stephaneww 19 Posted June 27, 2020 Share Posted June 27, 2020 (edited) 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 mass-energy 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] Edited June 27, 2020 by stephaneww Link to post Share on other sites

Mordred 1372 Posted June 29, 2020 Share Posted June 29, 2020 (edited) I'm not sure how your planning to apply solubility to the cosmological constant. You seem to be missing the scalar and vector aspects behind the equations you have given. Which has nothing to do with solubility. https://en.m.wikipedia.org/wiki/Solubility This is something I have stressed numerous times. Edited June 29, 2020 by Mordred Link to post Share on other sites

stephaneww 19 Posted June 29, 2020 Author Share Posted June 29, 2020 (edited) 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) : On 6/27/2020 at 6:01 AM, stephaneww said: which seems to be applicable in the case of perfect gases. 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. Edited June 29, 2020 by stephaneww Link to post Share on other sites

Mordred 1372 Posted June 29, 2020 Share Posted June 29, 2020 Well it's accurate to describe the cosmological constant as a perfect fluid as a negative vacuum. However it isn't something that is applicable to a medium with which solubility could be applied to. I would say your looking down the wrong garden path. Link to post Share on other sites

stephaneww 19 Posted June 29, 2020 Author Share Posted June 29, 2020 Thank you. I'll look for other cases where we'd have square root energy densities Link to post Share on other sites

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