stephaneww

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

ooops, I made an error direction trash please

Um, many things are badly said ... to be more clear : Planck force is the common point between the volumetric energy density of the cosmological constant and that of Planck vacuum catastrophe appears by omitting to multiply by their respective surfaces (except for factor 2) is it wright please ?

it's the same with lp^2 and Λ in m^-2 so we have an area divided by an area, but it's not relevant : It is only now that I understood that it is very likely the multiplication of an area and the inverse of an area : before making a detour by the pressures of a perfect fluid (that of the cosmological constant) and the mechanical definition of a pressure: [latex]P_{\Lambda}=-\frac{c^4*\Lambda}{8\pi*G}[/latex] [latex]P=\frac{F}{S}[/latex] with [latex]P[/latex] : pressure [latex]F[/latex] : force and [latex]S [/latex] : area now by isolating the Force we have [latex]F=P*S[/latex] assume that the Planck pressure is the same behavior as that of the cosmological constant, ie that of a perfect fluid, as seen above, a sign near. in terms of units, for [latex]P[/latex] is in Pascal (symbol Pa: [latex] kg*m^{-1}*s^{-2}[/latex] that is identical in terms of unity to Joules / m^3 ( in other words, a volume energy density) the Planck Force [latex]F_p[/latex] = volumetric energy density of Planck * (Planck length)2 it can be seen that the Planck Force =(c4/G) is also present in [latex]P_{\Lambda}[/latex] isolate the Planck Force in this last equality we have: [latex]F_p[/latex]= [latex]P_{\Lambda}*2*(4*\pi/\Lambda)[/latex] or in terms of dimensions [latex]\Lambda[/latex] is in [latex]m^{-2}[/latex] so [latex]1/\Lambda[/latex] is in [latex]m^2[/latex]. [latex]4*\pi/\Lambda[/latex] probably due to the fact that in the case of the cosmological constant the surface to be considered is that of a sphere and not a plan like in the quantum version factor [latex]2[/latex] is already present for relationships in cosmology and Planck dimensions, as in Planck mass flow and Hubble radius mass flow, for example. or maybe it comes from the fact that it takes 2 surfaces to see the energy of the vacuum via the Casimir effect [latex]F=P*S[/latex] is probably (within a factor of 2) a formula common to quantum mechanics and cosmology from their volume energy density and their respective area multiply both values in [latex]m^2[/latex] and [latex]m^{-2}[/latex] would be meaningless, count only the question of the Planck force in both cases, from the volumetric energy densities

oops sorry, this is only another writing of the equation of the vacuum catastrophe = 1/tp^2 * 1/Λ * 8pi. does this mean tant 1/tp^2 can't be considered as an energy in quantum mechanics as we do for Λ in the relativity? I don't know. only the begining is perhaps interesting, but I'm not sure of that

hello in my opinion, there is really something auround squares or square roots and 8 pi about the vacuum catastrophe : [latex]\frac{\hbar^2*\Lambda}{L_p^3* E_p} /8\pi=[/latex] the exact value of volumic energy density of cosmological constant exprimed in Joules/m^3 [latex]\hbar[/latex] reduced Planck constant (Joules*s) [latex]\Lambda[/latex] cosmological constant (s^-2) [latex]L_p[/latex] Planck length (meters) and [latex]E_p[/latex] Planck energy (Joules) thank you in advance if you can explain it to me or correct me please.

there is no question finaly : it's an another form of equations I posted before, sorry

oops I thougt I had find something interresting but not finaly, sorry finaly I let this : when we make energy density per volume of the cosmological constant (in Joule / m ^ 3) / Planck force (in Newton) * 8 * pi, I find exactly the numerical value of the cosmological constant in m ^ -2. can you explain to me why, confirm or correct me please?

and so Mordred please : is the factor 3 link to a question of pression (states equations with [latex]w[/latex])?

=[latex]\frac{3*c^2*H_0^2}{8\pi G}=\rho_c[/latex] to use the formula... (with LFRW equations)... it will be clearer

Hello. Certainly I didn't succed to prove ma first proposition with the equations of state, but using my method for the vacuum catastrophe applied to the Hubble constant, a parallel of the same type can be made from a point of view of quantum mechanics relativistic deterministic [latex]H_0=67,9 Km/s/Mpc = 2,2005*10^{-18} s^{-1}[/latex], [latex]T(Gly)=13,787*10^9[/latex] [latex]\frac{\hbar*H_0}{l_p^3}=5,4965*10^{52}\text{ Joules/m^3} \text{, }(H_0 \text{ in }s^{-1})[/latex] to follow the method of the vacuum catastrophe used in this tread we raise this volumetric energy density at square : [latex](\frac{\hbar*H_0}{l_p^3})^2=3,0211*10^{105}\text{ Joules^2/m^6}[/latex] [latex]\frac{E_p}{l_p^3}=4,6332*10^{113}\text{ Joules/m^3}[/latex] [latex](\frac{\hbar*H_0}{l_p^3})^2/\frac{E_p}{l_p^3}/8/\pi*3=7,7834*10^{-10}\text{ Joules/m^3}=[/latex] [latex]\text{exactly critical density of energy per volume at Hubble radius (edit : not necessarily at Hubble radius) in }\Lambda \text{CDM model}[/latex] the difference with the vacuum catastrophe is a factor 3 Is it enough to validate the method? ________________________ notes: Hubble radius based on [latex]T(Gly)=13,787*10^9[/latex] convert in seconds =[latex]4,3509*10^{17}s[/latex] is [latex]R=T(s)*c=4,3509*10^{17}s*299792458m/s=1,3044*10^{26}m[/latex] [latex]c= \text{light speed in vacuum}[/latex] for calculation of critical densité, I use [latex]H_0=67,9 Km/s/Mpc = 2,2005*10^{-18} s^{-1}[/latex]

That's perfectly true ... ... And so, I think that it's not me who can finish that work Of course, other contributions, in addition to what has been written to confirm or invalidate the first message of the thread, are welcome ...