Discussion:
the AP Equations that replace the Maxwell Equations of EM, using the product rule upon Ohm's law
Archimedes Plutonium
2017-06-14 08:48:24 UTC
Raw Message
Newsgroups: sci.physics
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.

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

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.

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
b***@gmail.com
2017-06-14 10:33:30 UTC
Raw Message
Confusing up electric network stuff with electric
field as usual! Got your brain fart backpack shouldered?

Work harder, AP! Express for example Joule heating, so
that we explain your mommies feet warmer in winter.

And lets for example explain the change of resistivity

voltage divider.
Post by Archimedes Plutonium
Newsgroups: sci.physics
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
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
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.
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
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
Post by Archimedes Plutonium
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.
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
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',
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',
Post by Archimedes Plutonium
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.
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
Archimedes Plutonium
2017-06-14 15:08:28 UTC
Raw Message
Alright, I need to let this soak in for several days. And the new equations do predict much that is new-- motion that varies from R to 1/R to 1/R^2.
Post by Archimedes Plutonium
Newsgroups: sci.physics
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
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
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.
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
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
Post by Archimedes Plutonium
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.
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
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',
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',
Post by Archimedes Plutonium
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.
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
Archimedes Plutonium
2017-06-15 05:24:04 UTC
Raw Message
Newsgroups: sci.physics
Date: Wed, 14 Jun 2017 21:42:24 -0700 (PDT)

Subject: finalized in my mind, electric-field = voltage= force, magnetic field
is a acceleration Re: arguing Voltage is electric field Re: Magnetic Field
comes nearest to being Force
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Thu, 15 Jun 2017 04:42:24 +0000

finalized in my mind, electric-field = voltage= force, magnetic field is a acceleration Re: arguing Voltage is electric field Re: Magnetic Field comes nearest to being Force

- hide quoted text -
Alright let us say volt = m^3/s^2 = electric field = force
Where mass = meter^2
Then, that causes Magnetic Field to have to be acceleration = m/ s^2
Now that does allow for consistency in the charge follows along inside of mass, whereas no magnetic monopole requires no mass but rather, Space to loop back around upon itself.
This is the first time i have heard of this theorem-- meter^2 = mass
What is the history of this theorem??
AP
I invented the mass=area theorem in 2015. It due to continuum
fluid aether.
Well, a theorem is mathematics, not physics. In physics, you would want to call it a concept, so like a cross-section mass.

But then again, this is what Rutherford was doing a lot of-- studies of particle scattering, and what can be called cross-section-mass. It is a good concept to sort of pry into units, simplify units.

I would not say you invented cross-section-mass concept, for that has been around since at least Rutherford.
Magnetic flux density B = 1/second
Alan, too much of an idealist, not enough clear thinking. Why invent something for frequency = 1/s.

With some silly obnoxious magnetic flux density. Tell, me, what passes through that mind of yours when you say magnetic flux density. What passes through my mind at this moment is I wish there was an organic beer flux flow through my mouth for it is too hot in this room.

You see, I worry about you Alan, as well as the thousands in colleges and universities that experience what you experienced-- an education of cloud nine idealisms, but little to no clear thinking, ever going on. It is as if you went to school and instead of learning the physics of how things work, you learned how to put long string of words together which really are fantasies-- the inductance power magnetic low frequency coil capacitor.
Magnetic field H = meter/second
Alright, Alan, that is really silly. For stop and think for a moment. Use common sense, not that wasted 8 years of Ivy league education. Just common sense. You have a two magnets on a table near you, and you know they are having a magnetic field, quite strong field, and if they ever get closer, they immediately bond together. So you are going to classify that motion as a m/s

Just commonsense should tell you that a Magnetic Field, at minimum is an acceleration, not a speed.

Now you called voltage a potential. Here again, what is "potential" in view of Physics. Is it just some word of language but no science meaning. You could just as well call voltage a reservoir, or perhaps a storage. So you call it a potential as if that means something, when it means nothing. The units m^2/s^2 do mean something, and you chose to call those units a potential. I call it something else, that m^2/s^2 is simply meters accelerated, a hyper form of acceleration, much like what I am faced with momentarily with the derivative of Resistance as a more orderly new resistance.

So skip the naming nonsense that goes on in science and try always to stick with what you know and then when m^2/s^2 comes up, it is different from acceleration but a form of acceleration as meters*acceleration

So you have Volts as m^2/s^2 and you have electric field as m/s^2 and you have magnetic field as m/s

That is all silly. For it should be that starting with Magnetic Field that it accelerates nearby iron filings, accelerates those filings means at minimum the Magnetic Field is m/s^2

And since the Electric Field involves mass particles with electric charge, means that at minimum the Electric field is a force of F= mass*accel or in your notation m^3/s^2

Now the Magnetic Field cannot have mass involved of its acceleration, for then it would not be a monopole. The Magnetic Field has to have massless particles-- photons, neutrinos to do the acceleration.

So we have emerging this picture::

Magnetic Field = acceleration  meter/second^2

Electric Field = force = Volt = mass* acceleration which in your notation is m^3/s^2

Your notation helps in getting to the finish-line faster and more clear, without the hemming and hawing over peripheral terms.

There are many terms in physics, that have no physical meaning other than an enhancement of known units.

There is your term m^2/s^2 which I just explained is enhanced acceleration

Then there are terms like 1/s^3, which to a novice, would never realize it is just enhanced frequency. Then there are m^3/s^3 or there is m^4/s^3, or 1/m^3,etc etc.

So when you have philosophers trying to do physics, you come up with all these laughable comedy  concepts. Never realizing it is time to come back to Earth and do physics.

So, this sideroad discussion has made it clear to me, that Magnetic Field is an acceleration, while Electric Field is the Voltage and is a force.

Now, thousands and thousands of experts in chemistry and physics have all known, full well, that voltage is Electromotive Force. The word "force" is not there to pass the time of day, but is there because it is well known that the events of the physical demonstration was due to the Voltage being a Force. There is no electric field, never was, never will be, because there is always Voltage. The two are one and the same thing.

One of the hugest and most lousy understanding and education in physics today, is the teaching that Volt and Electric Field are two different things. They are not. They are one and the same thing.

Voltage is a force, and magnetic field is an acceleration. Electricity is a force, voltage is a force, magnetism is a force, but magnetic field is a acceleration.

AP
Archimedes Plutonium
2017-06-15 11:33:10 UTC
Raw Message
Newsgroups: sci.physics
Date: Wed, 14 Jun 2017 21:42:24 -0700 (PDT)

Subject: finalized in my mind, electric-field = voltage= force, magnetic field
is a acceleration Re: arguing Voltage is electric field Re: Magnetic Field
comes nearest to being Force
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Thu, 15 Jun 2017 04:42:24 +0000

finalized in my mind, electric-field = voltage= force, magnetic field is a acceleration Re: arguing Voltage is electric field Re: Magnetic Field comes nearest to being Force

- hide quoted text -
Alright let us say volt = m^3/s^2 = electric field = force
Where mass = meter^2
Then, that causes Magnetic Field to have to be acceleration = m/ s^2
Now that does allow for consistency in the charge follows along inside of mass, whereas no magnetic monopole requires no mass but rather, Space to loop back around upon itself.
This is the first time i have heard of this theorem-- meter^2 = mass
What is the history of this theorem??
AP
I invented the mass=area theorem in 2015. It due to continuum
fluid aether.
Well, a theorem is mathematics, not physics. In physics, you would want to call it a concept, so like a cross-section mass.

But then again, this is what Rutherford was doing a lot of-- studies of particle scattering, and what can be called cross-section-mass. It is a good concept to sort of pry into units, simplify units.

I would not say you invented cross-section-mass concept, for that has been around since at least Rutherford.
Magnetic flux density B = 1/second
Alan, too much of an idealist, not enough clear thinking. Why invent something for frequency = 1/s.

With some silly obnoxious magnetic flux density. Tell, me, what passes through that mind of yours when you say magnetic flux density. What passes through my mind at this moment is I wish there was an organic beer flux flow through my mouth for it is too hot in this room.

You see, I worry about you Alan, as well as the thousands in colleges and universities that experience what you experienced-- an education of cloud nine idealisms, but little to no clear thinking, ever going on. It is as if you went to school and instead of learning the physics of how things work, you learned how to put long string of words together which really are fantasies-- the inductance power magnetic low frequency coil capacitor.
Magnetic field H = meter/second
Alright, Alan, that is really silly. For stop and think for a moment. Use common sense, not that wasted 8 years of Ivy league education. Just common sense. You have a two magnets on a table near you, and you know they are having a magnetic field, quite strong field, and if they ever get closer, they immediately bond together. So you are going to classify that motion as a m/s

Just commonsense should tell you that a Magnetic Field, at minimum is an acceleration, not a speed.

Now you called voltage a potential. Here again, what is "potential" in view of Physics. Is it just some word of language but no science meaning. You could just as well call voltage a reservoir, or perhaps a storage. So you call it a potential as if that means something, when it means nothing. The units m^2/s^2 do mean something, and you chose to call those units a potential. I call it something else, that m^2/s^2 is simply meters accelerated, a hyper form of acceleration, much like what I am faced with momentarily with the derivative of Resistance as a more orderly new resistance.

So skip the naming nonsense that goes on in science and try always to stick with what you know and then when m^2/s^2 comes up, it is different from acceleration but a form of acceleration as meters*acceleration

So you have Volts as m^2/s^2 and you have electric field as m/s^2 and you have magnetic field as m/s

That is all silly. For it should be that starting with Magnetic Field that it accelerates nearby iron filings, accelerates those filings means at minimum the Magnetic Field is m/s^2

And since the Electric Field involves mass particles with electric charge, means that at minimum the Electric field is a force of F= mass*accel or in your notation m^3/s^2

Now the Magnetic Field cannot have mass involved of its acceleration, for then it would not be a monopole. The Magnetic Field has to have massless particles-- photons, neutrinos to do the acceleration.

So we have emerging this picture::

Magnetic Field = acceleration  meter/second^2

Electric Field = force = Volt = mass* acceleration which in your notation is m^3/s^2

Your notation helps in getting to the finish-line faster and more clear, without the hemming and hawing over peripheral terms.

There are many terms in physics, that have no physical meaning other than an enhancement of known units.

There is your term m^2/s^2 which I just explained is enhanced acceleration

Then there are terms like 1/s^3, which to a novice, would never realize it is just enhanced frequency. Then there are m^3/s^3 or there is m^4/s^3, or 1/m^3,etc etc.

So when you have philosophers trying to do physics, you come up with all these laughable comedy  concepts. Never realizing it is time to come back to Earth and do physics.

So, this sideroad discussion has made it clear to me, that Magnetic Field is an acceleration, while Electric Field is the Voltage and is a force.

Now, thousands and thousands of experts in chemistry and physics have all known, full well, that voltage is Electromotive Force. The word "force" is not there to pass the time of day, but is there because it is well known that the events of the physical demonstration was due to the Voltage being a Force. There is no electric field, never was, never will be, because there is always Voltage. The two are one and the same thing.

One of the hugest and most lousy understanding and education in physics today, is the teaching that Volt and Electric Field are two different things. They are not. They are one and the same thing.

Voltage is a force, and magnetic field is an acceleration. Electricity is a force, voltage is a force, magnetism is a force, but magnetic field is a acceleration.

AP

Newsgroups: sci.physics
Date: Thu, 15 Jun 2017 02:24:49 -0700 (PDT)

Subject: first physics laws that multiply forces? Re: 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: Thu, 15 Jun 2017 09:24:50 +0000

first physics laws that multiply forces? Re: 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
Letting these soak in for several days

The magnificent heights these 4 equations bring to all of science. They compose the bulk of Physics other than the Atomic Theory, the EM physics are the two foundations of all science, anything else is details of Atoms and EM.
(iR +V) / R^2 = i'
Now I think that is Faraday's law with a produced current and the R term would be Lenz's law
2 i_accel * R_ordered, basically 2iR = V'
Now this is a supremely delightful law of physics where you have the math form of 2iR. I do not recall any other law of physics so neat, so compact, so vibrant so clean, two times two parameters. There simply should be some law of physics of that math form. Of course, I believe it is Ampere-Maxwell Law but not certain of it-- let it soak in to see of correct. There are two currents, the initial and displacement. There are two magnetic fields-- of course, since there are 2 currents.
(i^2 + V) / i^2 = R'
This is a law that Maxwell Equations never saw, never realized. In EM theory, it is based on watching the phenomenon of a thrusting bar magnet in coil producing electricity and of a current in copper wire producing a magnetic field. Maxwell got those, but failed to have a law that puts the coil of wire or the electric copper wire system into his equations. Why, he failed to inject those physical parameters is unknown, perhaps, he felt like so many other physicists, that laws of physics should be stripped of all
modeling that went into deriving the laws-- only the copper wires were not models but essential ingredients in forming the laws.

Now, what this third new law says is quite astounding for it is the motion law and why gravity is EM and why gravity has a range of motion varying from R to 1/R to 1/R^2.

So instead of electrons flowing in wire or coil, here we have stars flowing around galaxies and planets flowing around Sun. The law of the derivative of Resistance is that Space is Magnetic Field, lines of force forming TRACKS in Space and the planets and stars follow along those tracks, just as electrons follow along the copper wire.
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.
So let me soak this in, slowly.
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.
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.
And, already, I have a question, puzzling me. For the equations above for EM theory involve equations of terms that are forces themselves. Voltage is a force, and in a sense, resistance is a force-- like Lenz's law and in a sense current can be seen as electrons in motion as a force.

So are the Equations of EM supposed to be forces as terms, where we multiply forces. I can picture adding several forces, but here we multiply forces. So is this the first time in physics that we have forces as the terms in the laws. Usually the laws are terms that build up to being a force, but here we have multiplication and division of forces.

AP
Archimedes Plutonium
2017-06-15 21:37:53 UTC
Raw Message
Newsgroups: sci.physics
Date: Thu, 15 Jun 2017 14:30:29 -0700 (PDT)

Subject: Can I ditch the Quotient Rule and forever use just the Product Rule
Re: first physics laws that multiply forces?
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Thu, 15 Jun 2017 21:30:29 +0000

Can I ditch the Quotient Rule and forever use just the Product Rule Re: first physics laws that multiply forces?

Alright, I am letting this soak in, slowly.

But already have found a huge stumbling block. One I encountered when in college calculus, 47 years ago, my, time flys.

It was in Calculus class, and learning the Product Rule along with the Quotient Rule.

The one is (fg)' = f'g + fg'

the second is (f/g)' = (f'g - fg')/g^2

And, so, in college, what my mind said, later on, not during the tests, but after the tests and later on, my mind said, "make it simple" whenever you have something like V = iR and then have V/R = i, to make that V*R^-1 = i

But I think my simplicity of just converting R into R^-1 and ditching the Quotient Rule and only ever using the Product Rule, has caught up with me.

AP
b***@gmail.com
2017-06-15 22:12:22 UTC
Raw Message
A simple(r) form is:

(f/g)' = (f' - (f/g)g')/g

As you see, the input (f/g) is reused unchanged.
In auto diff, you get some speed-up through this.

This form of the quotient rule, is for example mentioned
in this old paper here, on page 4, 2.1 Rationale Komposition

Algorithmisches Differenzieren - Herbert Fischer, 1971
http://www-m1.ma.tum.de/foswiki/pub/Lehrstuhl/HFischerAlgorithmischesDifferenzieren1/Fi-Vorlesung-03.pdf
Post by Archimedes Plutonium
the second is (f/g)' = (f'g - fg')/g^2
"make it simple"
b***@gmail.com
2017-06-15 22:15:57 UTC
Raw Message
Corr.:
Sorry the link is a new script from 2006
Post by b***@gmail.com
Algorithmisches Differenzieren - Herbert Fischer, 1971
http://www-m1.ma.tum.de/foswiki/pub/Lehrstuhl/HFischerAlgorithmischesDifferenzieren1/Fi-Vorlesung-03.pdf
Archimedes Plutonium
2017-06-16 06:19:05 UTC
Raw Message
Newsgroups: sci.physics
Date: Thu, 15 Jun 2017 23:05:08 -0700 (PDT)

Subject: Re: Can I ditch the Quotient Rule and forever use just the Product
Rule Re: first physics laws that multiply forces?
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Fri, 16 Jun 2017 06:05:08 +0000

Re: Can I ditch the Quotient Rule and forever use just the Product Rule Re: first physics laws that multiply forces?
Post by Archimedes Plutonium
Alright, I am letting this soak in, slowly.
But already have found a huge stumbling block. One I encountered when in college calculus, 47 years ago, my, time flys.
It was in Calculus class, and learning the Product Rule along with the Quotient Rule.
The one is (fg)' = f'g + fg'
the second is (f/g)' = (f'g - fg')/g^2
Alright, so let us check these two equations with the Quotient Rule to see if something different emerges other than the subtraction sign.

(V/R)' = i'

and

(V/i)' = R'

So we have (V/R)' = (V'R - VR')/R^2

And (V/i)' = (V'i - Vi')/i^2

Now applying these::

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.

So we have (V/R)' = (V'R - VR')/R^2 = (iR - VR_ordered) / R^2

And (V/i)' = (V'i - Vi')/i^2 = (ii - Vi_accelerated )/ i^2

Now is that much different from what was obtained by Product Rule?

(V* R^-1)' = V'R^-1 + V*(R^-1)' = i*R^-1 - V*R^-2 = (iR - V) / R^2

2 i_accel * R_ordered, basically 2iR = V'

(V* i^-1)' = V'i^-1 + V*(i^-1)' = i*i^-1 - V*i^-2 = (i^2 - V) / i^2

Yes, that is very much so different, ignoring my negative sign error and other errors in Product Rule.

But the difference is this

Product Rule begets (iR - V) / R^2 while Quotient Rule begets (iR - VR_ordered) / R^2

So product rule has V while quotient rule has VR_ordered

Product Rule begets (i^2 - V) / i^2 while quotient rule begets    (i^2 - Vi_accelerated )/ i^2

Here again, the Quotient Rule provides a Vi_accel whilst the Product Rule provides just V.

So I think the Quotient Rule is vital, and the error of thinking the Product Rule can take over, is the error of g/f is not the same as g*f^-1 for g and f as function.

AP
Archimedes Plutonium
2017-06-16 07:12:17 UTC
Raw Message
On Friday, June 16, 2017 at 1:19:20 AM UTC-5, Archimedes Plutonium wrote:
Alright, here I am in trouble if the math comes out to be (kAA - jBB) / d^2 because that is the geometry of a hyperbola, and I need a closed loop geometry of addition (kAA + jBB)/ d^2
Post by Archimedes Plutonium
Post by Archimedes Plutonium
The one is (fg)' = f'g + fg'
the second is (f/g)' = (f'g - fg')/g^2
Alright, so let us check these two equations with the Quotient Rule to see if something different emerges other than the subtraction sign.
(V/R)' = i'
and
(V/i)' = R'
So we have (V/R)' = (V'R - VR')/R^2
And (V/i)' = (V'i - Vi')/i^2
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.
So we have (V/R)' = (V'R - VR')/R^2 = (iR - VR_ordered) / R^2
And (V/i)' = (V'i - Vi')/i^2 = (ii - Vi_accelerated )/ i^2
So here I am in trouble, with a negative sign and need to get a ellipse positive sign
Post by Archimedes Plutonium
Now is that much different from what was obtained by Product Rule?
(V* R^-1)' = V'R^-1 + V*(R^-1)' = i*R^-1 - V*R^-2 = (iR - V) / R^2
2 i_accel * R_ordered, basically 2iR = V'
(V* i^-1)' = V'i^-1 + V*(i^-1)' = i*i^-1 - V*i^-2 = (i^2 - V) / i^2
Yes, that is very much so different, ignoring my negative sign error and other errors in Product Rule.
Now one way to get a positive sign, is that signage in EM means direction of motion
Post by Archimedes Plutonium
But the difference is this
Product Rule begets (iR - V) / R^2 while Quotient Rule begets (iR - VR_ordered) / R^2
So that can be written as this, (iR + V(-R_ordered)) / R^2 where you have a direction of motion of R_ordered
Post by Archimedes Plutonium
So product rule has V while quotient rule has VR_ordered
Product Rule begets (i^2 - V) / i^2 while quotient rule begets    (i^2 - Vi_accelerated )/ i^2
And that can be written as this (i^2 + V(-i_accelerated))/ i^2
Post by Archimedes Plutonium
Here again, the Quotient Rule provides a Vi_accel whilst the Product Rule provides just V.
So I think the Quotient Rule is vital, and the error of thinking the Product Rule can take over, is the error of g/f is not the same as g*f^-1 for g and f as function.
So it is shaping up to be this (kAA + (-j)BB) / d^2

AP
Archimedes Plutonium
2017-06-16 09:43:56 UTC
Raw Message
Alright what is next?

Next i have to parse or resolve what these terms are

iR

VR

R^2

i^2

Vi

We know from Ohms law V= iR, so that eliminates the first leaving:

VR

R^2

i^2

Vi

Now is i^2 just accelerated i, and is R^2 just more ordered resistance?

AP
b***@gmail.com
2017-06-16 12:26:03 UTC
Raw Message

Tesla coil music. Sail by Awolnation

Post by Archimedes Plutonium
Alright what is next?
Next i have to parse or resolve what these terms are
iR
VR
R^2
i^2
Vi
VR
R^2
i^2
Vi
Now is i^2 just accelerated i, and is R^2 just more ordered resistance?
AP
Archimedes Plutonium
2017-06-16 20:45:49 UTC
Raw Message
Newsgroups: sci.physics
Date: Fri, 16 Jun 2017 02:47:26 -0700 (PDT)

Subject: Parsing or resolving new terms like VR, R^2, i^2, Vi
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Fri, 16 Jun 2017 09:47:27 +0000

Parsing or resolving new terms like VR, R^2, i^2, Vi

Alright what is next?

Next i have to parse or resolve what these terms are

iR

VR

R^2

i^2

Vi

We know from Ohms law V= iR, so that eliminates the first leaving:

VR

R^2

i^2

Vi

Now is i^2 just accelerated i, and is R^2 just more ordered resistance?

AP

Newsgroups: sci.physics
Date: Fri, 16 Jun 2017 03:32:36 -0700 (PDT)

Subject: Parsing or resolving new terms like VR, R^2, i^2, Vi
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Fri, 16 Jun 2017 10:32:36 +0000

Parsing or resolving new terms like VR, R^2, i^2, Vi

Alright looks like Power enters the picture. And would that be so delightful to say the Sun powers Earth to revolve around the Sun.

So i have Power in the terms Vi and i^2

The term V^2 is Resistance

The strangest term is VR with units kg^2*m^4/ A^3*s^6

AP

Newsgroups: sci.physics
Date: Fri, 16 Jun 2017 13:25:50 -0700 (PDT)

Subject: was there a huge flaw in Maxwell Equations-- silent on Power-- wattage
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Fri, 16 Jun 2017 20:25:51 +0000

was there a huge flaw in Maxwell Equations-- silent on Power-- wattage

I do not recall, Power = watts as any part of the Maxwell theory. Was this a grave error of the theory? Of course, there were no electrical gadgets around when Maxwell made the Equations, and time to reflect on Power, and so, Power was missed in the equations and theory.

Magnetic Field =  kg /A*s^2 = kg /C*s

Charge = C = A*s

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

Pressure = kg/m*s^2
Force = kg*m/s^2
Power = kg*m^2/s^3

Resistance = kg*m^2 /A^2*s^3  = kg*m^2 /C *A*s^2

Notice that multiply R by A we get V

Now I was looking on the web for discussion of Power in Maxwell theory. A Rutgers Univ. site says this

" We will find that electromagnetic energy flowing into a region will partially increase the stored energy in that region and partially dissipate into heat ..."

Earlier in that discussion is mentioned "ohmic power losses per unit volume"

Talking about the Lorentz force of Maxwell theory.

Now, what I am concerned with, is that Power is not really a part of the Maxwell theory, nor of the equations.

So as I derive the Equations for EM theory as these::

1) Ohm's law V = iR

2) V' = 2 i_accel * R_ordered which contains Faraday's Law

3) i' = (iR + V* (-R_ordered)) / R^2 which contains Ampere-Maxwell law

4) R' = (i^2 + V* (-i_accel)) / i^2 which is an equation that bespeaks of Power from Resistance

AP
Archimedes Plutonium
2017-06-17 01:57:38 UTC
Raw Message
Newsgroups: sci.physics
Date: Fri, 16 Jun 2017 18:50:44 -0700 (PDT)

Subject: New Physics the Sun Power-wattage moves Earth around the Sun-- that
is how gravity works, as wattage power
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Sat, 17 Jun 2017 01:50:44 +0000

New Physics the Sun Power-wattage moves Earth around the Sun-- that is how gravity works, as wattage power
Post by Archimedes Plutonium
Post by Archimedes Plutonium
Alright looks like Power enters the picture. And would that be so delightful to say the Sun powers Earth to revolve around the Sun.
So i have Power in the terms Vi and i^2
The term V^2 is Resistance
The strangest term is VR with units kg^2*m^4/ A^3*s^6
AP
I do not recall, Power = watts as any part of the Maxwell theory. Was this a grave error of the theory? Of course, there were no electrical gadgets around when Maxwell made the Equations, and time to reflect on Power, and so, Power was missed in the equations and theory.
Electric Power in watts is kg*m^2/ s^3

Electric Volts is kg*m^2 / A*s^3

Electric Resistance is kg*m^2 / A^2 * s^3

Do you see the pattern? Goes from W to W/A to W/A^2

So now, in gravity in spiral Galaxies we have motion of gravity going from R to 1/R to that of 1/R^2 where R in gravity is radius from center.

So what Newton said for Gravity was we have Sun and Earth

-------->    the Sun

-------->   the Earth

Both in linear momentum flying straight and straight by one another, until the force of gravity pulls on E so that it is no longer a parallel straight line path, but bending Earth in towards the Sun and thus, an orbit

General Relativity looks at it the same only with a added twist, by saying the Sun mass bends space and so the Earth mass can no longer be in a path straightline parallel but is now trapped into following the bent space like inside a ash-tray and going around the Sun.

What EM gravity says, is that is nonsense. What is happening is that Sun is a massive body of many charges, which makes Sun a Electrical and Magnetic body and the Earth is also a electrical magnetic body, and the Sun is the thrusting bar magnet and Earth is on a wire track formed by the magnetism of the Sun, and the Sun bar magnet causes by Electric Power to circle around the Sun, just as the bar magnet in Faraday's law causes the electron in the copper wire to circle around as current.

So in New Physics the Sun Power-wattage moves Earth around the Sun.

Now, sounds hard to believe, but then ask yourself, can a Newton gravity or General Relativity, can that force create solid body rotation? No, but a electric wattage can make things go around in solid body rotation.

AP
Archimedes Plutonium
2017-06-17 22:19:59 UTC
Raw Message
Now, many times it is shown in TV or documentaries of where Faraday or Maxwell are in a study with the electric motor invention of a copper wire going around in a mercury bath, with battery attached -- Faraday's invention of the first electric motor.

In The Mechanical Universe series by Cal Tech this scene is shown.

But what I want to draw attention to, is, instead of a copper wire going around in circles, envision that as planet Earth.

Better yet, a electric motor can be built from a copper wire, magnet attached to a battery, and the copper wire keeps spinning around and around-- seen ample times on YouTube.

So, replace the magnet with the Sun, and replace the copper wire with Earth. The battery is what?

Well, the battery is the force of gravity, powering the planet Earth as copper wire to revolve around the Sun as magnet.

What I want to emphasize, is that Gravity as a force is not that meek and miniscule force tauted by Old Physics. Gravity is EM and that means a very strong force, that is Powering Earth to orbit Sun. The Sun does not timidly or meekly adjust Earth's motion to go around the Sun, no, the Sun is with WATTAGE and POWER forcing Earth to go around the Sun.

AP
Archimedes Plutonium
2017-06-18 05:07:48 UTC
Raw Message
Re: New Physics the Sun Power-wattage moves Earth around the Sun-- that is how gravity works, as wattage power

So, what the physicists today think of gravity, they think it is Newton and General Relativity gravity, where objects in space like stars, planets all have linear momentum and when two objects are on parallel courses

------------>  Sun

------------> Earth

That the larger mass body pulls on the smaller mass so that it orbits.

General Relativity says the Sun bends space around the Sun and then when hapless Earth approaches that bent space, follows in the curvature of the bent space, the ashtray analogy.

Trouble with both explanations is that linear momentum seems to never dissipate, and we have signs of some galaxies actually accelerating not losing any momentum.

So in EM Gravity, the Sun is creating a electromagnetic track around the Sun and the Sun as a magnetic force not only attracts Earth, but is like a motor pushing Earth to follow this track the Sun created.

So gravity as EM is not a weak pulling force but is rather, a Wattage Power pushing Earth around the Sun.

Take a YouTube glimpse of how you connect a battery to a magnet and then craft a piece of copper wire so that it touches on top of the battery and touches slightly on the magnet base. You have a electric motor here. As the wire constantly spins around the battery + magnet. Well, the Earth is the copper wire and the Sun is the Magnet.

Now, Newton was a very smart guy, and the physicists after Newton were very smart, but they were also, very naive, very dumb, in that they never got worried about one huge important aspect of both Newton gravity and General Relativity. A gravity based on Newton and General Relativity is a force that is constantly dissipating, slowing down. And the solar system could not survive for billions of years with gravity like that.

But EM gravity is strong, and no slowing down, provided the Cosmos is itself electrical, a Atom Totality.

The electrons in atoms are in perpetual motion, they never slow down.

If our Solar System were under Newton gravity or General Relativity, each year should have spotted a slowing down, quite recognizable slowing down, but there is none.

AP
Archimedes Plutonium
2017-06-18 18:33:10 UTC
Raw Message
So in this view of Gravity, EM Gravity, is that it is a force that does not just attract a body, but pushes the body along a Track.

It is as if the Sun creates a roadway, a track, a path in the solar system for each planet, and pushes that planet along in that pathway.

And that makes a lot of sense, for how can a tiny force of gravity, keep planets in orbit for billions of years, if all the Sun did was deflect the planets a little towards the Sun. Deflection or as in General Relativity, bending of Space is not strong nor stable enough for billions of years of orbits. And the General Relativity theory is so stupid in mechanics, because how can a planet move in front of the Sun without getting run over by the Sun, if gravity were just bending of space. For the Sun to not run over Jupiter as Jupiter steps in front of Sun's forward motion, what has to happen is EM Gravity is a Power Wattage that pushes Jupiter in its path, whether in front of the Sun or in the rear of the Sun's forward motion.

Now, someone is going to say, well, where is the battery pack of the Sun to do the pushing of planets. And, my answer is to say, the entire Cosmos is a single atom of 231Pu, and the electrons are the battery pack. Atoms are electricity and magnetism.

Now, recently the Juno probe on Jupiter discovered Jupiter had twice as much magnetism as was thought, and caused the valves of Juno to break down. And recently Cassini is doing a flyby of Saturn Rings. What NASA needs to do is park Cassini smack dab in the middle of those Rings rather than have Cassini death dive. If NASA does not, well, the fools will only have to put another probe some 10 years later to Saturn to park into the Rings. Better the fools park Cassini rather than do it 10 years from now, the same thing.

AP
Archimedes Plutonium
2017-06-19 05:25:07 UTC
Raw Message
Newsgroups: sci.physics
Date: Sun, 18 Jun 2017 10:59:15 -0700 (PDT)

Subject: Earth Sun mechanics//SFFM-stepping in front of forward motion & SUNFM
& WakeFM & PearShape concepts
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Sun, 18 Jun 2017 17:59:15 +0000

Earth Sun mechanics//SFFM-stepping in front of forward motion & SUNFM & WakeFM & PearShape concepts

Alright, a lot of concepts to get through today.

Stepping in Front of Forward Motion was already discussed. It is the accounting of Sun revolving in galaxy and the idea the Sun is moving in Space at 230km/sec, a huge speed and Earth moving in orbit at 30km/sec. The question here is, when the Sun in plowing ahead at 230km/sec, do any of the planets cross in front of Sun's forward motion at any time?

This question has never been resolved.

And I pointed to a possible answer by looking at the Saturn Rings. Do the ice particles ever revolve around Saturn's forward motion? Here the answer is that the Rings do step in front of forward motion but at a tilt of 27 degrees.

So, what does that say for the Sun and its planets as a Sun Ring? It says that all the planets Step in Front of the Sun's forward motion at some time in their orbits.

And it means that only if Gravity was EM gravity, can this sort of arrangement carry on for billions of years. Only if gravity can Power Wattage the planets around the Sun can the Solar System be stable.

So, SUNFM is Sun's Forward Motion and it is a Vector of magnitude 230 km/second, yet I do not know where the vector's direction is if we could place an arrow.

Is this SUNFM vector, is it like Saturn and its Ring, a tilt of 27 degrees? I do not know.

And, this mystery of about 27 degrees for Earth is 23 degrees tilt. Is there some magical angle in EM to give 27 degrees?

Now the next concept I have dubbed WakeFM, the Wake of Forward Motion Vector and it basically asks, since planets have forward motion and since Sun has forward motion, that is the wake have a debris field since these bodies are moving at least 230km/second. Now a rock solid planet like Earth should have a small debris field, but a gas giant or the Sun should have quite a large debris field in space. Sort of like in a motor boat, you see the wake in back of forward motion. Only the wake of the Sun and planets should be hydrogen and helium particles and other particles. All of this, as a result of the fast motion in space. I doubt anyone has observed this Wake, but for the reason, no-one ever thought it existed.

Lastly, today, I want to talk about a feature of the Sun and planets due to this 230km/sec motion in space. That fast speed should leave a huge mark on our planets and sun. A mark I call the PearShape due to this fast motion. So, if you move at 230km/sec in Space, the forward part of the body should be the largest part of a oblate spheroid, the bottom of a pear. Whereas the backend of the forward motion would be the smallest surface area. The physics is that the forward motion flattens the front end and elongates the backend. Now, has anyone seen whether the Sun is oblate spheroid-- pear shaped? And is the flattened portion the vector direction of Sun's Forward Motion?

AP

Newsgroups: sci.physics
Date: Sun, 18 Jun 2017 16:08:03 -0700 (PDT)

Subject: Earth Sun mechanics//SFFM-stepping in front of forward motion & SUNFM
& WakeFM & PearShape concepts
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Sun, 18 Jun 2017 23:08:03 +0000

So is there some sort of favoring a 27 degree axis tilt in EM theory as a body moves in a EM field?

Is the Sun tilted 27 degree in space as it orbits the Milky Way galaxy. I have forgotten the name of the star the Sun is headed for-- was it Vega?? And would that put the disc of planets at a 27 degree tilt with the Sun? If so, this would mean the Saturn Rings is a duplicate miniature model of the Sun and its planets.

AP

Newsgroups: sci.physics
Date: Sun, 18 Jun 2017 21:32:24 -0700 (PDT)

Subject: indeed, Sun ecliptic is tilted 60 degrees moving in galactic plane
Re: Earth Sun mechanics//SFFM-stepping in front of forward motion & SUNFM &
WakeFM & PearShape concepts
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Mon, 19 Jun 2017 04:32:24 +0000

indeed, Sun ecliptic is tilted 60 degrees moving in galactic plane Re: Earth Sun mechanics//SFFM-stepping in front of forward motion & SUNFM & WakeFM & PearShape concepts
So is there some sort of favoring a 27 degree axis tilt in EM theory as a body moves in a EM field?
Is the Sun tilted 27 degree in space as it orbits the Milky Way galaxy. I have forgotten the name of the star the Sun is headed for-- was it Vega?? And would that put the disc of planets at a 27 degree tilt with the Sun? If so, this would mean the Saturn Rings is a duplicate miniature model of the Sun and its planets.
Alright, from a website I was able to get some data:: Physicsforums.com
"Motion of Earth and Sun around the Milky Way

Galactic Plane

Ecliptic Plane

And the Ecliptic Plane is tilted 60 degrees from the Galactic Plane

This is called the Apex of the Sun's Way

The Apex of the Sun's Way is what I was asking for earlier in the Sun's forward motion vector. Either one is the same, the Apex of Sun's Way or the Sun's Motion Vector.

Turns out, there is a 60 degree angle of the Ecliptic tilt.

Now, Saturn Rings are tilted about 30 degrees in its orbit around the Sun

The 30 and 60 degrees are complements of 90 degrees. Depending on how you view the Rings of Saturn, you get either 30 or 60 degrees. Depending on how you view the Planets as a Ring around the Sun, they are tilted 30 or 60 degrees.

Now from that picture, looking at the North Ecliptic Pole in the Apex of Sun's Way, the Sun's Motion Vector, is that the Sun is moving in the galactic plane at 30 degrees from the Sun's north pole. So out of convenience, let us assume the Sun is Earth, then the Motion Vector of the Sun would be a spot such as the northern point of Scotland, Oslo, Stockholm, Helsinki would be the Front Edge of the Sun's Forward Motion in Space. So if Earth were the Sun, it would be flying through Space at 230km/second at a spot like Oslo.

So, if gravity is EM gravity, there must be something in EM theory that causes a tilt of 30-60 degrees. What could it be???

I am not aware of any special angle of either 30 or 60 degrees.

AP

Newsgroups: sci.math
Date: Sun, 18 Jun 2017 11:33:10 -0700 (PDT)

Subject: Re: gravity is as strong as EM, for it is EM Re: when gravity is
EM-force, the Sun Power Wattage moves Earth around the Sun
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Sun, 18 Jun 2017 18:33:11 +0000

Re: gravity is as strong as EM, for it is EM Re: when gravity is EM-force, the Sun Power Wattage moves Earth around the Sun

So in this view of Gravity, EM Gravity, is that it is a force that does not just attract a body, but pushes the body along a Track.

It is as if the Sun creates a roadway, a track, a path in the solar system for each planet, and pushes that planet along in that pathway.

And that makes a lot of sense, for how can a tiny force of gravity, keep planets in orbit for billions of years, if all the Sun did was deflect the planets a little towards the Sun. Deflection or as in General Relativity, bending of Space is not strong nor stable enough for billions of years of orbits. And the General Relativity theory is so stupid in mechanics, because how can a planet move in front of the Sun without getting run over by the Sun, if gravity were just bending of space. For the Sun to not run over Jupiter as Jupiter steps in front of Sun's forward motion, what has to happen is EM Gravity is a Power Wattage that pushes Jupiter in its path, whether in front of the Sun or in the rear of the Sun's forward motion.

Now, someone is going to say, well, where is the battery pack of the Sun to do the pushing of planets. And, my answer is to say, the entire Cosmos is a single atom of 231Pu, and the electrons are the battery pack. Atoms are electricity and magnetism.

Now, recently the Juno probe on Jupiter discovered Jupiter had twice as much magnetism as was thought, and caused the valves of Juno to break down. And recently Cassini is doing a flyby of Saturn Rings. What NASA needs to do is park Cassini smack dab in the middle of those Rings rather than have Cassini death dive. If NASA does not, well, the fools will only have to put another probe some 10 years later to Saturn to park into the Rings. Better the fools park Cassini rather than do it 10 years from now, the same thing.

AP

Newsgroups: sci.physics
Date: Sun, 18 Jun 2017 21:56:01 -0700 (PDT)

Subject: explaining the origin of Earth's own 23 degree tilt //SFFM-stepping
in front of forward motion & SUNFM & WakeFM & PearShape concepts
From: Archimedes Plutonium <***@gmail.com>
Injection-Date: Mon, 19 Jun 2017 04:56:01 +0000

explaining the origin of Earth's own 23 degree tilt //SFFM-stepping in front of forward motion & SUNFM & WakeFM & PearShape concepts

Might that angle of 60 versus 30 degrees in EM theory come from the electroscope demonstration? The maximum angle of tilt in a electroscope is 60 degrees. Reasoning: maximum interaction angle is 60 degrees where charge cannot occupy same space.

So I think the electroscope answers the 60 or 30 degree tilt.

Now Earth tilt is 23 degrees but that is close enough to 30. The 60 and 30 are ideal cases.

Now in astronomy, many have tried to explain how Earth got its tilt whereas Mercury, Venus, Mars, Jupiter have none. And the answer is not the crazy bolide hitting Earth and being the Moon. The answer is, like Saturn's Ring, the answer is EM gravity and the Earth's magnetic field is so enormous for its size that it is a Magnetic field tilt. The Earth's own magnetism causes it to tilt. Now if the Earth had just some more magnetism, it could tilt all the way to 30 degrees rather than 23 degrees.

Now you may ask why not Jupiter tilting on its axis. And the answer is that Jupiter is just too massive.

But now I want to ask about the Wake of forward motion of Sun since its Vector Motion is 30 degrees off Sun's north pole. Can we see, or observe a Debris Field Wake from the South Pole of the Sun? Does the Sun leave a stream of particle debris starting at the South Pole? And, is the North Pole of the Sun have a larger surface area versus south pole of sun? A pear shaped Sun, oblate spheroid in favor of the north pole of sun?

Is any of that true?

AP