stephaneww

oops I missed that

um, nothing new in fact, it's under key : https://en.wikipedia.org/wiki/Planck_units#Base_units and it must be taken into account that: [latex]\Large{F_p=\frac{c^4}{G}}[/latex] so, we can deduce that it takes at least a relativistic context

Hello Strange, of course "p" is for "Planck" I only try to find an element of quantum gravity for a context deterministic of quantum mechanics and submit it to a friend who works on this question and as you say, it's not a usual way to approach the problem thank you for your answer edit : do the calculation, numeric value and dimension are ok

Hello I would like to know if the following calculation is allowed in quantum mechanics or if it's only the definition : [latex] \Large{F_p=\frac{G*m_p*m_p}{l_p^2}}[/latex] the numerical values are ok thanks in advanced

Hi, You have this also, for example : https://www.scienceforums.net/topic/114217-multiple-black-holes-at-the-center-of-our-galaxy/ Black holes seem to be much more numerous than expected within galaxies. They are good candidates to explain some (all?) of the dark matter and they have the cool features for dark matter: radiation, invisible, gravitational effect and hidden within galaxies…

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)

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

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] ???

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 ?

I propose a simplified version here : https://www.scienceforums.net/topic/115191-a-simply-way-to-understand-the-vacuum-catatastrophe/

…. in a cosmological quantum mechanics context.

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]

Thank you, it's obvious but I didn't think that before you say it .... [latex]L_(t)=\frac{a_0}{a_t}L_0[/latex] [latex]L[/latex] : length radius universe [latex]a_0 = 1[/latex] [latex]a_t[/latex] : scale factor it's correct please ? ok so my first post can't be right

Hi, I have in the idea that the decrease of the temperature of the universe and the expansion of the universe are compensated exactly in terms of energies and propose the following calculations age of the universe (recombination) 378 000 years, with temperature CMB T1 = 3000 K, redshift : z1= 1 100, radius observable universe : R1 = 3.96*10^23 m age of the universe today : 13.797.000.000 years, with temperature CMB T0 = 2.725 K, reshift z0=1, radius observable universe R0 = 4.36*10^26 m and [latex]H_0=67.74[/latex], z1=T1/T0 z1=3000K/2.725K=1100 R0 = (1+z) * R1 , V1 = 2.59 * 10^71 m^3, V0 = 3.47 * 10^80 m^3. (Vt = volume of observable universe ) [latex]H_{t1}=H_0*\sqrt{\Omega_m(1+z)^3+\Omega_{\text{rad}}(1+z)^4+\Omega_{\Lambda} }=158284\text{ km/s/Mpc}[/latex] with : [latex]\Omega_m=0.3089[/latex] [latex]\Omega_{\text{rad}}=9*10^{-5}[/latex] [latex]\Omega_{\Lambda}=0.6911[/latex] __________________________________________________________ [latex]V_t [/latex] : Volume of observable universe Boltzmann constant : [latex]k_B = 1.380 648 52 * 10^{-23}J/K[/latex] [latex]X1=k_B*T_1/V_1=1.38*10^{-23}*3000/(2.59*10^{71})=1.60*10^{-91}J/m^3[/latex] [latex]X0=k_B*T_0/V_0=1.38*10^{23}*2.725/(3.47*10^{80})=1.09*10^{-103}J/m^3[/latex] [latex](X1/X0)/(\Omega_m(1+z)^3+\Omega_{\text{rad}}(1+z)^4+\Omega_{\Lambda})=(1.47*10^{12})/(5.46*10^8)=2697[/latex] dimensionless [latex]M_t[/latex] : "total mass" of observable universe (for its energy do [latex]M_t* c^2[/latex], the next ratio is the same for mass or energy) [latex]M_1=1.22*10^{54}kg[/latex] [latex]M_0=2.99*10^{54}kg[/latex] [latex]M_0/M_1=2.45[/latex] dimensionless and we finally find : [latex]z_1=2697/2.45=1100 [/latex] exactly (the first value) _________________________________________________________ I haven't lookked for the demonstration yet, and I'm not even sure it makes sense Thank you in advance for your opinion

do you think that this is enought to accept the model of my first post ? (without equations state ) [latex]\frac{\Lambda_0}{3*H_0^2}=\Omega_{\Lambda_0}=0.69[/latex] source : http://www.cnrs.fr/publications/imagesdelaphysique/couv-PDF/IdP2008/03-Bernardeau.pdf (page 10) The cosmological constant appears here as a numeric multiple of a pulse expressed in [latex] s^{-1} [/latex] (squared) more precisely that of [latex] (H_0) [/latex] : Hubble constant (SI unit) thanks in advance for yours answers CNRS* = C : center, N : national, R : research, S: scientific