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How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here

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  How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 15:07 -0700
    How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 17:25 -0700
    Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Serigo <invalid@invalid.com> - 2016-09-18 20:32 -0500
      Magnetic Field in most-everything Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 19:38 -0700
        Re: Magnetic Field in most-everything Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-09-19 15:51 +0000
          people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 13:07 -0700
            Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors Serigo <invalid@invalid.com> - 2016-09-19 16:15 -0500
              Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 14:29 -0700
                Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors Serigo <invalid@invalid.com> - 2016-09-19 19:51 -0500
                  Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 19:41 -0700
                    Re: Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant Serigo <invalid@invalid.com> - 2016-09-20 09:55 -0500
                      Re: Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant "hanson" <hanson@quick.net> - 2016-09-20 09:03 -0700
            Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-09-19 22:26 +0000
              This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 19:23 -0700
                Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand Serigo <invalid@invalid.com> - 2016-09-19 21:41 -0500
                  (A*R)' = A'*R  +  R'*A and derivative (V*R^-1)' = V'*R^-1  +  V*(R^-1)' Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 22:12 -0700
                    derivative of magnetic field Re: (A*R)' = A'*R  +  R'*A and derivative (V*R^-1)' = V'*R^-1  +  V*(R^-1)' Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 23:45 -0700
                    Re: (A*R)' = A'*R + R'*A and derivative (V*R^-1)' = V'*R^-1 + V*(R^-1)' Serigo <invalid@invalid.com> - 2016-09-20 10:09 -0500
                Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-09-20 04:08 +0000
                  Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand Serigo <invalid@invalid.com> - 2016-09-20 10:28 -0500
                    Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-09-20 16:40 +0000
      Someone in Old Physics trying to flesh out Faraday's law Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 20:03 -0700
        Re: Someone in Old Physics trying to flesh out Faraday's law Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Serigo <invalid@invalid.com> - 2016-09-18 22:31 -0500
          Old Physics Faraday law math does not match the experiment Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 20:51 -0700
            math that matches is the Product Rule in Differentiation that gives a Lenz law inside of Faraday law Re: Old Physics Faraday law math does not match the experiment Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-19 02:47 -0700
      trying to teach a brainwashed that velocity on Magnetic Field is just another Magnetic Field Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2016-09-18 20:20 -0700

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#597449 — How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here

Voltage is the (1) Electric Potential, the (2) Potential Difference and (3) Electromotive Force and all with the Units of W/A = kg*m^2/A*s^3 

Voltage in Revised Maxwell theory is the Electric Field, for there is no electric field apart from voltage. When you speak of Voltage, you are speaking of Electric Field. 

Capacitance = farad = C/V = A^2*s^4 / kg*m^2 

Electrical Resistance = ohm = kg*m^2 /A^2*s^3 

Conductance = A/V = A^2*s^3 / kg*m^2 

Magnetic Flux = V*s = kg*m^2 /A*s^2 

Magnetic Field = tesla = kg /A*s^2 

Inductance =  kg*m^2 /A^2*s^2 

In Old Physics, the Faraday law is - curlxE = dB     

And one can immediately sense how wrong that is because it does not even give a opposing Magnetic Field to the changing magnetic field, as if a negative sign is sufficient to cover Lenz law.

So let us check this out in Units.

In differental Calculus we have a Thrusting Bar Magnet. The Magnetic Field is 

kg /A*s^2 

To get Current you need V=iR, you need i = V/R

So you need differention upon V and R^-1

V is kg*m^2/A*s^3 

and R is kg*m^2 /A^2*s^3

Notice that in all of these parameters listed that the Magnetic Field is a component thereof

For example, in R, it can be broken down into kg/A*s^2 then factor of m^2/A*s

V is kg/A*s^2 then factor of m^2/s

So, as we differentiate VR^-1 we end up with at least two terms on the rightside of the equation for Faraday's law and thus, we have the Lenz's Opposing Magnetic Field.

            . \ .  . | .   /. 
           . . \. . .|. . /. . 
              ..\....|.../... 
               ::\:::|::/:: 
---------------      ------------- 
--------------- (Y) ------------- 
---------------      -------------- 
               ::/:::|::\:: 
              ../....|...\... 
           . . /. . .|. . \. . 
            . / .  . | .   \ . 


Research 
Disclaimer: Due to the unconventional and speculative nature of this posting and thread, it would be inadvisable for students to apply any of the contents to their school course work.
 
http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies 

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.      

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe         
Archimedes Plutonium 

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#597477

The product rule of differentiation gives us three terms for the right side of Faraday law, a current, a Lenz magnetic field, and a constant for a spin term.

iPhone post 

AP

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#597484

On 9/18/2016 5:07 PM, Archimedes Plutonium wrote:
>
> Voltage is the (1) Electric Potential, the (2) Potential Difference
> and (3) Electromotive Force and all with the Units of W/A =
> kg*m^2/A*s^3
>
> Voltage in Revised Maxwell theory is the Electric Field, for there is
> no electric field apart from voltage. When you speak of Voltage, you
> are speaking of Electric Field.

nope.
Electric field is a vector field.
Voltage is a scalar number.

Electric Field units are V/m
   Voltage units are V

you have apples and oranges, mixed up.

>
> Capacitance = farad = C/V = A^2*s^4 / kg*m^2
>
> Electrical Resistance = ohm = kg*m^2 /A^2*s^3
>
> Conductance = A/V = A^2*s^3 / kg*m^2
>
> Magnetic Flux = V*s = kg*m^2 /A*s^2
>
> Magnetic Field = tesla = kg /A*s^2
>
> Inductance =  kg*m^2 /A^2*s^2
>
> In Old Physics, the Faraday law is - curlxE = dB

nope, Maxwell-Faraday equation is  =>  ∇ × E = -dB/dt


> And one can immediately sense how wrong that is because it does not
> even give a opposing Magnetic Field to the changing magnetic field,
> as if a negative sign is sufficient to cover Lenz law.

no, that is the negative sign in Faraday's law of induction.

Volt generated = -N (d BA)/dt    see? no curl.



> So let us check this out in Units.
>
> In differental Calculus we have a Thrusting Bar Magnet. The Magnetic
> Field is
>
> kg /A*s^2
>
> To get Current you need V=iR, you need i = V/R
>
> So you need differention upon V and R^-1
>
> V is kg*m^2/A*s^3
>
> and R is kg*m^2 /A^2*s^3
>
> Notice that in all of these parameters listed that the Magnetic Field
> is a component thereof
>
> For example, in R, it can be broken down into kg/A*s^2 then factor of
> m^2/A*s
>
> V is kg/A*s^2 then factor of m^2/s

> So, as we differentiate VR^-1 we end up with at least two terms on
> the rightside of the equation for Faraday's law and thus, we have the
> Lenz's Opposing Magnetic Field.
>
> . \ .  . | .   /. . . \. . .|. . /. . ..\....|.../... ::\:::|::/::
> ---------------      ------------- --------------- (Y) -------------
>  ---------------      -------------- ::/:::|::\:: ../....|...\... . .
> /. . .|. . \. . . / .  . | .   \ .
>
>
> Research Disclaimer: Due to the unconventional and speculative nature
> of this posting and thread, it would be inadvisable for students to
> apply any of the contents to their school course work.
>
> http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one
> big atom 
where dots of the electron-dot-cloud are galaxies
>
> I re-opened the old newsgroup PAU of 1990s and there one can read my
> recent posts without the hassle of spammers, off-topic-misfits,
> front-page-hogs, stalking mockers, suppression-bullies, and
> demonizers.
>
> https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe
>  Archimedes Plutonium
>

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#597493 — Magnetic Field in most-everything Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here

On Sunday, September 18, 2016 at 8:33:03 PM UTC-5, Serigo wrote:
> On 9/18/2016 5:07 PM, Archimedes Plutonium wrote:
> >
> > Voltage is the (1) Electric Potential, the (2) Potential Difference
> > and (3) Electromotive Force and all with the Units of W/A =
> > kg*m^2/A*s^3
> >
> > Voltage in Revised Maxwell theory is the Electric Field, for there is
> > no electric field apart from voltage. When you speak of Voltage, you
> > are speaking of Electric Field.
> 
> nope.
> Electric field is a vector field.
> Voltage is a scalar number.

Wrong again. The Magnetic Field is a vector Space: kg /A*s^2
And since that vector space is in all the below listed parameters, some of them inverted, means all of them are vectors.

Voltage is  W/A = kg*m^2/A*s^3 

Capacitance = farad = C/V = A^2*s^4 / kg*m^2 

Electrical Resistance = ohm = kg*m^2 /A^2*s^3 

Conductance = A/V = A^2*s^3 / kg*m^2 

Magnetic Flux = V*s = kg*m^2 /A*s^2 

Magnetic Field = tesla = kg /A*s^2 

Inductance =  kg*m^2 /A^2*s^2 

All of the above are multiples of the vector space Magnetic Field, hence, all are vectors.

You got bamboozled by Old Physics

AP

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#597542 — Re: Magnetic Field in most-everything Re: How to gather the missing terms in Faraday Law, and yes, Maxwell goofed here

Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:

>On Sunday, September 18, 2016 at 8:33:03 PM UTC-5, Serigo wrote:

>> nope.
>> Electric field is a vector field.
>> Voltage is a scalar number.

>Wrong again. The Magnetic Field is a vector Space: kg /A*s^2
>And since that vector space is in all the below listed parameters, some of them inverted, means all of them are vectors.

...

>All of the above are multiples of the vector space Magnetic Field, hence, all are vectors.

No, whether a derived unit is a vector or a scalar depends on
how it is derived. If the dot product is taken between two
vectors, you get a scalar.  The cross product of two vectors
is another vector.  For example, voltage is the dot product
of the distance vector and the electric field vector, thus it
is a scalar.  A (3D) vector has a direction, while a scalar is
just a value.

>You got bamboozled by Old Physics

>AP
>

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#597598 — people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors

On Monday, September 19, 2016 at 10:52:42 AM UTC-5, Michael Moroney wrote:
> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
> 
> >On Sunday, September 18, 2016 at 8:33:03 PM UTC-5, Serigo wrote:
> 
> >> nope.
> >> Electric field is a vector field.
> >> Voltage is a scalar number.
> 
> >Wrong again. The Magnetic Field is a vector Space: kg /A*s^2
> >And since that vector space is in all the below listed parameters, some of them inverted, means all of them are vectors.
> 
> ...
> 
> >All of the above are multiples of the vector space Magnetic Field, hence, all are vectors.
> 
> No, whether a derived unit is a vector or a scalar depends on
> how it is derived. If the dot product is taken between two
> vectors, you get a scalar.  The cross product of two vectors
> is another vector.  For example, voltage is the dot product
> of the distance vector and the electric field vector, thus it
> is a scalar.  A (3D) vector has a direction, while a scalar is
> just a value.
> 
> >You got bamboozled by Old Physics
> 
> >AP

The Magnetic Field is always a vector, regardless of any operation placed upon it. That is the meaning of dipole. Something like a parrot of understanding cannot grasp for themselves until told. Dipole means Magnetism requires a Field to mediate the two poles, and hence a vector is involved.

And no matter what operations you place on magnetism you never get rid of the vector field unless you get rid of the magnetism itself.

You still cling to the stupidity of a Electric Field which would be in your imagination (m/s) times kg /A*s^2. Unable to fathom that your electric-field is just a Magnetic Field times speed. A vector times speed.

And you are unable to fathom that voltage can never be scalar but always vector since voltage is kg*m^2/A*s^3 which is kg/A*s^2 times m^2/s, which is Magnetic Field vector times the factor m^2/s.
You are a parrot of Old Physics. You failed physics because never can you tell if something is right or wrong, and constantly look to what is in print as your Absolute Truth Guidance, not experiments to guide.

All these terms are vectors because they all are multiples of the Magnetic Field Vector

Voltage is  W/A = kg*m^2/A*s^3 

Capacitance = farad = C/V = A^2*s^4 / kg*m^2 

Electrical Resistance = ohm = kg*m^2 /A^2*s^3 

Conductance = A/V = A^2*s^3 / kg*m^2 

Magnetic Flux = V*s = kg*m^2 /A*s^2 

Magnetic Field = tesla = kg /A*s^2 

Inductance =  kg*m^2 /A^2*s^2 

When you do the thrusting bar magnet into a closed loop wire for Faraday, you as parrot will beep beep that the Maxwell Equations cover that experiment, but you have missing the Lenz opposing magnetic field. You are not capable of even recognizing that the experiment does not match the math, and that is why you are a failure of physics. As Abian used to often say, you accept only that which Mother Superior tells you to believe and accept.

AP

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#597609 — Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors

On 9/19/2016 3:07 PM, Archimedes Plutonium wrote:
> On Monday, September 19, 2016 at 10:52:42 AM UTC-5, Michael Moroney
> wrote:
>> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
>>
>>> On Sunday, September 18, 2016 at 8:33:03 PM UTC-5, Serigo wrote:
>>
>>>> nope. Electric field is a vector field. Voltage is a scalar
>>>> number.
>>
>>> Wrong again. The Magnetic Field is a vector Space: kg /A*s^2 And
>>> since that vector space is in all the below listed parameters,
>>> some of them inverted, means all of them are vectors.
>>

>>
>>> All of the above are multiples of the vector space Magnetic
>>> Field, hence, all are vectors.
>>
>> No, whether a derived unit is a vector or a scalar depends on how
>> it is derived. If the dot product is taken between two vectors, you
>> get a scalar.  The cross product of two vectors is another vector.
>> For example, voltage is the dot product of the distance vector and
>> the electric field vector, thus it is a scalar.  A (3D) vector has
>> a direction, while a scalar is just a value.
>>
>>> You got bamboozled by Old Physics
>>
>>> AP
>
> The Magnetic Field is always a vector, regardless of any operation
> placed upon it.

for our readers;

  dot product and cross product are operations used in Vector Analysis 
(vector calculus), which are in quite a few branches of science.

there are also the differential operators or del operator of
	grad
	curl
	div
	vector laplician
	Laplician
	
There is also Tensor Calculus, more difficult, it extends vector 
calculus to tensor fields, which can very over time, manifolds

most physicist and EEs should have some of these courses,


wiki has a lot on the math of Maxwell
https://en.wikipedia.org/wiki/Maxwell%27s_equations

and free book;

https://en.wikipedia.org/wiki/Book:Maxwell%27s_equations

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#597613 — Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors

On Monday, September 19, 2016 at 4:15:57 PM UTC-5, Serigo wrote:

Serigo, now, let us say you never learned Ohm's law, but you did learn Maxwell Equations. Would you be able to get Ohm's law or some equivalent out of the Maxwell Equations?

AP

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#597642 — Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors

On 9/19/2016 4:29 PM, Archimedes Plutonium wrote:
> On Monday, September 19, 2016 at 4:15:57 PM UTC-5, Serigo wrote:
>
> Serigo, now, let us say you never learned Ohm's law, but you did
> learn Maxwell Equations. Would you be able to get Ohm's law or some
> equivalent out of the Maxwell Equations?
>
> AP
>

you can't derive Ohm's Law from Maxwell Equations because of one 
important thing - you need to know how to treat a bunch of charges 
moving at random, which isn't contained in Maxwell equation.

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#597648 — Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant

On Monday, September 19, 2016 at 7:51:53 PM UTC-5, Serigo wrote:
> On 9/19/2016 4:29 PM, Archimedes Plutonium wrote:
> > On Monday, September 19, 2016 at 4:15:57 PM UTC-5, Serigo wrote:
> >
> > Serigo, now, let us say you never learned Ohm's law, but you did
> > learn Maxwell Equations. Would you be able to get Ohm's law or some
> > equivalent out of the Maxwell Equations?
> >
> > AP
> >
> 
> you can't derive Ohm's Law from Maxwell Equations because of one 
> important thing - you need to know how to treat a bunch of charges 
> moving at random, which isn't contained in Maxwell equation.

Well, then try explaining that of course whenever you have a electric current, you must have a voltage and whenever you have voltage, you must have resistance.

So, are you saying that Maxwell Equations cannot handle either voltage nor resistance nor both.

Seems a far cry from Feynman's assessment that all of EM to date is explained by the Maxwell Equations, yet here it cannot even explain voltage and resistance.

In my program, what I am doing is saying that all the Forces of Physics are a EM force, no other forces except EM exist, and that means, further, that Resistance in Ohm's law is the same as Lenz law opposing magnetic field.

All friction in Nature, all resistance, all impedance are all coming from Lenz's opposing magnetic field. As a car coasting and finally comes to rest by friction, every one of those friction acts is a Lenz law magnetic field rearing its ugly head to oppose further motion.

So, instead or your opinion that Resistance is too chaotic, my opinion is that the Maxwell Equations as given by Maxwell are seriously flawed and need the trashcan, with a set of new equations taking place.

Lenz law, when represented in Maxwell Equations is Resistance in Ohm's law and is imbedded in the Maxwell equations once those equations are proper and correct.

Look, here is an outline form of Faraday and Ampere laws with R, resistance in them:


The Faraday Law should look like this: 

(fg)' = f'g + g'f + Constant 

The Ampere Law should look like this also, where the rightside of equation has 3 terms. 

For Faraday's law it should look like this: 

(V*R^-1)' = A*R + B + constant where A is current, B is magnetic field and R is resistance and V is volts. 

The Ampere Law should look like this: 

(A*R)' = B + V*A^-1 + constant 

AP

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#597693 — Re: Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant

On 9/19/2016 9:41 PM, Archimedes Plutonium wrote:
> On Monday, September 19, 2016 at 7:51:53 PM UTC-5, Serigo wrote:
>> On 9/19/2016 4:29 PM, Archimedes Plutonium wrote:
>>> On Monday, September 19, 2016 at 4:15:57 PM UTC-5, Serigo wrote:
>>>
>>> Serigo, now, let us say you never learned Ohm's law, but you did
>>> learn Maxwell Equations. Would you be able to get Ohm's law or
>>> some equivalent out of the Maxwell Equations?
>>>
>>> AP
>>>
>>
>> you can't derive Ohm's Law from Maxwell Equations because of one
>> important thing - you need to know how to treat a bunch of charges
>>  moving at random, which isn't contained in Maxwell equation.
>
> Well, then try explaining that of course whenever you have a electric
> current, you must have a voltage and whenever you have voltage, you
> must have resistance.

      but those are not the same type of electric fields.

 From our analysis it appears that the electric field in Ohm’s law (7) 
must be something else than the electric field defined by Maxwell’s 
equations (1 – 6). In particular, it should not be confused with the 
static electric field as defined by (1) and (5) which is very obvious from 
our result in Sec. 2.

The microscopic picture of current flow suggests that electric currents 
should be conceived as moving electrons which are accelerated by an 
average field and stopped again by collisions. Each electron produces a 
magnetic field according to (2) and (6) which varies in time. As a result 
fluctuating electric fields are induced according to (3). On
average these fields seem to add up to a quasistatic macroscopic field 
which ultimately enters into Ohm’s law and may be expressed as the 
gradient of a potential. This is, however, not the electrostatic 
potential which is produced by a charge density according to (1) and 
(5). The divergence of the average macroscopic field in Ohm’s law
should, of course, vanish – in agreement with (10) – as it is a 
rotational field created microscopically by induction

https://www.researchgate.net/publication/282672796_Ohm%27s_Law_and_Maxwell%27s_Equations

>
> So, are you saying that Maxwell Equations cannot handle either
> voltage nor resistance nor both.

are you talking fields or circuits ?  the resistance or impeadance of 
free space is 377 ohms, Zo = sqroot(muo/eo)  Maxwell uses muo and eo, in 
his equations, but that does not apply to electrons jamming into 
particals and restarting up again like in a resistor

>
> Seems a far cry from Feynman's assessment that all of EM to date is
> explained by the Maxwell Equations, yet here it cannot even explain
> voltage and resistance.

you are applying EM to resistors, no can do.
if Feynman said that he is wrong.


>
> In my program, what I am doing is saying that all the Forces of
> Physics are a EM force, no other forces except EM exist, and that
> means, further, that Resistance in Ohm's law is the same as Lenz law
> opposing magnetic field.

gravity is not EM

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#597697 — Re: Faraday law looks like (V*R^-1)' = A*R + B + constant; Ampere law looks like (A*R)' = B + V*A^-1 + constant

"Serigo" <invalid@invalid.com> wrote in message
news:nrrim1$1d9v$1@gioia.aioe.org...
> On 9/19/2016 9:41 PM, Archimedes Plutonium wrote:
>> On Monday, September 19, 2016 at 7:51:53 PM UTC-5, Serigo wrote:
>>> On 9/19/2016 4:29 PM, Archimedes Plutonium wrote:
>>>> On Monday, September 19, 2016 at 4:15:57 PM UTC-5, Serigo wrote:
>>>>
>>>> Serigo, now, let us say you never learned Ohm's law, but you did
>>>> learn Maxwell Equations. Would you be able to get Ohm's law or
>>>> some equivalent out of the Maxwell Equations?
>>>>
>>>> AP
>>>>
>>>
>>> you can't derive Ohm's Law from Maxwell Equations because of one
>>> important thing - you need to know how to treat a bunch of charges
>>>  moving at random, which isn't contained in Maxwell equation.
>>
>> Well, then try explaining that of course whenever you have a electric
>> current, you must have a voltage and whenever you have voltage, you
>> must have resistance.
>
>      but those are not the same type of electric fields.
>
> From our analysis it appears that the electric field in Ohm’s law (7) must
> be something else than the electric field defined by Maxwell’s equations
> (1 – 6). In particular, it should not be confused with the static electric
> field as defined by (1) and (5) which is very obvious from our result in
> Sec. 2.
>
> The microscopic picture of current flow suggests that electric currents
> should be conceived as moving electrons which are accelerated by an
> average field and stopped again by collisions. Each electron produces a
> magnetic field according to (2) and (6) which varies in time. As a result
> fluctuating electric fields are induced according to (3). On
> average these fields seem to add up to a quasistatic macroscopic field which
> ultimately enters into Ohm’s law and may be expressed as the gradient of a
> potential. This is, however, not the electrostatic potential which is
> produced by a charge density according to (1) and (5). The divergence of
> the average macroscopic field in Ohm’s law
> should, of course, vanish – in agreement with (10) – as it is a rotational
> field created microscopically by induction
>
> https://www.researchgate.net/publication/282672796_Ohm%27s_Law_and_Maxwell%27s_Equations
>
>>
>> So, are you saying that Maxwell Equations cannot handle either
>> voltage nor resistance nor both.
>
> are you talking fields or circuits ?  the resistance or impeadance of free
> space is 377 ohms, Zo = sqroot(muo/eo)  Maxwell uses muo and eo, in his
> equations, but that does not apply to electrons jamming into particals and
> restarting up again like in a resistor
>
>>
>> Seems a far cry from Feynman's assessment that all of EM to date is
>> explained by the Maxwell Equations, yet here it cannot even explain
>> voltage and resistance.
>
> you are applying EM to resistors, no can do.
> if Feynman said that he is wrong.
>
>
>>
>> In my program, what I am doing is saying that all the Forces of
>> Physics are a EM force, no other forces except EM exist, and that
>> means, further, that Resistance in Ohm's law is the same as Lenz law
>> opposing magnetic field.
>
> gravity is not EM
>
hanson wrote:
Serge, check out Electro-Gavity.
__ G = (e/m_e)^2  /  [3 * (pi^2) * (a^3) * (N_A)^2)] __
>
<http://tinyurl.com/Electrogravity>
>
enjoy
>

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#597624 — Re: people that do not understand that if you have a Magnetic Field which is always a vector, that multiples thereof are also vectors

Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:

>On Monday, September 19, 2016 at 10:52:42 AM UTC-5, Michael Moroney wrote:
>> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:

>> >All of the above are multiples of the vector space Magnetic Field, hence,
>> >all are vectors.

>> No, whether a derived unit is a vector or a scalar depends on
>> how it is derived. If the dot product is taken between two
>> vectors, you get a scalar.  The cross product of two vectors
>> is another vector.  For example, voltage is the dot product
>> of the distance vector and the electric field vector, thus it
>> is a scalar.  A (3D) vector has a direction, while a scalar is
>> just a value.
>>=20
>> >You got bamboozled by Old Physics
>>=20
>> >AP

>The Magnetic Field is always a vector, regardless of any operation placed
>upon it. That is the meaning of dipole. Something like a parrot of
>understanding cannot grasp for themselves until told. Dipole means
>Magnetism requires a Field to mediate the two poles, and hence a vector
>is involved.

The magnetic field is a vector, yes, but not for the reasons you give.
I suspect you really don't understand what a vector is, or what the dot
products and cross products of vectors are.

>And no matter what operations you place on magnetism you never get rid of
>the vector field unless you get rid of the magnetism itself.

For the cross product, yes, not the dot product.  Again, do you even know
what they are?

>You still cling to the stupidity of a Electric Field which would be in your
> imagination (m/s) times kg /A*s^2. Unable to fathom that your electric-field
>is just a Magnetic Field times speed. A vector times speed.

You can't just "times" two vectors. You can take the dot product, or the cross
product, or create a 2d matrix/tensor from them.

The electric field is a vector, the electric potential is a scalar.
They are different. Electrodynamics requires and uses both.

>All these terms are vectors because they all are multiples of the Magnetic
>Field Vector

Some are scalars, some vectors, depending on how they are derived.
The base units are an oversimplification, they don't reveal how the
unit was derived.  In the list of base units, both dot products
and cross products are represented by multiplication, so information
is lost.  For example, torque and energy both have the same units,
(kg*m^2/s^2) and both can be derived from force and distance, but one
is a vector and the other a scalar. And they are very different.

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#597647 — This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand

On Monday, September 19, 2016 at 5:27:16 PM UTC-5, Michael Moroney wrote:
> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
> 
(snip)
> 
> >The Magnetic Field is always a vector, regardless of any operation placed
> >upon it. That is the meaning of dipole. Something like a parrot of
> >understanding cannot grasp for themselves until told. Dipole means
> >Magnetism requires a Field to mediate the two poles, and hence a vector
> >is involved.
> 
> The magnetic field is a vector, yes, but not for the reasons you give.

Oh, was it Serigo that whispered in your ear that you were dumb wrong by thinking a magnetic field is a vector only on how it was derived. Eh, what stupidity you have, and why you failed physics.

> I suspect you really don't understand what a vector is, or what the dot
> products and cross products of vectors are.
> 

So, when you are found stupid wrong, your comeback is to say "I think you do not understand such and such." You are a stupid twisted fool, a fool that failed physics, and no more belongs in sci.physics than do the spammers.

Hope you are not a science teacher in school.

What a fruitcake,,,,,,,,,,

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#597649 — Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand

On 9/19/2016 9:23 PM, Archimedes Plutonium wrote:
> On Monday, September 19, 2016 at 5:27:16 PM UTC-5, Michael Moroney
> wrote:
>> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
>>
> (snip)
>>
>>> The Magnetic Field is always a vector, regardless of any
>>> operation placed upon it. That is the meaning of dipole.
>>> Something like a parrot of understanding cannot grasp for
>>> themselves until told. Dipole means Magnetism requires a Field to
>>> mediate the two poles, and hence a vector is involved.
>>
>> The magnetic field is a vector, yes, but not for the reasons you
>> give.
>
> Oh, was it Serigo that whispered in your ear that you were dumb wrong
> by thinking a magnetic field is a vector only on how it was derived.
> Eh, what stupidity you have, and why you failed physics.

electric field, magnetic field, are vector fields, if you dont know 
that, you have no chance at all with Maxwells equations,

Your old posts from 2012 show you were working on maxwell back then, how 
did it go ?

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#597655 — (A*R)' = A'*R + R'*A and derivative (V*R^-1)' = V'*R^-1 + V*(R^-1)'

On Monday, September 19, 2016 at 9:41:53 PM UTC-5, Serigo wrote:
> On 9/19/2016 9:23 PM, Archimedes Plutonium wrote:
> > On Monday, September 19, 2016 at 5:27:16 PM UTC-5, Michael Moroney
> > wrote:
> >> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
> >>
> > (snip)
> >>
> >>> The Magnetic Field is always a vector, regardless of any
> >>> operation placed upon it. That is the meaning of dipole.
> >>> Something like a parrot of understanding cannot grasp for
> >>> themselves until told. Dipole means Magnetism requires a Field to
> >>> mediate the two poles, and hence a vector is involved.
> >>
> >> The magnetic field is a vector, yes, but not for the reasons you
> >> give.
> >
> > Oh, was it Serigo that whispered in your ear that you were dumb wrong
> > by thinking a magnetic field is a vector only on how it was derived.
> > Eh, what stupidity you have, and why you failed physics.
> 
> electric field, magnetic field, are vector fields, if you dont know 
> that, you have no chance at all with Maxwells equations,
> 
> Your old posts from 2012 show you were working on maxwell back then, how 
> did it go ?

So, what happens when you take the derivative of V*R^-1, V is voltage, R is resistance

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

Likewise what happens with

(A*R)' = A'*R  +  R'*A  where A is current and R is resistance.

Note: I will make R be that of the Lenz B field, so in the end, R is a magnetic field, pure magnetic field.

Can you get any traction Serigo?

AP

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#597661 — derivative of magnetic field Re: (A*R)' = A'*R + R'*A and derivative (V*R^-1)' = V'*R^-1 + V*(R^-1)'

On Tuesday, September 20, 2016 at 12:12:08 AM UTC-5, Archimedes Plutonium wrote:
(snipped)
> 
> So, what happens when you take the derivative of V*R^-1, V is voltage, R is resistance
> 
> (V*R^-1)' = V'*R^-1  +  V*(R^-1)'
> 
> Likewise what happens with
> 
> (A*R)' = A'*R  +  R'*A  where A is current and R is resistance.
> 
> Note: I will make R be that of the Lenz B field, so in the end, R is a magnetic field, pure magnetic field.

Alright, I made considerable progress in just one day.

If we use these as data:

(1) current through a capacitor is the derivative of the voltage across the capacitor with respect to time

(2) derivative of current is just current

(3) derivative of resistance is a constant K since resistance is independent of V or A

(4) derivative of B is what?

So now apply these I have:

Faraday Law (V*R^-1)' = V'*R^-1  +  V*(R^-1)' = A*R^-1 + Vk  = A*B  + Vk

Ampere Law (A*R)' = A'*R  +  R'*A  =  A*R  + kA  = A*B  + kA

Looks promising for a first stab chance. I need derivative of magnetic field.

Serigo, what is Derivative of Magnetic Field?

AP

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#597694 — Re: (A*R)' = A'*R + R'*A and derivative (V*R^-1)' = V'*R^-1 + V*(R^-1)'

On 9/20/2016 12:12 AM, Archimedes Plutonium wrote:
> On Monday, September 19, 2016 at 9:41:53 PM UTC-5, Serigo wrote:
>> On 9/19/2016 9:23 PM, Archimedes Plutonium wrote:
>>> On Monday, September 19, 2016 at 5:27:16 PM UTC-5, Michael Moroney
>>> wrote:
>>>> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
>>>>
>>> (snip)
>>>>
>>>>> The Magnetic Field is always a vector, regardless of any
>>>>> operation placed upon it. That is the meaning of dipole.
>>>>> Something like a parrot of understanding cannot grasp for
>>>>> themselves until told. Dipole means Magnetism requires a Field to
>>>>> mediate the two poles, and hence a vector is involved.

>>>> The magnetic field is a vector, yes, but not for the reasons you
>>>> give.
>>>
>>> Oh, was it Serigo that whispered in your ear that you were dumb wrong
>>> by thinking a magnetic field is a vector only on how it was derived.
>>> Eh, what stupidity you have, and why you failed physics.
>>
>> electric field, magnetic field, are vector fields, if you dont know
>> that, you have no chance at all with Maxwells equations,
>>
>> Your old posts from 2012 show you were working on maxwell back then, how
>> did it go ?
>
> So, what happens when you take the derivative of V*R^-1, V is voltage, R is resistance



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

dV/R - V*dR/r^2    now what ?

with respect to time ?




>
> Likewise what happens with
>
> (A*R)' = A'*R  +  R'*A  where A is current and R is resistance.
>
> Note: I will make R be that of the Lenz B field, so in the end, R is a magnetic field, pure magnetic field.
>
> Can you get any traction Serigo?
>
> AP
>

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#597653 — Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand

Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:

>On Monday, September 19, 2016 at 5:27:16 PM UTC-5, Michael Moroney wrote:
>> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
>> 
>(snip)
>> 
>> >The Magnetic Field is always a vector, regardless of any operation placed
>> >upon it. That is the meaning of dipole. Something like a parrot of
>> >understanding cannot grasp for themselves until told. Dipole means
>> >Magnetism requires a Field to mediate the two poles, and hence a vector
>> >is involved.
>> 
>> The magnetic field is a vector, yes, but not for the reasons you give.

>Oh, was it Serigo that whispered in your ear that you were dumb wrong 
>by thinking a magnetic field is a vector only on how it was derived.

Nope, I always knew that the magnetic field and the electric field were
vectors.  They have a magnitude and a direction, which is how vectors
appear to us in 3 dimensional space.

>Eh, what stupidity you have, and why you failed physics.

Nope, I have a degree in electrical engineering, which requires that I
pass physics (which I aced).  Not just the simple fact that you need to
pass certain mandatory physics courses in order to qualify for the degree,
you simply have to actually know the physics in order to have any chance
of passing all the electrical engineering courses you need for the degree.

Meanwhile, you have no degree, much less a physics degree, so if anyone
is a failure at physics, it is you, with all your pretending.

>> I suspect you really don't understand what a vector is, or what the dot
>> products and cross products of vectors are.

>So, when you are found stupid wrong, your comeback is to say "I think 
>you do not understand such and such." You are a stupid twisted fool, a 
>fool that failed physics, and no more belongs in sci.physics than do the 
>spammers.

Instead of addressing my comment, you launch into a personal attack.
That proves to me that you do not, in fact, know what a vector, a
dot product or a cross product are.

>Hope you are not a science teacher in school.

>What a fruitcake,,,,,,,,,,

Meanwhile, while you babble here about corrupting Maxwell's Equations,
scientists and engineers will continue to use the *real* Maxwell's
Equations and physics laws derived from the real Maxwell's Equations in
whole or part, to design and build all kinds of electromagnetic wizardry.
Just as they have been doing very successfully for some 150 years now.
Soon, you'll move on and start babbling about something else, and your
corrupted Maxwell's Equations babble will quickly be forgotten. Meanwhile,
the scientists and engineers, who have never heard of you, will continue
to use the actual Maxwell's Equations and their derivitives to develop
even more electromagnetic wizardry.  The actual Maxwell's Equations are
one of the most successful developments in physics ever.

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#597695 — Re: This moron thought a magnetic field is only a vector dependent on how it was derived Re: people that do not understand

On 9/19/2016 11:08 PM, Michael Moroney wrote:
> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
>
>> On Monday, September 19, 2016 at 5:27:16 PM UTC-5, Michael Moroney wrote:
>>> Archimedes Plutonium <plutonium.archimedes@gmail.com> writes:
>>>
>> (snip)
>>>

>
> Nope, I have a degree in electrical engineering, which requires that I
> pass physics (which I aced).  Not just the simple fact that you need to
> pass certain mandatory physics courses in order to qualify for the degree,
> you simply have to actually know the physics in order to have any chance
> of passing all the electrical engineering courses you need for the degree.


right on! undergrad physics was hard, full time homework and class ate 
up 20hrs a week, others would party, but we did physics homework all 
Saturday and night and all Sunday too, (along with the other ee HW) got 
only about 6 hours off on weekends.  Berkeley Physics, you had to get 
A+Bs or you were washed out, C was bad.  3 men enter, one man graduate. 
NO slackers. Yes, I have shingles. I may have lost out on my Engrlish, 
but gain it back in spades in Maths...

it is fun going back and looking at the equations again in different 
areas.  Maxwell is cool.

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