A metric current and sequel

Discussion in 'Alternative Theories' started by TIMO MOILANEN, Oct 1, 2020.

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Should they redefine the kg

  1. Real physics

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  2. I'm stubborn

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  1. TIMO MOILANEN Registered Member

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    Just ask

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    . I use to "incubate" stuff for myself first (feel like getting a better understanding), but I'm a bad reader too

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    .
     
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  3. TIMO MOILANEN Registered Member

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    Sorry I remembered wrong the fine structure constant is not included in Planck const. but electric const. ai=2*(3/16pi)^2 , E0i=3/(8c)*2(3/16pi)^2. This give ei^2=2*ai*E0i*hi*c <=> ai^2=ai^2
     
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  5. exchemist Valued Senior Member

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    Ah I see what you are getting at. But if they were "the same", that would mean that k = ε = μ. What you mean is not that they are "the same", but that they are related to one another, by the relations k =1/4πε and μ = 1/c²ε. Yes?

    I don't know why you keep putting the "i" suffix on these symbols. It does not seem to add anything.

    Also it would help if you could be consistent about upper and lower case symbols. Anybody reading will assume that K and k, or c and C, represent different quantities. I suggest you use lower case for :
    - k, the Coulomb constant,
    - c, the speed of light,
    - ε the dielectric permittivitty (i.e Greek epsilon, to avoid confusion with the charge on the electron, e) and
    -μ, the magnetic permeability.

    The "0" subscripts often attached to the last two indicate their values in free space, so we can leave those out, I think, to make it simpler.

    But thanks, I now understand what you meant by your first sentence. I will now try to read the rest of your post in the light of that.

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  7. TIMO MOILANEN Registered Member

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    The i suffix is very necessary , the only way to tell (new metric) from CODATA when calculating with both . What is worse all electrical units are slightly off too Ai,Ci and about 10 more , I listed them somewhere?
     
  8. TIMO MOILANEN Registered Member

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    K and k mean(trad.) Coulomb's constant, I write Ki for new(metric) . C upper case always the unit coulomb (As) , c always speed of light. The only Greece letter on my keyboard is µ , few years ago I had them all on office but from keyboard the can differ in languages. I do typing errors too

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  9. exchemist Valued Senior Member

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    OK now that I understand your first sentence, can you explain the second sentence, which I have marked in red?

    It is not clear what you are doing with Faraday's Law. Are you referring to Faraday's Law of Induction?

    If so, I can't immediately see what that has to do with Faraday's Constant, which is something one uses in electrochemistry and has nothing to do with electromagnetism, as far as I can see.

    Can you take us through your working, with explanations of the various steps?
     
  10. TIMO MOILANEN Registered Member

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    Sorry I mean Faraday's constant (Faraday's law of electrolysis). I have not written down Fi, (faradays metric constant) but the kind of Avogadro number I use NAi is two ways different. It is for mol of electrons("electrolysed") only, to avoid varying atom masses, and applied to Ai .Kind of Avogadro's number for electrolysis only. I think I have "forgotten" to write the formula for elementary charge ei=1/((1/Ki)^0.5*NAi) . So done

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    Last edited: Oct 24, 2020
  11. exchemist Valued Senior Member

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    9,970
    Yes I know what Faraday's constant is. I'm a chemist. It is the charge of a mole of electrons, ~96485 Coulombs. It does not depend in any way on atomic masses.

    I will await your answer to post 66.
     
  12. TIMO MOILANEN Registered Member

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    Yes that's why I used Faraday electrolysis law , but I needed n of protons for Ex. ei=3/8*(Mpi/(pi^3*c)^0.5 and ei=1/((1/E0i)^0.5*NAi) that give the exact decimal value of the Ai compatible proton mass. CODATA proton mass is derived from its energies in SI units (not what I'm calculating) and by the way Fi=94498.16785 Ci/mol(NAi) , not the all-purpose CODATA NA
     
  13. TIMO MOILANEN Registered Member

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    I hope I have covered these 3 first paragraphs, since I originally said "Faradays law"

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    not constant. Let me know if there is any "misunderstanding" or rendition unsolved.
     
    Last edited: Oct 25, 2020
  14. TIMO MOILANEN Registered Member

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    I have not forgotten , just pondering of a non-mathematical way to write, like some whish.
     
  15. exchemist Valued Senior Member

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    9,970
    Can you please clarify when you say "Faraday's Law", do you mean his law of induction or what you are calling his "electrolysis law"?
     
  16. TIMO MOILANEN Registered Member

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    "Faradays law of electrolysis" (First and Second )

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    Look Google ,Wikipedia, anywhere . I know I'm a chemist (bsc) and you should know too colleague. I did not mean "Faradays law of induction", as a chemist I happened to forget that one, and thus the deficient precision.
     
  17. exchemist Valued Senior Member

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    9,970
    OK. Now can you explain what the sentence I was asking about means? This was where you said: " With Coulomb's law I put in 1Ci ( with no definition) and via Faradays law I get Coulomb's const.Ki , elementary charge ei and Avogadro's const.NA tied together."

    How can you apply Coulomb's Law ( F= k.q1.q2/r²) without defining values for q1, q2 and r?
     
  18. TIMO MOILANEN Registered Member

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    That is the main idea all along the "math. process " having q1,q2 = 1Ci , distances =1m and most importantly F =1N , later Ai =1Ai in all formulas . This way I always have 1=k*1^2/1^2 and likewise with other equations , this way I can operate with the components of the constants that are 3, 4 , pi and the always self invited 2 from c^2/2. I'm still searching for a "good" definition of Ci and Ai , since comparing to old A SI definition is not exact and n elementary charges in 1s measuring 1 ampere is not yet done. Worst problem for scientists doin this is that n=(SI e^-1) * (e SI) is not 1A of any kind, but will be 0.9895 Ai. 6.3748*10^18 counts of ei will be 1A SI. n=6.3078*10^18 is roughly the count for Ai definition. While n for A SI def. =6.24150765*10^18 electrons. They will probably blame the apparatus , I hope they not "fit " in the theoretical e by conviction .
     
    Last edited: Oct 27, 2020
  19. exchemist Valued Senior Member

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    9,970
    OK I see, maybe....

    This article on the fine structure constant seems to help: https://en.wikipedia.org/wiki/Fine-structure_constant
    Nowadays, it seems, the value of the Coulomb constant, k, depends on the experimentally measured value of the fine structure constant, α = μ₀.e²c/2h = 2πk.e²/hc.
    It appears that, whereas μ₀ was previously defined as exactly 4π x 10⁻⁷ H/m, since the change you refer to it is determined experimentally from the fine structure constant and now has the value 4π x 1.000 000 000 82 H/m. So the difference is minuscule.

    But I don't see why any of this affects the value of the Coulomb or the Ampere. The Coulomb is defined as the charge on a set number of electrons and the Ampere follows from this.

    It's all set out here:
    https://en.wikipedia.org/wiki/2019_redefinition_of_the_SI_base_units#Ampere
     
  20. TIMO MOILANEN Registered Member

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    This have only to do with the comparison to old (present SI) values. The metric values origin from c,pi and natural numbers (2,3,4).
    After couching a few days I was able to settle on logical SI values for volt and amp.
    Ai=0.996787735 A(SI) , Vi=0.993585788 V(SI) =>Wi=0.990394127 W(SI) , since ampere is in I^2 F/L=i^2*µ0i/(2pi*r^2) . Electric field formula E=Q/4pi*E0i*r^2) and unit for [E]=N/C=V/m .This much for comparison, definition is 6.3078*10^18 =(64/9*pi^2*c^2) s^-1 elementary charges per second times 1.585336*10^-19 =(9/(64*pi^2*c^2) Ci .
     
    Last edited: Oct 30, 2020
  21. TIMO MOILANEN Registered Member

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  22. TIMO MOILANEN Registered Member

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  23. exchemist Valued Senior Member

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    9,970
    Sorry, I've enjoyed revising the theory of this stuff, up to a point, but my patience on this topic is now exhausted.

    There is no issue here, since if there were, physicists would be making a fuss, which they are not doing.

    I'll leave you to your eccentric obsessions.
     

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