User:Euclid

Forces on Detached Aether (The Lorentz Force Equation)

Figure 7. Representation of the Undisturbed, Attached Aether. Squares represent small portions of attached negative aether. Circles represent small portions of attached positive aether. Half-column numbers are just an arbitrary labeling of horizontal position to aid in description.

In order to derive the aetherial force equations it is useful to have a picture of the aether in mind. Figure 7 is therefore a useful construct. In Figure 7, the squares represent the attached negative aether, while the circles represent the attached positive aether. Recall that it is postulated above that each aether species is attached to itself and to its counterpart.

Figure 8. Representation of an infinite plane of Detached Aether within the Attached Aether. Grey circles represent detached, positive aether. Presence of detached positive aether causes displacement of the attached positive aether from its equilibrium position. Arrows show the magnitude of the attached positive aetherial displacements.

Now consider Figure 8, which depicts the situation wherein an infinite thin sheet of detached, positive aether is introduced into the attached aether. In such a case, the attached positive aether is pushed outward by the introduced detached aether. In this case, the circles of positive aether now have their centers upon the boundaries of the squares of negative aether. This leaves a condition where the displacement is P = (1/2)i for x > 1/2 and P = -(1/2)i for x < -1/2. (Here x is the horizontal coordinate of Figure 8, i is a unit vector in the x direction, and the unit for length is arbitrary.) In the case shown in Figure 8 there will be no net force on the detached aether, since the situation is completely symmetric. (Recall that it is postulated that the aether is incompressible in the presence of electromagnetic effects. Hence, when a sheet of detached positive aether is introduced into the attached positive aether it must push aside the attached positive aether in order to retain the constant overall aetherial density.)

Next, Figure 3 depicts the situation wherein two infinite sheets of detached, positive aether are introduced into the attached aether. In this case, the outward forces on the attached positive aether add for the attached positive aether outside of both sheets, and the forces cancel for the attached positive aether between the sheets. Hence, on the outside of each sheet we have attached aether that is displaced from its negative counterpart, while between the sheets there is no such displacement. With a linear restoring force, the energy of each displacement is proportional to the square of the displacement, $E = (1/2)k_1x^2$, which is the familiar equation for the energy stored in a spring. It is therefore useful to think of the situation as one of a field of springs. The springs between the sheets are in their relaxed, non-extended, state and they have no extra energy. The springs outside of the sheets are extended and thus have have extra internal stored energy.

Since the detached aether is free to move, the situation shown in Figure 9 is one that will lead to motion of the detached aether. This is because, should the detached aether move outward, the outward springs will relax to their un-extended lengths, releasing their energy. That released energy will go into the sheets of detached aether as kinetic energy. The force on the sheets comes from the flow of the attached aether toward the position occupied by the detached aether, as the in-rushing attached aether pushes the detached aether outward. Hence it is seen that two detached aether sheets will repel each other.

As a next step toward the more general case, the density of detached aether within the sheets can be assumed to be less than the attached aetherial density, and the densities of the two sheets might be different from one another. Defining $\rho_R$ to be the amount of detached “red” aether per unit volume ($\rho_R$ is the “red” aetherial density) and defining the thickness of the sheet as $\delta x$, the displacement (the length of the red arrows) will become

$x_R$ = $\rho_R \delta x$/$\rho_0$

where $\rho_0$ is the density of the undisturbed attached positive aether. Similarly,

$x_G$ = $\rho_G \delta x$/$\rho_0$, where $\rho_G$ is the “green” aetherial density. (The displacements will be reduced from what is shown in Figure 9 if the density of the detached aether is less than $\rho_0$, since there will be less detached aether pushing upon the neighboring attached aether. The reduction in the displacement will be proportional to the reduction in the density of the detached aether within the sheet.)

If $x_R$ is not equal to $x_G$ the displacement of the attached aether would no longer be zero between the sheets. Rather, the displacement would be $x_R - x_G$ between the sheets, where $x_R$ is the displacement caused by the detached aether on the left, and $x_G$ is the displacement caused by the detached aether on the right. In this unbalanced case, the displacement of the attached aether outside of the sheets will be $x_R + x_G$. In this unbalanced case, the energy of the “springs” between the sheets is $E_1 = (1/2)k_1(x_R - x_G)^2$, while the energy of the “springs” outside of the sheets is $E_2 = (1/2)k_1(x_R + x_G)^2$. Hence, in this unbalanced case, the energy liberated from the “springs” as the sheets move is given by Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \delta E} = E_2 - E_1 = (1/2)k_1(x_R + x_G)^2 - (1/2)k_1(x_R - x_G)^2 = (1/2)k_1(x_R^2 + 2x_Rx_G + x_G^2) - (1/2)k_1(x_R^2 - 2x_Rx_G + x_G^2) = 2k_1x_Rx_G . (Note that in the balanced case, $x_R = x_G = x/2$ where is the total stretching of the spring and hence in that case the energy release is $(1/2)k_1x^2$ as in the balanced case discussed above.)

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, μ_P by μ_N, 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, N_T = P_T, 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 N_T = P_T = 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.