25m views== 271st book of science by AP// Derivative, velocity, New Ohm's law recalibrated to one another// physics-math by Archimedes Plutonium

[Archimedes Plutonium]

271st book of science by AP// Derivative, velocity, New Ohm's law recalibrated to one another// physics-math by Archimedes Plutonium
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Archimedes Plutonium<***@gmail.com>
Dec 31, 2023, 2:58:46 PM (7 days ago)
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I was caught doing superconductivity in my 270th book of science. And in that research saw a dire need to unify Derivative with velocity (some like to call it speed) and with Ohm's law, either the V = i *R or the New Ohm's law V = i*B*E.

This project is long overdue for physics for most think of voltage as some electrical pressure. And so we have to examine that proposition very carefully and examine what current actually means.

AP

271st book of science by AP// Derivative, velocity, New Ohm's law recalibrated to one another// physics-math by Archimedes Plutonium
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Archimedes Plutonium
Dec 31, 2023, 3:27:24 PM (7 days ago)
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271st book of science by AP// Derivative, velocity, New Ohm's law recalibrated to one another//
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Archimedes Plutonium
Dec 31, 2023, 9:05:36 PM (7 days ago)
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In Calculus math and physics the calibration is easy to make. We have calculus derivative as dy/dx
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Archimedes Plutonium
Dec 31, 2023, 9:44:16 PM (7 days ago)
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On Sunday, December 31, 2023 at 9:05:36 PM UTC-6 Archimedes Plutonium wrote: (snipped) So if we take
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Archimedes Plutonium
Jan 3, 2024, 12:43:38 AM (5 days ago)
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Alright let me review units here, for a moment. Pressure is Force/area or kg/(meters*sec^2) Angular
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Archimedes Plutonium
Jan 3, 2024, 3:23:33 PM (4 days ago)
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Alright this is excellent progress, for UNITS are some of the toughest and major problems in all of
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Archimedes Plutonium
Jan 4, 2024, 12:39:21 AM (4 days ago)
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If I make no more progress than that of Voltage is the 2nd derivative of angular momentum which is
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Archimedes Plutonium
Jan 4, 2024, 3:50:38 PM (3 days ago)
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On Thursday, January 4, 2024 at 12:39:21 AM UTC-6 Archimedes Plutonium wrote: But why is mathematics
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Archimedes Plutonium
Jan 4, 2024, 11:39:45 PM (3 days ago)
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Apparently the physics literature already has AP's idea that the vector dot and vector products
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Archimedes Plutonium
Jan 5, 2024, 2:23:18 PM (2 days ago)
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On Thursday, January 4, 2024 at 11:39:45 PM UTC-6 Archimedes Plutonium wrote: Apparently the physics
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Archimedes Plutonium
12:55 AM (18 hours ago)
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Alright, I got most of what I wanted, a recalibration of units of Physics, the Calculus derivative
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Archimedes Plutonium<***@gmail.com>
5:09 PM (2 hours ago)
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Alright, I need an example to clarify that what was learned about vector dot and vector cross product.

Vector cross product is all summarized as Area of general triangle is 1/2 a*b (sine C).

All I need to do is slip into the picture, the Area of general triangle with a cosine angle.

AP
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Archimedes Plutonium<***@gmail.com>
7:09 PM (now)
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Alright, I looked in the math literature to see if any mathematician recalibrated triangle area to be that of cosine instead of sine. And apparently Archimedes Plutonium is the first to do so. And the reason I am the first is because I am the first to realize that Vector Dot Product is merely the inverse of Vector Cross Product. What I mean by that is that sine 60 degrees is 0.866 and cosine 30 degrees is 0.866 where the two are compliments to 90 degrees.

So let us start with the most simple example of all a equilateral triangle with all angles 60 degrees and all three sides the same length, say 4.

So we have area of Triangle is 1/2 Base * Height. We have base as 4 and now compute height from Pythagorean theorem is sqrt12 is 3.46. So the area is 1/2 (4)(3.46) = 6.92 square area.

Now let us use the Sine Area formula on this same triangle we have 1/2(4)(4) (sine 60degrees) = 1/2(16)(.866) = 8(0.866) = 6.92.

So the traditional formula of area of triangle as half of base times height matches the sine formula.

Lastly, we need to see where a cosine formula matches traditional formula, and matches sine formula.

Area of triangle from a cosine formula would be, since cosine is inverse sine, is 1/2(4)(4)(cosine 30 degrees) = 1/2(16)(.866) = 8(0.866) = 6.92.

The only difference between sine and cosine of triangle area is that we need to find the inverse angle for sine in the perpendicular height of triangle in general.

And how does that translate into Physics, the physical rotation of angular momentum?

AP


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Archimedes Plutonium

25m views== 271st book of science by AP// Derivative, velocity, New Ohm's law recalibrated to one another// physics-math by Archimedes Plutonium

[Archimedes Plutonium]

Archimedes Plutonium<***@gmail.com>
6:46 PM (2 hours ago)
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Recalibrating ampere with coulomb.

--- quoting Wikipedia on ampere ---
The 2019 redefinition of the SI base units defined the ampere by taking the fixed numerical value of the elementary charge e to be 1.602176634×10−19 when expressed in the unit C, which is equal to A⋅s, where the second is defined in terms of ∆νCs, the unperturbed ground state hyperfine transition frequency of the caesium-133 atom.

The SI unit of charge, the coulomb, "is the quantity of electricity carried in 1 second by a current of 1 ampere". Conversely, a current of one ampere is one coulomb of charge going past a given point per second:

1A = 1 C/s

In general, charge Q is determined by steady current I flowing for a time t as Q = I t.

Constant, instantaneous and average current are expressed in amperes (as in "the charging current is 1.2 A") and the charge accumulated (or passed through a circuit) over a period of time is expressed in coulombs (as in "the battery charge is 30000 C"). The relation of the ampere (C/s) to the coulomb is the same as that of the watt (J/s) to the joule.
--- end quoting Wikipedia on ampere ---


In University when I studied physics, I was presented with these definitions of Ampere and Coulomb

--- quoting Halliday & Resnick, Fundamentals of Physics, 3rd ed. 1988, Appendixes---

(1) electric current :: ampere :: A :: "... that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2 x 10^-7 newton per meter of length." (1946)

(2) quantity of electricity :: coulomb :: C :: A*seconds

(3) force :: newton :: N :: kg *(meter/second^2)
--- end quoting H & R textbook of 1988 ---

Sorry, I view the above as a mess. Nothing for which a commonsense person can wrap their head around. And I believe the confusion is because of these new facts presenting in the past two decades.

1) There is a Dirac magnetic monopole-- the 0.5MeV particle.

2) True electron of atoms is the muon, not the 0.5MeV particle. And the muon is stuck inside a 840MeV proton torus doing the Faraday law producing new magnetic monopoles.

3) Light Waves do not move as a straightline arrow with a front tip and rear end, but move as a closed loop pencil ellipse circuit.

4) Complimentary duality of particle and wave, electricity and magnetism.

Without these concepts when defining ampere and coulomb, no wonder they are a garbled mess.

AP


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Archimedes Plutonium<***@gmail.com>
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The direction AP is headed for is to make equal the ultimate units of 3D in geometry is volume making the ultimate units of volts in physics also be 3D.

That unification of volume to voltage requires Volts = current*(B*E) as similar to Volume = Length * Width * Depth.

The way I do that in this book is consider volume as a length of area, the area is width* depth and multiplied by length is volume. For physics, B*E is similar to width*depth and when multiplied the B*E is a parallelogram whose area is two triangles that compose the parallelogram. The area of one of these two triangles follows the formula Area = A cross B (sine angle) while the other complimentary triangle follows Area = A dot B (cosine of compliment angle of sine). So if the sine angle is 30 degrees the compliment angle is 60 degrees for cosine.

What I am heading for is the final units formula of Voltage is Volt = A*(magnetic field * electric field) written as A ((1/A)(meters/sec) * (A)(meters/sec). Which leaves me with Voltage = A(meters^2/seconds^2) a final configuration of energy.

In Geometry, volume is 3D and is the final configuration. In physics we need to end at 3D and be energy, and Voltage = A(meters^2/seconds^2 does just that.

But in New Physics, I need to recalibrate Ampere A and Coulomb C so that this final configuration makes sense.

AP

AP made the first of the above posts to sci.physics many hours ago. I passed a reCAPTcha test. I was approved, I posted it and it registered, but since then some miscreant deleted it.

[Archimedes Plutonium]

In this recalibration of physics units, the Magnetic field then becomes B = (1/Amp)(meters/second) while the Electric field becomes Amp(meters/second). And when one does the vector cross product of (B X E) we end up with energy units of meters^2/seconds^2. New Ohm's law is Voltage = A * (B X E) = A(m^2/s^2) of energy.

None of this was possible in Old Physics for the 6 ideas listed below.

1) There is a Dirac magnetic monopole-- the 0.5MeV particle.

2) True electron of atoms is the muon, not the 0.5MeV particle. And the muon is stuck inside a 840MeV proton torus doing the Faraday law producing new magnetic monopoles.

3) Light Waves do not move as a straightline arrow with a front tip and rear end, but move as a closed loop pencil ellipse circuit.

4) Complimentary duality of particle and wave, electricity and magnetism.

5) When does EM spectrum have rest mass and when not as in E = mc^2.

6) When does a EM wave collapse into a magnetic monopole of 0.5MeV?

AP


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[Archimedes Plutonium]

I am attempting to define New Physics Ampere and Coulomb. I am not sure I even need the Coulomb as being A*seconds.

Old Physics had missing the Dirac magnetic monopole, instead they used a general term of "charge" a really obnoxious and silly term for which they swept everything under the carpet of trash if they came upon trash in physics. In Old Physics, whenever a person is at a loss for explanation in electromagnetism, the word "charge" would enter the discussion and supposedly explain things, but it never did, for charge is a cover-up-of-ignorance.

So what I am doing is attempting to use geometry volume of math to solve what Voltage as energy is in physics. I am guided by geometry math.

We know that in geometry math that volume = an area moved through a distance. And the formula of Volume in geometry is L * W*D and the formula of Voltage is similar in Current * (magnetic field * electric field). We take the vector cross product (or the dot product) of B*E as a area of volume geometry and then multiply it by current to get a volume of electricity in mathematics. The B*E turns out to use sine or cosine and is the area of a parallelogram. Once we multiply by current we obtain that volume of electricity. And it should be a formula for Voltage as this V = current ((1/A)(meter/sec)* (A)(meter/sec)). That results in current (m^2/s^2) a energy term.

In math geometry the pinnacle peak of geometry is volume, there is nothing further in geometry than 3rd dimension volume. In keeping with geometry, physics pinnacle peak is energy, there is nothing further in physics than energy. And so when I end up with Voltage = A (m^2/s^2), I end up with energy, not much different from E = mc^2 which is wave energy, but for my A(m^2/s^2) that is particle energy of the Dirac magnetic monopole of 0.5MeV.

So I must decipher how waves transform into 0.5MeV Dirac magnetic monopoles to see if my formula A(m^2/s^2) is the true final formula for Voltage.

AP


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Archimedes Plutonium

[Archimedes Plutonium]

It is appealing to have magnetic field units be (1/A)(meters/second) and for electric field units be (A)(meters/second) for that makes Voltage be units of energy in the V = A*(B*E).

And where we get the vector cross product using sine for BxE as the area of parallelogram (the parallelogram rule of physics) of (BxE), or using the angle complement of the sine angle we find the vector dot product of B*E to have the same value as cross product. This complementarity of magnetism to electricity; of sine to cosine; of vector cross product to vector dot product, compelling reason to see the A, Ampere in magnetic field is inverse to A in the electric field.

And this ties in well with the calculus of mathematics that the angular momentum derivative with respect to time is energy so that the highest dimension in both math and physics when we speak of geometry dimension is 3rd dimension. So that voltage units as Ampere current * meters^2/seconds^2 fits all logical points of interest.

And in turn, we then obtain superconductivity into the Voltage equation where the Resistance is the term (B*E) and when the sine angle is 90 degrees -- subsequently cosine is thus 0 degrees, both with a value of 1 or superconductivity capable. We interpret that as meaning the flow of electricity as magnetic monopoles is passing through a material where the magnetism and electricity are perfectly aligned at 90 degrees. The more the angle is not 90 degrees the more resistance is involved.

That leaves me to straighten out a very difficult problem of the units of Ampere and Coulomb. And it is here that is most saddening to Old Physics. For they used "charge" to quell all questions of theory, but never defined the two precisely. It is a current no doubt, but the trouble is, it is both wave and particle. How do you define electricity as a wave? How do you have an equation of how many waves of EM spectrum superpositioned on one another then collapse into a particle of 0.5MeV go to compose a unit of electricity. So you measure Ampere as particle or wave, same question for Coulomb.

Halliday and Resnick says of Ampere as the "constant current" and says of Coulomb "quantity of electricity". Does that mean that Coulomb is a variable current? Does it mean the Ampere is not a quantity of electricity? Does it mean we have two different types of current? Or, does it mean that Coulomb is gibberish in an era where no magnetic monopoles were believed to exist.

Funny, how Old Physics used "charge" to explain almost everything that was hard to explain, but when it comes to the basic unit of current, Old Physics lands into a crisis of defining what is Ampere versus Coulomb.
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On Tuesday, January 16, 2024 at 2:06:50 PM UTC-6 Archimedes Plutonium wrote:
I am attempting to define New Physics Ampere and Coulomb. I am not sure I even need the Coulomb as being A*seconds.

Old Physics had missing the Dirac magnetic monopole, instead they used a general term of "charge" a really obnoxious and silly term for which they swept everything under the carpet of trash if they came upon trash in physics. In Old Physics, whenever a person is at a loss for explanation in electromagnetism, the word "charge" would enter the discussion and supposedly explain things, but it never did, for charge is a cover-up-of-ignorance.

So what I am doing is attempting to use geometry volume of math to solve what Voltage as energy is in physics. I am guided by geometry math.

We know that in geometry math that volume = an area moved through a distance. And the formula of Volume in geometry is L * W*D and the formula of Voltage is similar in Current * (magnetic field * electric field). We take the vector cross product (or the dot product) of B*E as a area of volume geometry and then multiply it by current to get a volume of electricity in mathematics. The B*E turns out to use sine or cosine and is the area of a parallelogram. Once we multiply by current we obtain that volume of electricity. And it should be a formula for Voltage as this V = current ((1/A)(meter/sec)* (A)(meter/sec)). That results in current (m^2/s^2) a energy term.

In math geometry the pinnacle peak of geometry is volume, there is nothing further in geometry than 3rd dimension volume. In keeping with geometry, physics pinnacle peak is energy, there is nothing further in physics than energy. And so when I end up with Voltage = A (m^2/s^2), I end up with energy, not much different from E = mc^2 which is wave energy, but for my A(m^2/s^2) that is particle energy of the Dirac magnetic monopole of 0.5MeV.

So I must decipher how waves transform into 0.5MeV Dirac magnetic monopoles to see if my formula A(m^2/s^2) is the true final formula for Voltage.

AP kindly asks Google to let AP run all three, sci.math, sci.physics, PAU as he runs PAU, now--- all pure science, no spam and no govt b.s.

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Archimedes Plutonium

[Archimedes Plutonium]

Post by Archimedes Plutonium
It is appealing to have magnetic field units be (1/A)(meters/second) and for electric field units be (A)(meters/second) for that makes Voltage be units of energy in the V = A*(B*E).
And where we get the vector cross product using sine for BxE as the area of parallelogram (the parallelogram rule of physics) of (BxE), or using the angle complement of the sine angle we find the vector dot product of B*E to have the same value as cross product. This complementarity of magnetism to electricity; of sine to cosine; of vector cross product to vector dot product, compelling reason to see the A, Ampere in magnetic field is inverse to A in the electric field.
And this ties in well with the calculus of mathematics that the angular momentum derivative with respect to time is energy so that the highest dimension in both math and physics when we speak of geometry dimension is 3rd dimension. So that voltage units as Ampere current * meters^2/seconds^2 fits all logical points of interest.
And in turn, we then obtain superconductivity into the Voltage equation where the Resistance is the term (B*E) and when the sine angle is 90 degrees -- subsequently cosine is thus 0 degrees, both with a value of 1 or superconductivity capable. We interpret that as meaning the flow of electricity as magnetic monopoles is passing through a material where the magnetism and electricity are perfectly aligned at 90 degrees. The more the angle is not 90 degrees the more resistance is involved.
That leaves me to straighten out a very difficult problem of the units of Ampere and Coulomb. And it is here that is most saddening to Old Physics. For they used "charge" to quell all questions of theory, but never defined the two precisely. It is a current no doubt, but the trouble is, it is both wave and particle. How do you define electricity as a wave? How do you have an equation of how many waves of EM spectrum superpositioned on one another then collapse into a particle of 0.5MeV go to compose a unit of electricity. So you measure Ampere as particle or wave, same question for Coulomb.
Halliday and Resnick says of Ampere as the "constant current" and says of Coulomb "quantity of electricity". Does that mean that Coulomb is a variable current? Does it mean the Ampere is not a quantity of electricity? Does it mean we have two different types of current? Or, does it mean that Coulomb is gibberish in an era where no magnetic monopoles were believed to exist.
Funny, how Old Physics used "charge" to explain almost everything that was hard to explain, but when it comes to the basic unit of current, Old Physics lands into a crisis of defining what is Ampere versus Coulomb.

The crisis occurs in Old Physics electromagnetism for they never treated electricity as being both a wave and a particle. When we measure current as to how many 0.5MeV particles-- the Dirac magnetic monopoles are moving we are having Coulomb quantity, but when we measure current as amperes A we are measuring current as a wave. A large superpositioning of many types of EM waves that have not collapsed into 0.5MeV magnetic monopoles.

This is why we would have a 1/A current in a magnetic field and why we would have a A current in electric field.

This ties into logic as one is the reverse or inverse of the other. The parallelogram rule issues the B X E as being 2 triangles forming a parallelogram, one of the triangles uses sine while the compliment of sine is the cosine. When the angle for sine is 90 degrees, it is 0 degrees for cosine and both have value 1.

So the definition of Ampere and Coulomb in Old Physics never overcame the reality of electricity being also a wave. Their definition was all based on particles in motion.

AP

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