stephaneww 19 Posted June 29, 2018 Share Posted June 29, 2018 (edited) I suggest a simple way to understand the "problem of the cosmological constant" or "the vacuum catastrophe" .Yours opinions are welcome. first, the volumetric density of energy of the quantum vacuum equals for example: [latex]\frac{E_p}{l_p^3}=\frac{\hbar /t_p}{l_p^3}=4,6*10^{113}Joules/m^3[/latex] [latex]t_p=c/l_p[/latex] [latex]\Large{\frac{\hbar*(c/l_p)}{l_p^3}=\frac{\frac{\hbar^2*(c^2/l_p^2)}{l_p^{3*2}}}{\frac{\hbar*(c/l_p)}{l_p^3}}}=4.6*10^{113}Joules/m^3[/latex] (1) [latex]\Lambda=1.11*10^{-52}m^{-2}[/latex] [latex]\text{Planck force} = F_p= c^4/G=1.2*10^{44}Newtons[/latex] the volumetric density of energy of the dark energy equals : [latex]F_p*\Lambda/8\pi=5.4*10^{-10}Joules/m^3[/latex] now we remplace [latex]1/l_p^2[/latex] by [latex]\Lambda[/latex] in (1) and divide by 8pi [latex]\Large{\frac{\frac{\hbar ^2*(c^2 * \Lambda)}{l_p^{3*2}}}{\frac{ \hbar *(c /l_p)}{l_p^3}}/(8\pi)}=[/latex] [latex]F_p*\Lambda/8\pi=5.4*10^{-10}Joules/m^3[/latex] , the volumetric density of energy of the dark energy it's as if the inverse of Planck's squared length was not the proper value to compute quantum vacuum energy the problem of the cosmological constant appears when we do [latex]\frac{4.6*10^{113}}{5.4*10^{-10}}=8*10^{122}adimensionless[/latex] Edited June 29, 2018 by stephaneww latex Link to post Share on other sites

stephaneww 19 Posted July 1, 2018 Author Share Posted July 1, 2018 (edited) On 29/06/2018 at 11:25 PM, stephaneww said: it's as if the inverse of Planck's squared length was not the proper value to compute quantum vacuum energy …. in a cosmological quantum mechanics context. Edited July 1, 2018 by stephaneww Link to post Share on other sites

Mordred 1372 Posted July 4, 2018 Share Posted July 4, 2018 (edited) Isn't the question of the vacuum catastrophe specifically "Why is the cosmological constant so small while the quantum Vacuum so large ?" I fail to see how this addresses that question. In our other thread I posted the following article, He shows a very accurate examination addressing this question. https://arxiv.org/pdf/1703.00543.pdf What you need to consider is the Cosmological constant is a macro size scale while the quantum vacuum is at the quantum scale the fluctuations occur at each coordinate in the latter case, where the cosmological constant is an average over a macro volume. In effect the former is smeared out on the macro scale. Edited July 4, 2018 by Mordred 1 Link to post Share on other sites

stephaneww 19 Posted July 5, 2018 Author Share Posted July 5, 2018 (edited) Thank you Mordred for your opinion. I was told that [latex]\Lambda \text{ exprimed in m}^{-2}[/latex] is an energy. Is it true please? (and why if it's easy to explain it, please) Moreover, is [latex]1/l_p^2[/latex] is of the same physical nature than [latex]\Lambda \text{ exprimed in m}^{-2}[/latex] please ? Edited July 5, 2018 by stephaneww latex Link to post Share on other sites

stephaneww 19 Posted July 8, 2018 Author Share Posted July 8, 2018 (edited) Hi, Another question comes to my mind : Isn't it curious to find the exact value of the energy volumic density of the cosmological constant from the volumic density of the quantum vacuum, to the nearest factor of [latex]8\pi[/latex], when we make appears value in [latex]m^{-2}[/latex] [latex],e.i\text{ : }1/l_p^2[/latex], in the formula of the volumic density of the quantum vacuum and that we replace it with [latex]\Lambda \text{ exprimed in }m^{-2}[/latex] ??? Edited July 8, 2018 by stephaneww Link to post Share on other sites

stephaneww 19 Posted July 8, 2018 Author Share Posted July 8, 2018 (edited) On 04/07/2018 at 3:34 PM, Mordred said: ...In our other thread I posted the following article, He shows a very accurate examination addressing this question. https://arxiv.org/pdf/1703.00543.pdf What you need to consider is the Cosmological constant is a macro size scale while the quantum vacuum is at the quantum scale the fluctuations occur at each coordinate in the latter case, where the cosmological constant is an average over a macro volume... I'm just starting to study the paper and I already have a problem:Indeed, by convention [latex]c=\hbar=1[/latex] but in the formula (8), if I do not make mistake,we should have without this convention : [latex]\Large{ \rho_{\text{eff}}^{\text{vac}}=\rho^{\text{vac}}*c^2+\frac{\lambda_b*c^4}{8*\pi*G}}=5.96*10^{-27}*c^2+1.11*10^{-52}*c^4/(8*\pi*G) = [/latex] [latex] \rho_ {\text {eff}}^{\text{vac}}=5.35*10^{-10} \text{ Joules/}/m^3 +5.35*10^{-10}\text{ Joules}/m^3=[/latex] [latex]2* \rho^{\text{vac}}=2* \Large{\frac{\lambda_b*c^4}{8*\pi*G}}[/latex] and so I, with this, I can't understand the following.. where did I make a mistake please, or can you explain me why it's necessary to use that ? edit : I think I have found my mistake : [latex] \lambda_b \text { is not } \Lambda[/latex] another possibility : [latex] \rho^{vac} \text{ is not "Einstein vacuum" }[/latex] but it 's the quantum vacuum Edited July 8, 2018 by stephaneww Link to post Share on other sites

Mordred 1372 Posted July 9, 2018 Share Posted July 9, 2018 (edited) Right each vacuum is its own field the quantum field is [latex]\rho_{vac}[/latex] this is based on zero point energy (QFT harmonic oscillator). [math]\Lambda[/math] is based on calc from commoving volume via critical density extrapolation over (a(t). Vacuums are specific to the fields being applied, for example the VeV Higg's (vacuum to vacuum) field involves a bosonic gauge and fermionic field interaction to determine. (goldstone bosons representing the guage). This later statement is telling. "That’s because the vacuum is not an eigenstate of the local energy density operator [math]T_{00}, although it is an eigenstate of the global Hamiltonian operator [math]H =\int Rd^3xT_{00}[/math]. This implies that the total vacuum energy all over the space is constant but its density ﬂuctuates at individual points. " (key note to consider it mean lifetime is shortened under HuP so each fluctuation is incredibly short lived.) Edited July 9, 2018 by Mordred Link to post Share on other sites

stephaneww 19 Posted July 10, 2018 Author Share Posted July 10, 2018 (edited) OK, thanks. I'm trying to rephrase for my approach: The energy density of the quantum vacuum is treated as a constant based on the inverse of the Planck length (see the first post of this thread) in the usual formulation of the cosmological constant. While this is true for the expected energy density, it is not true for the actual value. Indeed, one can consider that the quantum vacuum and the cosmological constant fluctuate locally in concert (see our first post on the subject with the approach of the pulsations: link to the thread ). Consequently, the value of the real quantum energy density must be calculated in a real quantum approach with [latex]\Lambda[/latex], of dimension [latex]m ^ -2[/latex], which is only a variant of the version of the pulsations.For this we must raise the energy density of the quantum vacuum squared to give the dimension of the inverse of the length of Plank, the dimension of the cosmological constant.This removes the problem of the cosmological constant since the formula of the modified quantum vacuum because it gives exactly the value of the energy density density of the cosmological constant of the stantard model. (to a factor of 8pi) Edited July 10, 2018 by stephaneww Link to post Share on other sites

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