Kinetic Energy

The following explanation of kinetic energy is not necessarily in agreement with the Standard Model. We present it in order to bring the understanding of kinetic energy into agreement with the Aether Physics Model.

There is not really such a "thing" as energy. Energy is a unit equal to the application of force across distance, or angular momentum at a frequency. Force and angular momentum are the active components of kinetic and potential energy. Force ultimately arises from the Gforce, and angular momentum ultimately 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 that is of interest, in physics we should account for the total transaction. With regard to 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 possessed by a body because of its motion, 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}



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 earned 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 is symbolic of 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, is indicative of the conservation of energy law.

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

A falling object has 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 were 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 do have direction, velocity, and ultimately energy, they must also have direction. Since half-spin subatomic particles only see the forward direction of quantum frequency, then all quantum frequency must yield positive time. But the length dimensions can be both positive and negative and thus yield both positive and negative distance.

In the financial analogy, the employer’s wealth is decreasing during the transaction while the employee’s wealth is increasing. 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 kinetic energy of the object is merely the instrument of the energy exchange between the objects. As in the financial transaction, the total change of energy state is equal to 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 available to sell at a profit.

The significance of tracking the wealth of both the employer and employee is the monitoring of the conservation of cash. 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, then the banks processing the checks would ultimately collapse. Maintaining the conservation of energy 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 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 so close to the ground. It vanishes due to the relative magnitudes of the mass of the Earth and the mass of an object. However, this acceleration midpoint occurs when the falling object has the mass of, say, 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 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 in relation to 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:


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 with respect to the falling object equal to:

\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 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 again reverse phase with a negative acceleration and negative kinetic energy. The result would be a positive phase kinetic energy with a change in direction of motion. Even the Earth experiences recoil, but due to its enormous mass compared to that of 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 it cannot erase from the physics.

In terms of 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. If for some reason the employee refuses the check (perfectly elastic collision) then the check reverts to the employer and the value of the funds reverses once again, thus returning them to their positive value.

A good 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 in relation to 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 up 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 it swinging will exactly equal the total energy lost due to friction as the balls eventually work back 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 is exactly equal to the kinetic energy that dissipates from the system.

As the ball lifts, the source of the lift stores energy in the gravitational field equal to the mass of the ball, 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 with respect to 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 kinetic energy of the ball. At the moment of contact, the positive phase potential energy that was converted to 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.

The moment the first swinging ball strikes the next ball in line, the first ball switches energy polarity with the next ball, which then collides with the middle ball while the first one comes to rest. 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 succession of balls until the second half of the cycle 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 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]


\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 the way 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}

In the case of 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 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 this means 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, and which has nothing to do with the potential energy required for balancing the kinetic energy of the impact. 

Furthermore, in the case of pushing an object for a period of time with a given force, the reaction force of Newton's Third Law of Motion is ignored; that is, 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 choose to ignore the reaction force of Newton's Third Law of Motion; the Earth is being worn down by the impact and thus giving up potential energy. The employee paycheck analysis above shows how to account for the full energy transaction. 

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. 

With regard to 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 to be insignificant in physics?" One cannot make calculations and measurements disappear out of convenience (an approximation), or to make them disappear based upon human choice of perception.

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

Eddy Current

Jean Bernard LeonFoucault investigated eddy current in the early 1800s. Eddy current is a unit that appeared as early as 1922[4]. For some reason though, scientists either ignored or lost its unit definition. Eddy current is an important unit and in Quantum Measurements Units 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 eddy current to magnetic field:

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

The eddy current is equal to 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 web site 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 goes on to describe 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 eddywbottom in opposite directions, producing eddy currents that are essentially rings about the tube, flowing in opposite directions at top and bottom, and moving with the falling magnet."

In an effort to test this theory, we dropped a NIB magnet down a copper tube. The magnet was 1" in diameter and nearly ¼" thick.eddymagnetsm

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 along 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.


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.


But as shown in the photos on the left, the magnet still dropped through at a slow rate, although slightly faster than the rate of drop 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 is a result of 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, and so the magnet tends to fall faster along this area. eddys3Nevertheless, 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 assure a good contact.

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

eddy c4

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 eddy current is an actual unit of electrical behavior. The current produced is within each atom and not within the macro structure of the atoms (copper tube in this case), at least not under normal conditions. The properties of angular momentum and resistance are capable of interacting 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 eddy current above, but with 1½” pipe and 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 at precisely the moment before the magnet drops and immediately after the magnet stops moving. The connections from the HP34970A DAQ unit are simple 2-wire setup since we are only looking for a general picture of the action.

EddyCurrentGraphThe resistance at the maximum is \(880.21m\Omega\) and at 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 \(521.42m\Omega\) both 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 the moment 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 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 angular momentum of the electron (Planck’s constant), then 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 \(mfld\) is equal to 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.


[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 

[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