Redefining units in terms of distributed charge and quantum measurements.

The Aether Physics Model constructs units with quantum measurements, as opposed to arbitrary or macro structure-based measurements such as meters, Earth revolutions, etc.  Quantum measurements provide a whole number of units for a quantum process or structure.  For example, the primary angular momentum of one electron moving at the speed of light determines the unit of one quantum photon.  Thus, there is a discrete relationship between the activity of electrons and the production of photons.

Constructing units from quantum measurements provides for easy comprehension of quantum processes.  Quantum physics, nanoscience, and chemistry would clearly benefit from this new system of units.

Quantum Units

There are essentially two stable forms of matter in our part of the Universe, the electron and the proton.  The neutron is a composite subatomic particle produced when a proton binds with an electron.  The photon comes into being when an atom absorbs excess primary angular momentum radiated from other atoms.  (see Photon Mechanics, page 223).

Since almost all controllable physical processes occur through interactions between the electron and photon, the quantum measurements of the electron usually define the quantum units.  As noted in Quantum Measurements on page 22, the electron quantum measurements are:

• Quantum Length: $${\lambda _C} = 2.426 \times {10^{ - 12}}m$$
• Quantum Frequency: $${F_q} = 1.236 \times {10^{20}}Hz$$
• Quantum Mass: $${m_e} = 9.109 \times {10^{ - 31}}kg$$
• Quantum Magnetic Charge: $${e_{emax}}^2 = 1.400 \times {10^{ - 37}}cou{l^2}$$
• Quantum Electrostatic Charge: $${e^2} = 2.567 \times {10^{ - 38}}cou{l^2}$$

The quantum length is equal to the Compton wavelength, the quantum frequency is equal to the speed of light divided by the Compton wavelength, quantum mass is the mass of the electron as measured by NIST, the quantum magnetic charge is the calculated magnetic charge, and the electrostatic charge is the elementary charge (as measured by NIST) squared.

Converting Charge Dimensions

There are two important differences between quantum measurement units and standard units with regard to the charge dimensions.  Charge dimensions always distribute (charge squared), and almost all charge dimensions express in terms of magnetic charge, as opposed to elementary charge.

Concerning distributed charge, the situation is somewhat complicated by the fact that five standard electrical units are already in the correct dimensions of distributed charge.  These units are permeability, permittivity, inductance, capacitance, and conductance.

Inductance is equal to the permeability of the Aether divided by length, and similarly, capacitance is equal to permittivity of the Aether divided by length.  (In the cgs system of units, units of length [cm] express inductance and capacitance).

So the units of inductance and capacitance already express in terms of distributed charge as follows:

$capc = 2.148 \times {10^{ - 23}}\frac{{se{c^2}cou{l^2}}}{{kg \cdot {m^2}}} \tag{6.1}$

$indc = 3.049 \times {10^{ - 18}}\frac{{kg \cdot {m^2}}}{{cou{l^2}}} \tag{6.2}$

All other electrically related units from Classical physics incorrectly express with single dimension of charge.  Further, the Standard Model has usually described the electrical units in terms of elementary charge.  Since the Aether donates electrostatic charge to subatomic particles, elementary charge has nothing to do with the action of subatomic particles in nearly all cases (magnetic moment is an exception).  In almost all cases, the magnetic charge of the subatomic particle is the active charge of the unit.

The magnetic charge is polar and behaves, in fact, like a tiny magnet.  The strong nuclear force, permanent magnetism, electromagnetism, the Casimir effect, Van der Waals forces… each of these is the action of the magnetic charge in a different situation.

In the case of resistance, where the standard unit in Classical physics appears to have distributed charge, there is a double distributed charge in the quantum measurements system of units, because resistance is a measurement of the action of two opposing subatomic particles colliding with each other.  Therefore, the magnetic charge is that of both subatomic particles experiencing the resistance.

The table below shows some units from Classical physics and the equivalent in the quantum measurement units.

 Aether Physics Model Classical Physics Resistance $$resn = \frac{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}}}{{{e_{emax}}^4}}$$ $$R = \frac{{kg \cdot {m^2}}}{{sec \cdot cou{l^2}}}$$ Potential $$potn = \frac{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}^2}}{{{e_{emax}}^2}}$$ $$V = \frac{{kg \cdot {m^2}}}{{se{c^2} \cdot coul}}$$ Current $$curr = {e_{emax}}^2 \cdot {F_q}$$ $$I = \frac{{coul}}{{sec}}$$ Magnetic Flux $$mflx = \frac{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}}}{{{e_{emax}}^2}}$$ $$\lambda = \frac{{kg \cdot {m^2}}}{{sec \cdot coul}}$$ Conductance $$cond = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}}}$$ $$G = \frac{{sec \cdot coul}}{{kg \cdot {m^2}}}$$

The usual rule for converting to quantum units from MKS units is to replace each dimension with its quantum measurement counterpart.  When it comes to the charge dimension, replace each single dimension of charge with $${{e_{emax}}^2}$$.  With the inductance, conductance, and capacitance units, the exponent of the charge dimensions remains unchanged.  The other exception is with magnetic moment; the charge involves both $${{e_{emax}}^2}$$ and $${e^2}$$.

Magnetic Moment

Magnetic moment is a unit that measures the influence of the Aether’s electrostatic charge against the magnetic charge of the subatomic particle.

The magnetic moment of the electron as defined by NIST is:

${\mu _e} = - 928.476362 \times {10^{ - 26}}J{T^{ - 1}}\tag{6.3}$

The NIST value of electron magnetic moment expresses in terms of quantum measurements as:

${\mu _e} = {g_e}{\lambda _C}^2{F_q}\frac{{e \cdot {e_{emax}}^2}}{{8\pi \cdot {e_{emax}}^2}} \tag{6.4}$

where $${g_e}$$ the is the electron g-factor as measured in the Lamb Shift.  In the electron unit of magnetic moment, the magnetic charge cancels out, since the electrons are acting on electrons.  Nevertheless, the magnetic charge terms belong in the equation in order to show that electrons are acting against other subatomic particles in the following NIST measured magnetic moment values.

The g-factor is an essential value related to the magnetic moment of the subatomic particle, as it corrects for the precession of the subatomic particle.

The NIST value for the proton magnetic moment is:

${\mu _e} = 1.410606633 \times {10^{ - 26}}J{T^{ - 1}} \tag{6.5}$

The NIST value of proton magnetic moment expresses in terms of quantum measurements as:

${\mu _p} = {g_p}{\lambda _C}^2{F_q}\frac{{e \cdot {e_{emax}}^2}}{{8\pi \cdot {e_{pmax}}^2}} \tag{6.6}$

where the proton g-factor is 5.58569 as listed on NIST. $${{e_{pmax}}^2}$$ is the magnetic charge of the proton, $${{e_{emax}}^2}$$ is the magnetic charge of the electron, and $$e$$ is the elementary charge.

The NIST value for the neutron magnetic moment notates as:

${\mu _n} = - 0.96623640 \times {10^{ - 26}}J{T^{ - 1}} \tag{6.7}$

and can be expressed as:

${\mu _n} = {g_{n - nist}}{\lambda _C}^2{F_q}\frac{{e \cdot {e_{emax}}^2}}{{8\pi \cdot {e_{pmax}}^2}} \tag{6.8}$

where $${g_{n - nist}}$$, the g-factor of the neutron, is -3.82608545 as defined by NIST, $${{e_{pmax}}^2}$$is the magnetic charge of the proton, $${{e_{emax}}^2}$$ is the magnetic charge of the electron, and $$e$$ is the elementary charge.  Notice that the equation balances by use of the magnetic charge of the proton instead of the neutron.  This is highly unlikely.

I am confident that the data used by NIST to produce these magnetic moment constants must be correct, as the equations above can be expressed in terms of quantum units (and g-factors).  However, it appears that the data for the neutron was misread, or the value for neutron g-factor was simply miscalculated.  That the neutron magnetic moment depends on the proton magnetic charge, and hence on the proton mass, seems impossible.

The above analysis also shows rather conclusively that all charge should distribute, including the elementary charge.  Based on Charles Coulomb’s observation that all charge must distribute in order for the force laws to work, and for consistency with the Aether Physics Model, we transpose the magnetic moment electrostatic charge dimensions.  The electron magnetic moment in the APM system is:

$emag = {g_e}{\lambda _C}^2{F_q}\frac{{e \cdot {e_{emax}}^2}}{{8\pi \cdot {e_{emax}}^2}} \tag{6.9}$

The proton magnetic moment in the APM system is:

$pmag = {g_p}{\lambda _C}^2{F_q}\frac{{e \cdot {e_{emax}}^2}}{{8\pi \cdot {e_{pmax}}^2}} \tag{6.10}$

And based on the NIST values for the neutron magnetic moment in the Standard Model, the neutron magnetic moment would be:

$nmag = {g_{n - nist}}{\lambda _C}^2{F_q}\frac{{e \cdot {e_{emax}}^2}}{{8\pi \cdot {e_{pmax}}^2}} tag{6.11}$

However, it is highly unlikely that nature has given a magnetic moment to the neutron, due to the magnetic charge of the proton.  So assuming the accuracy of the magnetic moment, correcting the quantum measurements changes the g-factor for the neutron:

$nmag = {g_n}{\lambda _C}^2{F_q}\frac{{e \cdot {e_{emax}}^2}}{{8\pi \cdot {e_{nmax}}^2}} \tag{6.12}$

The g-factor for the neutron must be -3.831359 if the neutron magnetic moment measurement is accurate.

From the expressions of magnetic moment in the Aether Physics Model, it appears that magnetic moment physically manifests by the interaction of the electrostatic and magnetic charges within each subatomic particle.  It is further apparent that the electron plays a key role in causing magnetic moment for each of the subatomic particles.

Comparing Magnetic Moments

Comparing the proportions of the NIST values to the APM values for the magnetic moment constants it appears that:

$emag = {\mu _e} \cdot e \tag{6.13}$

$pmag = {\mu _p} \cdot e \tag{6.14}$

$nmag = 0.999999{\mu _n} \cdot e \tag{6.15}$

Since the electron and proton magnetic moments can be calculated exactly by applying quantum measurements, and since the APM neutron magnetic moment is calculated using quantum measurements, it is likely that the above neutron magnetic moment error lies with the NIST value.

Based on the minor adjustments noted above, the actual value of neutron magnetic moment as calculated in the Standard model would be:

${\mu _n} = - 3.831359{\lambda _C}^2{F_q}\frac{{e \cdot {e_{emax}}^2}}{{8\pi \cdot {e_{nmax}}^2}} \tag{6.16}$

${\mu _n} = - 0.96623640 \times {10^{ - 26}}\frac{{{m^2} \cdot coul}}{{sec}} \tag{6.17}$

But whether this value of magnetic moment is useful or not would depend on how the equations used by NIST evolved.  If other adjustments compel compensation for the errors in the NIST constant, then the NIST formula will have to re-adjust as well.

Notice that each quantum measurement expression of magnetic moment includes the weak interaction constant of $$8\pi$$.  In addition, each magnetic moment unit resolves to a relationship between electrostatic and magnetic charge.  This indicates that the unit of magnetic moment directly relates to the dynamics of the weak nuclear interaction.

Changes

Some equations and laws need adjustment due to the new Aether Physics Model system of quantum measurement units, which bases on distributed charge.  For example, in the Standard Model, capacitance defines as charge divided by potential.

$C = \frac{Q}{V} \tag{6.18}$

However, in the Aether Physics Model all charge distributes, as Charles Coulomb pointed out.  Capacitance already has distributed units of charge in its dimensions, but charge and potential do not.  The effect is that when potential expresses in terms of distributed charge, Q disappears.  Therefore, it would be a prediction of the Aether Physics Model that capacitance is equal to the reciprocal of potential.

For capacitance to be related to charge, the Aether Physics Model dictates that charge is equal to capacitance times energy.

$Q = C\cdot E \tag{6.19}$

The charges specified in equations (6.18) and (6.19) are not elementary charge, rather they are magnetic charge.

Another important change regards the fundamental electromagnetic theories.  In modern electromagnetic theory, the $$B$$ field is magnetic flux density and the $$H$$ field is magnetic field intensity.  We learn from Clerk Maxwell that absolute permeability is equal to the ratio of $$B/H$$[1] as:

${\mu _0} = \frac{B}{H} \tag{6.20}$

But since the units of both magnetic flux density and magnetic field intensity should have distributed charge instead of single dimension charge:

$mfxd = \frac{{{m_e} \cdot {F_q}}}{{{e_{emax}}^2}} \tag{6.21}$

$mfdi = \frac{{{e_{emax}}^2 \cdot {F_q}}}{{{\lambda _C}}} \tag{6.22}$

The quantum measurement expression for equation yields:

$4\pi \cdot {\mu _0} = \frac{{mfxd \cdot chrg}}{{mfdi}} \tag{6.23}$

which suggests that the actual ratio of magnetic flux density to magnetic field intensity does not equal permeability.

Further, electromagnetic theory sees magnetic fields in terms of energy.

The total energy in any finite region of a magnetic field where the permeability is constant is the integral of the energy density over the volume or: $$W = \frac{1}{2}\int\limits_V {\mu {H^2}} dv$$[2]

The fact that the basic relation underlying modern electromagnetic theory does not fit into the Aether Physics Model does not negate over 100 years of electromagnetic theory.  However, if the Aether Physics Model is correct, all of electrodynamic theory needs reworking.

Instead of seeing magnetic fields in terms of energy, the Aether Physics Model sees them in terms of rotating magnetic field.  The Aether unit is itself the magnetic field.

New Units

After clarifying the definitions of dimension, measurement, and unit, it becomes possible to develop a system of quantum measurements, which allows for further development of quantum measurement analysis.

Ideally, quantum measurement analysis would mirror the physical processes of the observed physical world.  If this were true, we should be able to find a quantum measurement representation for every physical phenomenon.  Conversely, we should be able to find a physical process that matches any combination of quantum measurements.

In this section, we identify various new units.  The discovery of some units, like eddy current, actually occurred early in modern physics history and were either overlooked or discarded.  Other units have appeared unnoticed in modern physics equations all along, such as the photon.

A fully developed treatment of quantum measurement analysis would require another publication entirely.  Below is a small sampling of the new units utilized in the Aether Physics Model.  In most cases, the units could apply immediately to our understanding of physics.  In other cases, such as in understanding resonance, we need to review our measuring techniques.

Photon

In the Standard Model, the photon quantifies indirectly.  Instead of the photon, physicists quantify an energy packet and then treat it as though it were the photon itself.  This poor accounting creates many problems for the Standard Model.

In the Aether Physics Model, the photon defines in terms of the electron that produced it.  The electron is primary angular momentum and quantifies by Planck’s constant.  The photon then defines as the primary angular momentum of the electron times the speed of photons.

$phtn = h \cdot c \tag{6.24}$

Thus, the photon expands outward at the speed of photons and has the angular momentum of an electron.  As proposed by Cynthia Whitney[3], the photon remains connected to its source, even as it expands with cardioid geometry (see image page 158).

In the APM, there are two types, or “sizes,” of photons.  There is the electron/positron photon, and then there is a proton/antiproton photon.  The proton/antiproton photon hypothesizes to occur in fusion reactions and to generate via the same mechanics as the Casimir effect.  The quantification of the proton/antiproton photon is:

$pht{n_p} = {h_p} \cdot c \tag{6.25}$

where $${h_p}$$ is the APM value for proton angular momentum.

Light

Light comprises of quantum photons.  In the Standard Model, the photon packet of green light has a different frequency than the photon packet of red light; the different frequency means that each photon packet has a different energy from every other photon packet.  Further, if the mass/energy paradigm is used, the relativistic mass of each photon packet is different for each frequency of electromagnetic radiation.  Therefore, the photon packet of the Standard Model is not truly quantum.  The Standard Model presents an infinite number of various “sized” photon packets, one for each frequency. Unlike in the Standard Model, there is only one quantum photon in the Aether Physics Model.

In the Aether Physics Model, the photon is a true quantum.  To get light, photons produce in rapid succession at the frequency of the light.  Therefore, the unit of light is equal to photon times frequency.

$ligt = phtn \cdot freq \tag{6.26}$

An introduction to the mechanics of photons and light is on page 192.

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 is equal to magnetic flux squared.

$eddy = mfl{x^2} \tag{6.27}$

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:

$eddy = h \cdot resn \tag{6.28}$

Equation (6.28) 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.

$resn = time \cdot gyro \tag{6.29}$

Eddy current is also equal to potential times inductance.

$eddy = potn \cdot indc \tag{6.30}$

Eddy current is equal to inductance divided by capacitance:

$eddy = \frac{{indc}}{{capc}} = \frac{{{\mu _0}}}{{{\varepsilon _0}}} \tag{6.31}$

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

$eddy = mfld\frac{{momt}}{{chrg}} \tag{6.32}$

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 bottom 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.

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.  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 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.

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 (6.30).

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 Aether units in which they reside.

Gyration

The unit of gyration is equal to potential per charge:

$gyro = \frac{{potn}}{{chrg}} \tag{6.33}$

We discussed the eddy current unit on page 121.

Friction

Friction is a unit, which is equal to resistance times velocity.

$fric = resn \cdot velc \tag{6.34}$

Friction times charge is equal to rotating magnetic field.

$fric \cdot chrg = rmfd \tag{6.35}$

Understanding the friction unit helps in understanding the nature of resistance.  Take two objects, such as your hands, and press them together as though you were going to rub them.  As long as the two objects have lateral pressure but do not move, then only resistance is in effect.  When the objects are actually moving against each other, then friction is in effect.  Friction is resistance in motion.

In the discussion above concerning eddy current, eddy current is also equal to the friction applied to the ligamen circulatus of the subatomic particle.

Drag

The unit of drag is equal to the resistance times length.

$drag = resn \cdot leng \tag{6.36}$

When visualizing the unit of drag we would think of friction, except that instead of focusing on the moving resistance, we focus on the contact surface itself.  When charge drags against the Aether, it produces a magnetic field:

$drag \cdot chrg = mfld \tag{6.37}$

When angular momentum drags, it produces eddy current through a length:

$h \cdot drag = eddy \cdot leng \tag{6.38}$

Resonance

Distributed frequency is equal to resonance.  Viewing resonance in just one dimension of frequency is like viewing area in just one dimension of length.  The true meaning of resonance is lost when we change its dimensions.  The unit of resonance indicates there are two distinct dimensions of frequency involved.

$rson = fre{q^2} \tag{6.39}$

Modern physics does not measure capacitance and inductance as square roots, yet the resonance equation usually expresses as:

$F = \frac{1}{{2\pi \sqrt {LC} }} \tag{6.40}$

where $$F$$ is the “resonant frequency,” $$L$$ is the inductance and $$C$$ is the capacitance. (“Resonant frequency” is redundant and incorrect.  It is like saying “surface length.”)  Equation (6.40) loses much of its meaning by making it appear the inductance and capacitance measurements are square roots and expressing the resonance in terms of frequency.  It is as though modern physics has not yet discovered the unit of resonance.

To make the math of resonance compatible with the rest of physics, the correct expression would keep the natural measurements of inductance and capacitance and notate the result as frequency squared.  In the Aether Physics Model, the dimensions of resonance are equal to:

$rson = \frac{1}{{4\pi \cdot indc \cdot capc}} \tag{6.41}$

Equations (6.41) through (6.45) and the associated text were removed as they are in error.

The quantum realm exists in a five-dimensional space-resonance as opposed to a four-dimensional space-time.  If physicists wish to understand quantum existence properly, then we must design measurement equipment to measure directly in the resonance domain.  Presently, Fourier analysis attempts to account for this shortcoming by mathematically converting time domain measurements into frequency domain data.

The Aether Physics Model provides other ways to see resonance.  Earlier we demonstrated that $$potn$$ has the reciprocal dimensions of capacitance $$\left( {capc} \right)$$.  Therefore, resonance is equal to potential per inductance:

$rson = \frac{{potn}}{{indc}} \tag{6.46}$

The above equation manifests when winding a flat spiral secondary coil and then covering it with epoxy or some other dielectric.  If we seal the coil from electron leaks, the potential rises and so does the resonance.  When the coil is fully sealed, then the added dielectric increases the capacitance and the resonance decreases as in equation (6.47).

$rson = \frac{{curr}}{{capc \cdot h}} \tag{6.47}$

Capacitance times angular momentum is the product of the coil’s capacity to hold electrons times the number of electrons on one of the plates, or charge intensity.  Resonance is thus proportional to current and inversely proportional to the charge intensity.

Resonance relates to spherical geometry in the Aether unit.  The distributed frequency unit applies at the quantum level to produce space-resonance.  In the Aether unit graphic seen on the cover of this book, the two frequency dimensions are a source of space curvature.  Indeed, in acoustics, two longitudinal waves bounce through each other to produce a string of spheres.

The physics of resonance as distributed frequency extends to the macro realm of existence.  We can analyze a cylindrical pot of water with a vibration applied to its bottom.

Let us choose a 12” diameter pot and fill it with water.  The depth of the water is not important to this analysis, but we will choose six inches for the depth.  Applying a variable mechanical vibration to the bottom of the pot, we empirically discover maximum standing waves forming at 14.7Hz.  We then discover the distributed velocity of the water waves moving horizontally from the wall of the pot towards its center:

${\left( {14.7Hz} \right)^2} \cdot 2\pi {\left( {6in} \right)^2} = 31.534{\left( {\frac{m}{{sec}}} \right)^2} \tag{6.48}$

The resonance times the surface area is equal to the distributed velocity.  The distributed velocity is the average velocity of the water from the pot wall toward the center.  The distributed velocity is the product of the velocity in two orthogonal vectors and relates directly to the temperature of the water.

In quantum measurement units, however, the temperature of the water relates directly to the maximum temperature of quantum structures, as explained a little later.  Since the temperature of water involves distributed velocity far below the distributed speed of light, the value of the temp unit is very low.

$31.534{\left( {\frac{m}{{sec}}} \right)^2} = 3.509 \times {10^{ - 16}}temp \tag{6.49}$

The temperature scale at the macro level of our human existence depends upon the relative velocities of molecules, which are of a more complex order of existence than subatomic particles.  The reason that seemingly unrelated temperature units developed within physics is due to this complexity disparity between macro and quantum existence.  Further research must determine the scale factors between the various levels of complexity.  For now, we will simply refer to the result of equation (6.48) as “distributed velocity.”

The average distributed velocity of the water directly relates to the specific volume and average pressure of the water.

$vel{c^2} = spcv \cdot pres \tag{6.50}$

Empirically, we know the specific volume of water is equal to $$0.01602\frac{{f{t^3}}}{{lb}}$$, which in quantum measurement units equals $$63.781spcv$$.  Since we now have the average distributed velocity and specific volume of the water, we can determine the average pressure:

$\frac{{3.509 \times {{10}^{ - 16}}vel{c^2}}}{{63.781spcv}} = 5.589 \times {10^{ - 18}}pres = 3.204 \times {10^4}Pa \tag{6.51}$

Distributed velocity also relates to resonance in acoustics.  According to standard physics, the resonance of a vibrating string is equal to:

$F = \frac{1}{{2L}}\sqrt {\frac{T}{\rho }} \tag{6.52}$

where is the “resonant frequency”, is the length of the string, is the force applied to the string, and is the density of the string.[7]  Once again, it is obvious that resonance is not dependent upon the square root of force and density.  The quantum measurement units expression for the resonance of a string is:

$rson = \frac{{forc}}{{len{g^2} \cdot rbnd}} \tag{6.53}$

where $${rbnd}$$ (rebound) is the unit equal to mass per length in the Aether Physics Model.  Mass per length is also equal to line density.  Rebound is a measure of the strength for which an object with mass will reflect off an inelastic surface.  The greater the mass per length, the more intense the rebound will be.

Since we are dealing with resonance, there are two orthogonal frequencies involved; there is a wave of string traveling a velocity in one direction, and a wave of string traveling in the opposite direction.  In the fundamental quarter resonance there is one-half cycle between the ends of the string moving one direction and one-half cycle moving the opposite direction, which is inversely proportional to one-quarter of the total distributed wavelength.

$\frac{{rson}}{4} = \frac{{vel{c^2}}}{{4 \cdot len{g^2}}} \tag{6.54}$

The distributed velocity of the string depends upon the physical properties of the string and its environment.

It is clear that where equations show resonance as equal to the square root of measurements, the equations should express instead as distributed frequency.  Although such a change may meet initial resistance, it is essential to simplify physics by making it consistent throughout.  We will just have to get used to saying, “the resonance of an electrical circuit is equal to x [frequency unit] squared.”

Q Factor

The so-called “Q factor” of a coil indicates the “sharpness” of a resonance curve.  The Q factor is a dimensionless value derived from the following formula:

$Q = \frac{{\omega L}}{R} \tag{6.55}$

where $$\omega$$ is the frequency, $$L$$ is the inductance, and $$R$$ is the resistance.  In the APM, the unit represented by $$R$$ is actually magnetic flux.  The magnetic flux is a measure of the coil’s reactance, not its resistance.  In the APM, equation (6.55) expresses as:

$freq\cdot indc = mflx \tag{6.56}$

If the value for $$R$$ were measured as magnetic flux and with the correct charge dimensions, Q would always be equal to 1.  Since $$R$$ incorrectly measures as resistance, the Q factor is really an error factor, which also happens to coincide with the sharpness of the resonance.

The Aether Physics Model shows there is a balance between matter and environment and that minimizing the eddy current in the coil results in sharper resonance.  An identity arises from equations (6.46) and (6.47):

$\frac{{potn}}{{indc}} = \frac{{curr}}{{capc \cdot h}} \tag{6.57}$

We can transpose the identity such that:

$\frac{{potn \cdot h}}{{curr}} = \frac{{indc}}{{capc}} \tag{6.58}$

The value of $$h$$ is Planck’s constant.  The potential, current, and Planck’s constant are characteristics of the electron (matter), and inductance, and capacitance are characteristics of the Aether (environment).  Each side of equation (6.58) quantifies eddy current:

$\begin{array}{l} \frac{{potn \cdot h}}{{curr}} = eddy \\ \frac{{indc}}{{capc}} = eddy \\ \end{array} \tag{6.59}$

Minimizing the eddy current by changing the material and environmental characteristics of the coil increases the sharpness of the resonance.

Diverging Electric Field

The diverging electric field has a unit of its own and it is equal to electric field strength per length:

$dvef = \frac{{elfs}}{{leng}} \tag{6.60}$

Diverging electric field is also equal to electromagnetism (mass to magnetic charge ratio) times resonance:

$dvef = mchg \cdot rson \tag{6.61}$

Irradiance is expressed as diverging electric field times current:

$irrd = dvef \cdot curr \tag{6.62}$

Traditionally, irradiance is equal to power per area:

$irrd = \frac{{powr}}{{area}} \tag{6.63}$

Temperature

In the Standard Model, temperature appears as a dimension of its own and unrelated to the dimensions of length, time, mass and charge.  However, in the Aether Physics Model temperature is equal to velocity squared.  This makes sense since temperature defines as motion among colliding bodies.

$temp = {\lambda _C}^2 \cdot {F_q}^2 \tag{6.64}$

Defining temperature as “molecules in motion” is not enough, however.  Because there are different orders of reality, and molecules are just one order, distributed velocity must manifest in slightly different ways for each order of existence.  An electron exists in one fourth of the total available spin positions in the Aether, yet Aether directly encapsulates it.  The Aether exists in five-dimensional reality even though the electron only manifests four dimensions.  If we define a unit such as temperature as “molecules in motion,” we are missing key aspects of reality relevant to quantum existence.

Molecules, although composed of subatomic particles, exist on a larger scale.  There are new dimensions of existence added as complexity increases.  For example, the perception of color does not exist at the quantum level, but does exist at the level of animals, plants and minerals.  It is in this sense that temperature does not exist at the quantum level.  Although electrons and protons experience distributed velocity, they do not change state among gas, liquid, and solid, but produce plasma, instead.

Radiation is a case of distributed velocity moving in only one direction, namely outward from its source.  The case of standing waves is a case of distributed velocity moving one direction and then reflecting in the opposite direction.  The case of temperature specifically relates to the orders of atoms and molecules, which produce standing waves by bouncing off each other.

We developed our temperature scales of Celsius, Kelvin, and Fahrenheit specifically for measuring the distributed velocity within atoms and molecules bouncing off each other, which is why temperature seems to both relate to, and be in conflict with, our concept of radiation.  There is really no single term available having the same meaning as the phrase “distributed velocity,” and which applies to all of its manifestations.

The relationship of temperature to energy is:

$enrg = mass \cdot temp \tag{6.65}$

Knowing that 273.15K times 1.2929 kg/m3 equals one atmosphere, we can calculate the conversion factor for Kelvin to temp units:

$K = \frac{{\frac{{atm}}{{1.2929\frac{{kg}}{{{m^3}}}}}}}{{273.15}} \tag{6.66}$

$K = 286.91Sv \tag{6.67}$

$K = 3.19 \times {10^{ - 15}}temp \tag{6.68}$

Nevertheless, the unit for measuring molecules in motion does not directly apply to the unit for unidirectional radiation.  It is necessary to account for scaling factors.

Units Grid

Sometimes the lack of something speaks volumes.  In all of modern physics, nobody has made the effort to systematize all the known units.  This is understandable since modern physics has the wrong dimensions for charge, which makes it difficult to find meaningful patterns in unit structure.

The following tables show several groups of units in both their obverse and inverse expressions.  All of the known units are included.  Many of the units presented remain absent in modern physics.  Even with the addition of many new units, it is apparent that we have not even come close to identifying all the different manifestations of non-material existence.  The unit of eddy current does not fit into the table structure.  Also, there are at least two electromagnetic tables not included since they have no entries.

Some units have multiple expressions, but only one is given.  We present merely a beginning of the topic in this chapter and the tables below.

Supportive Electromagnetic Units

Obverse Units

Rotating Magnetic Field

Magnetic Field

Magnetic Volume

$${A_u} = \frac{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}^2}}{{{e_{emax}}^2}}$$  $$mfld = \frac{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}}}{{{e_{emax}}^2}}$$  $$mvlm = \frac{{{m_e} \cdot {\lambda _C}^3}}{{{e_{emax}}^2}}$$

Electric Potential

Magnetic Flux

Inductance

$$potn = \frac{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}^2}}{{{e_{emax}}^2}}$$  $$mflx = \frac{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}}}{{{e_{emax}}^2}}$$  $$indc = \frac{{{m_e} \cdot {\lambda _C}^2}}{{{e_{emax}}^2}}$$

Electric Field Strength

Magnetic Momentum

Permeability

$$elfs = \frac{{{m_e} \cdot {\lambda _C} \cdot {F_q}^2}}{{{e_{emax}}^2}}$$  $$emgm = \frac{{{m_e} \cdot {\lambda _C} \cdot {F_q}}}{{{e_{emax}}^2}}$$  $$perm = \frac{{{m_e} \cdot {\lambda _C}}}{{{e_{emax}}^2}}$$

Diverging Electric Field

Magnetic Flux Density

Electromagnetism

$$dvef = \frac{{{m_e} \cdot {F_q}^2}}{{{e_{emax}}^2}}$$  $$mfxd = \frac{{{m_e} \cdot {F_q}}}{{{e_{emax}}^2}}$$  $$mchg = \frac{{{m_e}}}{{{e_{emax}}^2}}$$

Inverse Units

Permittivity

$$? = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C}^3}}$$  $$? = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}}}$$  $$ptty = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}^2}}$$

Conductance

Capacitance

$$? = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C}^2}}$$  $$cond = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}}}$$  $$capc = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}^2}}$$

Curl?

Conductance Momentum

$$? = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C}}}$$  $$cmom = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C} \cdot {F_q}}}$$  $$? = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {\lambda _C} \cdot {F_q}^2}}$$

Exposure

Conductance Density

$$expr = \frac{{{e_{emax}}^2}}{{{m_e}}}$$  $$cden = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {F_q}}}$$  $$? = \frac{{{e_{emax}}^2}}{{{m_e} \cdot {F_q}^2}}$$

Opposing Electromagnetic Units

Obverse Units

Friction

Drag

Vorticular Opposition

$$fric = \frac{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}^2}}{{{e_{emax}}^4}}$$  $$drag = \frac{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}}}{{{e_{emax}}^4}}$$ $$vopp = \frac{{{m_e} \cdot {\lambda _C}^3}}{{{e_{emax}}^4}}$$

Rub

Resistance

Angular Opposition

$$rub = \frac{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}^2}}{{{e_{emax}}^4}}$$  $$resn = \frac{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}}}{{{e_{emax}}^4}}$$  $$aopp = \frac{{{m_e} \cdot {\lambda _C}^2}}{{{e_{emax}}^4}}$$

Plow

Skid

Linear Opposition

$$plow = \frac{{{m_e} \cdot {\lambda _C} \cdot {F_q}^2}}{{{e_{emax}}^4}}$$  $$skid = \frac{{{m_e} \cdot {\lambda _C} \cdot {F_q}}}{{{e_{emax}}^4}}$$  $$lopp = \frac{{{m_e} \cdot {\lambda _C}}}{{{e_{emax}}^4}}$$

Hold

Stop

Electromagnetic Opposition

$$hold = \frac{{{m_e} \cdot {F_q}^2}}{{{e_{emax}}^4}}$$  $$stop = \frac{{{m_e} \cdot {F_q}}}{{{e_{emax}}^4}}$$  $$eopp = \frac{{{m_e}}}{{{e_{emax}}^4}}$$

Inverse Units

$$? = \frac{{{e_{emax}}^4}}{{{m_e} \cdot {\lambda _C}^3}}$$  $$? = \frac{{{e_{emax}}^4}}{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}}}$$  $$? = \frac{{4\pi \cdot {e_{emax}}^4}}{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}^2}}$$

$$? = \frac{{{e_{emax}}^4}}{{{m_e} \cdot {\lambda _C}^2}}$$  $$? = \frac{{{e_{emax}}^4}}{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}}}$$  $$? = \frac{{4\pi \cdot {e_{emax}}^4}}{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}^2}}$$

$$? = \frac{{{e_{emax}}^4}}{{{m_e} \cdot {\lambda _C}}}$$  $$? = \frac{{{e_{emax}}^4}}{{{m_e} \cdot {\lambda _C} \cdot {F_q}}}$$  $$? = \frac{{4\pi \cdot {e_{emax}}^4}}{{{m_e} \cdot {\lambda _C} \cdot {F_q}^2}}$$

$$? = \frac{{{e_{emax}}^4}}{{{m_e}}}$$  $$? = \frac{{{e_{emax}}^4}}{{{m_e} \cdot {F_q}}}$$  $$? = \frac{{{e_{emax}}^4}}{{{m_e} \cdot {F_q}^2}}$$

Electric Units 1

Obverse Units

$$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}^3 \cdot {F_q}^3}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}^3 \cdot {F_q}^2}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}^3 \cdot {F_q}}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}^3}}$$

$$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}^2 \cdot {F_q}^3}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}^2 \cdot {F_q}^2}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}^2 \cdot {F_q}}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}^2}}$$

$$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C} \cdot {F_q}^3}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C} \cdot {F_q}^2}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C} \cdot {F_q}}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {\lambda _C}}}$$

$$? = \frac{1}{{{e_{emax}}^2 \cdot {F_q}^3}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {F_q}^2}}$$  $$? = \frac{1}{{{e_{emax}}^2 \cdot {F_q}}}$$  $$? = \frac{1}{{{e_{emax}}^2}}$$

Inverse Units

Charge Volume

$$? = {e_{emax}}^2 \cdot {\lambda _C}^3$$  $$? = {e_{emax}}^2 \cdot {\lambda _C}^3 \cdot {F_q}$$  $$? = {e_{emax}}^2 \cdot {\lambda _C}^3 \cdot {F_q}^2$$  $$? = {e_{emax}}^2 \cdot {\lambda _C}^3 \cdot {F_q}^3$$

Surface Charge

Magnetic Moment

Plasma

Ball Lightning?

$$? = {e_{emax}}^2 \cdot {\lambda _C}^2$$  $$? = {e_{emax}}^2 \cdot {\lambda _C}^2 \cdot {F_q}$$  $$? = {e_{emax}}^2 \cdot {\lambda _C}^2 \cdot {F_q}^2$$  $$? = {e_{emax}}^2 \cdot {\lambda _C}^2 \cdot {F_q}^3$$

Charge Length

(Charge Displacement)

Charge Velocity

Charge Acceleration

$$? = {e_{emax}}^2 \cdot {\lambda _C}$$  $$? = {e_{emax}}^2 \cdot {\lambda _C} \cdot {F_q}$$  $$? = {e_{emax}}^2 \cdot {\lambda _C} \cdot {F_q}^2$$  $$? = {e_{emax}}^2 \cdot {\lambda _C} \cdot {F_q}^3$$

Charge

Current

Charge Resonance

(Electric Coupling)

$$? = {e_{emax}}^2$$  $$? = {e_{emax}}^2 \cdot {F_q}$$  $$? = {e_{emax}}^2 \cdot {F_q}^2$$  $$? = {e_{emax}}^2 \cdot {F_q}^3$$

Electric Units 2

Obverse Units

Specific Charge

$$? = \frac{{{\lambda _C}^3}}{{{e_{emax}}^2 \cdot {F_q}^3}}$$  $$? = \frac{{{\lambda _C}^3}}{{{e_{emax}}^2 \cdot {F_q}^2}}$$  $$? = \frac{{{\lambda _C}^3}}{{{e_{emax}}^2 \cdot {F_q}}}$$  $$spch = \frac{{{\lambda _C}^3}}{{{e_{emax}}^2}}$$

Charge Distribution

$$? = \frac{{{\lambda _C}^2}}{{{e_{emax}}^2 \cdot {F_q}^3}}$$  $$? = \frac{{{\lambda _C}^2}}{{{e_{emax}}^2 \cdot {F_q}^2}}$$  $$? = \frac{{{\lambda _C}^2}}{{{e_{emax}}^2 \cdot {F_q}}}$$  $$chds = \frac{{{\lambda _C}^2}}{{{e_{emax}}^2}}$$

$$? = \frac{{{\lambda _C}}}{{{e_{emax}}^2 \cdot {F_q}^3}}$$  $$? = \frac{{{\lambda _C}}}{{{e_{emax}}^2 \cdot {F_q}^2}}$$  $$? = \frac{{{\lambda _C}}}{{{e_{emax}}^2 \cdot {F_q}}}$$  $$chgr = \frac{{{\lambda _C}}}{{{e_{emax}}^2}}$$

Inverse Units

Charge Density

$$chgd = \frac{{{e_{emax}}^2}}{{{\lambda _C}^3}}$$  $$? = \frac{{{e_{emax}}^2 \cdot {F_q}}}{{{\lambda _C}^3}}$$  $$? = \frac{{{e_{emax}}^2 \cdot {F_q}^2}}{{{\lambda _C}^3}}$$  $$? = \frac{{{e_{emax}}^2 \cdot {F_q}^3}}{{{\lambda _C}^3}}$$

Electric Flux Density

Current Density

$$efxd = \frac{{{e_{emax}}^2}}{{{\lambda _C}^2}}$$  $$cdns = \frac{{{e_{emax}}^2 \cdot {F_q}}}{{{\lambda _C}^2}}$$  $$? = \frac{{{e_{emax}}^2 \cdot {F_q}^2}}{{{\lambda _C}^2}}$$  $$? = \frac{{{e_{emax}}^2 \cdot {F_q}^3}}{{{\lambda _C}^2}}$$

Magnetic Field Intensity

$$? = \frac{{{e_{emax}}^2}}{{{\lambda _C}}}$$  $$mfdi = \frac{{{e_{emax}}^2 \cdot {F_q}}}{{{\lambda _C}}}$$  $$? = \frac{{{e_{emax}}^2 \cdot {F_q}^2}}{{{\lambda _C}}}$$  $$? = \frac{{{e_{emax}}^2 \cdot {F_q}^3}}{{{\lambda _C}}}$$

Electric Field Units

Obverse Units

Varying Electric Field

Electric Field

Specific Charge

$$? = \frac{{{\lambda _C}^3 \cdot {F_q}^3}}{{{e_{emax}}^2}}$$  $$vefd = \frac{{{\lambda _C}^3 \cdot {F_q}^2}}{{{e_{emax}}^2}}$$  $$efld = \frac{{{\lambda _C}^3 \cdot {F_q}}}{{{e_{emax}}^2}}$$  $$spch = \frac{{{\lambda _C}^3}}{{{e_{emax}}^2}}$$
Charge Temperature  Charge Sweep
$$? = \frac{{{\lambda _C}^2 \cdot {F_q}^3}}{{{e_{emax}}^2}}$$  $$chgt = \frac{{{\lambda _C}^2 \cdot {F_q}^2}}{{{e_{emax}}^2}}$$  $$chgs = \frac{{{\lambda _C}^2 \cdot {F_q}}}{{{e_{emax}}^2}}$$  $$? = \frac{{{\lambda _C}^2}}{{{e_{emax}}^2}}$$
Charge Acceleration  Charge Velocity

$$? = \frac{{{\lambda _C} \cdot {F_q}^3}}{{{e_{emax}}^2}}$$  $$chga = \frac{{{\lambda _C} \cdot {F_q}^2}}{{{e_{emax}}^2}}$$  $$chgv = \frac{{{\lambda _C} \cdot {F_q}}}{{{e_{emax}}^2}}$$  $$chgr = \frac{{{\lambda _C}}}{{{e_{emax}}^2}}$$
Charge Resonance  Charge Frequency  Charge
$$? = \frac{{{F_q}^3}}{{{e_{emax}}^2}}$$  $$chgr = \frac{{{F_q}^2}}{{{e_{emax}}^2}}$$  $$chgf = \frac{{{F_q}}}{{{e_{emax}}^2}}$$  $$chrg = \frac{1}{{{e_{emax}}^2}}$$

Inverse Units

Charge Density

$$chgd = \frac{{{e_{emax}}^2}}{{{\lambda _C}^3}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C}^3 \cdot {F_q}}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C}^3 \cdot {F_q}^2}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C}^3 \cdot {F_q}^3}}$$

Electric Flux Density

$$efxd = \frac{{{e_{emax}}^2}}{{{\lambda _C}^2}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C}^2 \cdot {F_q}}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C}^2 \cdot {F_q}^2}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C}^2 \cdot {F_q}^3}}$$

$$? = \frac{{{e_{emax}}^2}}{{{\lambda _C}}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C} \cdot {F_q}}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C} \cdot {F_q}^2}}$$  $$? = \frac{{{e_{emax}}^2}}{{{\lambda _C} \cdot {F_q}^3}}$$

Charge

$$chrg = {e_{emax}}^2$$  $$? = \frac{{{e_{emax}}^2}}{{{F_q}}}$$  $$? = \frac{{{e_{emax}}^2}}{{{F_q}^2}}$$  $$? = \frac{{{e_{emax}}^2}}{{{F_q}^3}}$$

Inertial Units 1

Obverse Units

Light

Photon

Rotation

Vortex

$$ligt = {m_e} \cdot {\lambda _C}^3 \cdot {F_q}^3$$  $$phtn = {m_e} \cdot {\lambda _C}^3 \cdot {F_q}^2$$  $$rota = {m_e} \cdot {\lambda _C}^3 \cdot {F_q}$$  $$vrtx = {m_e} \cdot {\lambda _C}^3$$

Power

Energy

Angular Momentum

Moment of Inertia

$$powr = {m_e} \cdot {\lambda _C}^2 \cdot {F_q}^3$$  $$enrg = {m_e} \cdot {\lambda _C}^2 \cdot {F_q}^2$$  $$h = {m_e} \cdot {\lambda _C}^2 \cdot {F_q}$$  $$minr = {m_e} \cdot {\lambda _C}^2$$

Shock Frequency or Light Intensity

Force

Momentum

Torque

$$shkf = {m_e} \cdot {\lambda _C} \cdot {F_q}^3$$  $$forc = {m_e} \cdot {\lambda _C} \cdot {F_q}^2$$  $$momt = {m_e} \cdot {\lambda _C} \cdot {F_q}$$  $$torq = {m_e} \cdot {\lambda _C}$$

Surface Tension

Intensity

Mass

$$irrd = {m_e} \cdot {F_q}^3$$  $$sten = {m_e} \cdot {F_q}^2$$  $$ints = {m_e} \cdot {F_q}$$  $$mass = {m_e}$$

Inverse Units

$$? = \frac{1}{{{m_e} \cdot {\lambda _C}^3}}$$  $$? = \frac{1}{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}}}$$ $$? = \frac{1}{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}^2}}$$   $$? = \frac{1}{{{m_e} \cdot {\lambda _C}^3 \cdot {F_q}^3}}$$

$$? = \frac{1}{{{m_e} \cdot {\lambda _C}^2}}$$  $$? = \frac{1}{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}}}$$  $$? = \frac{1}{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}^2}}$$  $$? = \frac{1}{{{m_e} \cdot {\lambda _C}^2 \cdot {F_q}^3}}$$

$$? = \frac{1}{{{m_e} \cdot {\lambda _C}}}$$   $$? = \frac{1}{{{m_e} \cdot {\lambda _C} \cdot {F_q}}}$$  $$? = \frac{1}{{{m_e} \cdot {\lambda _C} \cdot {F_q}^2}}$$ $$? = \frac{1}{{{m_e} \cdot {\lambda _C} \cdot {F_q}^3}}$$

$$? = \frac{1}{{{m_e}}}$$ $$? = \frac{1}{{{m_e} \cdot {F_q}}}$$   $$? = \frac{1}{{{m_e} \cdot {F_q}^2}}$$ $$? = \frac{1}{{{m_e} \cdot {F_q}^3}}$$

Inertial Units 2

Obverse Units

Mass Density

$$? = \frac{{{m_e} \cdot {F_q}^3}}{{{\lambda _C}^3}}$$  $$? = \frac{{{m_e} \cdot {F_q}^2}}{{{\lambda _C}^3}}$$  $$? = \frac{{{m_e} \cdot {F_q}}}{{{\lambda _C}^3}}$$  $$masd = \frac{{{m_e}}}{{{\lambda _C}^3}}$$

Force Density fdns

Surface Density

$$? = \frac{{{m_e} \cdot {F_q}^3}}{{{\lambda _C}^2}}$$  $$? = \frac{{{m_e} \cdot {F_q}^2}}{{{\lambda _C}^2}}$$  $$? = \frac{{{m_e} \cdot {F_q}}}{{{\lambda _C}^2}}$$  $$sfcd = \frac{{{m_e}}}{{{\lambda _C}^2}}$$

Pressure

Viscosity

Rebound

$$? = \frac{{{m_e} \cdot {F_q}^3}}{{{\lambda _C}}}$$ $$pres = \frac{{{m_e} \cdot {F_q}^2}}{{{\lambda _C}}}$$  $$visc = \frac{{{m_e} \cdot {F_q}}}{{{\lambda _C}}}$$  $$rbnd = \frac{{{m_e}}}{{{\lambda _C}}}$$

Inverse Units

Specific Volume

$$spcv = \frac{{{\lambda _C}^3}}{{{m_e}}}$$  $$? = \frac{{{\lambda _C}^3}}{{{m_e} \cdot {F_q}}}$$   $$? = \frac{{{\lambda _C}^3}}{{{m_e} \cdot {F_q}^2}}$$ $$? = \frac{{{\lambda _C}^3}}{{{m_e} \cdot {F_q}^3}}$$

$$? = \frac{{{\lambda _C}^2}}{{{m_e}}}$$   $$? = \frac{{{\lambda _C}^2}}{{{m_e} \cdot {F_q}}}$$ $$? = \frac{{{\lambda _C}^2}}{{{m_e} \cdot {F_q}^2}}$$   $$? = \frac{{{\lambda _C}^2}}{{{m_e} \cdot {F_q}^3}}$$

$$? = \frac{{{\lambda _C}}}{{{m_e}}}$$  $$? = \frac{{{\lambda _C}}}{{{m_e} \cdot {F_q}}}$$ $$? = \frac{{{\lambda _C}}}{{{m_e} \cdot {F_q}^2}}$$   $$? = \frac{{{\lambda _C}}}{{{m_e} \cdot {F_q}^3}}$$

Inertial Units 3

Obverse Units

$$? = \frac{{{m_e}}}{{{\lambda _C}^3 \cdot {F_q}^3}}$$  $$? = \frac{{{m_e}}}{{{\lambda _C}^3 \cdot {F_q}^2}}$$  $$? = \frac{{{m_e}}}{{{\lambda _C}^3 \cdot {F_q}}}$$

$$? = \frac{{{m_e}}}{{{\lambda _C}^2 \cdot {F_q}^3}}$$ $$? = \frac{{{m_e}}}{{{\lambda _C}^2 \cdot {F_q}^2}}$$ $$? = \frac{{{m_e}}}{{{\lambda _C}^2 \cdot {F_q}}}$$

$$? = \frac{{{m_e}}}{{{\lambda _C} \cdot {F_q}^3}}$$ $$? = \frac{{{m_e}}}{{{\lambda _C} \cdot {F_q}^2}}$$ $$? = \frac{{{m_e}}}{{{\lambda _C} \cdot {F_q}}}$$

Inverse Units

$$? = \frac{{{\lambda _C}^3 \cdot {F_q}}}{{{m_e}}}$$ $$? = \frac{{{\lambda _C}^3 \cdot {F_q}^2}}{{{m_e}}}$$  $$? = \frac{{{\lambda _C}^3 \cdot {F_q}^3}}{{{m_e}}}$$

$$? = \frac{{{\lambda _C}^2 \cdot {F_q}}}{{{m_e}}}$$ $$? = \frac{{{\lambda _C}^2 \cdot {F_q}^2}}{{{m_e}}}$$ $$? = \frac{{{\lambda _C}^2 \cdot {F_q}^3}}{{{m_e}}}$$

$$? = \frac{{{\lambda _C} \cdot {F_q}}}{{{m_e}}}$$ $$? = \frac{{{\lambda _C} \cdot {F_q}^2}}{{{m_e}}}$$ $$? = \frac{{{\lambda _C} \cdot {F_q}^3}}{{{m_e}}}$$

Length/Frequency Units 1

Obverse Units

Volume-Resonance

Flow

Volume

$$dtrd = {\lambda _C}^3 \cdot {F_q}^2$$  $$flow = {\lambda _C}^3 \cdot {F_q}$$ $$volm = {\lambda _C}^3$$

Radiation Dose or Temperature

Sweep or Angular Velocity

Area

$$temp = {\lambda _C}^2 \cdot {F_q}^2$$ $$swep = {\lambda _C}^2 \cdot {F_q}$$   $$area = {\lambda _C}^2$$

Acceleration

Velocity

Line

$$accl = {\lambda _C} \cdot {F_q}^2$$   $$velc = {\lambda _C} \cdot {F_q}$$  $$line = {\lambda _C}$$

Resonance

Frequency

$$rson = {F_q}^2$$  $$freq = {F_q}$$

Inverse Units

$$? = \frac{1}{{{\lambda _C}^3}}$$ $$? = \frac{1}{{{\lambda _C}^3 \cdot {F_q}}}$$  $$? = \frac{1}{{{\lambda _C}^3 \cdot {F_q}^2}}$$

$$? = \frac{1}{{{\lambda _C}^2}}$$  $$? = \frac{1}{{{\lambda _C}^2 \cdot {F_q}}}$$  $$? = \frac{1}{{{\lambda _C}^2 \cdot {F_q}^2}}$$

Wavenumber

$$wavn = \frac{1}{{{\lambda _C}}}$$  $$? = \frac{1}{{{\lambda _C} \cdot {F_q}}}$$ $$? = \frac{1}{{{\lambda _C} \cdot {F_q}^2}}$$

Time

Orbit

$$time = \frac{1}{{{F_q}}}$$  $$orbt = \frac{1}{{{F_q}^2}}$$

Length/Frequency Units 2

Obverse Units

Volume-Time

$$? = \frac{{{\lambda _C}^3}}{{{F_q}^3}}$$ $$? = \frac{{{\lambda _C}^3}}{{{F_q}^2}}$$   $$vlmt = \frac{{{\lambda _C}^3}}{{{F_q}}}$$

Active Area

$$? = \frac{{{\lambda _C}^2}}{{{F_q}^3}}$$  $$? = \frac{{{\lambda _C}^2}}{{{F_q}^2}}$$ $$acta = \frac{{{\lambda _C}^2}}{{{F_q}}}$$

Dynamic Length

$$? = \frac{{{\lambda _C}}}{{{F_q}^3}}$$ $$? = \frac{{{\lambda _C}}}{{{F_q}^2}}$$  $$dynl = \frac{{{\lambda _C}}}{{{F_q}}}$$

Inverse Units

$$? = \frac{{{F_q}}}{{{\lambda _C}^3}}$$  $$? = \frac{{{F_q}^2}}{{{\lambda _C}^3}}$$  $$? = \frac{{{F_q}^3}}{{{\lambda _C}^3}}$$

$$? = \frac{{{F_q}}}{{{\lambda _C}^2}}$$ $$? = \frac{{{F_q}^2}}{{{\lambda _C}^2}}$$  $$? = \frac{{{F_q}^3}}{{{\lambda _C}^2}}$$

Scalar Wave

$$sclw = \frac{{{F_q}}}{{{\lambda _C}}}$$  $$? = \frac{{{F_q}^2}}{{{\lambda _C}}}$$  $$? = \frac{{{F_q}^3}}{{{\lambda _C}}}$$

More Example Calculations

We 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.

The 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 $$14.142cm$$.

$15.759\frac{{cm}}{{sec}} \cdot 897.4msec = 14.142cm \tag{6.69}$

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 adjusting for the different charge dimensions.

$620m\Omega \cdot \frac{{3.382 \times {{10}^{40}}}}{{cou{l^2}}} = .620resn \tag{6.70}$

Notice that the value for $$\Omega$$ is the same as the value for $$resn$$.  It will always be so.  The total averaged electrons dragged at any moment along the magnet’s fall are:

$.620resn \cdot 14.142cm = 3.614 \times {10^{10}}drag \tag{6.71}$

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

$3.614 \times {10^{10}}drag \cdot chrg = 3.614 \times {10^{10}}mfld \tag{6.72}$

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 $$3.614 \times {10^{10}}$$ dragging electrons and $$3.614 \times {10^{10}}$$ Aether units.

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

$\frac{{3.614 \times {{10}^{10}}mfld}}{{14.142cm}} = .62mflx \tag{6.73}$

Converting $$mflx$$ to $$weber$$:

$.62mflx \cdot 2.112 \times {10^{ - 4}}coul = .62weber \tag{6.74}$

Notice that once again the value of magnetic flux is exactly equal to the value of $$weber$$.  This holds true for potential, current, and most other units as well.

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.

Other tests for the accuracy of quantum measurement units are easily verifiable.  For example:

$5A \cdot 2\Omega = 5v \tag{6.75}$

$5curr \cdot 2resn = 5potn \tag{6.76}$

It makes sense that if the quantum measurements are accurate for Ohm’s law, then they will also be accurate for the newly identified quantum measurements presented above.

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 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:

${E_k} = \frac{{m{v^2}}}{2} \tag{6.77}$

If $${E_k}$$ is a unit of energy, then equation (7.77) 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:

$\frac{E}{2} = \frac{{m{v^2}}}{2} \tag{6.78}$

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 $$300, thus the economy has a net gain of zero dollars, is indicative of the conservation of energy law. According to the standard explanation of kinetic energy, it has no direction, being a scalar quantity. Nevertheless, since dimensions comprise all units, and since dimensions have a more primary nature than units, the units must obtain their characteristics from the dimensions. A falling object has direction toward the ground, which sees a falling body directed toward it. From the perspective of the ground, it is as though the ground were moving toward the falling object. Length and frequency have direction, nullifying the arbitrary statement that “kinetic energy has no direction.” Since length and frequency dimensions do have direction, velocity, and ultimately energy, they must also have direction. Since half-spin subatomic particles only see the forward direction of frequency, then all quantum frequency must yield positive time. But the length dimensions can be both positive and negative and thus yield both positive and negative distance. In the financial analogy, the employer’s wealth is decreasing during the transaction while the employee’s wealth is increasing. This is true even though the paycheck remains the same value throughout the transaction and moves unidirectionally from employer to employee. The paycheck is merely an instrument of exchange. The employer and employee are the real parties to the transaction. Similarly, kinetic energy is always associated with moving objects, such as electrons, photons, or swinging balls. The kinetic energy of the object is merely the instrument of the energy exchange between the objects. As in the financial transaction, the total change of energy state is equal to twice the kinetic energy. One might ask, “What does the employee care about the employer’s wealth decreasing by$$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 a mass of 1kg hangs a distance of 10m above the Earth.  The gravitational 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:

$1kg \cdot 10m \cdot g = 98.066joule \tag{6.79}$

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:

$\frac{{98.066joule}}{{5.98 \times {{10}^{24}}kg}} = 1.672 \times {10^{ - 24}}m \tag{6.80}$

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 kinetic energy of the Earth with respect to the falling object equal to:

$- 1.672 \times {10^{ - 24}}m \cdot 5.98 \times {10^{24}}kg = - 98.052joule \tag{6.81}$

So not only does the object have a positive phase kinetic energy of about $$98joule$$, but the Earth also has a negative phase kinetic energy of about $$- 98joule$$ at the time of impact.  And since the energy is a vector quantity, the Earth’s negative phase kinetic 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 kinetic 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:

$E = \frac{{{E_O}}}{2} + \frac{{{E_E}}}{2} \tag{6.82}$

where $${{E_O}}$$ is the potential energy of the object and $${{E_E}}$$ is the kinetic energy of the Earth.  The net energy gain of the system is equal to:

$E = \frac{{{E_O}}}{2} + \frac{{{E_E}}}{2} = 0 \tag{6.83}$

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.

$E = \frac{{{E_O}}}{2} + \frac{{{E_f}}}{2} \tag{6.84}$

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 force constant of the Earth.

$- \frac{E}{2} = m \cdot - h \cdot g \tag{6.85}$

Equation (6.85) 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.

Because of the conservation of energy law for any full cycle of motion, the positive phase kinetic energy minus the negative phase kinetic energy minus the friction loss will equal zero:

$\frac{{{E_p}}}{2} - \frac{{{E_n}}}{2} - \frac{{{E_f}}}{2} = 0 \tag{6.86}$

where $${{E_p}}$$ is positive phase kinetic energy and $${{E_n}}$$ is negative phase kinetic energy.

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.  The net kinetic 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:

$\frac{{m{v^2}}}{2} + 0 = \frac{{m{v^2}}}{2} + \frac{{m{v^2}}}{2}\tag{6.87}$  [9]

or

$\frac{{{E_O}}}{2} + 0 = \frac{{{E_O}}}{2} + \frac{{{E_f}}}{2} \tag{6.88}$

and therefore it is assumed that:

$\frac{{m{v^2}}}{2} = E \tag{6.89}$

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.

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

[1] Warren B. Boast Principles of Electric and Magnetic Fields (Harper & Brothers, New York, 1948) 173

[2] Warren B. Boast Principles of Electric and Magnetic Fields (Harper & Brothers, New York, 1948) 179

[3] Whitney, Cynthia Kolb, Essay 1: This is Not Einstein’s Postulate (Galilean Electrodynamics, Space Time Analysis LTD, Winter 2005) pp 43-44

[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 41

[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