TIMO MOILANEN
Registered Member
Just askI won't attempt to comment on anything else until this is resolved.
Just askI won't attempt to comment on anything else until this is resolved.
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^2included in both ei and Planck constant hi
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?Coulombs const. Ki , electric const. E0i and magnetic const. µ0i are one and same constant just "turned over". E0i=1/(4pi*Ki) , µ0i=1/(E0i*c^2) , just fitted into different "geometry"
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?I don't know why you keep putting the "i" suffix on these symbols. It does not seem to add anything.
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 tooI suggest you use lower case for :
To start Coulomb's ,electric E0i and magnetic constants µ0i are one and the same. 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. Faradays constant F is included via the eq. to get out ei . Not surprisingly the fine structure constant ai is included in both ei and Planck constant hi . ei=2*ai/c^2 that upside down is clearer c^2/(2*ai) is both a kinetic energy and equally large potential energy. Planck constant is also 4 times the electric const. in nuclear units (/c^2). Then I just check the constants with eq. ei^2=2*ai*E0i*hi*c so all ad up . Here I also derived the mass of a proton Mpi since I used E0i=3*Mpi*c^2 /(16*pi) to begin with . Then just testing the constants on most equations I found and transforming half a dozen "minor" constants to apply metric ampere Ai. Ai have a similar definition as the SI A (=stupid) but can be compared to the SI ampere via the magnetic constant µ0i.
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 doneIf 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
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.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.
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 NAYes 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 hope I have covered these 3 first paragraphs, since I originally said "Faradays law"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.
I have not forgotten , just pondering of a non-mathematical way to write, like some whish.I will await your answer to post 66
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"?I hope I have covered these 3 first paragraphs, since I originally said "Faradays law"not constant. Let me know if there is any "misunderstanding" or rendition unsolved.
"Faradays law of electrolysis" (First and Second )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"?
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.""Faradays law of electrolysis" (First and Second )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.
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 .How can you apply Coulomb's Law ( F= k.q1.q2/r²) without defining values for q1, q2 and r?
OK I see, maybe....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 .
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).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.
Sorry, I've enjoyed revising the theory of this stuff, up to a point, but my patience on this topic is now exhausted.https://1drv.ms/u/s!ArSE2R4ReZrzijSLOPWZHlAUXF0x
Any better luck?