stephaneww
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 Birthday 10/02/1968
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I have no idea where I'm going.I don't even know if it's applicable, although this encourages me to say yes ( cosmological constant seen as a perfect gas with a negative pressure) : But it's going to be very complicated and the finalization of this project is more than uncertain. I don't know the solubility issues at all and I don't know if scalar and vector aspects are involved in this potential approach. I just want to know if at first glance it seems completely silly or if it's worth trying to find out more about it.

What is this "fundamental pure number" please ?
stephaneww replied to stephaneww's topic in Astronomy and Cosmology
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What is this "fundamental pure number" please ?
stephaneww replied to stephaneww's topic in Astronomy and Cosmology
I've got it. It's just [math]\frac{1}{t_p^2\Lambda_{s^{2}}}[/math] or [math]\frac{1}{l_p^2\Lambda_{m^{2}}}[/math], whatever you want. The factor is [math]8\pi[/math]. 
The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
done here : https://www.scienceforums.net/topic/122453anattempttoapproachanotionofsolubilityincosmologytoexplainthecosmologicalconstant/ 
What is this "fundamental pure number" please ?
stephaneww posted a topic in Astronomy and Cosmology
Hi, I have read here under (2.1) that [math]c^5/(G \hbar \Lambda_{s^{2}}[/math] is a "fundamental pure number" It's the word "fundamental" that questions my mind. What is that fondamental notion please? 
Hi, Planck's mass density = [math]\frac{m_p}{l_p^3}=\frac{c^5}{G^2 \hbar} \text{ }\frac{kg}{m^3} [/math] is assumed to be the maximum density in relativity with the massenergy equivalence of GR, we have: Planck's energy density = [math]A=\frac{m_p c^2}{l_p^3}=\frac{m_p l_p^2}{l_p^3 t_p^2}=\frac{m_p}{l_p t_p^2}=\frac{c^7}{G^2 \hbar} \text{ }\frac{kg}{m s^2} [/math] [math]A = \frac{m_p c^2}{l_p^3}=\hbar c l_p^{4}\text{ }J/m^3[/math] [math]A =\hbar c (l_p^{2})^2=\frac{c^7}{G^2 \hbar} \text{ }J/m^3[/math] (1) By dimensional analysis, we make the hypothesis that [math]l_p^{2}[/math] can be replaced in (1) by [math]\Lambda_{m^{2}}[/math] to obtain [math]B[/math] a "Planck's energy density of the cosmological constant" and we add a divisor [math]\frac{1}{(8 \pi)^2}[/math] : [math]B = \frac{1}{(8 \pi)^2} \hbar c( \Lambda_{m^{2}} )^2 \text{ }J/m^3[/math] (2) It turns out that the geometric mean of the squared root of [math]A[/math] and [math]B[/math] is the energy density of the cosmological constant [math]C[/math]. [math]\Large{C=\sqrt{A}\sqrt{B} =\sqrt{\frac{c^7}{G^2 \hbar}} \sqrt{\frac{\hbar c ( \Lambda_{m^{2}})^2 }{(8 \pi)^2}} \text{ }J/m^3}[/math] [math]\Large{C=\sqrt{A B} =\sqrt{\frac{c^7 \text{ }c \text{ }( \Lambda_{m^{2}} )^2 \hbar}{(8 \pi)^2 G^2 \hbar} } \text{ }J/m^3} [/math] [math]\Large{C=\frac{c^4 \Lambda_{m^{2}} }{(8 \pi) G } \text{ }J/m^3}[/math] (3) which is the exact formula for the energy density of the cosmological constant. We must now try to explain the physical reason for this relationship: A square root of energy density could be related to the notion of solubility. Indeed we find the dimension of [math]\sqrt{A}[/math] or [math]\sqrt{B}[/math] is [math]=\sqrt{J/m^3}=J^{1/2}m^{3/2}[/math] which exists in physics with the Hildebrand solubility parameter (https://en.wikipedia.org/wiki/Hildebrand_solubility_parameter#Definition) which seems to be applicable in the case of perfect gases. [math]A[/math] could "dissolve" in [math]B[/math] with the effect of a constant cosmological constant energy density (= a constant cosmological vacuum energy density) causing an acceleration of expansion Something close to the Hansen solubility parameters (https://en.wikipedia.org/wiki/Hansen_solubility_parameter) in cosmology would be more appropriate? The exact equations of this supposed dissolution effect are yet to be determined if the above makes sense. Finally, we can note for the anecdote, that this presentation is a proposed explanation to the number in 10^122 of the equality (2.1) of this paper (https://royalsocietypublishing.org/doi/full/10.1098/rspa.2007.0370) Indeed: [math]A/C=C/B=\frac{8\pi}{l_p^2 \Lambda_{m^{2} }} = 8,... * 10 ^{122}[/math]

The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
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The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
Hi swantsont Yes ! I saw the word "vacuum" in the penultimate paragraph of the PDF. My level of English is not enough to explain why it appears and how the author relates it to the subject of this article. Can you help me understand by telling me what the connections are between the word an the Planck density in the paper, please? Hi Mordred c^7 comes from the transformation of Planck's density in mass/m^3 into energy density with the multiplication by c^2 (equality (1) = [math]c^5/ (G^2 hbar)[/math] of the pdf of this message : https://www.scienceforums.net/topic/118858thesolutionofthecosmologicalconstantproblem/?do=findComment&comment=1145564 
The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
No, I did not address the problem of the cosmological constant but another question: (Planck density of matter/density of the cosmological constant) in 10^122. I admit that the fact that this number in 10^122 is close to the value in 10^122 of the cosmological constant problem is only a coincidence. I have sketched a physical explanation of the problem that I have solved mathematically, I will come back to it later, I don't have time to develop it now. 
The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
What is the IOW, please? Otherwise, very quickly, I didn't really deal with the problem of the cosmological constant but another point that shows a ratio in 10^122 
The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
file attached , the English version. I can't get the translation: the PDF is an image file. source : https://ui.adsabs.harvard.edu/ I'll get back to you on the rest later. article_20160.pdf 
The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
Which one? Thank you, I did my best but I didn't realize the importance of the contribution of all frequencies. I will try to go deeper with what I find on the web. Of course ! Because it was the one who was consistent with [math]t_p[/math]. 1. being wrong I don't know if it's still relevant. Indeed, they are'nt derived from the QM, but by dimensional analysis (which I have sometimes been reproached for using in this thread, unless I'm mistaken) and in their formula, [math]\hbar[/math], the minimum energy value of a quanta appears each time. But that doesn't take away from the fact that they are conceptually linked to a fundamental physical level : edit : that's why I thought that the energy density of Planck could be used to make an approach to the problem of the cosmological constant as in the document I used as a reference. But this is not going to stop my research with the zero point energy approach. https://en.wikipedia.org/wiki/Planck_units#Significance edit 2 : click on "show" of Table 5: Interpretations of the Planck units [21] and take a look at the interpretation of "density". It does seem to have a physical meaning, doesn't it? Now that you have put your finger on the question of the contribution of all frequencies to zeropoint energy, this constancy becomes much less obvious now for me As far as the constancy of the energy density of the cosmological constant you will find plenty of references that assert this within the framework of the standard cosmological model. 
The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
The numerical value of the vacuum catastrphe X is dimensionless and must be a ratio in SI units, for example : [math]X=\frac{t_p^{2}}{\Lambda_{s^{2}}}=\frac{l_p^{2}}{\Lambda_{m^{2}}}=\frac{\text{quantum energy density} (J/m^3)}{\text{cosmological constant energy density} (J/m^3)}[/math] I have sketched in this jumble of thread a physical meaning with the meaning of a square root of an energy density. I put it back if necessary if what follows makes sense, otherwise there's no point in looking at it. Question : 1. Is [math]1/t_p[/math] the value of [math]\nu[/math] for calculating zeropoint energy? 2. And divide it by [math]l_p^3[/math] to get its energy density?  If yes, my solution remains valid to within one constant and not only with numbers but with formulas.  If it is not the case, I remain in the unknown for the moment. The values [math]t_p[/math] and [math]l_p[/math] are not only a system of units in my approach but minimum physical values that make sense (smallest possible time value for known physics, smallest possible length value in QM). That's why I don't understand the criticism about the unit system. Not at all; I hadn't paid attention to it. Vacuum energy (quantum and cosmological) is a constant. The problem is that we expect a ratio of 1 and that it is in 10^122 or 10^123. Moreover the energy density of the cosmological constant is fixed while the age of the universe varies. The link made and the two seem more than suspicious 
The solution of the Cosmological constant problem ?
stephaneww replied to stephaneww's topic in Speculations
I wonder if you're not looking for lice in my head : Here (https://en.wikipedia.org/wiki/Planck_units#In_cosmology) we have: Cosmological constant 5.6 Ã— 10^122 tp^ 2 which is just another way of presenting the cosmological constant problem with a numerical factor (which depends among other things on the precision of the data) and a factor pi close to my presentation of the problem