Dubbelosix 152 Posted April 15, 2019 Report Share Posted April 15, 2019 (edited) Since the late cosmology just asserts the physics of the Firemdann equation, we will stick with the main result which was based on strict scientific reason: 1. Cold big bangs start at near absolute zero temperature By modification of the Friedmann equation for reversible and irreversible dynamics, the Helmholtz-Gibbs thermodynamic phase change of the Friedmann equation (which incidently, merges the physics of the opening post with Motz and Kraft's version of the pre big bang phase) takes a very simple form:[math]\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = -\frac{1024 G}{360}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \frac{dU_2}{dV}[/math] This last equation, I call the unification equation, not because it tells us about why pre-big bang states exist, but unifies the notion of the expanding universe under thermal pressure. The term on the right is a cosmological interpretation of the zero point fluctuations When matter appears, the a above pre-big bang phase transcends into the Friedmann equation. We will cover implications soon. Edited April 18, 2019 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 17, 2019 Author Report Share Posted April 17, 2019 Ok, so let's start a quiz. The left hand side tells you the state of an ''almost'' Friedmann term you would find in his famous equation, with a small addition of a time derivative. The right hand side tells us about a condensate not so much different to a zero point field. These are now questions anyone can try and answer and I will tell you if you are close to the mark. 1. If the left hand side says something about expansion, what is it telling us about the condensate on the right hand side? 2. What significance of this third derivative on the left hand, possibly say about the right hand side? Quote Link to post Share on other sites

Dubbelosix 152 Posted April 17, 2019 Author Report Share Posted April 17, 2019 No takers? Quote Link to post Share on other sites

Dubbelosix 152 Posted April 18, 2019 Author Report Share Posted April 18, 2019 The equation in the OP has been selected to reveal an equality, but I should present the full equation in which the final equality presents a rather attractve solution involving a ratio of the zeta function, with a complete thermodynamic coefficient per unit particle which has the time derivative attached to it, meaning the number of particles are allowed to be non-conserved. The real task is to find the boundary in which this can no longer happen, leading to the ordinary Friedmann equation. The full OP equation is [math]\int\ \frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2})\ dV = \frac{3072 G}{60}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ PV = \frac{1024 G}{360}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \mathbf{U}V = \frac{3072 G}{60} \frac{\zeta(4)}{\zeta(3)}\dot{N}k_BT[/math] This is a bit complicated to understand, which was why I reduced it to the equation in the OP. I won't give any answers away yet, but maybe this fuller equation could give insight into the questions I provided in the OP. Quote Link to post Share on other sites

Bradpitt4 1 Posted April 18, 2019 Report Share Posted April 18, 2019 Since the late cosmology just asserts the physis of the Firemdann equation, we will stick with the main result which was based on strict scientific reason: 1. Cold big bangs start at near absolute zero temperature By modification of the Friedmann equation for reversible and irreversible dynamics, the Helmholtz-Gibbs thermodynamic phase change of the Friedmann equation (which incidently, merges the physics of the opening post with Motz and Kraft's version of the pre big bang phase) takes a very simple form:[math]\frac{\dot{T}}{T}(\frac{\ddot{T}}{T} + \frac{kc^2}{a^2}) = -\frac{1024 G}{360}(\frac{k^3_B \zeta(3)}{\hbar^3c^3}\frac{dV}{dt}T^3)\ \frac{dU_2}{dV}[/math] This last equation, I call the unification equation, not because it tells us about why pre-big bang states exist, but unifies the notion of the expanding universe under thermal pressure. The term on the right is a cosmological interpretation of the zero point flucuations When matter appears, the a above pre-big bang phase transcends into the Friemdann equation. We will cover implications soon. Ok, so let's start a quiz. The left hand side tells you the state of an ''almost'' Friedmann term you would find in his famous equation, with a small addition of a time derivative. The right hand side tells us about a condensate not so much different to a zero point field. These are now questions anyone can try and answer and I will tell you if you are close to the mark. 1. If the left hand side says something about expansion, what is it telling us about the condensate on the right hand side? 2. What significance of this third derivative on the left hand, possibly say about the right hand side?1. I had to really dissect your equation to get at the heart of the matter of that left hand side is saying that expansion is c squared and the right hand side seems to indicate that it happens when the very first particles in the universe were plus charge for a little over one Planck time but less than a nanosecond but that expansion came when the first down quarks emerged is just gravity. 2. I had to really sink my teeth into the implications of that equation, but it seems that the lagrangian produces the plus up quarks in which a single down quark on the left side can be paired with multiple up quarks. OceanBreeze 1 Quote Link to post Share on other sites

OceanBreeze 425 Posted April 18, 2019 Report Share Posted April 18, 2019 1. I had to really dissect your equation to get at the heart of the matter of that left hand side is saying that expansion is c squared and the right hand side seems to indicate that it happens when the very first particles in the universe were plus charge for a little over one Planck time but less than a nanosecond but that expansion came when the first down quarks emerged is just gravity. 2. I had to really sink my teeth into the implications of that equation, but it seems that the lagrangian produces the plus up quarks in which a single down quark on the left side can be paired with multiple up quarks. It would be great if you could explain how you came to see all that, because Dubbelosix never explains any of his equations. Quote Link to post Share on other sites

fahrquad 83 Posted April 18, 2019 Report Share Posted April 18, 2019 "Thoery"? Quote Link to post Share on other sites

OceanBreeze 425 Posted April 18, 2019 Report Share Posted April 18, 2019 "Thoery"? It didn't take you long to spot the Hole in his theory! Quote Link to post Share on other sites

exchemist 732 Posted April 18, 2019 Report Share Posted April 18, 2019 1. I had to really dissect your equation to get at the heart of the matter of that left hand side is saying that expansion is c squared and the right hand side seems to indicate that it happens when the very first particles in the universe were plus charge for a little over one Planck time but less than a nanosecond but that expansion came when the first down quarks emerged is just gravity. 2. I had to really sink my teeth into the implications of that equation, but it seems that the lagrangian produces the plus up quarks in which a single down quark on the left side can be paired with multiple up quarks.I have difficulty in seeing how something with the dimensions L²/T² can be a rate of expansion. The expansion rate is usually quoted in km/sec/Mpc, i.e. dimensions of L/T². Quote Link to post Share on other sites

OceanBreeze 425 Posted April 18, 2019 Report Share Posted April 18, 2019 Bradpitt4, on 18 Apr 2019 - 1:10 PM, said: 1. I had to really dissect your equation to get at the heart of the matter of that left hand side is saying that expansion is c squared and the right hand side seems to indicate that it happens when the very first particles in the universe were plus charge for a little over one Planck time but less than a nanosecond but that expansion came when the first down quarks emerged is just gravity. 2. I had to really sink my teeth into the implications of that equation, but it seems that the lagrangian produces the plus up quarks in which a single down quark on the left side can be paired with multiple up quarks. I have difficulty in seeing how something with the dimensions L²/T² can be a rate of expansion. The expansion rate is usually quoted in km/sec/Mpc, i.e. dimensions of L/T². Not only that, but I looked and looked again, but I still can't see those damn quarks! Quote Link to post Share on other sites

exchemist 732 Posted April 18, 2019 Report Share Posted April 18, 2019 Not only that, but I looked and looked again, but I still can't see those damn quarks! Yes I think he is probably pulling Doublesox's leg. In fact the dimensions of the two terms in the bracket on the left side of the equation seem to be different anyway: the first terms has dimensions of T⁻² while the second has dimensions of (E/Temp) x L²/T² , i.e. ML⁴/T⁴Temp. So the equation looks as if it is trying to add apples and oranges, unless the unexplained "a" in the denominator has dimensions of √M.L²/T²Temp, which seems implausible. Par for the course, I suppose. OceanBreeze 1 Quote Link to post Share on other sites

Dubbelosix 152 Posted April 18, 2019 Author Report Share Posted April 18, 2019 "Thoery"? A very rare typo :) Quote Link to post Share on other sites

Dubbelosix 152 Posted April 18, 2019 Author Report Share Posted April 18, 2019 1. I had to really dissect your equation to get at the heart of the matter of that left hand side is saying that expansion is c squared and the right hand side seems to indicate that it happens when the very first particles in the universe were plus charge for a little over one Planck time but less than a nanosecond but that expansion came when the first down quarks emerged is just gravity. 2. I had to really sink my teeth into the implications of that equation, but it seems that the lagrangian produces the plus up quarks in which a single down quark on the left side can be paired with multiple up quarks. I promised myself that if anyone really attempted to understand this, I would remain nice. But this is just hogwash, even other posters can see this. Quote Link to post Share on other sites

OceanBreeze 425 Posted April 18, 2019 Report Share Posted April 18, 2019 Yes I think he is probably pulling Doublesox's leg. I see what you did there. :good: Quote Link to post Share on other sites

Dubbelosix 152 Posted April 18, 2019 Author Report Share Posted April 18, 2019 I see what you did there. :good: Yes, was quite inventive. But let's try and leave some jokes sparse from now on, despite the ridiculous statement made by... well I know who it is, I'd like to keep this as serious as possible. Quote Link to post Share on other sites

exchemist 732 Posted April 18, 2019 Report Share Posted April 18, 2019 (edited) I see what you did there. :good:Actually I've just had a look at the Friedmann equations and so now I know what a and k are. I find I was wrongly taking k to be Boltzmann's constant (seeing as Doublesox's equation contains temperature so is implicitly thermodynamic) but it is isn't that. In the context of Friedmann's equations, k is a dimensionless parameter, set to be between -1 and +1, according to whether a 3-spherical, flat or hyperbolic universe is being considered. So it is a sort of "shape" factor. a, on the other hand is the "scale factor", again dimensionless. So I need to re-do my dimension check on the second term in the LHS bracket. I now get something a lot simpler, viz. L²/T². So it is still a nonsense, as one cannot add terms with different dimensions, but at least we don't have to worry about daft things like √M. Maybe if Doublesox can add something to the first term with dimensions L², he can make it look more superficially plausible, at least. Edited April 18, 2019 by exchemist Quote Link to post Share on other sites

Dubbelosix 152 Posted April 18, 2019 Author Report Share Posted April 18, 2019 I have difficulty in seeing how something with the dimensions L²/T² can be a rate of expansion. The expansion rate is usually quoted in km/sec/Mpc, i.e. dimensions of L/T². What are you on about? You're the last person I would expect to be an expert on dimensional analysis, I do it everyday. Try again, or don't. But don't prance about these posts trying to ''uncover'' an error when I have presented the equations to physicists who very clearly understand it. Just accept that you are wrong, or at least learn some more dimensional analysis that forgoes your own basic knowledge of it. Quote Link to post Share on other sites

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