stephaneww

for the relationship to work correctly in terms of dimensions, it is also necessary to remember that we have: [math]P_p \Lambda / 4 = 1. kg/s^3[/math] and [math]P_p = 3.62831 .kg.m^2/s^3[/math]

Hello. I don't have much free time to continue studying electromagnetism right now. But I'll get back to it as soon as I can. Just a remark on the orders of magnitude. We are often surprised by the small value of the cosmological constant. It is directly linked to Planck's power by the relationship : [math]\Lambda=\frac{4}{P_p}[/math] (remember the relationship of power surface density power expressed in [math]W/m^2[/math] or [math]kg/s^3[/math])

after an exchange with an electromagnetism specialist, this is false: [math]2 \pi[/math] is only a scalar so whatever point in the universe is considered, and if I don't say something stupid again, each point in the universe follows the perimeter of a circle. as : its direction is tangent to the circle so "it is undergoing an acceleration."

Sorry please read in the post https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1119086 so : [math]\frac{c^4 \Lambda}{8 \pi G}.c .2. \pi=1 W/m^2/[/math] i.e.

Indeed, this presentation is far too difficult for me...

I started to take a closer look.... Beforehand we will assume that [math]\frac{1}{2} W/m^2[/math] is the value that accurately replaces [math]\frac{1}{1,994} W/m^2[/math]. We have the energy density of the cosmological constant =[math]\frac{c^4 \Lambda}{8 \pi G}. J/m^3[/math] Here we have Let's follow the "intensity" link. We find : By following the velocity link, we have : We had found: so : [math]\frac{c^4 \Lambda}{8 \pi G} .c.2.\pi=1. W/m^2[/math] i.e where [math]c[/math] is the speed and [math]2 \pi[/math] is the .. so whatever point in the universe is considered, and if I don't say something stupid again, each point in the universe follows the perimeter of a circle. as : its direction is tangent to the circle so "it is undergoing an acceleration." I don't know if this first approach is correct. If anyone can help me, thank you in advance edit : Um, there may be a problem with the

I'm going back to work to fill in the blanks and also because there's a lot of nonsense… 1* Okay, if I understand correctly, we have two scalar fields' ( I have revised the notion of partial derivative and its modern notation) What can we do with these scalar fields [math]\phi_p \phi[/math] ? I don't understand their usefulness for calculations. ... and I don't know how to use gradients. 2* In this post, https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1116338 there's been 2 big mistakes. -Indeed, if [math]\Lambda_{s^{-2}}=\Lambda_{m^{-2}}.c^2[/math] then [math](\Lambda_{s^{-2}})^2/c^4=(\Lambda_{m^{-2}})^2[/math] The last line "[math]D=[/math]", of the first post, [math]=B[/math] (demonstration of geometric mean page 2) becomes : [math]D=B=h.(\Lambda_{s^{-2}})^2/c^3/(8∗\pi)^2=6.11∗10−133.kg.m^{−1}.s^{−2}[/math] -the second big error is on the value of [math]H_{p,minimum}[/math] , we do not fond a minimum but only [math]\sqrt{\Lambda_{s^{-2}}}=3.152*10^{-18}.s^{-1}[/math] 3* consequences for this post https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1116465 [math]R[/math], [math]V[/math], [math]\rho_c[/math] are no longer extreme values, and [math]\Omega_{\Lambda}=\frac{1}{3}[/math]. One might think that the cyclical universe is collapsing, but there is still this indicator that ,perhaps, remains valid : [math]\frac{V.\rho_c.\Omega_{\Lambda}.c.\pi}{V}=\frac{1}{1.994}kg/s^3 \text{ or W/}m^2[/math] [math]=\frac{c^4.\Lambda.c.\pi}{8 . \pi G}=\frac{1}{1.994}kg/s^3[/math] [math]=\frac{F_p.\Lambda.c.\pi}{8 . \pi }=\frac{F_p.\Lambda.c}{8}=\frac{1}{1.994}kg/s^3[/math] But even with that, it is not certain that the universe is cyclical. 4* I agree with you on that point. 5* I still have to learn electromagnetism to try to understand this power surface density...

...And with power of Planck we have : [math]\Lambda=\frac{0.5*8}{c F_p}=1.102422 *10^{-52}m^{-2}[/math] because the unit of 0.5 is [math]kg.s^{-3}[/math] edit : Um, I didn't notice: source : https://en.wikipedia.org/wiki/Planck_force#Other_derivations

Thank you very much. I'm so happy that I put +1 edit : Indeed, if we name X, the value of 1,994 the variation of the cosmological constant from datas, give X in the 3 equations. Second indice, 2pi is the divisor for the Planck reduced constant. edit 2 So [math]\Lambda[/math] would be equal to: [math]\Lambda=0.5/0.5\frac{t_p^2.8 \pi}{h}=1.10242*10^{-52}m^{-2}[/math]

Do you think that if we take 2 instead of 1,994 in post 3 on page 5, we could deduce the exact value, from the Planck units, of the cosmological constant value? I am very tempted to answer "Yes" in the case of a flat universe

However, if we do this simple calculation by dimensional analysis: [math]m_p/t_p^3*2 \pi =8.72*10^{122}.kg/s^{3}[/math] that is the value of the vacuum catastrophe at less than 1% But I don't know what to conclude from that... and if I do, in my spreadsheet: [math]m_p/t_p^3*1.994 \pi =8.72*10^{122}.kg/s^{3}[/math] it is the exact numerical value of the vacuum catastrophe

Yes, generally. Even for Planck's mass?

I don't know if it's a maximum power. I'm not a theorist. What I know about on this subject Planck's force is this : source : https://en.wikipedia.org/wiki/Planck_force#General_relativity But I don't know if it's the same thing for the power of Planck.

I just saw here that: [math] F_p=\frac{c^4 }{G}[/math] [math] P_p=\frac{c^5 }{G}= 3.62831×10^{52} W[/math] … and I have realized the calculation with [math] P_p=c.F_p [/math]

Oops. I knew it was the base, but we have to divide by 8 in more. . Do [math]c.F_p.\Lambda/8=0.50144.kg/s^3[/math] value of posts 2 and 3 page 5 : https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1117499 https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1117557