Secrets of the Aether

In complex physical systems, the Aether Physics Model is consistent for the most part with Classical Mechanics. However, some physics equations need correction due to the error in the Standard Model with units based on single-dimension charge. Also, the concept of kinetic energy is somewhat deficient in the Standard Model, and we provide our alternative presentation. 

Ohm's Law in APM Classical Mechanics

Ohm's law involves the units of potential, current, and resistance. Ohm's law works in all systems of units.  

\begin{equation}potn=curr\cdot resn\end{equation}

Ohm's law pertains to electric current, which describes the physics of the electrostatic charge in an electric circuit. There is a second law, which has so far been missed by mainstream physics, which describes the physics of magnetic charge in a magnetic circuit: 

\begin{equation}\label{magnetic_current}mcur\cdot indc=resn\end{equation}.

The magnetic current law (\ref{magnetic_current}) pertains to the magnetic current of the circuit. 

Maxwell's Equations in Classical Mechanics

The cgs system of units originally expressed charge as a distributed dimension (charge squared). The advent of the MKS and SI systems of units expressed charge as a single dimension (charge). This results in many errors in mainstream physics.

Let us look at how gravitational force should relate to the electric nature of space (Aether) utilizing the Classical Mechanics values for the Sun and Earth:

\begin{equation}\label{gravity_electric}G\frac{m_{sun}}{au}\frac{m_{earth}}{au}=1.05\times 10^{24}efxd\cdot A_{u}\end{equation}

where au is distance given as an astronomical unit.

The gravitational force between the Sun and the Earth equals \(1.05\times 10^{24}forc\), the same as \(3.543\times 10^{22}newton\). In QMU, the unit of force also expresses as:

\begin{equation}\label{forc_electric}forc=efxd\cdot A_{u}\end{equation}.

Equation (\ref{gravity_electric}) shows that the gravitational force involves a tensor between the length density of two massive objects and electric flux density. When the correct system of units is applied, new predictions about gravity and the electric force become apparent. The surface tension (force per length) of space per electric flux density generates a potential:

\begin{equation}\frac{sten}{efxd}=potn\end{equation}.

And so, as two massive bodies in space move toward each other, the potential between them increases, creating a potential discharge. Such discharges are witnessed as electrical flashes when satellites are smashed into asteroids and comets collide with the Sun and Jupiter.

We provide a more detailed analysis of Maxwell's equations by stepping through Maxwell's work step by step.

Electric Flux Density and Electric Field Strength (Electric Field Intensity)

The mainstream explanation for electric flux density and electric field strength (aka electric field intensity) presents in terms of single-dimension charge. This creates problems in mainstream theory concerning the actual interpretation of the units. 

Electric flux density (D) is given as:

\begin{equation}\label{efxd_D}D=\frac{Q}{4\pi r^{2}}\end{equation}

And electric field strength (E) is given as:

\begin{equation}\label{elfs_E}E=\frac{Q}{4\pi\epsilon_{0}\epsilon_{r}r^{2}}\end{equation}

Due to the charge needing to be notated as charge squared rather than just as charge, the correct relationship of electric flux density per electric field strength should be:

\begin{equation}\frac{elfs}{efxd}=fric\end{equation}

Where \(fric\) is the unit of friction (resistance times velocity). It is believed by mainstream physicists that:

\begin{equation}\frac{D}{E}=\epsilon_{0}\epsilon_{r}\end{equation}

Which is incorrect. Furthermore, the correct relationship of efxd and permittivity (permittivity is the reciprocal of the Aether unit) is \(efxd\cdot A_{u}=forc\), as shown in equation (\ref{forc_electric}).

Kinetic Energy

The following Classical Mechanics explanation of kinetic energy does not necessarily agree with the Standard Model. We present it to bring the understanding of kinetic energy into an agreement with the Aether Physics Model.

There is not really such a "thing" as energy. Energy is a unit equal to applying force across a distance or angular momentum at a frequency. Force and angular momentum are the active components of kinetic and potential energy. Force arises from the Gforce, and angular momentum arises from dark matter.

When we understand that energy is just a unit of convenience, one can think of all processes in the physical Universe as energy transactions. Although one can choose to see only that portion of a transaction of interest, we should account for the total transaction in physics. Concerning kinetic energy, it is not actually a unit. Kinetic energy is the positive phase of an energy transaction.

According to Newtonian physics, kinetic energy is:

The energy a body possesses because of its motion is equal to one-half the mass of the body times the square of its speed.[8]

The kinetic energy equation thus notates as:

\begin{equation}\label{Ek}{E_k} = \frac{{m{v^2}}}{2} \end{equation}

If \({E_k}\) is a unit of energy, then equation (\ref{Ek}) is not a true equation because the two sides do not equal each other. The left side would have twice the value of the right side. Kinetic energy is, therefore, not a unit but rather a component of an equation removed from its true context. A proper equation using kinetic energy is:

\begin{equation}\frac{E}{2} = \frac{{m{v^2}}}{2} \end{equation}

or

\begin{equation}E=E_{k}+E_{p}\end{equation}

where \({E_{k}}\) is kinetic energy and \({E_{p}}\) is potential energy.

Thus, kinetic energy is just half the energy transaction.

Comprehending kinetic energy is easy when compared to a financial transaction. An employee earns a paycheck. The employer pays the employee. Let us say the paycheck is \$300. The total change in wealth between the employer and employee at the moment the check is handed over is \$600 (the employer is \$300 poorer, and the employee is \$300 richer.) However, despite the total change of wealth being \$600, only \$300 changes hands. The \$300 paycheck is tangible to the employer before paying the employee but becomes intangible to the employer after giving it away. Likewise, the employee’s wages were intangible before getting paid but tangible after receipt of the check.

Symbolically, the paycheck is kinetic energy. Kinetic energy is tangible, as it is the work done. The employee’s accrued wages could be symbolic of potential energy. The potential energy is intangible, being unusable. In the transaction, the total change in wealth symbolizes total energy. The fact that the employer’s wealth decreases by \$300 and the employee’s wealth increases by \$300. Thus, the economy has a net gain of zero dollars, indicating the energy law conservation.

According to the standard explanation of kinetic energy, it has no direction, being a scalar quantity. Nevertheless, since dimensions comprise all units and have a more primary nature than units, they must obtain their characteristics from the dimensions.

A falling object has a direction toward the ground, which sees a falling body directed toward it. From the perspective of the ground, it could be as though the ground was moving toward the falling object.

Length and frequency have direction, nullifying the arbitrary statement that “kinetic energy has no direction.” Since length and frequency dimensions have direction, velocity, and energy, they must also have direction. Since half-spin subatomic particles only see the forward direction of quantum frequency, all quantum frequencies must yield positive time. But the length dimensions can be positive and negative and thus yield both positive and negative distances.

In the financial analogy, the employer’s wealth decreases during the transaction while the employee’s wealth increases. This is true even though the paycheck remains the same value throughout the transaction and moves unidirectionally from employer to employee. The paycheck is merely an instrument of exchange. The employer and employee are the real parties to the transaction.

Similarly, kinetic energy is always associated with moving objects, such as electrons, photons, or the swinging balls of Newton's cradle. The object's kinetic energy is merely the instrument of the energy exchange between the objects. As in the financial transaction, the total change of energy state equals twice the kinetic energy.

One might ask, “What does the employee care about the employer’s wealth decreasing by \$300?” After all, the employee earned the paycheck, and the employer has marketable goods to sell at a profit.

The significance of tracking the employer's and employee's wealth is monitoring cash conservation. The conservation of cash is important to the economy in which the transaction takes place. If employers wrote checks for \$300, but employees cashed the checks and received \$450 per check, the banks processing the checks would ultimately collapse. Maintaining energy conservation in our physics transactions is just as important, not because the Universe would collapse, but because the Universe will not allow it to be otherwise.

Despite the common assumption in mainstream Classical Mechanics that an object on Earth falls toward the ground while the ground remains stationary, there is an acceleration midpoint between the object and the ground. The acceleration midpoint is the point on a line segment between two objects where they will collide.

The acceleration midpoint commonly vanishes from equations because it is close to the ground. It vanishes due to the relative magnitudes of the Earth's mass and an object's mass. However, this acceleration midpoint occurs when the falling object has the mass of the Moon. The mass of a very small object merely indicates a different scale. The Earth moves a very tiny distance toward the falling object, while the falling object moves practically all the distance toward the Earth.

Let us assume an object with a mass of 1kg hangs a distance of 10m above the Earth. The gravitational acceleration constant of the Earth is \(g = 9.8066\frac{m}{{se{c^2}}}\). The potential energy stored in the Earth’s gravitational field concerning the object is then:

\begin{equation}1kg \cdot 10m \cdot g = 98.066 joule \end{equation}

The mass of the Earth is \(5.98 \times {10^{24}}kg\). Since the falling object travels nearly all the distance, we can calculate the distance that the Earth will traverse as:

\begin{equation}\frac{1kg\cdot10m}{5.98\times{{10}^{24}}kg}=1.672\times{10^{-24}}m\end{equation}

However, the distance traveled by the Earth to the acceleration midpoint is a negative length compared to the distance traveled by the falling object. This would make the total potential energy of the Earth concerning the falling object equal to the following:

\begin{equation}{5.98\times{10^{24}}kg}\cdot{-1.672\times{10^{-24}}m}\cdot g =-98.052joule\end{equation}

So not only does the object have a positive phase kinetic energy of about \(98joule\), but the Earth also has a negative phase potential energy of about \(-98joule\) at the time of impact. And since the energy is a vector quantity, the Earth’s negative phase potential energy is 180º out of phase with the falling object. Thus, at the moment of impact, the positive kinetic energy of the falling object becomes negative (it decelerates to a stop), and the Earth's negative phase potential energy becomes positive (and the friction caused by the Earth’s immovability generates \(98joule\) of heat.) The total energy exchange of the system is equal to:

\begin{equation}E = \frac{{{E_k}}}{2} + \frac{{{E_p}}}{2} \end{equation}

where \({{E_k}}\) is the kinetic energy of the object and \({{E_p}}\) is the potential energy of the Earth. The net energy gain of the system is equal to:

\begin{equation}\frac{{{E_k}}}{2} - \frac{{{E_p}}}{2} = 0 \end{equation}

which is the conservation of energy.

When the two objects collide, the energy phases reverse polarity. If the collision were perfectly elastic, the positive phase kinetic energy, made negative at the collision, would reverse phase with negative acceleration and negative kinetic energy again. The result would be a positive phase kinetic energy with a change in the direction of motion. Even the Earth experiences recoil, but due to its enormous mass compared to the falling object, it is on the scale of \({10^{ - 24}}m\), which is considerably smaller than the quantum length. The recoil is extremely small but cannot be erased from the physics.

Regarding the financial analogy, while the employer possesses the check, the funds the check represents have a positive value in the bank account. However, when the employer transfers the check to the employee, its value must subtract. Therefore, the check transaction reverses the polarity of the funds. Suppose the employee refuses the check (perfectly elastic collision). In that case, the check reverts to the employer, and the value of the funds reverses once again, thus returning them to their positive value.

An example of energy phase exchange is the swinging ball demonstration known as “Newton’s cradle." If one ball lifts and drops, it has positive kinetic energy concerning the four stationary balls. The positive phase kinetic energy will change to negative phase kinetic energy and eventually transfer the positive phase kinetic energy to the ball at the opposite end, which will cause it to swing up and in the same direction as the first ball. Since the balls are all the same mass, the ball on the end would swing to the same height as the first ball, assuming no frictional loss.

With all the balls at rest, the energy needed to raise the first ball and start swinging will equal the energy lost due to friction as the balls eventually return to the rest state.

\begin{equation}E = \frac{{{E_k}}}{2} + \frac{{{E_f}}}{2} \end{equation}

where \({{E_f}}\) is the energy lost to friction. In other words, the frictional loss equals the kinetic energy dissipating from the system.

As the ball lifts, the source of the lift stores energy in the gravitational field equal to the ball's mass, times the height raised, times the gravitational acceleration force constant of the Earth.

\begin{equation}\label{potnenrg} - \frac{E}{2} = m \cdot - h \cdot g \end{equation}

Equation (\ref{potnenrg}) is the correct form for the potential energy equation since the energy phase is negative concerning kinetic energy. The height is negative because length has direction, and the ball moves away from the Earth.

When the ball releases, it swings toward the next ball in line. Until impact, the energy stored in the gravitational field increasingly converts into the ball's kinetic energy. At the moment of contact, the positive phase potential energy converted to the motion now manifests as positive phase kinetic energy in the collision. Also, at the moment of collision, the next ball in line sees an oncoming mass with a velocity but a velocity of the opposite polarity, so it has a negative phase kinetic energy.

When the first swinging ball strikes the next ball in line, the first ball switches energy polarity with the next ball, colliding with the middle ball while the first one rests. Since the distance between the second and the middle ball is zero, the energy polarity instantaneously exchanges between them. The middle ball has the same exchange with the fourth ball, and the fourth ball has the same exchange with the ball on the opposite end, which, because it is the last ball, retains the positive energy, transferring it to the gravitational field as the ball moves up and away from the Earth.

As the positive kinetic energy exchanges from ball to ball, and as the end balls move through the air, the balls give up some of the positive phase kinetic energy in the form of friction, similar to a free-falling ball striking the Earth in an inelastic collision, but spread out over time.

Eventually, the rising ball on the end stores all its positive phase kinetic energy in the gravitational field as positive phase potential energy, thus giving up its motion. The ball comes to rest, and due to the Earth's gravitational force, the energy polarity reverses relative to the original motion as it begins moving in the opposite direction. When the ball swings back toward a collision, it transfers the negative phase kinetic energy along the balls' succession until the cycle's second half is complete. Again, some of the negative phase kinetic energy is lost to friction.

The importance of the energy phase concept is especially apparent when we look at the Standard Model Classical Mechanics explanation of kinetic energy. In that model, the kinetic energy of a falling object collides with the ground, which is assumed to have zero kinetic energy, which it does, but which is irrelevant. The ground has potential energy, and it is the potential energy that balances the kinetic energy of the falling object. In the Standard Model explanation, the net energy of the two is supposed to be equal to the kinetic energy of the falling object plus the energy converted to friction from the collision. So the equation for kinetic energy in the Standard Model expresses as:

\begin{equation}\frac{{m{v^2}}}{2} + 0 = \frac{{m{v^2}}}{2} + \frac{{m{v^2}}}{2}\end{equation}[9]

or

\begin{equation}\frac{{{E_k}}}{2} + 0 = \frac{{{E_k}}}{2} + \frac{{{E_f}}}{2} \end{equation}

and therefore, it is assumed that:

\begin{equation}\frac{{m{v^2}}}{2} = E \end{equation}

which is an unbalanced equation.

However, Newton’s cradle demonstrates the actual physics of collisions. Positive phase kinetic energy reverses phase with negative phase kinetic energy at the moment of collision, thus conserving energy. This presents a potential flaw in how the Standard Model explains kinetic energy.

Another argument used by the Standard Model to incorrectly imply that the Earth has zero kinetic energy is to present the equation:

\begin{equation}E_{k}=\frac{1}{2}\cdot\frac{time^2\cdot forc^2}{mass}\end{equation}

If pushing the Earth for ten seconds and with a force of ten newtons, the Earth will develop only \(8.489\times 10^{-22}joule\) of kinetic energy. Compare the same pushing for ten seconds at ten newtons to a \(1 kg\) object, and the \(1 kg\) object develops \(5000  joule\) of kinetic energy.

The Standard Model Classical Mechanics interpretation of this scenario is that the Earth has near zero kinetic energy, but it is the potential energy the physicists should be looking at. Physicists correctly note the Earth has near zero kinetic energy, assume that the potential energy is also near zero, and then irrationally eliminate the energy dimensions and the tiny kinetic energy value of the Earth. Thus the precision of physics is then replaced with an approximation based on human perception, which has nothing to do with the potential energy required for balancing the kinetic energy of the impact. 

Furthermore, when pushing an object for some time with a given force, the reaction force of Newton's Third Law of Motion is ignored; something is providing the force that is doing the pushing. In the case of an object impacting the Earth due to gravitational acceleration, Standard Model physicists again ignore the reaction force of Newton's Third Law of Motion; the impact is wearing down the Earth, thus giving up potential energy. The employee paycheck analysis above shows how to account for the full energy transaction. 

According to the Standard Model, the tiny amount of kinetic energy developed within the Earth is nothing more than a distraction that has no bearing on the transfer of potential energy from the Earth to the kinetic energy of the impacting object. Concerning the "tiny kinetic energy" argument being nearly equal to zero, one might ask a Standard Model physicist, "At what value is kinetic energy considered to be meaningful if the Earth's kinetic energy is considered insignificant in physics?" One cannot make calculations and measurements disappear out of convenience (an approximation) or make them disappear based on the human choice of perception.

In conclusion, physics equations invoking kinetic energy must account for both positive kinetic energy and negative potential energy phases to conserve energy properly.

Eddy Current

Jean-Bernard Leon Foucault investigated eddy currents in the early 1800s. Eddy Current is a unit that appeared as early as 1922[4]. For some reason, scientists either ignored or lost its unit definition. Eddy current is an important Classical Mechanics unit, and in Quantum Measurements Units, it is equal to magnetic flux squared.

\begin{equation}\label{eddy1}eddy = mfl{x^2} \end{equation}

Eddy current also has other expressions and relates to Ohm’s law. According to the Aether Physics Model, eddy current is also equivalent to angular momentum times resistance:

\begin{equation}\label{eddy2}eddy = angm \cdot resn \end{equation}

Equation (\ref{eddy2}) represents the measurement of electron relaxation times by eddy current damping. When the external magnetic field from a primary coil switches off, it releases the induced magnetic field in a secondary coil. The electrons in the secondary coil quantified by their angular momentum are then relaxed[5]. Depending on the material of the secondary coil, the electrons will gyrate to a magnetic realignment. Due to the geometrical structure of the atoms and free electrons, the time it takes to gyrate back to stable magnetic realignment will vary from material to material. This unit of time times gyration toward magnetic realignment is the unit of resistance.

\begin{equation}resn = time \cdot gyro \end{equation}

Eddy current is also equal to potential times inductance.

\begin{equation}\label{eddy3}eddy = potn \cdot indc \end{equation}

Eddy current is equal to inductance divided by capacitance:

\begin{equation}eddy = \frac{{indc}}{{capc}} = \frac{{{\mu _0}}}{{{\varepsilon _0}}} \end{equation}

Significantly, eddy current is equal to Aether per curl:

\begin{equation}eddy = \frac{{A_{u}}}{{curl}} \end{equation}

Another observation of interest is the relationship of an eddy current to the magnetic field:

\begin{equation}eddy = mfld\frac{{momt}}{{chrg}} \end{equation}

The eddy current equals the magnetic field times momentum per magnetic charge. Thus, the eddy current is dependent upon a moving magnetic field.

According to many experts, eddy current is a complete path electrical current that flows through the conductor as the magnetic flux changes.

According to a website by Dr. James B. Calvert[6]:

"A magnet produces a pure magnetic field in its rest frame. Anything moving with respect to the magnet sees an electric field in addition to the magnetic field that is roughly proportional to the relative velocity. An electron free to move, as in copper, will be set into motion by the electric field it sees.  ...  This current is called the eddy current, since it flows in closed loops in a conducting plate like eddying water."

Dr. Calvert describes the physical eddy current within a copper tube. A neodymium-iron-boron (NIB) magnet drops through. "The magnetic field passes through the tube walls at top and Eddy Current experiment in Classical Mechanicsbottom in opposite directions, producing eddy currents that essentially ring about the tube, flowing in opposite directions at top and bottom, and moving with the falling magnet."

We dropped a NIB magnet down a copper tube to test this theory. The magnet was 1" in diameter and nearly ¼" thick.Small magnet eddy current experiment in Classical Mechanics.

As the magnet dropped, it dropped at a much slower velocity than it would in free space, as Dr. Calvert explained it would.

The plane of the magnet was almost perfectly perpendicular to the length of the tube during its descent.

According to Dr. Calvert, the magnetic field of the magnet moving through the copper tube made the copper tube see an electric current. This electric current flowed in one direction near the top of the magnet and in the opposite direction near the bottom of the magnet.

To test the theory, we slit a section of copper pipe along its length, thus preventing any current flow around the periphery of the tube.

Figure 3. Copper tube with slit along length.

Eddy Current tube side slit in Classical Mechanics

We then dropped the magnet into the slit tube. If the eddy currents were propagating through the periphery of the tube, they would not form in this experiment and would drop straight through.

Figures 4 & 5 Magnet falling down slit tube.

Falling magnet in Classical Mechanics slit tube experiment.Falling magnet in Classical Mechanics slit tube experiment.

But as shown in the photos on the left, the magnet still dropped through at a slow rate, although slightly faster than the drop rate through the un-slit tube. In addition, the magnet did not fall perpendicular to the length of the tube. Instead, it fell with a noticeable tilt toward the slit.

The interpretation of this experiment is that the eddy current results from the angular momentum of the electrons (cut by the magnetic field) times the resistance of the electrons (cut by the magnetic field). Along the slit, there are no electrons and thus no eddy currents, so the magnet tends to fall faster along this area. Eddy current connections in Classical Mechanics slit tube experiment.Nevertheless, the angular momentum in the atoms along the path of the magnetic field still contributes to eddy currents, and thus this portion of the magnet tends to fall slower. This results in the tilt of the magnet as it falls.

We attached an HP 34970A data acquisition switch with a built-in digital multimeter to test for resistance. Two terminals were soldered mid-length, one on each side of the slit, as in the image to the right. We cleaned the terminals to ensure good contact.

The magnet dropped down the tube while measuring resistance at the terminals. Several tests ran, each producing the same graph, as shown below.

Resistance in copper slit tube

The spike at the beginning of the drop occurred at the beginning of each test. Apparently, resistance increases as the magnet approaches the test leads and then abruptly decreases just before passing. Then the resistance gradually returns to normal as the magnet moves away.

The preliminary conclusion is that an eddy current is an actual unit of electrical behavior in Classical Mechanics. The current produced is within each atom and not within the macrostructure of the atoms (copper tube in this case), at least not under normal conditions. The properties of angular momentum and resistance can interact to produce a combined effect that we call eddy currents.

This, of course, is not the standard explanation for eddy current. The normal explanation is that the magnet generates a potential on the leads, and thus the ohmmeter, expecting no potential, is “fooled” into seeing less (or more) resistance. This is, of course, true, as measurement does show an increase in potential at the edges of the pipe as the magnet passes by. However, the induced potential reacting to the inductance of the copper is also a way of seeing eddy current, as in equation (\ref{eddy3}).

The difference between the understandings of eddy current presented here and the standard interpretation of eddy current is the standard interpretation considers resistance a characteristic of a material rather than an effect of electricity. According to the APM, the eddy current develops because subatomic particles interact with the fabric of Aether units in which they reside.

More Example Calculations

EddyCurrentPipesWe will repeat the slit tube experiment for the above eddy current with a 1½” pipe and a 1½” magnet. The length of the pipe is 11.875” (30.162cm), and the magnet is .375” thick with a .5” diameter hole. The data screen below represents the resistance of the pipe at the terminal while the magnet drops through the slit tube.

The markers are the green vertical lines in the graph and are set precisely before the magnet drops and immediately after the magnet stops moving. The connections from the HP34970A DAQ unit are simple 2-wire setups since we are only looking for a general picture of the action.

EddyCurrentGraphThe resistance at the maximum is \(880.21m\Omega\), and at the minimum is \( - 162.63m\Omega \) with a reference resistance of \(358.79m\Omega \). Therefore, at first, we see that the change in resistance is exactly \(521.42m\Omega\) above and below the reference resistance. The interval from the beginning of the magnet drop to the maximum resistance was \(897.4msec\). The interval from the minimum resistance to the moment the magnet stopped moving was \(915.8msec\). Between the maximum and minimum moments, \(100.8msec\) elapsed.

The magnet fell \(30.162cm\) in \(1.914sec\). The velocity of the magnet was \(15.759\frac{{cm}}{{sec}}\). Between the moments the magnet started falling and the maximum resistance, the magnet traveled \(14.142cm\).

\begin{equation}15.759\frac{{cm}}{{sec}} \cdot 897.4msec = 14.142cm \end{equation}

The mean resistance from when the magnet started falling to the maximum resistance was \(620m\Omega \), so we can calculate the average drag during that interval. First, we need to convert the unit of \(\Omega \) to the unit of \(resn\) by applying the charge conversion factor to the different charge dimensions. Since resistance has a charge to the fourth in APM units and charge squared in MKS units, the charge conversion factor must be squared:

\begin{equation}\frac{620m\Omega}{ccf^2} = 2.402\times 10^{-5}resn \end{equation}

The total averaged electrons dragged at any moment along the magnet’s fall are:

\begin{equation}2.402\times 10^{-5}resn \cdot 14.142cm = 1.4 \times {10^{6}}drag \end{equation}

Since the magnetic charge is directly proportional to the electron's angular momentum (Planck’s constant), the magnetic charge is also a constant of the electron. The magnetic charge represents as \({e_{emax}}^2\) or as its variable “\(chrg\),” so the averaged magnetic field in the first \(897msec\) of fall is:

\begin{equation}1.4 \times {10^{6}}drag \cdot chrg = 1.4 \times {10^{6}}mfld \end{equation}

The \(mfld\) unit is the Aether unit, but without accounting for its rotation. Therefore, the unit of \(mfld\) equals a unit of Aether. As the magnet falls from the start position to the point of maximum resistance, at any given moment along the fall, it involves the action of an average \(1.4 \times {10^{6}}\) dragging electrons and \(1.4 \times {10^{6}}\) Aether units.

Assuming an average magnetic field during the \(14.142cm\) of fall, the average magnetic flux would be:

\begin{equation}\frac{{1.4 \times {{10}^{6}}mfld}}{{14.142cm}} = 2.402\times 10^{-5}mflx \end{equation}

Converting \(mflx\) to \(weber\):

\begin{equation}2.402\times 10^{-5}mflx \cdot ccf = 9.934\times 10^{-20}weber \end{equation}

Of course, a test of the accuracy of this exercise would be the magnet’s magnetic flux measurement, which is not available at the time of this writing.

references

[4] A Course in Electrical Engineering Volume II - Alternating Currents, McGraw Hill Book Company, Inc., 1947 pg 259

[5] Arthur F. Kip Fundamentals of Electricity and Magnetism (McGraw Hill Book Company, New York, St. Louis, San Francisco, Toronto, London, Sidney, 1969) 316

[6] Dr. James B. Calvert, Associate Professor Emeritus of Engineering, University of Denver Registered Professional Engineer, State of Colorado No.12317 http://www.du.edu/~jcalvert/phys/eddy.htm 

[7] John  Backus, The Acoustical Foundations of Music (W.W. Norton & Company, New York, 1977) p 59

[8] The American Heritage® Dictionary of the English Language, Fourth Edition Copyright © 2003 by Houghton Mifflin Company.

[9] Edward R. McCliment, Physics (Orlando, Harcourt Brace Jovanovich, Inc., 1984) 150