Archimedes Plutonium

2017-06-14 08:48:24 UTC

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Newsgroups: sci.physicsRaw Message

Date: Tue, 13 Jun 2017 23:03:08 -0700 (PDT)

Subject: Taking the Product Rule, (f*g)' = f'g + fg', of the 3 permutations of

Ohm's law to achieve all of the Maxwell Equations

From: Archimedes Plutonium <***@gmail.com>

Injection-Date: Wed, 14 Jun 2017 06:03:08 +0000

Taking the Product Rule, (f*g)' = f'g + fg', of the 3 permutations of Ohm's law to achieve all of the Maxwell Equations

Alright, let us get going on this. It is called the Product Rule in math (f*g)' = f'g + fg' and we use it on three of the four listed below::

1) V= i*R

2) V' = (iR)'

3) i' = (V*R^-1)'

4) R' = (V*i^-1)'

Three are differential equations.

Now I do not know how many times Maxwell used the Product Rule for his original 20 differential equations or later, his 8 differential equations. Before his equations were corrupted by vector fields.

I like science to be simple, not the fancy schmanzy for you lose sight of what the underlying physical reality is. And it is a shame, darn shame Maxwell did not use Ohm's law as a framework, instead, he used all sorts of models, which is great if you are comfortable with models, but the trouble with models is that you invariably miss large sections of the physical science you are doing. Whereas if you pick a universal law of electricity, magnetism, as Ohm's law, you really cannot go wrong.

Ampere's law looks to be (3), and Faraday's law looks to be (2). The law of electricity and magnetism looks to be all wrapped up into (1) since R and V contain magnetic field and "i" is electric charge and current.

Now where do I get Coulomb law as not kAA/d^2 but rather as (kAA+jBB)/d^2 ? Where do I get that? Well, when you differentiate (f*g)' you get f'g + fg'. So you get a two term numerator.

Alright, let us do the Calculus Product Rule upon

(iR)' = i'*R + i*R', now that was fun and easy

next,

(V*R^-1)' = V'*R^-1 + V*(R^-1)'

next,

(V*i^-1)' = V'* (i^-1) + V* (i^-1)'

Now, the math calculus is really just about over and finished, hard to believe, when you see students across the world, around the world struggling with the nightmare of vector field calculus. Time to go out and have a coconut cream pie topped with vanilla ice cream, all organic.

Now we have to finish off the EM laws, and here is the only tricky part, the part that bedevils, not just the student but the experienced and greenhorn professors of math and physics. Here we have to interpret what the derivative of V, of i and of R mean.

Now here is a partial list of derivatives of physical concepts::

The derivative of voltage is current A or i.

The derivative of charge is current.

The derivative of current is accelerated current.

The derivative of Resistance R is more ordered more patterned resistance, let me call it R_ordered.

What is the derivative of Magnetism? Even though magnetism is not directly found in Ohm's law?

Years ago i came to a conclusion on what the derivative of magnetism was.

Magnetism is magnetic field kg/A*s^2 and Voltage= Electric Field is kg*m^2/A*s^3, and ohms = Resistance is kg*m^2/A^2*s^3

If you take the derivative of Magnetic Field, what I suspect comes out is the Ohm's Resistance. And this makes quite a lot of common sense. That if the world is just and only EM forces, what we view as friction, as resistance, as impedance, is all, all back to magnetic fields.

So taking the derivative of R, is going to be an ordered R as the outcome.

AP

On Wednesday, June 14, 2017 at 1:30:09 AM UTC-5, Archimedes Plutonium wrote:

Alright, I am just so anxious to see which one is Faraday's law and if I can get (kXX + jYY)/d^2 rather than kXX/d^2

1) V= i*R

2) V' = (iR)'

3) i' = (V*R^-1)'

4) R' = (V*i^-1)'

Three are differential equations.

Now I do not know how many times Maxwell used the Product Rule for his original 20 differential equations or later, his 8 differential equations. Before his equations were corrupted by vector fields.

I like science to be simple, not the fancy schmanzy for you lose sight of what the underlying physical reality is. And it is a shame, darn shame Maxwell did not use Ohm's law as a framework, instead, he used all sorts of models, which is great if you are comfortable with models, but the trouble with models is that you invariably miss large sections of the physical science you are doing. Whereas if you pick a universal law of electricity, magnetism, as Ohm's law, you really cannot go wrong.

Ampere's law looks to be (3), and Faraday's law looks to be (2). The law of electricity and magnetism looks to be all wrapped up into (1) since R and V contain magnetic field and "i" is electric charge and current.

Now where do I get Coulomb law as not kAA/d^2 but rather as (kAA+jBB)/d^2 ? Where do I get that? Well, when you differentiate (f*g)' you get f'g + fg'. So you get a two term numerator.

Alright, let us do the Calculus Product Rule upon

(iR)' = i'*R + i*R', now that was fun and easy

next,

(V*R^-1)' = V'*R^-1 + V*(R^-1)'

I am going to examine this one first for it is current derivative.2) V' = (iR)'

3) i' = (V*R^-1)'

4) R' = (V*i^-1)'

Three are differential equations.

Now I do not know how many times Maxwell used the Product Rule for his original 20 differential equations or later, his 8 differential equations. Before his equations were corrupted by vector fields.

I like science to be simple, not the fancy schmanzy for you lose sight of what the underlying physical reality is. And it is a shame, darn shame Maxwell did not use Ohm's law as a framework, instead, he used all sorts of models, which is great if you are comfortable with models, but the trouble with models is that you invariably miss large sections of the physical science you are doing. Whereas if you pick a universal law of electricity, magnetism, as Ohm's law, you really cannot go wrong.

Ampere's law looks to be (3), and Faraday's law looks to be (2). The law of electricity and magnetism looks to be all wrapped up into (1) since R and V contain magnetic field and "i" is electric charge and current.

Now where do I get Coulomb law as not kAA/d^2 but rather as (kAA+jBB)/d^2 ? Where do I get that? Well, when you differentiate (f*g)' you get f'g + fg'. So you get a two term numerator.

Alright, let us do the Calculus Product Rule upon

(iR)' = i'*R + i*R', now that was fun and easy

next,

(V*R^-1)' = V'*R^-1 + V*(R^-1)'

V' * R^-1 + V * (R^-1)'

derivative of Voltage is current A

derivative of Resistance is Ordered Resistance R_order

A/R + V/R^2 = AR/R^2 + V/R^2 = (AR + V)/R^2

Not bad, not bad at all

Now I have to see if the other two differential equations yield about the same thing

AP

Newsgroups: sci.physics

Date: Wed, 14 Jun 2017 01:40:56 -0700 (PDT)

Subject: let these soak in for some days to see if they are good Re: Taking

the Product Rule, (f*g)' = f'g + fg',

From: Archimedes Plutonium <***@gmail.com>

Injection-Date: Wed, 14 Jun 2017 08:40:57 +0000

let these soak in for some days to see if they are good Re: Taking the Product Rule, (f*g)' = f'g + fg',

Now doing all three, i end up with these

(iR +V) / R^2 = i'

2 i_accel * R_ordered, basically 2iR = V'

(i^2 + V) / i^2 = R'

Now the i' looks to be Faraday's law with current and where Lenz's law is the V term since magnetic field is within Resistance

And V' looks to be Ampere-Maxwell law two resultant currents and the magnetic fields are in the R term. The Old Maxwell Eq. predicted just one magnetic field, missing the displacement current magnetic field. Now let me just think about this for several days, for I may have them mixed up.

Finally the R' looks like the force of gravity as (kAA+jBB)/d^2 and is the copper wire in Faraday, Ampere laws. The R' equation is a new one, which Maxwell totally missed. Maxwell had nothing for the copper wires in Ampere and Faraday laws. It is a nasty habit of most scientists, that they leave things out, figuring they were not "science in general".

Now if I include Ohm's law itself along with the three differential equations, I believe I have captured all the laws of EM theory.(iR +V) / R^2 = i'

2 i_accel * R_ordered, basically 2iR = V'

(i^2 + V) / i^2 = R'

Now the i' looks to be Faraday's law with current and where Lenz's law is the V term since magnetic field is within Resistance

And V' looks to be Ampere-Maxwell law two resultant currents and the magnetic fields are in the R term. The Old Maxwell Eq. predicted just one magnetic field, missing the displacement current magnetic field. Now let me just think about this for several days, for I may have them mixed up.

Finally the R' looks like the force of gravity as (kAA+jBB)/d^2 and is the copper wire in Faraday, Ampere laws. The R' equation is a new one, which Maxwell totally missed. Maxwell had nothing for the copper wires in Ampere and Faraday laws. It is a nasty habit of most scientists, that they leave things out, figuring they were not "science in general".

Of course the idea of a magnetic monopole is forever silly, because magnetism is a closed loop, to be a monopole, means you can exist linearly, not closed but open. An electron can exist with charge, linearly, open, not closed for the charge need not emanate and loop back around.

So in the AP Equations of EM, the law of electricity and magnetism are contained in the Ohm's law. Then, the Faraday, Ampere-Maxwell and the Gravity-Resistance laws come out of the differentiation of Ohm's law.

And we see that two of the four laws of EM are of the Force math form of (kAA + jB)/d^2. Looks like I did not get exactly what I wanted of (kAA+jBB)/d^2 but rather got a lesser of (kAA + jB) / d^2.

AP