Jump to content

stephaneww

Senior Members
  • Content Count

    292
  • Joined

  • Last visited

  • Days Won

    1

stephaneww last won the day on September 15

stephaneww had the most liked content!

Community Reputation

17 Neutral

About stephaneww

  • Rank
    Atom
  • Birthday 10/02/1968

Profile Information

  • Location
    France
  • Favorite Area of Science
    cosmology

Recent Profile Visitors

3925 profile views
  1. 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]
  2. 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])
  3. 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."
  4. 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.
  5. Indeed, this presentation is far too difficult for me...
  6. 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
  7. 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...
  8. ...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
  9. 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]
  10. 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
  11. 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
  12. 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.
  13. 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]
  14. 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
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.