# Two Component Aether

The two component, solid aether theory is a proposed theoretical framework of physics, specifically applying to the field of Classical electromagnetism and relevant to particle physics. It was proposed by D. J. Larson in 1998.[1] The proposal involves some relatively simple models as well as two elementary force laws for the underlying aether, and then goes on to show how Maxwell's equations follow. The theory is not currently accepted as probable by most mainstream physicists. If true, however, it would represent a major paradigm shift.

The two component aether model proposes a Luminiferous aether that is made of two incompressible substances; one is called positive aether and the other is called negative aether. Each of these substances is in a solid state, so that the aether is capable of transmitting transversely polarized waves. (It is known that Light is transversely polarized, and that transversely polarized waves can typically form only in solids.) It is then assumed that under normal conditions each small volume of positive aether is attached to a corresponding small volume of negative aether. However, this attachment can be broken by putting enough energy into the aether, resulting in a situation where free (detached) aether is formed. Further, two Force laws are postulated. The first force law results from an assumption that each aetherial substance is under tension. The second force law is based on a proposal that forces result when one type of aether flows through another type. The tension and flow forces are postulated to be simple linear relationships.

With the above axiomatic basis in place, it is then shown by Vector calculus that the initial assumptions and force laws lead directly to Maxwell's equations. Many other explanations readily follow from the initial assumptions as well. Electric charge is identified as free (detached) aether. Therefore, the aether itself is modeled as two solid blocks of charge - one positive and the other negative - that are attached both internally and to each other. This idea of a "sea" of charge is similar to an idea originally put forward by Paul Dirac, although in the case of the two component aether model there are two seas, one of each sign, and not the single sea proposed by Dirac.

The two component aether model gives physical understandings for other things as well. The physical origins of the Electric field and the Magnetic field are understood in terms of the underlying aether. The law of Charge conservation is understood. It becomes clear why the Magnetic monopole does not exist. The mechanism for the Displacement current is readily apparent and it is unified with the mechanism for physical currents. One of the nagging problems of all aether models is solved, as it is explained how material bodies can freely move through the aether. Importantly, tests for the theory are proposed.

## Introduction and Relation to Special Relativity

After Albert Einstein's exposition of the theory of special relativity, the concept of an aether was soon considered by the majority of physicists to be a discredited idea. However, the luminiferous aether concept has many intuitive advantages. Classical physicists believed that an understanding of nature should enable physical laws to be stated in words as well as mathematics, and that an underlying physical model for the mathematics was crucial to that understanding. Waves were believed to be the result of a substance in motion. In the case of light this required the existence of a new substance, the aether. James Clerk Maxwell devoted a considerable amount of his life toward the pursuit of a mechanical model that would underlie his famous equations. However, a classical aether leads to a confrontation with special relativity. It is well known that relativity correctly predicts experimental results. What is less known is that the predecessor of relativity, Lorentz’s theory of electrons, predicts the results of experiments equally well. One school of thought is that Einstein greatly simplified the idea of space by abandoning the aether. However, special relativity replaces the simple aether concept by a triply infinite set of equivalent spaces and times throughout the universe. By reintroducing an aether, the simplicity of a single space and time can be recovered. Whether Einstein's relativity or Absolute Theory is correct is still an open question. No experiment has ever been conducted that would indicate that Absolute Theory is wrong. In fact, Bell's theorem tests by Alain Aspect provide some evidence that supports the Absolute Theory view over that of Einstein.

## Initial Axioms of the Two Component Aether Theory

One approach to science is to start with a set of axioms, and then determine if those axioms are a good representative of nature by developing those axioms into testable hypotheses. The initial axioms of the two component aether follow.

Axiom 1) Newton's Law, F=dp/dt is assumed to be correct, where F is the force, p is the momentum and t is time. The usual definitions are assumed, where the momentum is β$\gamma$mc, $\gamma$ = (1 - β2)-1/2, β = v/c, v is the distance traveled per unit time, and c is the speed of light.

Axiom 2) It is assumed that in the absence of sources and waves, a homogeneous, isotropic, two component solid incompressible aether occupies all space.

From the fact that electrostatics and the theory of atomic dislocations in a uniform solid state lattice are both governed by Poisson's equation, it follows that the scalar potential may result from the dislocation of one thing with respect to another. This fact motivates Axiom 3.

Axiom 3) It is postulated that the aether is made up of two components. One component is called positive aether, the other is called negative aether.

Of these axioms, only the two component nature of the aether is a departure from the classical luminiferous aether. But even on that score, most materials on earth consist of positive (proton) and negative (electron) components, so Axiom 3 can be regarded as an entirely natural assumption once one assumes an aether.

Moving on from those initial assumptions, the two component aether model makes some further assumptions.

Axiom 4) It is proposed that in the absence of sources and sinks each small volume of negative aether is attached to an equally small volume of positive aether. (Note that this is again similar to normal matter where electrons are attached to nuclei in atoms.)

Axiom 5) It is assumed that the attachment between positive and negative aether can be broken, resulting in free, detached aether. (Note that this is similar to the ionization of atoms.)

Drawing from the standard theory of dislocations within a solid, the presence of a source is evidence of some extra substance disturbing the otherwise homogeneous material. Sources are therefore proposed to be regions wherein there is some extra positive aether. For the two-component aether model this extra aether is postulated to no longer be attached to the solid - it is detached and free to move about within the remaining aether. Since the source and sink term in Maxwell’s use of the Poisson equation is identified as electric charge, detached positive aether is identified as positive electric charge, while detached negative aether is identified as negative electric charge. The two-component aether model then readily leads to an understanding of the law of electric charge conservation. Whenever a given amount of positive aether becomes detached (and positive charge is formed), an equal amount of negative aether becomes detached (and an equal amount of negative charge is formed).

### Pictorially Envisioning the Aether

Figure 1. A cube of bound (attached) aether along with free (detached) pieces of aether.

Figure 1 depicts a cube of aether, as well as a free piece of positive aether and a free piece of negative aether. The aether can be envisioned as a typical ionic solid, although in the case of the aether, positive aether plays the role of a typical positive ion, whereas negative aether plays the role of a typical negative ion. Free detached aether occurs when some of the bound aether becomes detached from the lattice of the solid. Note that this model has two relevant precursors in the history of science. In 1930, Paul Dirac postulated a theoretical model of the vacuum as an infinite sea of particles possessing negative energy in order to explain the anomalous negative-energy quantum states predicted by his equation for relativistic electrons. In the aether model proposed here, it is seen that there are two such infinite seas, one positive and one negative, and that they are bound together in a solid lattice. The second precursor comes from standard solid state physics of semiconductors. In typical silicon bonding, each silicon atom has four valance electrons that contribute to bonding within the lattice. By introducing doping atoms with five valence electrons, an extra electron is brought into the lattice structure. This extra electron is free to move within the lattice, which is what leads to the conductive nature of the semiconductor. Similarly, the two component aether model proposes that the detached aether components are allowed to flow within the lattice. The extra components can flow since the bound components already have a partner with which to bind, allowing the extra component to flow throughout the solid, similar to the electron flow in the n-type super conductor. Since it is postulated that the aether is incompressible, when a positive portion of free aether enters the cube, the other positive portions must bulge outward to make room for it.

### Aetherial Force Laws

From Maxwell’s equations it is known that a current (flowing charge) acts as a source term. Therefore, flows of detached aether (which has been identified as charge above) should lead to forces. This leads to the following simple flow force law:

Aetherial Flow Force Law: When aether moves with respect to attached aether of like (opposite) kind, a force is generated on the attached aether that is proportional to the rate of aether flow, and the direction of the force is in the same (opposite) direction as the flow.

Since the force generated by a moving charge - detached aether - on the attached negative aether is in the opposite direction as the force generated on the attached positive aether, the net reaction force back on the charged particle is zero, and there is no drag force felt by moving charges as they move through the aether. Since all matter is composed of elementary particles, this overcomes one of the classical problems of the aether - which is how planets and stars could move through an aether without slowing down. Since the elementary charged particles can traverse the aether without energy loss, so will larger bodies that are made up of those elementary particles.

It is known from experimental evidence that light exhibits transverse polarization. Further, it is known that waves in a solid are observed to exhibit transverse polarization, whereas waves in a fluid are not. This fact about ordinary materials lends itself to the conclusion that the aether must be a solid. Waves on a solid string exhibit many of the properties of light. Waves on a string exhibit transverse polarization. The underlying medium, the string, is a solid under tension. For this reason the two aetherial components are each proposed to be a solid under tension. The tension is postulated to obey the following simple force law:

Tension Force Law. When a small cube of aether is deformed, the magnitude of the tension perpendicular to each of any two originally parallel faces is proportional to the separation of those two faces, T = k1L. Here k1 is some arbitrary constant and L is the distance that the two faces have been separated from each other.

## Review of Relevant Vector Calculus

Figure 2. Arbitrary Vector P and its Cartesian Components.

The derivation of Maxwell's Equations that follows below will involve some vector calculus concepts, and hence a review of these concepts is useful here. First, recall that any vector can be decomposed into its Cartesian components. The Cartesian components are the projection that the vector has along three orthogonal axes. For instance, Figure 2 shows that the vector P has a projection in the width direction of Px, a projection in the height direction of Py, and a projection in the depth direction of Pz. Adding these projections together results in the full vector P. It is usual to define i as a vector of unit length in the x direction, j as a unit vector in the y direction, and k as a unit vector in the z direction. Other useful vector calculus relations include the dot or scalar product, which is defined for two vectors P and Q as:

$\mathbf{P} \cdot \mathbf{Q} = P_x Q_x + P_y Q_y + P_z Q_z$

The scalar product is equal to the magnitude of P times the magnitude of Q times the cosine of the angle made between the two vectors. It is a scalar result - it is just a number with no direction. The vector or cross product is defined as:

$\mathbf{P} \times \mathbf{Q} = \mathbf{i} \left ( P_y Q_z - P_z Q_y \right ) + \mathbf{j} \left ( P_z Q_x - P_x Q_z \right ) + \mathbf{k} \left ( P_x Q_y - P_y Q_x \right )$

From calculus the partial derivative of any function is defined as the function evaluated at x plus Δx minus the function evaluated at x divided by Δx, where Δx is allowed to become arbitrarily small:

$\frac{\partial P_x}{\partial x} = [P_x(x + \Delta x, y, z, t) - P_x(x, y, z, t)]/\Delta x$

All other variables within the function, such as y, z and the time t, are held constant during the evaluation. The above equation shows the definition of the partial derivative applied to the x component of the vector P. A useful quantity of vector calculus is the gradient operator:

$\nabla = \frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k}$

The gradient operator is a vector quantity that can be applied to functions. The gradient operator can also be combined with an arbitrary vector through the dot product operation, and the result is called the divergence of that vector. From the definition of the gradient operator it is easy to see that the divergence of P is:

$\nabla \cdot \mathbf{P} = \frac{\partial P_x}{\partial x} \mathbf{i} + \frac{\partial P_y}{\partial y} \mathbf{j} + \frac{\partial P_z}{\partial z} \mathbf{k}$

The gradient operator can also be combined with an arbitrary vector through the cross product operation, and the result is called the curl of that vector, which is expanded here.

$\nabla \times \mathbf{P} = ( \frac{\partial P_z}{\partial y} - \frac{\partial P_y}{\partial z} ) \mathbf{i} + ( \frac{\partial P_x}{\partial z} - \frac{\partial P_z}{\partial x} ) \mathbf{j} + ( \frac{\partial P_y}{\partial x} - \frac{\partial P_x}{\partial y} ) \mathbf{k}$

Continuing with the vector calculus review, it is possible to decompose any vector field into its longitudinal and transverse components in a process called Helmholtz decomposition. The longitudinal component of the vector field is a portion of the vector field that has zero curl, while the transverse component has zero divergence. Since the longitudinal component of the vector field has zero curl, it can be expressed as the gradient operator applied to a scalar field:

$\mathbf{P_L} = \nabla \Psi_P$

P; is just a simple function that has a single scalar value at every point in space and time. One can take the curl of PL obtained in the above fashion to easily show that the curl does indeed vanish.)

Since the divergence of the transverse component of P is zero, the divergence of P is equal to the divergence of its longitudinal portion. And now, since the longitudinal component of P is equal to the gradient of ΨP, the divergence of the longitudinal portion is the divergence of the gradient of ΨP. The divergence of the gradient operator is called the Laplace operator or Laplacian, and hence the Laplacian applied to ΨP is equal to the divergence of P:

$\nabla^2 \Psi_P = \nabla \cdot \mathbf{P}$

(1)

It will be important below to consider the case where the divergence of PL is zero. With P decomposed into its longitudinal and transverse components, P = PL + PT, if the divergence of PL is zero then the decomposition can be chosen such that PL is zero. This can be seen by first assuming that PL is not zero, but rather, it is some remaining function R. But since the divergence of PL is equal to zero, this means that the divergence of R is zero, and R is transverse. Hence, a new transverse vector P'T can be formed by making P'T equal to the original PT plus R. And a new P'L can be formed by setting it equal to the original PL minus R. (Here R, which has both zero curl and zero divergence, is subtracted from PL to form P'L and added to PT to form P'T so that the sum P'L + P'T remains P.) The divergence of P'T is zero, since the divergence of both PT and R are zero, and also P'L is also now zero. Hence, there is freedom to choose the longitudinal and transverse components in such a way as to arrange for the longitudinal component to be zero if the divergence of the longitudinal component is zero.

## Derivation of Maxwell's Equations

### Definitions of P and N

It is postulated (see above) that in the absence of disturbances an aether exists throughout all of space that is homogeneous, isotropic, and incompressible. In this rest condition it is possible to divide the aether into small volume elements (cubes) and label each cube with the coordinates of its center, r = (x, y, z). The motion of the aether can then be analyzed by looking at deviations from this equilibrium condition at each point over time. Both the cube of the negative aether and the cube of the positive aether can move independently. Here, the vector P is defined to be the position of the positive aether cube relative to its equilibrium point. The magnitude of P is the distance that the positive aether has moved away from its equilibrium position, while the direction of P is the direction it has moved away from its equilibrium position. For the negative aether, similar considerations apply, and the vector N is defined to be the position of the negative aether cube relative to its equilibrium point.

### A Cylinder of Aether

Figure 3. Undisturbed, attached, positive aether. Cylindrical boundaries arbitrarily assigned so that each cylindrical cross section has an equal area.

At this point it is again useful to attempt to picture what is meant by the mathematical constructs just discussed. Since the aether is proposed to be a solid lattice occupying all of space, it can of course be divided up for analysis into volume elements of any shape that is desired. In Figure 3, the aether is divided into cylindrical shells. The figure shows a cross section of the shells, and in the dimension perpendicular to the drawing nothing changes. Each shell shown has the same area. Without any sources or sinks, both the vector P and the vector N are zero everywhere, since none of the aether has been disturbed.

In Figure 4, a cylinder of detached positive aether (blue) is injected into the attached aether (green) to occupy the central cylinder. This action pushes the attached positive aether from cylinder 1 into cylinder 2, and from cylinder 2 into cylinder 3 and so on. An analogy to this situation would be placing a cylindrically shaped water balloon into a bathtub full of water. The water in the balloon will push the bathtub water out of the way, moving the bathtub water out radially. In the aether model, the figure envisions placing a cylinder of detached positive aether into the aether sea, and this will move the attached aether radially outward. In the two-component aether model there are two such seas, and this action leaves the negative aether fully in its original place, so the vector N will be everywhere zero as a result of adding the detached positive aether. But the vector P is now no longer zero, and in this case the vector P is inversely proportional to the distance from the center of the cylinder once the analysis is done outside of the cylinder. This result comes from the fact that each cylinder has the same cross sectional area. Since the area of the cylinder for a small difference in radius δr is 2$\pi$rδr, it can be seen that keeping the area constant means that δr must be proportional to the inverse of the radius. Putting a cylinder of free, detached aether into the attached aether thus moves the attached aether radially outward in a way that is inversely proportional to the distance from the disturbance.

Figure 4. Situation where detached, positive aether (shown in blue) is injected into the attached, positive aether (shown in green). Each cylinder of attached aether is moved radially outward by one position as a result of the injection.

For one specific example, lets look at the situation at the boundary of the inserted attached aether. At that point, the magnitude of P is equal to the radius of the inserted aether, since the attached aether at that point has been moved from the center. The direction of the displacement P is radially outward from the center. For aether that was originally at the edge of the first section, it moves radially outward to the edge of the second section, and so on. For this example, P is always directed radially outward and that its magnitude drops off as the inverse of distance. In general, the detached aether will not be a perfect cylinder, and there may be detached negative aether as well as detached positive aether. But the important point is that the attached positive aether will move by a distance P depending on the insertion of a sources of free positive aether, and the negative aether will move a distance N depending on the insertion of sources of free negative aether. Additionally, waves within the solid may also cause the aether to move away from its equilibrium position. To sum up, P(x, y, z, t) is the vector field that describes how far positive aether has moved away from its equilibrium position at time t. N(x, y, z, t) is the similar expression for the negative aether.

### Ramifications of Incompressibility - Poisson's Equation

Figure 5. Two cubes of aether. Cube on left shows undisturbed boundaries. Cube on right shows displaced boundaries.

Figure 5 shows two cubes of aether. On the left is an undistorted cube of attached aether, which is Δx wide by Δy high by Δz deep. Note that if a substance is incompressible, the density of the substance is always constant. Hence, if an amount of free, detached aether is added to the first cube, the size of the volume must expand in order to keep the density constant. Such an expanded volume is shown on the right. In that second cube, the width is increased by an amount δx, with the height and depth also changed by δy and δz respectively. Now the amount of attached, positive aether in the second cube is equal to the amount in the first cube, but the attached aether density will be less since it is the total density that remains constant. (The total density is the sum of the attached density and the detached density.) Since the added detached aether expands the boundary of the cube, the new attached aether density is:

$\rho_{AP} = \rho_0 (\Delta x \Delta y \Delta z) / [(\Delta x + \delta x)(\Delta y + \delta y)(\Delta z + \delta z)]$

$\approx \rho_0 (\Delta x \Delta y \Delta z) / [\Delta x \Delta y \Delta z + \delta x \Delta y \Delta z + \delta y \Delta x \Delta z + \delta z \Delta x \Delta y]$

(2)

$= \rho_0 / [1 + \delta x / \Delta x + \delta y / \Delta y + \delta z / \Delta z]$

$\approx \rho_0 [1 - \delta x / \Delta x - \delta y / \Delta y - \delta z / \Delta z]$

The density of anything is the amount of it divided by the volume it occupies. The original density, which is the density in the small cube that has no detached aether, is called $\rho_0$ in Eq. 2. The amount of the attached aether in the larger cube is the same amount that it had before the cube is expanded, which is $\rho_0 (\Delta x \Delta y \Delta z)$. The volume of the larger cube is simply $[(\Delta x + \delta x) + (\Delta y + \delta y) + (\Delta z + \delta z)]$. Now it will generally be the case that the amount of attached aether is far greater than the detached aether, and so the $\delta$ quantities are very much smaller than the $\Delta$ quantities in Eq. 2. This allows the approximations made in Eq. 2. In the first approximation, the denominator is expanded by keeping only the first order terms in small quantities. The second approximation uses the fact that one over (one plus something small) is about equal to one minus something small, resulting in a rather simple equation for the density of the attached aether in a region where detached aether is immersed within it.

To further refine Eq. 2, recall that the vector P is the displacement of the attached positive aether with respect to its equilibrium position. Hence, the x component of P at x is the displacement of the left yz face of the cube shown on the right in Fig. 5, and the x component of P at x+Δx is the displacement of the right yz face of that cube. Therefore $\delta x = P_x(x + \Delta x, y, z, t) - P_x(x, y, z, t)$. Dividing each side of the equation by $\Delta x$ leaves $\delta x / \Delta x = [P_x(x + \Delta x, y, z, t) - P_x(x, y, z, t)]/\Delta x \approx \frac{\partial P_x} {\partial x}$. Repeating the derivation for y and z will lead to similar expressions. This allows Eq. 2 to be re-expressed as:

$\rho_{AP} \approx \rho_0 (1 - \nabla \cdot \mathbf{P})$

(3)

Repeating the derivation for the negative aether results in a similar equation for N:

$\rho_{AN} \approx \rho_0 (1 - \nabla \cdot \mathbf{N})$

(4)

As a final step in analyzing the effect of incompressibility, note that incompressibility means that the density of the attached aether in any volume plus the density of the detached, free, aether will sum to $\rho_0$, which is the density of the attached aether when it is undisturbed by the sources and sinks of free aether. Hence, Eqs. (3) and (4) can be manipulated by bringing $\rho_0 \nabla \cdot \mathbf{P}$ to the left hand side, bringing $-\rho_{AP}$ to the left hand side, and then dividing through by $\rho_0$:

$\nabla \cdot \mathbf{P} = (\rho_0 - \rho_{AP})/\rho_0 = \rho_{dp}/\rho_0$

(5)

In Eq. 5, $\rho_0 - \rho_{AP} = \rho_{dp}$ where the detached aether density is $\rho_{dp}$. After all, $\rho_0$ is the undisturbed aether density, and $\rho_{AP}$ is the attached aether density, and by incompressibility the detached aether density plus the attached aether density must equal $\rho_0$. Lastly, recall Eq. 1, where it was shown that the Laplacian applied to a scalar function $\Psi_P$ is equal to the divergence of the vector P. Applying that to Eq. 5, and realizing that a similar derivation applies to N, results in the equations:

$\nabla^2 \Psi_P = \nabla \cdot \mathbf{P} = \rho_{dp}/\rho_0$

(6)

$\nabla^2 \Psi_N = \nabla \cdot \mathbf{N} = \rho_{dn}/\rho_0$

(7)

### Tension Forces

Figure 6. Tension forces in the aether.

With the ramifications of incompressiblity now in hand, it is time to turn to the affect that forces have on the aether. The first force to consider is the force of tension. Fig. 6 shows three cubes of aether as well as the tension forces acting on the central cube. Of course, in the real aether the cubes (which are arbitrarily specified for analysis only) will have adjoining faces. Here they are separated out only in an effort to make the drawing clear. It is known that tension forces in a solid lead to transversely polarized waves, with a familiar example being waves on a string. Since light is also known to be a transversely polarized wave, it is natural to assume that the aether is a solid under tension. Note from Fig. 6 that the longitudinal force will typically cancel out, as the longitudinal component of T2 is almost exactly opposite to the longitudinal component of T1. This leads to the observation that it is the transverse components of the tension that are responsible for the forces which cause the wave motion within the solid aether.

The tension force will be directed from the center of one cube toward the center of the adjacent one. Recall that the vector P is the relative position of the positive aether with respect to its undisturbed, equilibrium position. Hence, the cube with an equilibrium position of r is in general at r + P(x, y, z, t) and the cube with an equilibrium position of r + Δx is in general at r + Δx + P(x + Δx, y, z, t). Hence, the direction of the force is simply the direction between these two cube centers:

$\mathbf{D} = \mathbf{r} + \mathbf{\Delta x} + \mathbf{P}(x + \Delta x, y, z, t) - \mathbf{r} - \mathbf{P}(x, y, z, t) = \Delta x\mathbf{i} + \mathbf{P}(x + \Delta x, y, z, t) - \mathbf{P}(x, y, z, t)$

Assuming small oscillations, $[\mathbf{P}(x + \Delta x, y, z, t) - \mathbf{P}(x, y, z, t)]/\Delta x << 1$

Now form a vector in the direction of D that has a magnitude close to unity. First form the quantity d:

$\mathbf{d} = lim (\Delta x \rightarrow 0) \mathbf{D}/\Delta x = \mathbf{i} + \frac{\partial \mathbf{P}} {\partial x} = \mathbf{i} + \frac{\partial {P_x}} {\partial x} \mathbf{i} + \frac{\partial {P_y}} {\partial x} \mathbf{j} + \frac{\partial {P_z}} {\partial x} \mathbf{k}$

Now form the quantity u, keeping terms to first order in small quantities:

$|\mathbf{d}| = [(1 + \frac{\partial {P_x}} {\partial x})^2 + (\frac{\partial {P_y}} {\partial x})^2 + (\frac{\partial {P_z}} {\partial x})^2]^{1/2} \approx 1 + \frac{\partial {P_x}} {\partial x}$

$\mathbf{u} = \mathbf{d}/|\mathbf{d}| \approx [(1 + \frac{\partial {P_x}} {\partial x}) \mathbf{i} + \frac{\partial {P_y}} {\partial x} \mathbf{j} + \frac{\partial {P_z}} {\partial x} \mathbf{k}]/[ 1 + \frac{\partial {P_x}} {\partial x}] \approx \mathbf{i} + \frac{\partial {P_y}} {\partial x} \mathbf{j} + \frac{\partial {P_z}} {\partial x} \mathbf{k}$

With u now derived, the Tension force per unit area can be expressed as T times u, where T is the magnitude of the tension force per unit area. T on the left face of the cube is

$\mathbf{T}(x, y, z, t) = T(x, y, z, t)[\mathbf{i} + \frac{\partial {P_y}(x, y, z, t)} {\partial x} \mathbf{j} + \frac{\partial {P_z}(x, y, z, t)} {\partial x} \mathbf{k}]$.

while T on the right face is

$\mathbf{T}(x + \Delta x, y, z, t) = T(x + \Delta x, y, z, t)[\mathbf{i} + \frac{\partial {P_y}(x + \Delta x, y, z, t)} {\partial x} \mathbf{j} + \frac{\partial {P_z}(x + \Delta x, y, z, t)} {\partial x} \mathbf{k}]$

At this point the first order Taylor series expansion can be used: $T(x + \Delta x, y, z, t) \approx T(x, y, z, t) + \frac{\partial T} {\partial x} \Delta x$. Here, $\frac{\partial T} {\partial x}\Delta x << T(x, y, z, t)$. This leaves:

$\mathbf{T}(x + \Delta x, y, z, t) = [T(x, y, z, t) + \frac{\partial T} {\partial x}\Delta x][\mathbf{i} + \frac{\partial {P_y}(x + \Delta x, y, z, t)} {\partial x} \mathbf{j} + \frac{\partial {P_z}(x + \Delta x, y, z, t)} {\partial x} \mathbf{k}]$

Now the force on the side of the cube will be equal to the area of that face of the cube multiplied by the force per unit area (which is T) on that face, so therefore the force is equal to Δy times Δz multiplied by T, and, recognizing that the forces are acting in opposite directions, the total force on the cube due to the tension forces on the yz faces is:

$F_{TYZ} = \Delta y \Delta z [\mathbf{T}(x + \Delta x, y, z, t) - \mathbf{T}(x, y, z, t)]$,

or,

$F_{TYZ} = \Delta x \Delta y \Delta z [\mathbf{T}(x + \Delta x, y, z, t) - \mathbf{T}(x, y, z, t)]/\Delta x$

It is assumed that $\frac{\partial {P_y}} {\partial x}$ and $\frac{\partial {P_z}} {\partial x}$ are small quantities in comparison to 1, and it is further assumed that deviations of T from the equilibrium value T0 will always be small, leaving $T(x, y, z, t) \approx T_0$. Then, from the above expressions and keeping only terms first order in small quantities:

$F_{TYZ} = \Delta x \Delta y \Delta z [\frac{\partial T} {\partial x}\mathbf{i} + T_0\frac{\partial^2 P_y} {\partial x^2}\mathbf{j} + T_0\frac{\partial^2 P_z} {\partial x^2}\mathbf{k}]$

(8)

What remains is to further evaluate the term $\frac{\partial T} {\partial x}$. Recall that the tension force law stated above that the tension is proportional to the amount of distance that the aether is stretched, which can be expressed mathematically as:

$T = K_1 L$

where K1 is some arbitrary constant, and L is the distance the aether is stretched. With no additional stretching, a small cube of aether in its normal equilibrium state will have an equilibrium length of Δx, and the tension will be equal to some universal tension T0 that exists in the undisturbed aether. Hence T0 is equal to K1Δx, or, K1 is T0 divided by Δx. If the aether is displaced, the x displacement of the right side of the cube is Px(x + Δx, y, z, t) while the displacement at the left side is Px(x, y, z, t). The amount of stretching is the difference between these two displacements, and so the tension force that results from the Tension force law is

$T = K_1 L = K_1[\Delta x + P_x(x + \Delta x, y, z, t) - P_x(x, y, z, t)]$

and with K1 = T0/Δx

$T = T_0[\Delta x + P_x(x + \Delta x, y, z, t) - P_x(x, y, z, t)]/\Delta x = T_0[1 + \frac{\partial P_x} {\partial x}]$

from which

$\frac{\partial T} {\partial x} = T_0[\frac{\partial^2 P_x} {\partial x^2}]$

Substituting this into Eq. 8 leaves:

$F_{TYZ} = \Delta x \Delta y \Delta z [T_0\frac{\partial^2 P_x} {\partial x^2}\mathbf{i} + T_0\frac{\partial^2 P_y} {\partial x^2}\mathbf{j} + T_0\frac{\partial^2 P_z} {\partial x^2}\mathbf{k}]$

The above expression is the tension force on a small cube of aether that results from the tension forces that are present on the yz faces of the cube. By doing exactly the same derivation for the xy and xz faces analogous expressions for the forces are derived in those directions. As can be seen, the only thing that changes is that the partial derivatives are evaluated with respect to y and z, instead of x. The sum of these three force components result in the total force expression, and recalling that P is just the sum of its components, and that the Laplacian is the sum of the second partial derivatives, the equation for the tension force can be expressed as:

$\mathbf{F_T} = \Delta x \Delta y \Delta z T_0 \nabla^2\mathbf{P}$

(9)

### Flow Forces

Figure 1. A cube of bound (attached) aether along with free (detached) pieces of aether.

With the tension forces now evaluated, the forces that result due to flows through the aether will be investigated. To aid in visualization, Figure 1 is repeated nearby. There are three possible flows that can occur within the two component solid aether that will affect the equation of motion of the attached positive aether. Free detached positive aether can flow through the representative sample cube, as can free detached negative aether. Lastly, the attached negative aether can flow with respect to the attached positive aether.

Each flow rate is easily stated as the density of what is flowing multiplied by its velocity relative to the attached positive aether. For the case of negative detached aether, the rate is:

$\rho_{dn} (\mathbf{v_{dn}} - \frac{\partial \mathbf{P}} {\partial t})$

Note that the partial derivative of P with respect to time is simply the velocity of the attached positive aether within the analysis cube, and hence the term in parentheses is the relative velocity between the detached negative aether and the attached positive aether.

For the case of negative attached aether, the rate is:

$(\rho_0 - \rho_{dn}) (\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t})$

In the above case, the density of the attached negative aether is equal to the undisturbed negative aether density minus the detached negative aether density, since the overall density of the aether is constant. And in this case, it is the partial derivatives of the position vectors with respect to time that yield the velocities of the attached aether components.

For the case of positive detached aether, the rate is:

$\rho_{dp} (\mathbf{v_{dp}} - \frac{\partial \mathbf{P}} {\partial t})$

Note that the total time derivative, rather than the partial time derivative, would have been more exact in the expressions above. However, since it is assumed that aetherial motions are small, the partial derivative of the aetherial position vectors with respect to x, y, or z are small. For that reason, the partial derivative with respect to time is the first order quantity in the expression. This analysis always keeps only first order terms in small quantities and it will then let the region of analysis tend to zero to arrive at the desired result, which is a common technique when using calculus to evaluate physical phenomena. Continuing in this approach, terms involving the density of the detached aether multiplied by the motion of the attached aether are neglected as these terms are of second order in small quantities under most conditions. This is because the density of the detached aether is always much less than the nominal aether density and the speed of the attached aether is much less than the speed of light, while the velocity of the detached aether (the velocity of charged particles) is often comparable to the speed of light. (Terms involving $\rho_{dn}\frac{\partial \mathbf{P}} {\partial t}$, $\rho_{dn} \frac{\partial \mathbf{N}} {\partial t}$ and $\rho_{dp}\frac{\partial \mathbf{P}} {\partial t}$ are neglected as being second order in small quantities.) Next, the analysis must recognize that the total force will be proportional to the the volume of the cube, since the size of the cube will determine the size of the flows for any given density. The flow force can then be formulated as the volume of the cube multiplied by some constant multiplied by the rates of flows, once care is taken to use the proper sign for the direction of the force as determined by the flow force law that was postulated above. Recall that the postulate was that flowing positive aether would result in a force on the positive aether in the same direction as the flow, and flowing negative aether would result in a force in the opposite direction of the flow. Hence, the force on an arbitrary cube is:

$\mathbf{F_F} = k_2 \Delta x \Delta y \Delta z [\rho_{dp} \mathbf{v_{dp}} - \rho_{dn} \mathbf{v_{dn}} - \rho_0(\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t})]$

(10)

### Aetherial Equation of Motion

Now that all of the forces acting on the positive attached aether have been identified and evaluated, it is possible to determine the equation of motion for the attached, positive, aether. It will be assumed here that the velocity of the attached aether is small compared to the velocity of light, and hence the equation F=ma will be applied. Now the mass of a cube of aether is equal to its mass density times its volume Δx Δy Δz and the acceleration is equal to the second partial derivative of the position of the cube with respect to time, $\frac{\partial^2 \mathbf{P}} {\partial t^2}$. Putting this all together, and recalling the forces for the flow and tension leaves:

$\mathbf{F} = m\mathbf{a} = \mu_P \rho_0 \Delta x \Delta y \Delta z \frac{\partial^2 \mathbf{P}} {\partial t^2} = \mathbf{F_T} + \mathbf{F_F}$

$= \Delta x \Delta y \Delta z T_0 \nabla^2\mathbf{P} + k_2 \Delta x \Delta y \Delta z [\rho_{dp} \mathbf{v_{dp}} - \rho_{dn} \mathbf{v_{dn}} - \rho_0(\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t})]$

or, by eliminating the factor Δx Δy Δz from each term:

$\mu_P \rho_0 \frac{\partial^2 \mathbf{P}} {\partial t^2} = T_0 \nabla^2\mathbf{P} + k_2[\rho_{dp} \mathbf{v_{dp}} - \rho_{dn} \mathbf{v_{dn}} - \rho_0(\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t})]$

Now divide each side of the equation by T0:

$\frac{\mu_P \rho_0} {T_0} \frac{\partial^2 \mathbf{P}} {\partial t^2} = \nabla^2\mathbf{P} + \frac{k_2} {T_0}[\rho_{dp} \mathbf{v_{dp}} - \rho_{dn} \mathbf{v_{dn}} - \rho_0(\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t})]$

And then multiply each side of the equation by minus one and take the Laplacian term over to the left hand side and rearrange the terms:

$\nabla^2\mathbf{P} - \frac{\mu_P \rho_0} {T_0} \frac{\partial^2 \mathbf{P}} {\partial t^2} = \frac{k_2} {T_0}[-\rho_{dp} \mathbf{v_{dp}} + \rho_{dn} \mathbf{v_{dn}} + \rho_0(\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t})]$

It is now useful to set some constants:

$\frac{1} {c^2} = \frac{\mu_P \rho_0} {T_0}$

and

$\frac{k_2} {T_0} = \frac{4 \pi a^2} {c}$

In each of the above expressions c is the speed of light, and a is some fundamental length. With these definitions and to first order in small quantities, the following equation presents the equation of motion of the positive, attached aether:

$\nabla^2\mathbf{P} - \frac{1} {c^2} \frac{\partial^2 \mathbf{P}} {\partial t^2} = \frac{4 \pi a^2} {c}[\rho_{dn} \mathbf{v_{dn}} + \rho_0(\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t}) - \rho_{dp} \mathbf{v_{dp}}]$

(11)

The derivation of the equation of motion for the attached negative aether will follow the exact same steps as does the derivation for the attached positive aether by just replacing P by N, N by P, μP by μN, and T0 by T'0:

$\nabla^2\mathbf{N} - \frac{\mu_N \rho_0} {T'_0} \frac{\partial^2 \mathbf{N}} {\partial t^2} = \frac{k_2} {T'_0}[-\rho_{dn} \mathbf{v_{dn}} + \rho_{dp} \mathbf{v_{dp}} + \rho_0(\frac{\partial \mathbf{P}} {\partial t} - \frac{\partial \mathbf{N}} {\partial t})]$

now make the following substitutions:

$\frac{1} {c^2} = \frac{\mu_N \rho_0} {T'_0}$

and

$\frac{k_2} {T'_0} = -\frac{4 \pi a^2} {c}$

With the above substitutions, the equation of motion for the negative attached aether is:

$\nabla^2\mathbf{N} - \frac{1} {c^2} \frac{\partial^2 \mathbf{N}} {\partial t^2} = \frac{4 \pi a^2} {c}[\rho_{dn} \mathbf{v_{dn}} + \rho_0(\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t}) - \rho_{dp} \mathbf{v_{dp}}]$

(12)

Note that $\frac{k_2} {T'_0}$ has now been set with a negative sign. This means that $T'_0$ is a negative quantity with respect to $T_0$. Note also that the substitution for $\frac{1} {c^2}$ involves both $T'_0$ and $\mu_N$, so this means that $\mu_N$ is now the negative of $\mu_P$. Hence, the derivation introduces the new concept of negative mass into physics. This makes the aetherial components true matter / antimatter partners in nature, since they have opposite mass and tension, as well as the flows affecting things in an opposing fashion.

### Arriving at Maxwell's Equations

With the analysis of compressibility and the aether equations of motion now done, the results from above can be used in a derivation of Maxwell's Equations. First, recall that the investigation concerning incompressibility of the aether led to Eqs. 6 and 7, which are repeated here:

$\nabla^2 \Psi_P = \nabla \cdot \mathbf{P} = \rho_{dp}/\rho_0$

(6)

$\nabla^2 \Psi_N = \nabla \cdot \mathbf{N} = \rho_{dn}/\rho_0$

(7)

Poisson's equation can be derived quite simply by subtracting these two equations:

$\nabla^2 \Psi_P - \nabla^2 \Psi_N = \nabla^2 (\Psi_P - \Psi_N) = (\rho_{dp} - \rho_{dn})/\rho_0$

(13)

Next, define $\phi$ and $\rho_D$ through the following expressions:

$\Psi_P - \Psi_N = -\phi/4\pi\rho_0$

and

$\rho_D = \rho_{dp} - \rho_{dn}$

Substituting in $\phi$ and $\rho_D$ into Eq. 13 leaves:

$\nabla^2 \phi/4\pi\rho_0 = -\rho_D/\rho_0$

Rearranging and cancelling common terms leads to:

$\nabla^2 \phi = -4\pi\rho_D$

(14)

Eq. 14 is recognized as Poisson's Equation.

In order to simplify the mathematics, the vector J is now defined as the net current of detached aether that flows through the attached aether. J is the density of the detached positive aether times its velocity, minus the density of the detached negative aether times its velocity:

$\mathbf{J} = \rho_{dp}\mathbf{v_{dp}} - \rho_{dn}\mathbf{v_{dn}}$

And hence, Eq. 11 becomes:

$\nabla^2\mathbf{P} - (1/c^2)\frac{\partial^2 \mathbf{P}} {\partial t^2} = (4\pi a^2/c)[-\mathbf{J} + \rho_0(\frac{\partial \mathbf{N}} {\partial t} - \frac{\partial \mathbf{P}} {\partial t})]$

(15)

The review of vector calculus above has shown that it is always possible for any vector field to be divided into a transverse component that has zero divergence, and a longitudinal component that has zero curl. Manipulations of the transverse expression will be done further down; the longitudinal portion of Eq. 15 is:

$\nabla^2\mathbf{P_L} - (1/c^2)\frac{\partial^2 \mathbf{P_L}} {\partial t^2} = (4\pi a^2/c)[-\mathbf{J_L} + \rho_0(\frac{\partial \mathbf{N_L}} {\partial t} - \frac{\partial \mathbf{P_L}} {\partial t})]$

The review of vector calculus above has also shown that the longitudinal displacements can be represented as the gradient of a scalar:

$\mathbf{P_L} - \mathbf{N_L} = \nabla \Psi_P - \nabla \Psi_N = -\nabla \phi/(4\pi\rho_0)$

Using the above substitution leaves the longitudinal equation as:

$\nabla^2\mathbf{P_L} - (1/c^2)\frac{\partial^2 \mathbf{P_L}} {\partial t^2} = -(4\pi a^2/c)\mathbf{J_L} + (a^2/c)(\frac{\partial} {\partial t})\nabla \phi$

(16)

To manipulate this equation further, use is made of the continuity equation, which is:

$\frac{\partial \rho_D} {\partial t} = - \nabla \cdot \mathbf{J} = - \nabla \cdot \mathbf{J_L}$

(17)

In Eq. 17, the last expression results since $\nabla \cdot \mathbf{J_T}$ is zero by the definition of the transverse component. Also, recalling that Poisson's Equation is $\nabla^2\phi = -4\pi\rho_D$ the partial derivative of time can be taken on both sides to yield:

$(\frac{\partial} {\partial t})\nabla^2\phi = -4\pi(\frac{\partial \rho_D} {\partial t})$

(18)

The divergence of the right hand side of Eq. 16 can now be taken, and the results of Eqs. 17 and 18 substituted therein to yield:

$-(4\pi a^2/c)\mathbf{\nabla \cdot J_L} + (a^2/c)(\frac{\partial} {\partial t})\nabla^2 \phi = (4\pi a^2/c)(\frac{\partial \rho_D} {\partial t}) - (a^2/c)4\pi(\frac{\partial \rho_D} {\partial t}) = 0$

Therefore the divergence of the right hand side of Eq. 16 is zero, and since the curl is also zero by the definition of the longitudinal component, this means that any remaining value can be assigned to the transverse component leaving:

$(4\pi a^2/c)\mathbf{J_L} = (a^2/c)(\frac{\partial} {\partial t})\nabla \phi$

(19)

Now consider the transverse portion of Eq. 15:

$\nabla^2\mathbf{P_T} - (1/c^2)\frac{\partial^2 \mathbf{P_T}} {\partial t^2} = (4\pi a^2/c)[-\mathbf{J_T} + \rho_0(\frac{\partial \mathbf{N_T}} {\partial t} - \frac{\partial \mathbf{P_T}} {\partial t})]$

Note that the flow and tension forces have resulted above in identical equations of motion for P and N, while the displacements caused by the ramification of incompressibility (Poisson's Equation) are purely longitudinal. Hence the transverse aether disturbance will be the same for both the positive attached aether and the negative attached aether, or, NT = PT, and this means that the second term on the right hand side of the above equation is zero:

$\nabla^2\mathbf{P_T} - (1/c^2)\frac{\partial^2 \mathbf{P_T}} {\partial t^2} = (4\pi a^2/c)[-\mathbf{J_T}]$

Now, from Eq. 19 it is possible to add $-(4\pi a^2/c)\mathbf{J_L} + (a^2/c)(\frac{\partial} {\partial t})\nabla \phi$ to the right hand side (since this is adding zero):

$\nabla^2\mathbf{P_T} - (1/c^2)\frac{\partial^2 \mathbf{P_T}} {\partial t^2} = -(4\pi a^2/c)\mathbf{J_T} - (4\pi a^2/c)\mathbf{J_L} + (a^2/c)(\frac{\partial} {\partial t})\nabla \phi$

And now the current terms are combined:

$\nabla^2\mathbf{P_T} - (1/c^2)\frac{\partial^2 \mathbf{P_T}} {\partial t^2} = -(4\pi a^2/c)\mathbf{J} + (a^2/c)(\frac{\partial} {\partial t})\nabla \phi$

At this point the notation will be adjusted so that the equation will become more familiar by making the assignments NT = PT = a2A:

$\nabla^2\mathbf{A} - (1/c^2)\frac{\partial^2 \mathbf{A}} {\partial t^2} = -(4\pi/c)\mathbf{J} + (1/c)(\frac{\partial} {\partial t})\nabla \phi$

(20)

Further, since A is identified as a transverse vector:

$\nabla \cdot \mathbf{A} = 0$

(21)

It is also timely to recall Eq. 14:

$\nabla^2 \phi = -4\pi\rho_D$

(14)

Eqs. 14, 20 and 21 are readily recognized as Maxwell's Equations in the Coulomb gauge.

## Physical Understandings Enabled by the Two Component Aether Model

The two component aether model has now been shown to possess the ability to derive Maxwell's Equations in a mathematically rigorous way. However, the theory is capable of providing many more revelations with respect to the understanding of electrodynamics.

### Identifying Electric Charges as Free Aether

The first physical identification (which has already been mentioned above) comes from Poisson's Equation, shown above as Eq. 14. In Eq. 14 the source term, $\rho_D$, is identified as free, detached aether. That source term is more conventionally identified as electric charge. From this observation it is now understood that electric charge is in actuality an amount of aether that has become detached from the predominant attached aether. Furthermore, positive charge is one form of detached aether, while negative charge is the other form of detached aether. Also, this identification shows that the aether itself can be considered to be two infinite seas of charge - one negative, the other positive, which are internally attached and in the state of a solid. Lastly, since electric current is understood as the motion of electric charge, electric currents are now identified as moving detached aether.

### Identifying the Static Electric Field as an Aetherial Displacement

The scalar potential in the above analysis satisfies the equation:

$-\nabla \phi = 4 \pi \rho_0 (\mathbf{P_L} - \mathbf{N_L}) = 4 \pi \rho_0 (\mathbf{P} - \mathbf{N})$

Figure 3. Undisturbed, attached, positive aether. Cylindrical boundaries arbitrarily assigned so that each cylindrical cross section has an equal area.

The second equality above results from the relation P_T = N_T, so P_L - N_L = P - N. With the static electric field being the gradient of $\phi$, this reveals that static electric fields are nothing more than the distance that the positive attached aether is separated from its negative counterpart. Nearby, Figs. 3 and 4 (repeated from above) show two depictions of the positive, attached aether. Fig. 3 shows undisturbed aether, while the Fig. 4 shows the case where a cylinder of detached positive aether has been inserted. The attached aether is green, and the detached aether blue. Since the aether is incompressible, when detached aether is inserted it must push the attached aether out of the way. Vectors show some such displacements. As can be seen, the further away the original attached aether is from the center the less it needs to move out of the way. Several rings are drawn in the figures such that each annular region contains the same area. Since the area is $2 \pi r \delta r$, it is easy to see that $\delta r$ decreases as 1/r. Hence the displacement, P, caused by inserting a cylinder of detached aether, decreases as 1/r. In this aether model charge has been identified as detached aether, and the static electric field is identified as being proportional to the displacement of the attached aether. Hence, Figs. 3 and 4 show the familiar result that for an infinite cylinder of charge the electric field falls off as 1/r.

Figure 4. Situation where detached, positive aether (shown in blue) is injected into the attached, positive aether (shown in green). Each cylinder of attached aether is moved radially outward by one position as a result of the injection.

The two component aether model finally gives us an answer for what the static electric field is in physical terms. Under the two component aether model, the static electric field is recognized as a quantity that is proportional to the separation of the positive aether from the negative aether.

### Identifying the Vector Potential as Aetherial Displacement

The two component aether model also provides an answer to the question of what the vector potential is. The vector potential is the transverse component of the displacement of the attached aether. As discussed above, both the positive and negative aether have the same equation of motion for their transverse displacements.

### Identifying Light as an Aetherial Wave

Light can now be identified as an aetherial wave - just as the classical theory anticipated in the time of Maxwell.

### How Bodies Can Freely Move Inside the Aether

The two component aether model provides an answer to the question of how bodies can traverse through a solid aether without slowing down. The answer to that question comes from the flow force law that stipulates that flow forces are in one direction for positive aether and in the opposite direction for the negative aether. Hence, the reaction force on all charged particles will be zero as those particles traverse the aether. Since all bodies are constructed of charged particles such as electrons and (possibly) quarks, those composite bodies will also pass through the aether with no change in energy.

### Origin of the Law of Charge Conservation

The law of charge conservation is easily obtained from the two component aether model. Since charge is identified as detached aether, whenever positive aether becomes detached, it must become detached from negative aether, freeing an equal amount of negative aether in the process.

### Why Magnetic Monopoles Do Not Exist

The absence of magnetic monopoles also follows from the two component aether model. When written in terms of the electric and magnetic fields, Maxwell's equations have a certain symmetry that has led many people to propose that a magnetic charge should exist alongside the known electric charge. But by the two component aether model, it is seen that the physical entities underlying Maxwell's equations are entities that are described by the scalar and vector potentials in the Coulomb gauge, and in that form the symmetry argument for the existence of a magnetic monopole no longer exists. In fact, there is no room for a magnetic monopole in the two component aether model. Positive electric charge is identified as detached, positive aether, and negative charge as detached negative aether. There is no similar substance to form magnetic monopoles in the two component aether model.

### Unification of the Physical and Displacement Currents

The two component aether model unifies physical currents and the displacement current with a common physical understanding. What are understood as physical currents in the contemporary sense are charges in motion: physical currents are charge times the velocity of the charge. In the two component aether model, those physical currents are identified as detached aether times the velocity of the detached aether. But also, the displacement current, which is the time derivative of the electric field, can be shown to be the velocity of the attached aether.

The time derivative of the electric field is:

$(1/c)\frac{\partial \mathbf{E}} {\partial t} = -(1/c)\frac{\partial [(1/c)\partial \mathbf{A}/ \partial t + \nabla \phi]} {\partial t}$

$= (1/c)\frac{\partial [4\pi\rho_0(\mathbf{P} - \mathbf{N}) - (1/ca^2) \partial \mathbf{P_T} /\partial t]} {\partial t}$

Since P and N are the displacements of the attached aether, one component of the displacement current is simply the velocity of the attached aether. The other component involves the vector potential, which is also proportional to the displacement of the attached aether. Hence both the physical currents and displacement currents are unified in the two component aether model by the understanding that they both are simply the motion of quantities of aether. (Although it is of note that the displacement current involves a term containing the second partial derivative with respect to time.)

## Introduction of Negative Mass

An important aspect of the two component aether model is the introduction of negative mass into physical theory. It is through the presence of negative mass that the equation of motion for the negative attached aether becomes the same as the equation of motion for the positive attached aether, and that equality is a crucial aspect in arriving at the derivation of Maxwell's equations above. While one could possibly change the flow force laws so that negative mass is not required, such a change in the flow force laws would also need to retain the result that the reaction force back on the moving charge is zero. So far, no such flow force law has been found, and for this reason negative mass is preferred in the theory. Also, with one component of the aether having a positive mass, and the other having an equal and opposite negative mass, the total aether mass is zero, and hence the aether will not contribute to the gravitational field which affects the motion of the stars and the planets.

## Possible Future Tests

While the two component aether model has many very nice features, it is important for any new theory to be able to point to specific tests that will help to separate it from competing ideas. In order to be a difference, it must make a difference. And for the two component aether model, there are two tests that can be done.

It is known that as entities travel through space their processes slow down. (Clocks retard in an effect called time dilation.) In the two component aether model light is an oscillation of the aether and therefore light is aether in motion. Since clocks traveling through the aether slow down, by the two component aether model clocks immersed in moving aether (for instance, clocks within electromagnetic waves) should also slow down. Now the effect may be quite small, since the theory gives no guide as to how large the oscillations are, but the effect should be observable at some level.

Another test should also be possible for low velocity particles in electromagnetic wave environments. Recall that in the derivation of Maxwell's equations terms involving the density of the detached aether multiplied by the velocity of the attached aether were neglected since those quantities are typically small in comparison to the other terms in the equation. But for the case where the velocity of the detached aether is small, namely, when a charged particle velocity is small, and the intensity of the attached aether oscillations are large this approximation will begin to break down, and that should yield experimental differences under appropriate conditions.

## Identified Future Research

As with all possible advances in science, there is room for further study. Since the two component aether model shows how charges lead to forces that create electric and magnetic fields, it would in principle be possible to derive the Lorentz Force Equation in a way that is consistent with the model. It would also be a remaining task to experimentally evaluate the constants defined for the theory which are a, $\rho_0$, and T0.

More intriguing is the future possibility of isolating large portions of aether from the background and manipulating its flow, as this could lead to extremely important practical breakthroughs.

## References

1. D.J. Larson, Phys. Essays vol. 11, no. 4, 524 (1998).