EM laws have a natural order Re: AP's 151st book, Part 3, TEACHING TRUE PHYSICS, 1st year college, by Archimedes Plutonium.

Archimedes Plutonium<plutonium....@gmail.com>
Jun 28, 2021, 9:54:14 PM (11 days ago)
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This maybe just the right thing I need to teach Faraday constant in classrooms.

It comes from a website of Minneapolis Community and Tech College titled "Determination of Faraday's Constant"

They perform the electrolysis of water.

Now another website, LD Chemistry from a German company outlines the Faraday Constant experiment, very similar to the above Minneapolis experiment. So one can use both to help explain.

And here is where the instructor of the 1st year College Physics is crucial in explaining how Faraday constant and how mole number is obtained to determine the 1.6*10^-19 Coulomb of magnetic monopole.

AP

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Archimedes Plutonium<plutonium....@gmail.com>
Jun 29, 2021, 1:26:26 PM (10 days ago)
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Making sense of the Elementary Coulomb of 1.6*10^-19.

We discovered that the Magnetic Monopole is the particle of 0.5MeV rest mass and that the combining of two monopoles is a dipole of 1MeV. The Monopole is Static Electricity, and we learned that the dipole of combining two monopoles is a full loop circuit and is Dynamic Electricity. Dynamic Electricity is what flows in electricity in circuits.

We learned that the DC current is double transverse wave of dipoles, while the AC current is longitudinal waves of dipoles.

So, let us make some sense of this number that is called the Elementary Couloumb of 1.6* 10^-19.
And the sense we make of it is that sense we make of Avogadro's number or mole. The mole is about 6*10^23 neutrons (or protons) in 1 gram of matter. Every gram of matter has the same number of neutrons 6*10^23 neutrons (or some prefer protons). That was easy.

But now what is the common sense explanation of the Elementary Coulomb? And this looks difficult because it is a small number of 10^-19 whereas mole is a huge number of 10^23.

To explain Elementary Coulomb we look at the mass of the neutron or proton in terms of grams. And the mass of the single neutron is 1.6*10^-24 gram. We can call the neutron as the Elementary Mass of Matter. We can now make sense of the Elementary Coulomb of 1.6*10^-19 as being the Elementary particle of electricity.
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Archimedes Plutonium<plutonium....@gmail.com>
Jun 30, 2021, 12:41:05 AM (10 days ago)
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Alright, I am very happy with these results and the explanation of the Elementary Coulomb as equal to 1.60*10-19.

And we can also see, that the elementary mass for neutron is 1.6*10^-24 gram.

The neutron and proton and muon are particles with rest mass. The electricity dipole is not a particle of rest mass but a wave and its elementary energy is 1.6*10^-19 Coulomb.

But let us look some more at this fascinating number 1.60*10^-19, because mathematics has a fascinating geometry number constant called the Golden Ratio of A is to A+B as A is to B from the Golden Rectangle.

A B
-------------------
| | |
| | |
| | |
------------------- A+B

A+B = A
------- ----
A B

And this number is (1+sqrt5) / 2 = 1.618....

Now if we look at the prefix only of Elementary Coulomb as 1.60 and compare with 1.61 we have a sigma error of
161/160 = 0.6%. So close in Sigma Error we can call them equal.

Now I want to mention also, that the exponent power of Elementary Coulomb of 10^-19 appears in physics in the force strength of Coulomb law, the inverse square law, and the force of gravity, another inverse square law. The ratio of how strong Coulomb force is to gravity force is of the order 10^38 stronger which is got from squaring 10^19 or that of gravity is 10^-38 in strength to the Coulomb force.

AP
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Archimedes Plutonium<plutonium....@gmail.com>
Jul 1, 2021, 11:52:54 AM (8 days ago)
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On Monday, June 28, 2021 at 5:18:35 PM UTC-5 Archimedes Plutonium wrote:

On Monday, June 28, 2021 at 3:03:50 PM UTC-5 Archimedes Plutonium wrote:
By all means, we must teach the most direct experiment, the experiment that is aligned with theory as the best in teaching. Because other experiments are too much off course in learning. Those other experiments such as oil-drop are good for precision, but horrible in learning and teaching to young students.

So the Faraday constant is 96485 Amperes/mole

I do not know if Faraday had this number above by 1833, or even close to that number. Anyone know?

The Avogadro constant is 6.02*10^23 per mole

Chemistry probably had that number by 1833.

So let us assume Faraday could have computed the Coulomb constant by 1833 from his formula of F/N_A.

So this brings up the huge huge problem of why the vast gulf of ignorance there was in physics. Because, F/N_A leads to the logic that there must exist a magnetic monopole in order to have F/N_A.

I say-- must exist a magnetic monopole -- because otherwise you have Physics with a untenable definition of Coulomb as this nonsense---

--- quoting H&R ---
Appendix A, quoting "Electric current, Ampere, A = ...that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2*10^-7 newton per meter of length." (1946)
--- end quoting ---
Where C Coulomb = Ampere*seconds = A*s.

Certainly is nice to have an update on a definition of 1946.

--- quoting Wikipedia ---
The SI system defines the coulomb in terms of the ampere and second: 1 C = 1 A × 1 s.[4] The 2019 redefinition of the ampere and other SI base units fixed the numerical value of the elementary chargewhen expressed in coulombs, and therefore fixed the value of the coulomb when expressed as a multiple of the fundamental charge (the numerical values of those quantities are the multiplicative inverses of each other). The ampere is defined by taking the fixed numerical value of the elementary charge e to be 1.602176634×10−19 coulombs.[5]

Thus, one coulomb is the charge of approximately 6241509074460762607.776 elementary charges, where the number is the reciprocal of 1.602176634×10−19 C.[6] It is impossible to realize exactly 1 C of charge, since it is not an integer number of elementary charges.

--- end quoting ---
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Archimedes Plutonium<plutonium....@gmail.com>
Jul 2, 2021, 12:32:30 AM (8 days ago)
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3) Defining the units of Coulomb and Ampere as C = A*seconds; and the Elementary-Coulomb.

Before we define what is the Coulomb quantity of electricity we need to review Avogadro's number and some call it the mole. I actually prefer Avogadro's number but those in chemistry prefer mole.

Most students have come across mole in High School chemistry or physics class, so it is not new to them. Some have not learned it well enough to remember it. And some are learning it for the first time.

I am not going to teach Avogadro's number (or mole) here and one can learn it from my book:

World's first logical teaching of 6.02*10^23, Avogadro's number and "mole"; refurbished with "hyasys" and 5.98*10^23 // Chemistry series, book 6

Kindle Edition

by Archimedes Plutonium (Author)

Preface: I was worried I did not reach the target audience on this topic of mole and Avogadro's number. The cover picture is a High School physics textbook, a later edition of PSSC that I took when I was in High School circa 1967. It teaches mole and Avogadro's number. So I needed to write this textbook starting High School, because it is shameful to teach wrong science in either High School or University.

Cover Picture is page 448-9 of PSSC Physics, 3rd ed., 1971, Haber-Schaim, Cross, Dodge, Walter, explaining how Rutherford in 1919 weighed the mass of the proton in a experiment. It is disheartening to find I cannot see that experiment in a Google search because of another famous experiment of Rutherford of the gold leaf. And lucky for me that I retrieved this PSSC book I used in High School circa 1967-8 that explains how the proton mass was weighed in physics history. The mass of the proton is vital to understand the mole and Avogadro's number concept. Sad of course that every Old Chemistry and Old Physics textbooks that teaches the concept mole and Avogadro number never mentions the proton mass as crucial to understanding.

What I am going to do is review Avogadro's number because it is a similar idea to the "Elementary Coulomb".

So what was Avogadro's Number in most simple teaching? Avogadro's number, or mole, was taking a "elementary mass" and knowing that in all 1 gram of mass, there is a specific number of these elementary mass units. So if I have a 1 gram of iron or a 1 gram of silver, or a 1 gram of oxygen gas or a 1 gram of sodium or a 1 gram of hydrogen or a 1 gram of H2O or a 1 gram of CO2.. If I have 1 gram of anything of matter, it has a number of particles that is the same number in all 1 gram of matter. What is that elementary particle of mass? It is called nucleon in Old Physics and is either the neutron at 940MeV or the proton 840MeV + muon 105MeV = 945MeV.

Re: EM laws have a natural order Re: AP's 151st book, Part 3, TEACHING TRUE PHYSICS, 1st year college, by Archimedes Plutonium.

And so the 6 laws of EM theory, and, by the way, those would be all the laws of the entire Physics, where everything else is a "shade of those laws". And they have a natural ordering.

1st law magnetic field units defines the magnetic monopole
2nd law is New Ohm's law V= CBE

I was tempted to make the 3rd law be the differential equation V' = (CBE)'.

Where derivative is a creator function in physics, so we have the creation of voltage. But, no, I now realize due to many factors that V' has to be the last and 6th of the EM laws.

So the 3rd law is either Faraday or Ampere-Maxwell law. And because C=coulomb= quantity of electricity, I am placing Faraday law as 3rd law C' = (V/(BE))' with Ampere-Maxell law as 4th law B' = (V/(CE))'. And of course the derivative means a creator of C and a creator of B. Now the 5th law then is E' = (V/(CB))' leaving V' as the 6th law.

In the course of writing TEACHING TRUE PHYSICS, 1st year College I wanted the book to flow and it is not going to flow if laws are not in a ordered sequence.

And the laws pretty much follow an ordering of a increasing complexity of the units. Except for V= CBE for it generates the other 4 laws as all permutations of the derivative.

Now some may wonder about a integral or integration for the laws of EM theory. And this is unnecessary. For example in just pure math, say our starting function is Y= x^2 then the derivative is 2x and the integral of 2x is x^2. So the integral of Y=x^2 is 1/3* (x^3) and derivative of that is x^2.. In Old Physics they wrote their Maxwell equations in either derivative form or in integral form. But in New Physics, we can only write the EM laws in differential form because derivative means create, while integral means entire Space+Energy. We already have entire space plus energy in V= CBE so we cannot write the EM laws in integral form. We must write them in derivative form.

Now there are many factors for the reason that V' has to be the 6th law, and one of which is the definition of AC current, another is Meissner effect and superconductivity and diamagnetism.

AP
King of Science, especially Physics

Those 6 laws are these.

1) Magnetic monopole telling us what magnetism and electric current and magnetic field and electric field are.

2) New Ohm's law Voltage = quantity of current C times magnetic field times electric field. V= CBE.

3) Rate of change of C, quantity current, C' = (V/(BE))' Faraday law

4) Rate of change of B, magnetic field, B' = (V/(CE))' Ampere law

5) Rate of change of E, electric field, E' = (V/(CB))' Coulomb law & gravity

6) The rate of change of V= CBE as V' = (CBE)'. Capacitor law

I need to start with those 6 laws and on units of electromagnetism and the definition of magnetic monopoles that composes photons, and neutrinos.

Units

V = voltage
A = ampere current
s = seconds
C = Coulomb, amount of electricity = A*s
B = Magnetic field
E = Electric field
m = meters
kg = kilogram mass

Angular momentum L =  kg*m^2/s

Action = kg*m^2/s where angular momentum = action = electric field

Current = A  where the A represents Ampere

Quantity of Current = C = A*s where the C represents Coulomb

Magnetic Field B = kg/ C*s = kg/ A*s^2

Electric Field is E = kg*m^2/ A*s^2 = kg*m^2/ C*s

Voltage = C*B*E  = kg*m^2 /A*s^3 = kg*m^2 /C*s^2

Now the Derivative for the Differential Equations of EM theory is with respect to either seconds or to meters. So for instance C' derivative in Faraday law with respect to seconds is Ampere. B' as in Ampere-Maxwell law with respect to seconds would be Voltage without meter^2. E' in Coulomb-gravity law with respect to seconds would be voltage. But E' with respect to meters would be a Force = kg*acceleration.

Here we are going to define Quantum Duality as either the derivative with respect to meters or with respect to seconds.

AP
King of Science, especially Physics