stephaneww Posted June 7, 2020 Share Posted June 7, 2020 (edited) when you have a moment: enter in excel the exact values (from the formulas) of : - Planck's mass - Planck's length then the most accurate value [math]C [/math] of the Coulomb you can find. (6,24150962915265*10^18 elementary charge ) calculate the value of : [math]m_p.l_p/C^2*10^7[/math] estimate the numerical deviation in % from the inverse of the fine structure constant ([math]1/ \alpha[/math]) I find 0.00002%. Can you confirm please ? gauge the relevance in dimensional terms knowing that this calculation comes from this question: https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1118152 which also gives an acceptable approximation of the value of the vacuum catastrophe, but dimensioned, whereas the vacuum catastrophe is dimensionless. Please ___________________________________ basic working formula before simplification : [math](F_p^2/e^2)\text{ / Planck's surface density (in kg/s^3=W/^2)[]/math] - Fp: Planck force - e: elementary charge Edited June 7, 2020 by stephaneww Link to comment Share on other sites More sharing options...
stephaneww Posted June 7, 2020 Author Share Posted June 7, 2020 (edited) Hi I still have one more check to do before concluding that we can no longer simplify the units making alpha dimensionless ^^ Edited June 7, 2020 by stephaneww Link to comment Share on other sites More sharing options...
stephaneww Posted June 7, 2020 Author Share Posted June 7, 2020 (edited) the 1/10^-7 = 10^7 came from μ0 = 4π × 10^−7 kg m A^−2 s^-2, so no puzzle finaly for [math]\alpha[/math] , except for me , if I don't made a new no sense Edited June 7, 2020 by stephaneww Link to comment Share on other sites More sharing options...
swansont Posted June 7, 2020 Share Posted June 7, 2020 Post the formula for all of your terms, and simplify. It might be illuminating. Attacking this numerically is the wrong approach, IMO. It introduces opportunities for round-off errors Link to comment Share on other sites More sharing options...
stephaneww Posted June 7, 2020 Author Share Posted June 7, 2020 (edited) Okay, but not much of physical interest, except maybe something to do with the cosmological constant problem. [math]\Large{\frac {1}{\alpha}=\frac {F_p^2}{e^2}. \frac {t_p^3}{m_p.2.\pi} . \frac {8.\pi^2}{c} .\frac {1}{\mu_0} }[/math] - [math]\alpha[/math] : fine constant structure - [math]F_p[/math] : Planck's force - [math]e[/math] : elementary charge - [math] \Large{\frac {t_p^3}{m_p.2.\pi}} [/math] : inverse of Planck's surface power density * 1/(2pi) note 1 : units of surface power density = W/m^2 = kg/s^3 note 2 : possible link with the problem of the cosmological constant ? - [math]c[/math] : speed of light [math]\mu_0[/math] : vacuum permeability I would suggest the simplifications later in another message: too afraid to exceed the maximum editing time to correct the latex Edited June 7, 2020 by stephaneww Link to comment Share on other sites More sharing options...
swansont Posted June 7, 2020 Share Posted June 7, 2020 Try using formulas that only use fundamental constants. Not Planck units. So you can simplify. e.g. G is going to cancel out of the planck length and mass, as well as a factor of c. Link to comment Share on other sites More sharing options...
stephaneww Posted June 7, 2020 Author Share Posted June 7, 2020 (edited) 1 hour ago, swansont said: Try using formulas that only use fundamental constants. Not Planck units. So you can simplify. e.g. G is going to cancel out of the planck length and mass, as well as a factor of c. This will be simpler than replacing Planck's units with their formulas in [math] F_p^2[/math] ? Simplifying [math] F_p^2[/math] aside I believe it's just as simple no ? Besides, problem for me, I don't see how you do it.. Edit : I understand but I don't know which is the simplest method Edited June 7, 2020 by stephaneww Link to comment Share on other sites More sharing options...
swansont Posted June 7, 2020 Share Posted June 7, 2020 48 minutes ago, stephaneww said: This will be simpler than replacing Planck's units with their formulas in F2p ? I’m not sure why Fp made an appearance, but yes. 48 minutes ago, stephaneww said: Simplifying F2p aside I believe it's just as simple no ? Besides, problem for me, I don't see how you do it.. Edit : I understand but I don't know which is the simplest method It’s basic algebra. Link to comment Share on other sites More sharing options...
stephaneww Posted June 8, 2020 Author Share Posted June 8, 2020 3 minutes ago, swansont said: I’m not sure why Fp made an appearance, but yes. Me neither, but here it's more a question of the physical meaning of equality, and on this point I have a huge doubt. Link to comment Share on other sites More sharing options...
swansont Posted June 8, 2020 Share Posted June 8, 2020 Recasting things in terms of Planck units is just a circular argument https://en.wikipedia.org/wiki/Planck_units Look at table 2. If you multiply lp and mp you get ħ/c If you look at the formula for Planck charge, you see from the formula for qp (and you square it) that 4πε0* ħ c = e^2/alpha https://en.wikipedia.org/wiki/Vacuum_permittivity Does that help? Like I said, it looks like you have roundoff error, and you've shown that 1 = 1 Link to comment Share on other sites More sharing options...
joigus Posted June 8, 2020 Share Posted June 8, 2020 16 minutes ago, Mordred said: Nope the cosmological constant has no gravitational force or Coulomb force term. A force is a vector it has a magnitude and direction. The cosmological constant is a scalar quantity. It's value only has a magnitude. This is one critical detail you have to learn to seperate. The two types of fields will have different dynamics. I totally agree with this +1. I also concur with Swansont that it looks like you've proven 1=1 there. I'm not sure. I'd have to go over the details. I do have an argument of why the cosmological constant gives the order of 10120 times bigger than it should be when you apply QFT. It has nothing to do with the coupling constants, as Mordred suggests. Maybe I'll talk about it some day in Speculations so that everybody can slap me in the face openly for getting sloppy. Or maybe I won't have the guts. It's to do with how incredibly many particles you're producing when you take the Mp cutoff seriously (you plan to actually go to Mp collision energies.) Link to comment Share on other sites More sharing options...
swansont Posted June 8, 2020 Share Posted June 8, 2020 ! Moderator Note The thread’s topic is not the cosmological constant. I have moved posts into that thread Link to comment Share on other sites More sharing options...
stephaneww Posted June 8, 2020 Author Share Posted June 8, 2020 Swansont : It seems to me that I linked the two questions from the very beginning, didn't I On 6/7/2020 at 5:12 AM, stephaneww said: gauge the relevance in dimensional terms knowing that this calculation comes from this question: https://www.scienceforums.net/topic/118858-the-solution-of-the-cosmological-constant-problem/?do=findComment&comment=1118152 Link to comment Share on other sites More sharing options...
swansont Posted June 8, 2020 Share Posted June 8, 2020 6 minutes ago, stephaneww said: Swansont : It seems to me that I linked the two questions from the very beginning, didn't I Yes, but you made a new thread for it. Which defines the new discussion.We aren’t going to have multiple open threads on one topic. Link to comment Share on other sites More sharing options...
stephaneww Posted June 8, 2020 Author Share Posted June 8, 2020 (edited) erase I didn't post in the good thread Edited June 8, 2020 by stephaneww Link to comment Share on other sites More sharing options...
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