where the wave number is [math]k = \frac{2\pi}{\lambda}[/math] With [math]\Delta k \Delta x \geq \frac{1}{2}[/math] Which is an analogue of an uncertainty principle obtained by Fourier, well before Heisenberg was on the picture, that a picture stated that a superposition of waves could not have both a small size and also a small number of frequencies. It makes the entropy appear as [math]\mathbf{S} \leq k_B\frac{\Delta R}{4 \Delta \lambda} = k_B\frac{ \Delta k \Delta x}{4}[/math] This is where the Fourier transform part cones in... The momentum of the system is expressed as [math]p = (\frac{h}{2\pi})k[/math] By multiplying [math]\Delta k \Delta x \geq \frac{1}{2}[/math] By [math]\frac{h}{2\pi}[/math] allows us to obtain [math](\Delta k \frac{h}{2\pi}) \Delta x = \Delta p \Delta x \geq \frac{h}{2\pi} = \frac{\hbar}{2}[/math] To be fair, the first part really addresses a rebuttal of your statements about nothing being new here, the rest was just showcase. Edited February 21, 2023 by TheCosmologist Link to comment Share on other sites More sharing options...

exchemist Posted February 27, 2023 Share Posted February 27, 2023 Just in case of any doubts about authorship, here is Gareth's paper on Quora: https://www.quora.com/profile/Gareth-Lee-Meredith-1/On-The-Elastic-Paper-Part-II-A-Rebuttal-To-Accusation-and-Further-Insights-When-I-proposed-this-new-theory-a-critiqu?ch=15&oid=100930088&share=8daaadb7&srid=uUK54g&target_type=post Link to comment Share on other sites More sharing options...

Sensei Posted February 27, 2023 Share Posted February 27, 2023 (edited) On 2/20/2023 at 3:34 PM, joigus said: That's exactly what ChatGPT would say. Aren't you just bits on my storage? (let me see your code.. koti didn't survive this stage) On 2/20/2023 at 3:41 PM, TheCosmologist said: you're very strange. I am very much a fleshy human being. Where is your vat? Let me see.. https://en.wikipedia.org/wiki/Brain_in_a_vat If this makes you laugh, it's not your fault, I just stimulated your brain cells (the one in vat).. Edited February 27, 2023 by Sensei 1 Link to comment Share on other sites More sharing options...

joigus Posted February 27, 2023 Share Posted February 27, 2023 5 hours ago, Sensei said: Aren't you just bits on my storage? (let me see your code.. koti didn't survive this stage) LOL. Yeah, on second thought that's not that clear, what I said. Sometimes I say things just to see how people react. I suppose you need to probe the other one somehow. Especially online, where a lot of usual clues are missing. Link to comment Share on other sites More sharing options...

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