Function Conjunction → Electromagnetic Potentials

The basic idea here is that the electromagnetic potentials $\varphi$ and $A$ may be the underlying key to several inventions related to electromagnetic forces.

Introduction

From December 2016 to March 2017, I (S.H.O. ^talk) have been conducting electromagnetic simulations using JavaScript and the THREE.js script library (http://threejs.org). Based on these results, I have determined that the magnetic component of the Lorentz force:

$\mathbf{F_{mag}} = q\ \mathbf{v} \times \mathbf{B} = q\ \left[ \nabla_\mathbf{A}(\mathbf{v}\cdot\mathbf{A}) - (\mathbf{v}\cdot\nabla)\mathbf{A} \right]$

causes transfers of energy within the magnetic rotor assembly (such as electron kinetic energy and inductive storage energy into wire kinetic energy (wired rotor example), or atomic electron kinetic energy to magnetic domain kinetic energy (permanent magnet rotor example)) that, in the low-frequency approximation, pretty much matches the amount of energy transfer between fields and the driving (stator) coils which is due to the transformer induction electric field $- \frac{\partial \mathbf{A}}{\partial t}$ acting on currents in the coils from the relative motion of the magnetic rotor assembly.

As a result, the only way using conventional physics to explain various devices, such as the Marinov Motor, the Distinti Paradox2, and, especially, the Marinov Generator is to more accurately define the electric field $\mathbf{E}$, which is part of the full Lorentz Force equation, and then see if it leads to predictions that confirm observations, especially those found figure 9 of Cyril Smith's 2009 paper on the "Marinov Generator". S.H.O. ^talk 22:09, 5 March 2017 (PST)

Prior content in the "Comment Record" section:

• http://www.sho.wiki/index.php?title=Electromagnetic_Potentials&diff=1036&oldid=951
• http://www.sho.wiki/index.php?title=Electromagnetic_Potentials&diff=next&oldid=1165

Prior content in the "Background" section:

http://www.sho.wiki/index.php?title=Electromagnetic_Potentials&diff=next&oldid=1162

Novel Force laws proposed by various researchers

James Wesley's proposal

James Wesley proposed adding the "motional induction" on charge q. In SI Units, this can be expressed as:

$-q(\mathbf{v}\cdot\nabla)\mathbf{A}$

to the Lorentz force.

The idea behind this was to explain an observation in an experiment involving a "Marinov Motor" in which longitudinal induction forces were produced.

The extra term is equivalent to:

$-(\mathbf{v}\cdot\nabla)\mathbf{A} = \begin{matrix} - \left[ v_x \frac{∂A_x}{∂x} + v_y \frac{∂A_x}{∂y} + v_z \frac{∂A_x}{∂z} \right] \mathbf{e}_x \\ - \left[ v_x \frac{∂A_y}{∂x} + v_y \frac{∂A_y}{∂y} + v_z \frac{∂A_y}{∂z} \right] \mathbf{e}_y \\ - \left[ v_x \frac{∂A_z}{∂x} + v_y \frac{∂A_z}{∂y} + v_z \frac{∂A_z}{∂z} \right] \mathbf{e}_z \end{matrix}$

Where $\mathbf{v}$ is the velocity of the charge.

The Lorentz force is:

$\mathbf{F} = q\left[-\nabla \varphi- \frac{\partial \mathbf{A}}{\partial t}+ \nabla_\mathbf{A}(\mathbf{v}\cdot\mathbf{A})-(\mathbf{v}\cdot\nabla)\mathbf{A} \right]$

Therefore, adding the extra term proposed by Wesley results in:

$\mathbf{F} = q\left[-\nabla \varphi- \frac{\partial \mathbf{A}}{\partial t}+ \nabla_\mathbf{A}(\mathbf{v}\cdot\mathbf{A})-2(\mathbf{v}\cdot\nabla)\mathbf{A} \right]$

The problem with this modification:

In the case of an electrical charge approaching a wire, this additional term proposed by Wesley would double the force of deflection. This is not observed.

Consider the vector potential due a current-carrying wire on the x-axis. Both the current and the vector potential of this current point in the $+x$ direction. Now have a charge approaching this wire perpendicularly. Both the magnetic Lorentz force and the additional Wesley term predict the same force. The proposed additional term is superfluous and would double the force of deflection if added.

Cyril Smith's proposal

Cyril Smith proposed adding the following gradient to the Lorentz force:

$- \nabla_\mathbf{A}(\mathbf{v}\cdot\mathbf{A})$

This is equal to:

$- \nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A}) = \begin{matrix} - \left[ v_x \frac{∂A_x}{∂x} + v_y \frac{∂A_y}{∂x} + v_z \frac{∂A_z}{∂x} \right] \mathbf{e}_x \\ - \left[ v_x \frac{∂A_x}{∂y} + v_y \frac{∂A_y}{∂y} + v_z \frac{∂A_z}{∂y} \right] \mathbf{e}_y \\ - \left[ v_x \frac{∂A_x}{∂z} + v_y \frac{∂A_y}{∂z} + v_z \frac{∂A_z}{∂z} \right] \mathbf{e}_z \end{matrix}$

The idea behind this was to explain an observation in an experiment involving a "Marinov Generator" in which longitudinal induction forces were produced.

The Lorentz force is:

$\mathbf{F} = q \left[ -\nabla \varphi - \frac{\partial \mathbf{A}}{\partial t} + \nabla_\mathbf{A}(\mathbf{v}\cdot\mathbf{A}) - (\mathbf{v}\cdot\nabla)\mathbf{A} \right]$

Therefore, adding the extra term proposed by Cyril Smith results in:

$\mathbf{F} = q \left[ -\nabla \varphi - \frac{\partial \mathbf{A}}{\partial t} + 0\nabla_\mathbf{A}(\mathbf{v}\cdot\mathbf{A}) - (\mathbf{v}\cdot\nabla)\mathbf{A} \right]$

The problem with this modification:

In the case of two parallel current-carrying wires, this additional term proposed by Cyril Smith negates the magnetic forces between the currents. It turns out that the extra term may yield forces perpendicular to the velocity. The relevant field components are:

$- \nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A})_\bot = \begin{matrix} - \left[ v_y \frac{∂A_y}{∂x} + v_z \frac{∂A_z}{∂x} \right] \mathbf{e}_x \\ - \left[ v_x \frac{∂A_x}{∂y} + v_z \frac{∂A_z}{∂y} \right] \mathbf{e}_y \\ - \left[ v_x \frac{∂A_x}{∂z} + v_y \frac{∂A_y}{∂z} \right] \mathbf{e}_z \end{matrix}$

Explaining the Marinov Motor and Cyril Smith's "Marinov Generator" using Conventional Physics

See the online bulletin thread titled "Electromagnetism and relativity" for details on this video.

The Liénard–Wiechert electric field was derived from the Liénard–Wiechert potentials $\varphi$ and $A$ by Kirk T. McDonald, Professor Emeritus of Princeton University in New Jersey.

In the sub-sections below, Gaussian units are used unless otherwise noted. Also, do note that all electric fields due to source charges are, in the following sub-sections, evaluated in the rest frame of each target charge subject to them separately. This procedure has roots in an approach to electromagnetism introduced by Edward M. Purcell in Section 5.6 of the Berkeley Physics Course (Volume II) titled Electricity and Magnetism and is explained by a video by Veritasium titled "How Special Relativity Makes Magnets Work". This avoids having to perform calculations based on the magnetic field viewed by an arbitrary inertial observer. This procedure relies on the relative velocities between the charges. The calculations in the sub-sections below are valid for $v \ll c$.

The Liénard–Wiechert electric fields for electrically-neutral currents

From the paper titled "Onoochin's Paradox" by Kirk T. McDonald, we have following statement:

For calculations of the Lorentz force to be accurate to order $\frac{1}{c^2}$, it suffices to use eq. (4) for the magnetic field. However, to maintain the desired accuracy the electric field of a moving charge must also include effects of retardation, as can be obtained from an expansion of the Liénard–Wiechert fields (for details, see the appendix of ),

$\mathbf{E} \approx q\ \frac{\mathbf{\hat{r}}}{r^2} \left(1 + \frac{v^2}{2c^2} - 3\frac{\left(\mathbf{\hat{r}} \cdot \mathbf{v}\right)^2}{2c^2} \right) - \frac{q}{2c^2r}\left[\mathbf{a}+\left(\mathbf{a} \cdot \mathbf{\hat{r}} \right)\mathbf{\hat{r}}\right]$

where $\mathbf{a}$ is the acceleration $\mathbf{a}$ of the charge $q$ at the present time.

Let's consider the situation where the acceleration $\mathbf{a}$ of charge $q$ is negligible. The electric field at $\mathbf{r}$ due to source charge $q$ located at the origin is:

$\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \left(1 + \frac{v^2}{2c^2} - 3\frac{\left(\mathbf{\hat{r}} \cdot \mathbf{v}\right)^2}{2c^2} \right)$

In the Coulomb gauge, the first term in the parentheses comes from the electric scalar potential of a charge at rest in the observer's inertial frame. In event that the charge is contained within an electrically-neutral body, the electric field reduces to:

$\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \left(\frac{v^2}{2c^2} - 3\frac{\left(\mathbf{\hat{r}} \cdot \mathbf{v}\right)^2}{2c^2} \right)$

Let's consider a target charge $Q$ located at $\mathbf{r}$ at rest in the observer's inertial frame. The inertial observer and the charge $q$ agree on what the electric field $\mathbf{E}$ is, they agree that there is no magnetic force on $Q$, and finally, they agree on the acceleration of $Q$.

Another way to express this result is in terms of the angle $\theta$ between $\mathbf{v}$ and $\mathbf{r}$:

$\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \frac{v^2}{2c^2} \left(1 - 3\ cos^2\theta \right)$

The above equation can be broken up into two parts, one based on the relative azimuthal velocity $\mathbf{v}_{\theta}$ of the source, and one based on the relative radial velocity $\mathbf{v}_r$ of the source. First we rearrange the equation:

$\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \left(\frac{v^2 - \left(\mathbf{\hat{r}} \cdot \mathbf{v}\right)^2}{2c^2} - 2\frac{\left(\mathbf{\hat{r}} \cdot \mathbf{v}\right)^2}{2c^2} \right)$

Since the relative azimuthal velocity and the relative radial velocity are orthogonal, we can express the following:

$\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \left(\frac{v_{\theta}^2}{2c^2} - \frac{v_r^2}{c^2} \right)$

The relative azimuthal velocity can be split into orthogonal components in $x$ and $y$ according to the following Pythagorean relation:

$|\mathbf{v}_{\theta}|^2 = |\mathbf{v}_x|^2 + |\mathbf{v}_y|^2$

$v_{\theta}^2 = v_x^2 + v_y^2$

Therefore:

$\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \left(\frac{v_x^2}{2c^2} + \frac{v_y^2}{2c^2} - \frac{v_r^2}{c^2} \right)$

The variables $v_x$, $v_y$, and $v_r$ are mutually independent from each other. Therefore, the electric field $\mathbf{E}$ can be split into three co-radial contributions:

$\mathbf{E}_x = + q\ \frac{\mathbf{\hat{r}}}{r^2} \frac{v_x^2}{2c^2}$

$\mathbf{E}_y = + q\ \frac{\mathbf{\hat{r}}}{r^2} \frac{v_y^2}{2c^2}$

$\mathbf{E}_r = - q\ \frac{\mathbf{\hat{r}}}{r^2} \frac{v_r^2}{c^2}$

Consider the existence of four charges:

$q_-$ : the negative charge of loose electrons of the source current element

$Q_-$ : the negative charge of loose electrons of the target current element

$q_+$ : the positive charge of metallic atoms (excluding loose electrons) of the source current element

$Q_+$ : the positive charge of metallic atoms (excluding loose electrons) of the target current element

Consider the existence of their corresponding current elements:

$id\mathbf{l}$ : the source current element

$Id\mathbf{L}$ : the target current element

Where:

$i$ is the source current and $d\mathbf{l}$ is its length element.

$I$ is the target current and $d\mathbf{L}$ is its length element.

The current elements are equal to:

$id\mathbf{l} = q_- \mathbf{v}_d$

$Id\mathbf{L} = Q_- \mathbf{V}_d$

Where:

$\mathbf{v}_d$ is the drift velocity of the electrons of the source current element.

$\mathbf{V}_d$ is the drift velocity of the electrons of the target current element.

Say we want to calculate the force on the target current element due to the source current element. This requires us to analyze four different forces:

Force $\mathbf{F}_{--}$ on $Q_-$ by field $\mathbf{E}_{--}$ of $q_-$

Force $\mathbf{F}_{+-}$ on $Q_-$ by field $\mathbf{E}_{+-}$ of $q_+$

Force $\mathbf{F}_{-+}$ on $Q_+$ by field $\mathbf{E}_{-+}$ of $q_-$

Force $\mathbf{F}_{++}$ on $Q_+$ by field $\mathbf{E}_{++}$ of $q_+$

These four forces are dependent on four different relative velocities (source velocity w.r.t. target velocity):

Relative velocity $\mathbf{v}_{--} = \mathbf{v}_- - \mathbf{V}_-$

Relative velocity $\mathbf{v}_{+-} = \mathbf{v}_+ - \mathbf{V}_-$

Relative velocity $\mathbf{v}_{-+} = \mathbf{v}_- - \mathbf{V}_+$

Relative velocity $\mathbf{v}_{++} = \mathbf{v}_+ - \mathbf{V}_+$

The corresponding relative speeds are:

Relative speed $v_{x--} = \left(v_- - V_-\right)_x$

Relative speed $v_{x+-} = \left(v_+ - V_-\right)_x$

Relative speed $v_{x-+} = \left(v_- - V_+\right)_x$

Relative speed $v_{x++} = \left(v_+ - V_+\right)_x$

Relative speed $v_{y--} = \left(v_- - V_-\right)_y$

Relative speed $v_{y+-} = \left(v_+ - V_-\right)_y$

Relative speed $v_{y-+} = \left(v_- - V_+\right)_y$

Relative speed $v_{y++} = \left(v_+ - V_+\right)_y$

Relative speed $v_{r--} = \left(v_- - V_-\right)_r$

Relative speed $v_{r+-} = \left(v_+ - V_-\right)_r$

Relative speed $v_{r-+} = \left(v_- - V_+\right)_r$

Relative speed $v_{r++} = \left(v_+ - V_+\right)_r$

The two drift velocities are:

$\mathbf{v}_d = \mathbf{v}_- - \mathbf{v}_+$ : The drift velocity of the loose electrons of the source current element

$\mathbf{V}_d = \mathbf{V}_- - \mathbf{V}_+$ : The drift velocity of the loose electrons of the target current element

Let the effective velocities of the current elements be:

$\mathbf{v} = \left( \mathbf{v}_- + \mathbf{v}_+ \right)/2$

$\mathbf{V} = \left( \mathbf{V}_- + \mathbf{V}_+ \right)/2$

So the effective relative velocity between the current elements (source velocity w.r.t. target velocity) is:

$\mathbf{v}_{rel} = \mathbf{v} - \mathbf{V}$

The effective velocity of each current element is halfway between the velocity of the negative charges and the velocity of the positive charges, so one may rather define a new variable, the deviation velocity, to be one-half the drift velocity of the electrons:

$\mathbf{u} = \mathbf{v}_d / 2$ is the deviation velocity of the source current element.

$\mathbf{U} = \mathbf{V}_d / 2$ is the deviation velocity of the target current element.

All four relative velocities can be expressed in terms of the deviation velocities $\mathbf{u}$ and $\mathbf{U}$ together with the relative velocity $\mathbf{v}_{rel}$.

Relative velocity $\mathbf{v}_{--} = \mathbf{v}_- - \mathbf{V}_- = \mathbf{u} + \mathbf{v}_{rel} - \mathbf{U}$

Relative velocity $\mathbf{v}_{+-} = \mathbf{v}_+ - \mathbf{V}_- = -\mathbf{u} + \mathbf{v}_{rel} - \mathbf{U}$

Relative velocity $\mathbf{v}_{-+} = \mathbf{v}_- - \mathbf{V}_+ = \mathbf{u} + \mathbf{v}_{rel} + \mathbf{U}$

Relative velocity $\mathbf{v}_{++} = \mathbf{v}_+ - \mathbf{V}_+ = -\mathbf{u} + \mathbf{v}_{rel} + \mathbf{U}$

The corresponding relative speeds are:

Relative speed $v_{x--} = \left(v_- - V_- = u + v_{rel} - U\right)_x$

Relative speed $v_{x+-} = \left(v_+ - V_- = -u + v_{rel} - U\right)_x$

Relative speed $v_{x-+} = \left(v_- - V_+ = u + v_{rel} + U\right)_x$

Relative speed $v_{x++} = \left(v_+ - V_+ = -u + v_{rel} + U\right)_x$

Relative speed $v_{y--} = \left(v_- - V_- = u + v_{rel} - U\right)_y$

Relative speed $v_{y+-} = \left(v_+ - V_- = -u + v_{rel} - U\right)_y$

Relative speed $v_{y-+} = \left(v_- - V_+ = u + v_{rel} + U\right)_y$

Relative speed $v_{y++} = \left(v_+ - V_+ = -u + v_{rel} + U\right)_y$

Relative speed $v_{r--} = \left(v_- - V_- = u + v_{rel} - U\right)_r$

Relative speed $v_{r+-} = \left(v_+ - V_- = -u + v_{rel} - U\right)_r$

Relative speed $v_{r-+} = \left(v_- - V_+ = u + v_{rel} + U\right)_r$

Relative speed $v_{r++} = \left(v_+ - V_+ = -u + v_{rel} + U\right)_r$

The equation for the electric field contains squared values of the speed. As noted before:

$\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \left(\frac{v^2 - \left(\mathbf{\hat{r}} \cdot \mathbf{v}\right)^2}{2c^2} - 2\frac{\left(\mathbf{\hat{r}} \cdot \mathbf{v}\right)^2}{2c^2} \right)$

The electric fields on charges $Q_+$ and $Q_-$, in their own respective and distinct rest frames, are as follows:

$\mathbf{E}_- = \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( q_- \left(v_{x--}^2 + v_{y--}^2 - 2 v_{r--}^2\right) + q_+ \left(v_{x+-}^2 + v_{y+-}^2 - 2 v_{r+-}^2\right) \right)$

$\mathbf{E}_+ = \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( q_- \left(v_{x-+}^2 + v_{y-+}^2 - 2 v_{r-+}^2\right) + q_+ \left(v_{x++}^2 + v_{y++}^2 - 2 v_{r++}^2\right) \right)$

The Forces due to the Liénard–Wiechert electric fields for electrically-neutral currents

It may be more helpful to calculate the forces due to relative radial velocities separately from the forces due to relative azimuthal velocities. Therefore:

$\mathbf{F}_x = \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( Q_- q_- v_{x--}^2 + Q_- q_+ v_{x+-}^2 + Q_+ q_- v_{x-+}^2 + Q_+ q_+ v_{x++}^2 \right)$

$\mathbf{F}_y = \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( Q_- q_- v_{y--}^2 + Q_- q_+ v_{y+-}^2 + Q_+ q_- v_{y-+}^2 + Q_+ q_+ v_{y++}^2 \right)$

$\mathbf{F}_r = \frac{\mathbf{\hat{r}}}{r^2} \frac{- 2}{2c^2} \left( Q_- q_- v_{r--}^2 + Q_- q_+ v_{r+-}^2 + Q_+ q_- v_{r-+}^2 + Q_+ q_+ v_{r++}^2 \right)$

It follows that $\mathbf{F}_x$ and $\mathbf{F}_y$ are functions of currents perpendicular to radial vector $\mathbf{r}$ while $\mathbf{F}_r$ is a function of currents co-linear with radial vector $\mathbf{r}$

All charges ($Q_-$, $Q_+$, $q_-$, and $q_+$) may contribute simultaneously to the azimuthally-directed (transverse) currents in $x$ and $y$ and the radially-directed (longitudinal) currents in $r$.

It will be very advantageous to simplify these formulas. For electrically-neutral currents, we can recognize the following:

$q_+ = - q_-$

$Q_+ = - Q_-$

This allows us to factor out $Q_+ q_+$ with the following result:

$\mathbf{F}_x = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( v_{x--}^2 - v_{x+-}^2 - v_{x-+}^2 + v_{x++}^2 \right)$

$\mathbf{F}_y = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( v_{y--}^2 - v_{y+-}^2 - v_{y-+}^2 + v_{y++}^2 \right)$

$\mathbf{F}_r = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{- 2}{2c^2} \left( v_{r--}^2 - v_{r+-}^2 - v_{r-+}^2 + v_{r++}^2 \right)$

Next, we will work on simplifying the contents within the parentheses. To make matters simpler, we will move the subscript to the lower right corner of the parentheses.

$\mathbf{F}_x = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 \right)_x$

$\mathbf{F}_y = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 \right)_y$

$\mathbf{F}_r = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{- 2}{2c^2} \left( v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 \right)_r$

These can be distributed back to the contents within the parentheses after they are substituted for a different expression in terms of the currents. The terms in the parentheses (disregarding their sign) are as follows:

$v_{--}^2 = \left(u + v_{rel} - U\right)^2 = u^2 + v_{rel}^2 + U^2 + (2u - 2U) v_{rel} - 2 u U$

$v_{+-}^2 = \left(-u + v_{rel} - U\right)^2 = u^2 + v_{rel}^2 + U^2 + (-2u - 2U) v_{rel} + 2 u U$

$v_{-+}^2 = \left(u + v_{rel} + U\right)^2 = u^2 + v_{rel}^2 + U^2 + (2u + 2U) v_{rel} + 2 u U$

$v_{++}^2 = \left(-u + v_{rel} + U\right)^2 = u^2 + v_{rel}^2 + U^2 + (-2u + 2U) v_{rel} - 2 u U$

Therefore sum of the forces on $Q_-$ and $Q_+$ depends on:

$v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = ((2u - 2U) - (-2u - 2U) - (2u + 2U) + (-2u + 2U))v_{rel} + ((-2) - (2) - (2) + (-2)) u U$

$v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = ((2u - 2U) + (2u + 2U) - (2u + 2U) - (2u - 2U))v_{rel} - 8 u U$

$v_{--}^2 - v_{+-}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = ((-2U) + (2U) - (2U) - (- 2U))v_{rel} - 8 u U$

$v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = (0)v_{rel} - 8 u U$

$v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = - 8 u U$

$v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = - 8 (v_d/2)(V_d/2)$

$v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = - 2 (v_d)(V_d)$

The force on $Q_-$ depends on:

$v_{--}^2 - v_{+-}^2 = ((2u - 2U) - (-2u - 2U))v_{rel} + ((-2) - (2)) u U$

$v_{--}^2 - v_{+-}^2 = 4 u v_{rel} - 4 u U$

The force on $Q_+$ depends on:

$v_{++}^2 - v_{-+}^2 = ((-2u + 2U) - (2u + 2U))v_{rel} + (- (2) + (-2)) u U$

$v_{++}^2 - v_{-+}^2 = -4 u v_{rel} - 4 u U$

Recalling that:

$\mathbf{F}_x = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 \right)_x$

$\mathbf{F}_y = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 \right)_y$

$\mathbf{F}_r = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{- 2}{2c^2} \left( v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 \right)_r$

$v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = - 2 (v_d)(V_d)$

Substitution yields:

$\mathbf{F}_x = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( - 2 (v_d)(V_d) \right)_x$

$\mathbf{F}_y = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( - 2 (v_d)(V_d) \right)_y$

$\mathbf{F}_r = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{- 2}{2c^2} \left( - 2 (v_d)(V_d) \right)_r$

Simplified:

$\mathbf{F}_x = - \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_x$

$\mathbf{F}_y = - \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_y$

$\mathbf{F}_r = + \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( (v_d)(V_d) \right)_r$

As stated earlier, the current elements are equal to:

$id\mathbf{l} = q_- \mathbf{v}_d$

$Id\mathbf{L} = Q_- \mathbf{V}_d$

We can now substitute the currents into the equation. First we substitute $Q_- q_-$ for $Q_+ q_+$:

$\mathbf{F}_x = - \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_x$

$\mathbf{F}_y = - \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_y$

$\mathbf{F}_r = + \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( (v_d)(V_d) \right)_r$

Next, we assign each charge with their corresponding drift velocities:

$\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (Q_- V_d) (q_- v_d) \right)_x$

$\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (Q_- V_d) (q_- v_d) \right)_y$

$\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( (Q_- V_d) (q_- v_d) \right)_r$

Next, we make a substitution for the current elements:

$\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (Id\mathbf{L}) \cdot (id\mathbf{l}) \right)_x$

$\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (Id\mathbf{L}) \cdot (id\mathbf{l}) \right)_y$

$\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( (Id\mathbf{L}) \cdot (id\mathbf{l}) \right)_r$

This can be written as:

$\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( Ii (d\mathbf{L} \cdot d\mathbf{l}) \right)_x$

$\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( Ii (d\mathbf{L} \cdot d\mathbf{l}) \right)_y$

$\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( Ii (d\mathbf{L} \cdot d\mathbf{l}) \right)_r$

Or:

$\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( d\mathbf{L} \cdot d\mathbf{l} \right)_x$

$\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( d\mathbf{L} \cdot d\mathbf{l} \right)_y$

$\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2Ii}{c^2} \left( d\mathbf{L} \cdot d\mathbf{l} \right)_r$

Or:

$\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( d\mathbf{L}_x \cdot d\mathbf{l}_x \right)$

$\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( d\mathbf{L}_y \cdot d\mathbf{l}_y \right)$

$\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2Ii}{c^2} \left( d\mathbf{L}_r \cdot d\mathbf{l}_r \right)$

Adding the forces together, their sum is:

$\mathbf{F} = \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( 2 d\mathbf{L}_r \cdot d\mathbf{l}_r - d\mathbf{L}_x \cdot d\mathbf{l}_x - d\mathbf{L}_y \cdot d\mathbf{l}_y \right)$

$\mathbf{F} = \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( 2 d\mathbf{L}_r \cdot d\mathbf{l}_r - d\mathbf{L}_\theta \cdot d\mathbf{l}_\theta \right)$

In S.I. Units:

$\mathbf{F} = \frac{\mathbf{\hat{r}}}{4\pi\epsilon_0r^2} \frac{Ii}{c^2} \left( 2 d\mathbf{L}_r \cdot d\mathbf{l}_r - d\mathbf{L}_\theta \cdot d\mathbf{l}_\theta \right)$

Or:

$\mathbf{F} = \frac{\mu_0 I i \mathbf{\hat{r}}}{4\pi r^2} \left( 2 d\mathbf{L}_r \cdot d\mathbf{l}_r - d\mathbf{L}_\theta \cdot d\mathbf{l}_\theta \right)$

The second differential is:

$d^2\mathbf{F} = \frac{\mu_0 \mathbf{\hat{r}}}{4\pi r^2} \left( 2 \mathbf{I}_r \cdot \mathbf{i}_r - \mathbf{I}_\theta \cdot \mathbf{i}_\theta \right)$

Special Scenario: No relative motion between positive charges

In the case that the positive charges $q_+$ and $Q_+$ are essentially stationary, we can simplify the derivation of the field and force equations in the rest frame of the positive charges.

As stated in the parent section, the two drift velocities are:

$\mathbf{v}_d = \mathbf{v}_- - \mathbf{v}_+$ : The drift velocity of the loose electrons of the source current element

$\mathbf{V}_d = \mathbf{V}_- - \mathbf{V}_+$ : The drift velocity of the loose electrons of the target current element

These become:

$\mathbf{v}_d = \mathbf{v}_-$ : The drift velocity of the loose electrons of the source current element

$\mathbf{V}_d = \mathbf{V}_-$ : The drift velocity of the loose electrons of the target current element

As stated in the parent section, the effective velocities of the current elements are:

$\mathbf{v} = \left( \mathbf{v}_- + \mathbf{v}_+ \right)/2$

$\mathbf{V} = \left( \mathbf{V}_- + \mathbf{V}_+ \right)/2$

These become:

$\mathbf{v} = \mathbf{v}_- /2$

$\mathbf{V} = \mathbf{V}_- / 2$

As stated in the parent section, a new variable, the deviation velocity, is defined as one-half the drift velocity of the electrons:

$\mathbf{u} = \mathbf{v}_d / 2$ is the deviation velocity of the source current element.

$\mathbf{U} = \mathbf{V}_d / 2$ is the deviation velocity of the target current element.

Therefore:

$\mathbf{v} = \mathbf{u}$

$\mathbf{V} = \mathbf{U}$

The effective relative velocity between the current elements (source velocity w.r.t. target velocity) is:

$\mathbf{v}_{rel} = \mathbf{v} - \mathbf{V}$

Therefore:

$\mathbf{v}_{rel} = \mathbf{u} - \mathbf{U}$

As derived in the parent section, all four relative velocities can be expressed in terms of the deviation velocities $\mathbf{u}$ and $\mathbf{U}$ together with the relative velocity $\mathbf{v}_{rel}$.

Relative velocity $\mathbf{v}_{--} = \mathbf{v}_- - \mathbf{V}_- = \mathbf{u} + \mathbf{v}_{rel} - \mathbf{U}$

Relative velocity $\mathbf{v}_{-+} = \mathbf{v}_- - \mathbf{V}_+ = \mathbf{u} + \mathbf{v}_{rel} + \mathbf{U}$

Relative velocity $\mathbf{v}_{+-} = \mathbf{v}_+ - \mathbf{V}_- = -\mathbf{u} + \mathbf{v}_{rel} - \mathbf{U}$

Relative velocity $\mathbf{v}_{++} = \mathbf{v}_+ - \mathbf{V}_+ = -\mathbf{u} + \mathbf{v}_{rel} + \mathbf{U}$

By substituting for $\mathbf{v}_{rel}$, we get:

Relative velocity $\mathbf{v}_{--} = \mathbf{v}_- - \mathbf{V}_- = 2\mathbf{u} - 2\mathbf{U}$

Relative velocity $\mathbf{v}_{-+} = \mathbf{v}_- - \mathbf{V}_+ = 2\mathbf{u}$

Relative velocity $\mathbf{v}_{+-} = \mathbf{v}_+ - \mathbf{V}_- = -2\mathbf{U}$

Relative velocity $\mathbf{v}_{++} = \mathbf{v}_+ - \mathbf{V}_+ = 0$

As derived in the parent section, the following is a series sum of terms (a function of relative speeds) which will be used to help calculate the forces between currents:

$v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2$

Based on the above results, we have:

$v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = (2u - 2U)^2 - (2u)^2 - (-2U)^2 + 0$

$v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = (4u^2 + 4U^2 - 8uU) - 4u^2 - 4U^2 + 0$

$v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = - 8uU$

$v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = - 8 (v_{d}/2)(V_{d}/2)$

$v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = -2 (v_{d})(V_{d})$

This produces the same results as the parent section titled "Explaining the Marinov Motor and Cyril Smith's 'Marinov Generator' using Conventional Physics" and the previous section titled "The Forces due to the Liénard–Wiechert electric fields for electrically-neutral currents":

In S.I. Units:

$\mathbf{F} = \frac{\mathbf{\hat{r}}}{4\pi\epsilon_0r^2} \frac{Ii}{c^2} \left( 2 d\mathbf{L}_r \cdot d\mathbf{l}_r - d\mathbf{L}_\theta \cdot d\mathbf{l}_\theta \right)$

Or:

$\mathbf{F} = \frac{\mu_0 I i \mathbf{\hat{r}}}{4\pi r^2} \left( 2 d\mathbf{L}_r \cdot d\mathbf{l}_r - d\mathbf{L}_\theta \cdot d\mathbf{l}_\theta \right)$

The second differential is:

$d^2\mathbf{F} = \frac{\mu_0 \mathbf{\hat{r}}}{4\pi r^2} \left( 2 \mathbf{I}_r \cdot \mathbf{i}_r - \mathbf{I}_\theta \cdot \mathbf{i}_\theta \right)$

Deriving James Wesley's additional force term in the case of co-linear current elements

As pointed out above in the sub-section titled "James Wesley's proposal", James Wesley proposed adding the "motional induction" on charge $q$. In SI Units, this can be expressed as:

$-q(\mathbf{v}\cdot\nabla)\mathbf{A}$

How may we derive the same result for current elements located and oriented along the same line? Equation 2b of the paper titled "Observations of the Marinov Motor" can be adapted to the form above, resulting in:

$\mathbf{F} = -Q_- \left(\mathbf{V}_d \cdot \nabla \right) \mathbf{A} = Q_- \left(-\mathbf{V}_d \cdot \nabla \right) \mathbf{A} = - I \left( d\mathbf{L} \cdot \nabla \right) \mathbf{A}$

In the case that we are dealing only with co-linear current elements, the result can be expressed as:

$\mathbf{F}_r = \left( -Q_- \left(\mathbf{V}_d \cdot \nabla \right) \mathbf{A} = Q_- \left(-\mathbf{V}_d \cdot \nabla \right) \mathbf{A} = - I \left( d\mathbf{L} \cdot \nabla \right) \mathbf{A} \right)_r$

As stated in the parent section, the force on $Q_-$ depends on:

$v_{--}^2 - v_{+-}^2 = ((2u - 2U) - (-2u - 2U))v_{rel} + ((-2) - (2)) u U$

$v_{--}^2 - v_{+-}^2 = 4 u v_{rel} - 4 u U$

As stated in the parent section, The force on $Q_+$ depends on:

$v_{++}^2 - v_{-+}^2 = ((-2u + 2U) - (2u + 2U))v_{rel} + (- (2) + (-2)) u U$

$v_{++}^2 - v_{-+}^2 = -4 u v_{rel} - 4 u U$

In the case that that there is no relative velocity $v_{rel}$ between the current elements:

$v_{--}^2 - v_{+-}^2 = - 4 u U = -4 (v_d/2)(V_d/2) = - (v_d)(V_d)$

$v_{++}^2 - v_{-+}^2 = - 4 u U = -4 (v_d/2)(V_d/2) = - (v_d)(V_d)$

As stated in the section "The Forces due to the Liénard–Wiechert electric fields for electrically-neutral currents", the electric forces on charges $Q_-$ and $Q_+$ are:

$\mathbf{F}_{r-} = + \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( v_{--}^2 - v_{+-}^2 \right)_r$

$\mathbf{F}_{r+} = + \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( v_{++}^2 - v_{-+}^2 \right)_r$

From this and above, it follows that:

$\mathbf{F}_{r-} = + \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( - (v_d)(V_d) \right)_r$

$\mathbf{F}_{r+} = + \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( - (v_d)(V_d) \right)_r$

Therefore:

$\mathbf{F}_{r-} = + \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( - (v_d)(V_d) \right)_r$

$\mathbf{F}_{r+} = + \frac{Q_+ q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( + (v_d)(V_d) \right)_r$

The effective electric fields experience by each charge is:

$\mathbf{E}_{r-} = + \frac{q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( - (v_d)(V_d) \right)_r$

$\mathbf{E}_{r+} = + \frac{q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( + (v_d)(V_d) \right)_r$

Given that:

$Q_+ = - Q_-$

$\mathbf{U} = \mathbf{v}_d / 2$

$\mathbf{v}_d = \mathbf{U} - (-\mathbf{U})$

$Id\mathbf{L} = Q_- \mathbf{V}_d = Q_- (+ \mathbf{U}) + Q_+ (-\mathbf{U})$

It can be shown that:

$\mathbf{F}_{r+} = \left( - Q_+ \left((-\mathbf{U}) \cdot \nabla \right) \mathbf{A} \right)_r$

$\mathbf{F}_{r-} = \left( - Q_- \left((+\mathbf{U}) \cdot \nabla \right) \mathbf{A} \right)_r$

$\mathbf{F}_r = \left( - Q_- \left(\mathbf{V}_d \cdot \nabla \right) \mathbf{A} \right)_r = \mathbf{F}_{r-} + \mathbf{F}_{r+}$

This is based on the directional derivative of vector potential $\mathbf{A}$ along $Id\mathbf{L}$.

It is important to note that the total force $\mathbf{F}_r$ in $r$ remains unaffected by the relative motion between the target current element $I_rd\mathbf{L}_r$ and the source current element $i_rd\mathbf{l}_r$ responsible for vector potential $\mathbf{A}_r$. However, forces $\mathbf{F}_{r-}$ and $\mathbf{F}_{r+}$ are affected by the relative speed $(v_{rel})_r$ between the current elements in $r$, which leads to a kind of electromagnetic induction acting between co-linear current elements in relative motion in $r$. This feature may help to explain Cyril Smith's "Marinov Generator".

A New Idea: The Makerarc

The above result supports a development beyond the S.H.O. Drive. A much more powerful and compact permanent magnet system is now envisioned.

The preliminary name is Makerarc (MAY-KERR-ARK), which stands for:

• Magnetic
• Atom
• Kinetic
• Energy
• Reservoir
• And
• Resource
• Channel

Details pending. Stay tuned. Sincerely, S.H.O. ^talk 22:09, 5 March 2017 (PST)

What makes a magnet tick? Clues can be found in the video titled "MAGNETS: How Do They Work?" by Veritasium. S.H.O. ^talk 22:08, 23 March 2017 (PDT)

An attempt to produce the Makerarc back in March 2017 failed to demonstrate the predictions of the term $\mathbf{F}_r = \left( - Q_- \left(\mathbf{V}_d \cdot \nabla \right) \mathbf{A} \right)_r$ where I assumed that an external vector potential $\mathbf{A}$ would lead to a force co-linear to current elements. This may be due to the supercurrent nature of electrons bounded to atoms, allowing atoms to be capable of preserving or "fixing" the amount of magnetic flux that passes through them, just as "macroscopic" superconducting currents do for superconductors, and therefore, by extension be capable of resisting changes in the magnetic vector potential. This does not negate the possibility of a Marinov Generator, as designed by Cyril Smith, because in his case the currents receiving power were inside a conductor where an externally applied vector potential is not fully shielded against, permitting $\mathbf{F}_r = \left( - Q_- \left(\mathbf{V}_d \cdot \nabla \right) \mathbf{A} \right)_r$ to yield a force. At the moment, I am still trying to decide on the correct model, but hopefully the surviving concept is similar to what I have developed so far in this page. Sincerely, S.H.O. ^talk 19:55, 14 July 2017 (PDT)

Explaining "Altered" Lenz' Law Devices

It is anticipated that the longitudinal force described above in the section titled "Explaining the Marinov Motor and Cyril Smith's 'Marinov Generator' using Conventional Physics" may explain some types of purported Reduced-Lenz devices. A good example can be found in the video below ("Ray's No Back EMF Generator"), although in this example, the longitudinal force increases the drag, mainly in positions where pancake generator coil is mostly outside the cylindrical boundary of the permanent magnet. This creates an illusion of a "Reduced" Lenz' Law effect when the magnet is mostly within the cylindrical boundary of the permanent magnet:

Simulations in JavaScript and THREE.js have determined that in many other configurations of currents and magnets, the magnetic Lorentz forces $q\ \mathbf{v} \times \mathbf{B}$ will be opposed in part by the additional force. This makes certain types of magnetic circuit arrangements more efficient at electrical power generation (in terms of output power vs. input power) but less efficient as an electrical motor. Simulations of the S.H.O. Drive indicated that the additional force may vary between assisting or opposing the magnetic Lorentz force, which is helpful depending on whether the goal is to develop an efficient motor or an efficient generator, respectively. The simulated S.H.O. Drive designs appeared to switch between these two extremes every quarter cycle. The only way to make it work would have been to increase the inductive reactance (so that the current would be delayed by about a quarter cycle), and even then, simulations indicated that the extra term was typically insufficient to completely reverse the net force. However simulations of various rotor and S.H.O. coil curve modifications for the S.H.O. Drive showed that it was possible for the additional force to be a significant percentage of the magnetic Lorentz force. Per more recent simulations (early March), the Makerarc design (previous section) will improve upon this many fold. S.H.O. ^talk 23:39, 5 March 2017 (PST)

Explaining the Newman Motor

See also: Memory Lane

A Newman Motor-style coil and magnet arrangement, like that shown in the video below, have been simulated by me using JavaScript and THREE.js.

The extra electric field term predicts a significant opposition to the magnetic Lorentz force at angles slightly straying from the "top-dead-vertical" position, making it a better generator than a motor. However, when energy is discharged from the "generator coil" to the "motor coil" of Newman's motor, the rotor will have often changed position to the point where the magnetic Lorentz force becomes increasingly significant, helpful for motive purposes. Newman's motor operated at a high Q, which facilitated energy recovery. S.H.O. ^talk 00:00, 6 March 2017 (PST)

See also

• Magnetic Energy
• The Anatomy of a Physical Expression
• Functions composed of Physical Expressions

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