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| | :<math>Id\mathbf{L} = Q_- \mathbf{V}_d</math> | | :<math>Id\mathbf{L} = Q_- \mathbf{V}_d</math> |
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| − | We can now substitute the currents into the equation. First we substite <math>Q_+ q_+</math> for <math>Q_- q_-</math>: | + | We can now substitute the currents into the equation. First we substitute <math>Q_+ q_+</math> for <math>Q_- q_-</math>: |
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| | :<math>\mathbf{F}_x = - \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_x</math> | | :<math>\mathbf{F}_x = - \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_x</math> |
Revision as of 17:55, 23 March 2017
The basic idea here is that the electromagnetic potentials [math]\varphi[/math] and [math]A[/math] 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:
- [math]\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] [/math]
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 [math]- \frac{\partial \mathbf{A}}{\partial t}[/math] 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[1], the Distinti Paradox2[2], and, especially, the Marinov Generator[3][4] is to more accurately define the electric field [math]\mathbf{E}[/math], 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".[4] S.H.O. talk 22:09, 5 March 2017 (PST)
Prior content in the "Comment Record" section:
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 Paul Wesley's proposal
James Paul Wesley proposed adding the "motional induction" on charge q:[1]
- [math]-(\mathbf{v}\cdot\nabla)\mathbf{A}[/math]
to the Lorentz force.
The idea behind this was to explain an observation in an experiment involving a "Marinov Motor"[1] in which longitudinal induction forces were produced.
The extra term is equivalent to:[5]
- [math]-(\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}[/math]
Where [math]\mathbf{v}[/math] is the velocity of the charge.
The Lorentz force is:
[math]\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][/math]
Therefore, adding the extra term proposed by Stefan Marinov results in:
[math]\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][/math]
The problem with this modification:
In the case of an electrical charge approaching a wire, this additional term proposed by Marinov doubles 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 Marinov term predict the same force.
Now consider the case where the wire is moving toward the charge. In this case, both the transformer induction [math]- \frac{\partial \mathbf{A}}{\partial t}[/math] and the additional Marinov term predict the same force acting on the charge directed parallel to the vector potential of the current. The term proposed by Marinov doubles the observed force in these cases.
Cyril William Smith's proposal
Cyril Smith proposed adding the following gradient to the Lorentz force:[3]
[math]- \nabla_\mathbf{A}(\mathbf{v}\cdot\mathbf{A})[/math]
This is equal to:
- [math]- \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}[/math]
The idea behind this was to explain an observation in an experiment involving a "Marinov Generator"[3] in which longitudinal induction forces were produced.
The Lorentz force is:
[math]\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][/math]
Therefore, adding the extra term proposed by Cyril Smith results in:
[math]\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][/math]
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:
- [math]- \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}[/math]
Explaining the Marinov Motor and Cyril Smith's "Marinov Generator" using Conventional Physics
In the sub-sections below, Gaussian units are used unless otherwise noted.
The E-field derived from the Liénard–Wiechert Potentials
From the paper titled "Onoochin's Paradox" by Kirk T. McDonald[6][7], we have following statement:
For calculations of the Lorentz force to be accurate to order [math]\frac{1}{c^2}[/math], 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 [8][9] (for details, see the appendix of [10]),
- [math]\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][/math]
where [math]\mathbf{a}[/math] is the acceleration [math]\mathbf{a}[/math] of the charge [math]q[/math] at the present time.
Let's consider the situation where the acceleration [math]\mathbf{a}[/math] of charge [math]q[/math] is negligible:
- [math]\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)[/math]
Next, let's clarify that [math]q[/math] is the source of E field by attaching a prime mark or ' to it:
- [math]\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)[/math]
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:
- [math]\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)[/math]
Let's consider a target charge [math]q[/math] located at [math]\mathbf{r}[/math] at rest in the observer's inertial frame. The inertial observer and the charge [math]q[/math] agree on what the electric field [math]\mathbf{E}[/math] is, they agree that that there is no magnetic force on [math]q[/math], and finally, they agree on the acceleration of [math]q[/math].
Another way to express this result is in terms of the angle [math]\theta[/math] between [math]\mathbf{v}[/math] and [math]\mathbf{r}[/math]:
- [math]\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \frac{v^2}{2c^2} \left(1 - 3\ cos^2\theta \right)[/math]
The above equation can be broken up into two parts, one based on the relative azimuthal velocity of the source [math]\mathbf{v}_{\theta}[/math], and one based on the relative radial velocity of the source [math]\mathbf{v}_r[/math]. First we rearrange the equation:
- [math]\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)[/math]
Since the relative azimuthal velocity and the relative radial velocity are orthogonal, we can express the following:
- [math]\mathbf{E} = q\ \frac{\mathbf{\hat{r}}}{r^2} \left(\frac{v_{\theta}^2}{2c^2} - \frac{v_r^2}{c^2} \right)[/math]
The relative azimuthal velocity can be split into orthogonal components in x and y according to the following Pythagorean relation:
- [math]|\mathbf{v}_{\theta}|^2 = |\mathbf{v}_x|^2 + |\mathbf{v}_y|^2[/math]
- [math]v_{\theta}^2 = v_x^2 + v_y^2[/math]
Therefore:
- [math]\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)[/math]
Three relative orthogonal velocities contribute separate to the force between charges. Therefore, the electric field [math]\mathbf{E}[/math] can be split into three components:
- [math]\mathbf{E}_x = q\ \frac{\mathbf{\hat{r}}}{r^2} \frac{v_x^2}{2c^2}[/math]
- [math]\mathbf{E}_y = q\ \frac{\mathbf{\hat{r}}}{r^2} \frac{v_y^2}{2c^2}[/math]
- [math]\mathbf{E}_r = q\ \frac{\mathbf{\hat{r}}}{r^2} \frac{v_r^2}{c^2}[/math]
Consider the existence of four charges:
- [math]q_-[/math] : the negative charge of loose electrons of the source current element
- [math]Q_-[/math] : the negative charge of loose electrons of the target current element
- [math]q_+[/math] : the positive charge of metallic atoms (excluding loose electrons) of the source current element
- [math]Q_+[/math] : the positive charge of metallic atoms (excluding loose electrons) of the target current element
Consider the existence of their corresponding current elements:
- [math]id\mathbf{l}[/math] : the source current element
- [math]Id\mathbf{L}[/math] : the target current element
Where:
- [math]i[/math] is the source current and [math]d\mathbf{l}[/math] is its length element.
- [math]I[/math] is the target current and [math]d\mathbf{L}[/math] is its length element.
The current elements are equal to:
- [math]id\mathbf{l} = q_- \mathbf{v}_d[/math]
- [math]Id\mathbf{L} = Q_- \mathbf{V}_d[/math]
Where:
- [math]\mathbf{v}_d[/math] is the drift velocity of the electrons of the source current element.
- [math]\mathbf{V}_d[/math] 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 [math]\mathbf{F}_{--}[/math] on [math]Q_-[/math] by field [math]\mathbf{E}_{--}[/math] of [math]q_-[/math]
- Force [math]\mathbf{F}_{-+}[/math] on [math]Q_-[/math] by field [math]\mathbf{E}_{-+}[/math] of [math]q_+[/math]
- Force [math]\mathbf{F}_{+-}[/math] on [math]Q_+[/math] by field [math]\mathbf{E}_{+-}[/math] of [math]q_-[/math]
- Force [math]\mathbf{F}_{++}[/math] on [math]Q_+[/math] by field [math]\mathbf{E}_{++}[/math] of [math]q_+[/math]
These four forces are dependent on four different relative velocities (source velocity w.r.t. target velocity):
- Relative velocity [math]\mathbf{v}_{--} = \mathbf{v}_- - \mathbf{V}_-[/math]
- Relative velocity [math]\mathbf{v}_{-+} = \mathbf{v}_- - \mathbf{V}_+[/math]
- Relative velocity [math]\mathbf{v}_{+-} = \mathbf{v}_+ - \mathbf{V}_-[/math]
- Relative velocity [math]\mathbf{v}_{++} = \mathbf{v}_+ - \mathbf{V}_+[/math]
The corresponding relative speeds are:
- Relative speed [math]v_{x--} = \left(v_- - V_-\right)_x[/math]
- Relative speed [math]v_{x-+} = \left(v_- - V_+\right)_x[/math]
- Relative speed [math]v_{x+-} = \left(v_+ - V_-\right)_x[/math]
- Relative speed [math]v_{x++} = \left(v_+ - V_+\right)_x[/math]
- Relative speed [math]v_{y--} = \left(v_- - V_-\right)_y[/math]
- Relative speed [math]v_{y-+} = \left(v_- - V_+\right)_y[/math]
- Relative speed [math]v_{y+-} = \left(v_+ - V_-\right)_y[/math]
- Relative speed [math]v_{y++} = \left(v_+ - V_+\right)_y[/math]
- Relative speed [math]v_{r--} = \left(v_- - V_-\right)_r[/math]
- Relative speed [math]v_{r-+} = \left(v_- - V_+\right)_r[/math]
- Relative speed [math]v_{r+-} = \left(v_+ - V_-\right)_r[/math]
- Relative speed [math]v_{r++} = \left(v_+ - V_+\right)_r[/math]
The two drift velocities are:
- [math]\mathbf{v}_d = \mathbf{v}_- - \mathbf{v}_+[/math] : The drift velocity of the loose electrons of the source current element
- [math]\mathbf{V}_d = \mathbf{V}_- - \mathbf{V}_+[/math] : The drift velocity of the loose electrons of the target current element
Let the effective velocities of the current elements be:
- [math]\mathbf{v} = \left( \mathbf{v}_- + \mathbf{v}_+ \right)/2[/math]
- [math]\mathbf{V} = \left( \mathbf{V}_- + \mathbf{V}_+ \right)/2[/math]
So the effective relative velocity between the current elements (source velocity w.r.t. target velocity) is:
- [math]\mathbf{v}_{rel} = \mathbf{v} - \mathbf{V}[/math]
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:
- [math]\mathbf{u} = \mathbf{v}_d / 2[/math] is the deviation velocity of the source current element.
- [math]\mathbf{U} = \mathbf{V}_d / 2[/math] is the deviation velocity of the target current element.
All four relative velocities can be expressed in terms of the deviation velocities [math]\mathbf{u}[/math] and [math]\mathbf{U}[/math] together with the relative velocity [math]\mathbf{v}_{rel}[/math].
- Relative velocity [math]\mathbf{v}_{--} = \mathbf{v}_- - \mathbf{V}_- = \mathbf{u} + \mathbf{v}_{rel} - \mathbf{U}[/math]
- Relative velocity [math]\mathbf{v}_{-+} = \mathbf{v}_- - \mathbf{V}_+ = \mathbf{u} + \mathbf{v}_{rel} + \mathbf{U}[/math]
- Relative velocity [math]\mathbf{v}_{+-} = \mathbf{v}_+ - \mathbf{V}_- = -\mathbf{u} + \mathbf{v}_{rel} - \mathbf{U}[/math]
- Relative velocity [math]\mathbf{v}_{++} = \mathbf{v}_+ - \mathbf{V}_+ = -\mathbf{u} + \mathbf{v}_{rel} + \mathbf{U}[/math]
The corresponding relative speeds are:
- Relative speed [math]v_{x--} = \left(v_- - V_- = u + v_{rel} - U\right)_x[/math]
- Relative speed [math]v_{x-+} = \left(v_- - V_+ = u + v_{rel} + U\right)_x[/math]
- Relative speed [math]v_{x+-} = \left(v_+ - V_- = -u + v_{rel} - U\right)_x[/math]
- Relative speed [math]v_{x++} = \left(v_+ - V_+ = -u + v_{rel} + U\right)_x[/math]
- Relative speed [math]v_{y--} = \left(v_- - V_- = u + v_{rel} - U\right)_y[/math]
- Relative speed [math]v_{y-+} = \left(v_- - V_+ = u + v_{rel} + U\right)_y[/math]
- Relative speed [math]v_{y+-} = \left(v_+ - V_- = -u + v_{rel} - U\right)_y[/math]
- Relative speed [math]v_{y++} = \left(v_+ - V_+ = -u + v_{rel} + U\right)_y[/math]
- Relative speed [math]v_{r--} = \left(v_- - V_- = u + v_{rel} - U\right)_r[/math]
- Relative speed [math]v_{r-+} = \left(v_- - V_+ = u + v_{rel} + U\right)_r[/math]
- Relative speed [math]v_{r+-} = \left(v_+ - V_- = -u + v_{rel} - U\right)_r[/math]
- Relative speed [math]v_{r++} = \left(v_+ - V_+ = -u + v_{rel} + U\right)_r[/math]
The equation for the electric field contains square values of the speed. As noted before:
- [math]\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)[/math]
The electric fields on charges [math]Q_+[/math] and [math]Q_-[/math] are as follows:
- [math]\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)[/math]
- [math]\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)[/math]
It may be more helpful to calculate the forces due to relative radial velocities separately from the forces due to relative azimuthal velocities. Therefore:
- [math]\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)[/math]
- [math]\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)[/math]
- [math]\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)[/math]
It follows that [math]\mathbf{F}_x[/math] and [math]\mathbf{F}_y[/math] are functions of currents perpendicular to radial vector [math]\mathbf{r}[/math] while [math]\mathbf{F}_r[/math] is a function of currents co-linear with radial vector [math]\mathbf{r}[/math]
All charges ([math]Q_-[/math], [math]Q_+[/math], [math]q_-[/math], and [math]q_+[/math]) 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. To start, we can recognize the following:
- [math]q_+ = - q_-[/math]
- [math]Q_+ = - Q_-[/math]
This allows us to factor out [math]Q_+ q_+[/math] with the following result:
- [math]\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)[/math]
- [math]\mathbf{F}_t = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( v_{t--}^2 - v_{t+-}^2 - v_{t-+}^2 + v_{y++}^2 \right)[/math]
- [math]\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)[/math]
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.
- [math]\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[/math]
- [math]\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[/math]
- [math]\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[/math]
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:
- [math]v_{--}^2 = \left(u + v_{rel} - U\right)^2 = u^2 + v_{rel}^2 + U^2 + (2u - 2U) v_{rel} - 2 u U[/math]
- [math]v_{+-}^2 = \left(-u + v_{rel} - U\right)^2 = u^2 + v_{rel}^2 + U^2 + (-2u - 2U) v_{rel} + 2 u U[/math]
- [math]v_{-+}^2 = \left(u + v_{rel} + U\right)^2 = u^2 + v_{rel}^2 + U^2 + (2u + 2U) v_{rel} + 2 u U[/math]
- [math]v_{++}^2 = \left(-u + v_{rel} + U\right)^2 = u^2 + v_{rel}^2 + U^2 + (-2u + 2U) v_{rel} - 2 u U[/math]
Therefore sum of the forces on [math]Q_-[/math] and [math]Q_+[/math] would depend on:
- [math]v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = ((2u - 2U) - (-2u - 2U) - (2u + 2U) + (-2u + 2U))v_{rel} + ((-2) - (2) - (2) + (-2)) u U[/math]
- [math]v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = ((2u - 2U) + (2u + 2U) - (2u + 2U) - (2u - 2U))v_{rel} - 8 u U[/math]
- [math]v_{--}^2 - v_{+-}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = ((-2U) + (2U) - (2U) - (- 2U))v_{rel} - 8 u U[/math]
- [math]v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = (0)v_{rel} - 8 u U[/math]
- [math]v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = - 8 u U[/math]
The force on [math]Q_-[/math] would depend on:
- [math]v_{--}^2 - v_{+-}^2 = ((2u - 2U) - (-2u - 2U))v_{rel} + ((-2) - (2)) u U[/math]
- [math]v_{--}^2 - v_{+-}^2 = 4 u v_{rel} - 4 u U[/math]
The force on [math]Q_+[/math] would depend on:
- [math]v_{++}^2 - v_{-+}^2 = ((-2u + 2U) - (2u + 2U))v_{rel} + (- (2) + (-2)) u U[/math]
- [math]v_{++}^2 - v_{-+}^2 = -4 u v_{rel} - 4 u U[/math]
The force between electrically-neutral current elements
In the parent section, we had the following:
- [math]\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[/math]
- [math]\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[/math]
- [math]\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[/math]
As stated in the parent section:
- [math]v_{--}^2 - v_{+-}^2 - v_{-+}^2 + v_{++}^2 = - 2 (v_d)(V_d)[/math]
Substitution yields:
- [math]\mathbf{F}_x = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( - 2 (v_d)(V_d) \right)_x[/math]
- [math]\mathbf{F}_y = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{2c^2} \left( - 2 (v_d)(V_d) \right)_y[/math]
- [math]\mathbf{F}_r = \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{- 2}{2c^2} \left( - 2 (v_d)(V_d) \right)_r[/math]
Simplified:
- [math]\mathbf{F}_x = - \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_x[/math]
- [math]\mathbf{F}_y = - \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_y[/math]
- [math]\mathbf{F}_r = + \frac{Q_+ q_+ \mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( (v_d)(V_d) \right)_r[/math]
As stated earlier, the current elements are equal to:
- [math]id\mathbf{l} = q_- \mathbf{v}_d[/math]
- [math]Id\mathbf{L} = Q_- \mathbf{V}_d[/math]
We can now substitute the currents into the equation. First we substitute [math]Q_+ q_+[/math] for [math]Q_- q_-[/math]:
- [math]\mathbf{F}_x = - \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_x[/math]
- [math]\mathbf{F}_y = - \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (v_d)(V_d) \right)_y[/math]
- [math]\mathbf{F}_r = + \frac{Q_- q_- \mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( (v_d)(V_d) \right)_r[/math]
Next, we apportion each charge with their corresponding drift velocities:
- [math]\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (Q_- V_d) (q_- v_d) \right)_x[/math]
- [math]\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (Q_- V_d) (q_- v_d) \right)_y[/math]
- [math]\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( (Q_- V_d) (q_- v_d) \right)_r[/math]
Next, we make a substitution for the current elements:
- [math]\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (Id\mathbf{L}) \cdot (id\mathbf{l}) \right)_x[/math]
- [math]\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( (Id\mathbf{L}) \cdot (id\mathbf{l}) \right)_y[/math]
- [math]\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( (Id\mathbf{L}) \cdot (id\mathbf{l}) \right)_r[/math]
This can be written as:
- [math]\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( Ii (d\mathbf{L} \cdot d\mathbf{l}) \right)_x[/math]
- [math]\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{1}{c^2} \left( Ii (d\mathbf{L} \cdot d\mathbf{l}) \right)_y[/math]
- [math]\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2}{c^2} \left( Ii (d\mathbf{L} \cdot d\mathbf{l}) \right)_r[/math]
Or:
- [math]\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( d\mathbf{L} \cdot d\mathbf{l} \right)_x[/math]
- [math]\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( d\mathbf{L} \cdot d\mathbf{l} \right)_y[/math]
- [math]\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2Ii}{c^2} \left( d\mathbf{L} \cdot d\mathbf{l} \right)_r[/math]
Or:
- [math]\mathbf{F}_x = - \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( d\mathbf{L}_\theta \cdot d\mathbf{l}_x \right)[/math]
- [math]\mathbf{F}_y = - \frac{\mathbf{\hat{r}}}{r^2} \frac{Ii}{c^2} \left( d\mathbf{L}_\theta \cdot d\mathbf{l}_y \right)[/math]
- [math]\mathbf{F}_r = + \frac{\mathbf{\hat{r}}}{r^2} \frac{2Ii}{c^2} \left( d\mathbf{L}_r \cdot d\mathbf{l}_r \right)[/math]
Adding the forces together, their sum is:
- [math]\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)[/math]
- [math]\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)[/math]
In S.I. Units:
- [math]\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)[/math]
Or:
- [math]\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)[/math]
The second differential is:
- [math]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)[/math]
Special Scenario: No relative motion between positive charges
In the case that the positive charges [math]q_+[/math] and [math]Q_+[/math] 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:
- [math]\mathbf{v}_d = \mathbf{v}_- - \mathbf{v}_+[/math] : The drift velocity of the loose electrons of the source current element
- [math]\mathbf{V}_d = \mathbf{V}_- - \mathbf{V}_+[/math] : The drift velocity of the loose electrons of the target current element
These become:
- [math]\mathbf{v}_d = \mathbf{v}_-[/math] : The drift velocity of the loose electrons of the source current element
- [math]\mathbf{V}_d = \mathbf{V}_-[/math] : 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:
- [math]\mathbf{v} = \left( \mathbf{v}_- + \mathbf{v}_+ \right)/2[/math]
- [math]\mathbf{V} = \left( \mathbf{V}_- + \mathbf{V}_+ \right)/2[/math]
These become:
- [math]\mathbf{v} = \mathbf{v}_- /2[/math]
- [math]\mathbf{V} = \mathbf{V}_- / 2[/math]
As stated in the parent section, a new variable, the deviation velocity, is defined as one-half the drift velocity of the electrons:
- [math]\mathbf{u} = \mathbf{v}_d / 2[/math] is the deviation velocity of the source current element.
- [math]\mathbf{U} = \mathbf{V}_d / 2[/math] is the deviation velocity of the target current element.
Therefore:
- [math]\mathbf{v} = \mathbf{u} [/math]
- [math]\mathbf{V} = \mathbf{U} [/math]
The effective relative velocity between the current elements (source velocity w.r.t. target velocity) is:
- [math]\mathbf{v}_{rel} = \mathbf{v} - \mathbf{V}[/math]
Therefore:
- [math]\mathbf{v}_{rel} = \mathbf{u} - \mathbf{U}[/math]
As derived in the parent section, all four relative velocities can be expressed in terms of the deviation velocities [math]\mathbf{u}[/math] and [math]\mathbf{U}[/math] together with the relative velocity [math]\mathbf{v}_{rel}[/math].
- Relative velocity [math]\mathbf{v}_{--} = \mathbf{v}_- - \mathbf{V}_- = \mathbf{u} + \mathbf{v}_{rel} - \mathbf{U}[/math]
- Relative velocity [math]\mathbf{v}_{-+} = \mathbf{v}_- - \mathbf{V}_+ = \mathbf{u} + \mathbf{v}_{rel} + \mathbf{U}[/math]
- Relative velocity [math]\mathbf{v}_{+-} = \mathbf{v}_+ - \mathbf{V}_- = -\mathbf{u} + \mathbf{v}_{rel} - \mathbf{U}[/math]
- Relative velocity [math]\mathbf{v}_{++} = \mathbf{v}_+ - \mathbf{V}_+ = -\mathbf{u} + \mathbf{v}_{rel} + \mathbf{U}[/math]
By substituting for [math]\mathbf{v}_{rel}[/math], we get:
- Relative velocity [math]\mathbf{v}_{--} = \mathbf{v}_- - \mathbf{V}_- = 2\mathbf{u} - 2\mathbf{U}[/math]
- Relative velocity [math]\mathbf{v}_{-+} = \mathbf{v}_- - \mathbf{V}_+ = 2\mathbf{u}[/math]
- Relative velocity [math]\mathbf{v}_{+-} = \mathbf{v}_+ - \mathbf{V}_- = -2\mathbf{U}[/math]
- Relative velocity [math]\mathbf{v}_{++} = \mathbf{v}_+ - \mathbf{V}_+ = 0[/math]
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:
- [math]v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2[/math]
Based on the above results, we have:
- [math]v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = (2u - 2U)^2 - (2u)^2 - (-2U)^2 + 0[/math]
- [math]v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = (4u^2 + 4U^2 - 8uU) - 4u^2 - 4U^2 + 0[/math]
- [math]v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = - 8uU[/math]
- [math]v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = - 8 (v_{d}/2)(V_{d}/2)[/math]
- [math]v_{--}^2 - v_{-+}^2 - v_{+-}^2 + v_{++}^2 = -2 (v_{d})(V_{d})[/math]
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 force between electrically-neutral current elements":
In S.I. Units:
- [math]\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)[/math]
Or:
- [math]\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)[/math]
The second differential is:
- [math]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)[/math]
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)
Explaining "Altered" Lenz' Law Devices
It is anticipated that the longitudinal force described in the previous section 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 [math]q\ \mathbf{v} \times \mathbf{B}[/math] 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#The Energy Machine of Joseph Newman
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)
References
- ↑ 1.0 1.1 1.2 http://redshift.vif.com/JournalFiles/Pre2001/V05NO3PDF/v05n3phi.pdf
- ↑ http://www.distinti.com/docs/pdx/paradox2.pdf
- ↑ 3.0 3.1 3.2 http://overunity.com/14691/the-marinov-generator/
- ↑ 4.0 4.1 http://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=13897
- ↑ http://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=13908
- ↑ http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.299.8534
- ↑ freeweb.siol.net/markoor/onoochin.pdf
- ↑ https://docs.google.com/file/d/0B817m31MAj0wZTZjZmMwMjgtY2Y5YS00YTQ5LThjM2EtNzhjYTYzNzFlZDY0/edit?hl=en_GB&pli=1
- ↑ https://docs.google.com/file/d/0B817m31MAj0wMDI1YjllYjctY2NhOS00M2M2LWFlMTUtYjVmYTkyZmVlY2M2/edit?hl=en_GB
- ↑ http://web.archive.org/web/20170318210550/http://puhep1.princeton.edu/~mcdonald/examples/ph501/ph501lecture24.pdf
See also
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