Difference between revisions of "Electromagnetic Potentials"

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(→‎Explaining the Marinov Motor and Cyril Smith's "Marinov Generator" using Conventional Physics)
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==Explaining the Marinov Motor and Cyril Smith's "Marinov Generator" using Conventional Physics==
 
==Explaining the Marinov Motor and Cyril Smith's "Marinov Generator" using Conventional Physics==
  
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===The Longitudinal E-field derived from the Liénard–Wiechert Potentials===
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In the sub-sections below, [https://en.wikipedia.org/wiki/Centimetre%E2%80%93gram%E2%80%93second_system_of_units CGS units] are used unless otherwise noted.
  
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Believe it or not, the electromagnetic potentials formulated independently by Alfred-Marie Liénard<ref name="Liénard">https://docs.google.com/file/d/0B817m31MAj0wZTZjZmMwMjgtY2Y5YS00YTQ5LThjM2EtNzhjYTYzNzFlZDY0/edit?hl=en_GB&pli=1</ref> and Emil Wiechert<ref name="Wiechert">https://docs.google.com/file/d/0B817m31MAj0wMDI1YjllYjctY2NhOS00M2M2LWFlMTUtYjVmYTkyZmVlY2M2/edit?hl=en_GB</ref> anticipate an electric field with longitudinal components.
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===The E-field derived from the Liénard–Wiechert Potentials===
  
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The Liénard–Wiechert potentials <math>\varphi</math> (scalar potential field) and <math>\mathbf{A}</math> (vector potential field) are for a ''source point'' charge <math>q_s</math> at position <math>\mathbf{r}_s</math> traveling with velocity <math>\mathbf{v}_s</math>:
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From the paper titled "Onoochin's Paradox" by Kirk T. McDonald, we have following statement:
  
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:<math>\varphi(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \left(\frac{q_s}{(1 - \mathbf{n} \cdot \boldsymbol{\beta}_s)|\mathbf{r} - \mathbf{r}_s|} \right)_{t_r}</math>
+
<blockquote>''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 <ref name="Liénard">https://docs.google.com/file/d/0B817m31MAj0wZTZjZmMwMjgtY2Y5YS00YTQ5LThjM2EtNzhjYTYzNzFlZDY0/edit?hl=en_GB&pli=1</ref><ref name="Wiechert">https://docs.google.com/file/d/0B817m31MAj0wMDI1YjllYjctY2NhOS00M2M2LWFlMTUtYjVmYTkyZmVlY2M2/edit?hl=en_GB</ref> (for details, see the appendix of <ref>http://web.archive.org/web/20170318210550/http://puhep1.princeton.edu/~mcdonald/examples/ph501/ph501lecture24.pdf</ref>),
  
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and
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:<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>
  
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:<math>\mathbf{A}(\mathbf{r},t) = \frac{\mu_0c}{4 \pi} \left(\frac{q_s \boldsymbol{\beta}_s}{(1 - \mathbf{n} \cdot \boldsymbol{\beta}_s)|\mathbf{r} - \mathbf{r}_s|} \right)_{t_r} = \frac{\boldsymbol{\beta}_s(t_r)}{c} \varphi(\mathbf{r}, t)</math>
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''where <math>\mathbf{a}</math> is the acceleration <math>\mathbf{a}</math> of the charge <math>q</math> at the present time.</blockquote>
  
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where <math>\boldsymbol{\beta}_s(t) = \frac{\mathbf{v}_s(t)}{c}</math> and  <math>\mathbf{n} = \frac{\mathbf{r} - \mathbf{r}_s}{|\mathbf{r} - \mathbf{r}_s|}</math>.
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Let's consider the situation  where the acceleration of charge <math>q</math> is negligible:
  
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The electric field derived from this is:
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:<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>
  
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:<math>\begin{align}\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2(1 - \mathbf{n} \cdot {\boldsymbol \beta}_s)^3} \left[\left(\mathbf{n} - {\boldsymbol \beta}_s\right)(1-{\beta_s}^2) + |\mathbf{r} - \mathbf{r}_s|(\mathbf{n} \cdot \dot{\boldsymbol \beta}_s/c) (\mathbf{n} - {\boldsymbol \beta}_s) - |\mathbf{r} - \mathbf{r}_s|\big(\mathbf{n} \cdot (\mathbf{n} - {\boldsymbol \beta}_s)\big) \dot{\boldsymbol \beta}_s/c \right]\end{align}</math>
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Next, let's clarify that <math>q</math> is the source of E field by attaching a subscript ''s'' to it:
  
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which is equal to <math>-{\boldsymbol \nabla}\varphi - \frac{d\mathbf{A}}{dt}</math>
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:<math>\mathbf{E} = q_s\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>
  
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If acceleration <math>\dot{\boldsymbol \beta}_s</math> is negligible, then we have:
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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:
  
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:<math>\begin{align}\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2(1 - \mathbf{n} \cdot {\boldsymbol \beta}_s)^3} \left(\mathbf{n} - \hat{{\boldsymbol \beta}}_s\right)(1-{\beta_s}^2)\end{align}</math>
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:<math>\mathbf{E} = q_s\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>
  
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If <math>\mathbf{n}</math> is parallel to <math>{\boldsymbol \beta}_s</math>, then:
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Let's consider a receiver charge <math>q_r</math> located at <math>\mathbf{r}</math> at rest in the observer's inertial frame. The inertial observer and the charge <math>q_r</math> agree on what the electric field <math>E</math> is, they agree that that there is no magnetic force on <math>q_r</math>, and finally, they agree on the acceleration of <math>q_r</math>.
  
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:<math>\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2(1 - \beta_s)^3} \left(\mathbf{n} - {\boldsymbol \beta}_s\right)(1 - {\beta_s}^2)</math>
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The above equation can be broken up into two parts, one based on the ''transverse relative velocity'' of the source <math>\mathbf{v}_{s\bot}</math>, and one based on the ''longitudinal relative velocity'' of the source<math>\mathbf{v}_{s\parallel}</math>:
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:<math>\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2(1 - \beta_s)^3} \left(\mathbf{n} - {\boldsymbol \beta}_s\right)(1 - \beta_s)(1 + \beta_s)</math>
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−
:<math>\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2(1 - \beta_s)^2} \left(\mathbf{n} - {\boldsymbol \beta}_s\right)(1 + \beta_s)</math>
+
  
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If in addition to the aforementioned condition, the condition <math>|{\boldsymbol \beta}_s| \ll |\mathbf{n}|</math> is also satisfied, then we can approximate <math>\mathbf{\hat{n}} - \hat{{\boldsymbol \beta}}_s</math> as <math>\mathbf{\hat{n}}</math>. This affects only the direction of the calculated electric field, not its magnitude. Regarding the instance of <math>\left(\mathbf{n} - {\boldsymbol \beta}_s\right)</math>, what we can do is factor out the magnitude <math>|\mathbf{n} - {\boldsymbol \beta}_s|</math> from the vector, and the use the approximation from the unit vector. Note, remember that <math>\mathbf{n}</math> and <math>{\boldsymbol \beta}_s</math> were taken to be co-linear. Therefore:
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:<math>\mathbf{E} = q_s\frac{\mathbf{\hat{r}}}{r^2} \left(\frac{\mathbf{v}_{s\bot}^2}{2c^2} - 2\frac{\mathbf{v}_{s\parallel}^2}{2c^2} \right)</math>
  
−
:<math>\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2(1 - \beta_s)^2} \mathbf{\hat{n}} \left(1 - \beta_s\right)(1 + \beta_s)</math>
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===Scenario 1: All charges confined to a line===
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:<math>\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2(1 - \beta_s)^2} \mathbf{\hat{n}} (1 + \beta_s)</math>
+
−
:<math>\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2} \frac{1 + \beta_s}{1 - \beta_s} \mathbf{\hat{n}}</math>
+
  
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Recognizing that the above condition limits the situation to non-relativistic velocities, that is to say <math>{\boldsymbol \beta} \ll 1</math> or <math>|\mathbf{v}_s| \ll c</math>, then we can approximate <math>1 - \beta_s</math> as <math>1</math> without significantly affecting the numerator <math>1 + \beta_s</math>. Therefore:
+
We can consider in the surrounding environment a system of charges with arbitrary position and velocity which observe an electric field <math>\mathbf{E}</math> that is unique to each charge's rest frame.
  
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:<math>\mathbf{E}(\mathbf{r}, t) = \frac{q_s}{4 \pi \epsilon_0} \frac{1 + \beta_s}{|\mathbf{r} - \mathbf{r}_s|^2} \mathbf{\hat{n}}</math>
+
Let's start off by considering the simplest case where all the charges are distributed along a straight line and all of their velocities are tangent to this line. Therefore, the velocity-dependent electric field experienced by a charge <math>q_r</math> in its rest frame due a charge <math>q_s</math> is:
  
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The "rigid" contribution to the electric field is:
+
:<math>\mathbf{E} = - q_s\frac{\mathbf{\hat{r}}}{r^2} \frac{\left(\mathbf{v}_{s}-\mathbf{v}_{r}\right)^2}{c^2}</math>
 +
:<math>\mathbf{E} = - q_s\frac{\mathbf{\hat{r}}}{r^2} \frac{\left(\mathbf{v}_{s \to r}\right)^2}{c^2}</math>
  
−
:<math>\mathbf{E}(\mathbf{r}, t)_{rigid} = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2} \mathbf{\hat{n}}</math>
+
Let's consider that along this line there is an element of a wire at velocity <math>\mathbf{v}_{wire}</math> relative to charge <math>q_r</math> carrying loose electrons with drift velocity <math>\mathbf{v}_{drift}</math> relative to that wire. And let's say the total electrical charge of that wire <math>Q_{wire}=+Q</math> + loose electrons <math>Q_{drift}=-Q</math> is zero. What is the electric field created by this current element according to a receiver charge <math>q_{r}</math> moving at velocity <math>\mathbf{v}_r</math>?
  
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If we consider an electromagnetic device composed of electrically neutral objects, what remains is a "quasistatic" contribution to the electric field.
+
:<math>\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q \mathbf{v}_{wire \to r}^2}{c^2} + \frac{-Q \left(\mathbf{v}_{wire \to r}+\mathbf{v}_{drift}\right)^2}{c^2}\right)</math>
  
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:<math>\mathbf{E}(\mathbf{r}, t)_{quasistatic} = \frac{q_s}{4 \pi \epsilon_0} \frac{\beta_s}{|\mathbf{r} - \mathbf{r}_s|^2} \mathbf{\hat{n}}</math>
+
:<math>\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q \mathbf{v}_{wire \to r}^2}{c^2} - \frac{Q \left(\mathbf{v}_{wire \to r}+\mathbf{v}_{drift}\right)^2}{c^2}\right)</math>
  
−
Given that:
+
:<math>\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{2 Q \mathbf{v}_{wire \to r}\mathbf{v}_{drift} + \mathbf{v}_{drift}^2}{c^2}\right)</math>
  
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:<math>\mathbf{A}(\mathbf{r},t) = \frac{\boldsymbol{\beta}_s(t_r)}{c} \varphi(\mathbf{r}, t)</math>
+
To simplify this result, let's consider the average velocity of charges comprising the wire+current:
  
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We can substitute for <math>\beta_s</math>:
+
:<math>\mathbf{v}_{rel} = \mathbf{v}_{wire \to r} + \mathbf{v}_{drift}/2</math>
  
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:<math>|\mathbf{A}(\mathbf{r},t)|c = \beta_s \varphi(\mathbf{r}, t)</math>
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This is effectively the relative velocity of a current element <math>\mathbf{I}_s dl = Q_{drift} \mathbf{v}_{drift}</math> with respect to receiver charge <math>q_r</math>
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:<math>\beta_s = \frac{|\mathbf{A}(\mathbf{r},t)|c}{\varphi(\mathbf{r}, t)}</math>
+
  
−
Because this scenario considers non-relativistic velocities, that is to say <math>{\boldsymbol \beta} \ll 1</math> or <math>|\mathbf{v}_s| \ll c</math>, we can again approximate the denominator. Therefore:
+
From here, we can find the alternate expression for electric field at charge <math>q_{r}</math> due to the current element <math>\mathbf{I}_s dl</math>:
  
−
:<math>\beta_s = \frac{|\mathbf{A}(\mathbf{r},t)|c}{\frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|}}</math>
+
:<math>\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q \left(\mathbf{v}_{avg} - \mathbf{v}_{drift}/2\right)^2}{c^2} + \frac{-Q \left(\mathbf{v}_{avg} + \mathbf{v}_{drift}/2\right)^2}{c^2}\right)</math>
  
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Therefore, substitution for <math>\beta_s</math> gives:
+
:<math>\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q \left(\mathbf{v}_{avg} - \mathbf{v}_{drift}/2\right)^2}{c^2} - \frac{Q \left(\mathbf{v}_{avg} + \mathbf{v}_{drift}/2\right)^2}{c^2}\right)</math>
  
−
:<math>\mathbf{E}(\mathbf{r}, t)_{quasistatic} = \frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|^2} \frac{|\mathbf{A}(\mathbf{r},t)|c}{\frac{q_s}{4 \pi \epsilon_0} \frac{1}{|\mathbf{r} - \mathbf{r}_s|}} \mathbf{\hat{n}}</math>
+
:<math>\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{-Q \mathbf{v}_{avg} \mathbf{v}_{drift}}{c^2}\right)</math>
  
−
:<math>\mathbf{E}(\mathbf{r}, t)_{quasistatic} = \frac{|\mathbf{A}(\mathbf{r},t)|c}{|\mathbf{r} - \mathbf{r}_s|} \mathbf{\hat{n}}</math>
+
:<math>\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q_{drift} \mathbf{v}_{avg} \mathbf{v}_{drift}}{c^2}\right)</math>
  
−
Next, substituting for <math>\mathbf{A}(\mathbf{r},t)</math> gives:
+
:<math>\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{\mathbf{I}_s \mathbf{v}_{avg}}{c^2}\right)</math>
  
−
:<math>\mathbf{E}(\mathbf{r}, t)_{quasistatic} = \frac{\frac{\beta_s(t_r)}{c} \varphi(\mathbf{r}, t)c}{|\mathbf{r} - \mathbf{r}_s|} \mathbf{\hat{n}}</math>
+
====Special Scenario 1.1: Forces between co-linear current elements confined to a straight line====
  
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:<math>\mathbf{E}(\mathbf{r}, t)_{quasistatic} = -\beta_s(t_r) \nabla\varphi(\mathbf{r}, t)</math>
+
If we wish to consider the force on charge <math>q_s</math> due to the electric field of this current element, we simply multiply by target charge <math>q_r</math>.
  
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===Force is four-dimensional, not three dimensional===
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:<math>\mathbf{F} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{\mathbf{I}_s q_r \mathbf{v}_{avg}}{c^2}\right)</math>
  
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At the beginning of physics studies, students are normally treated to the famous equation <math>\mathbf{F} = m \mathbf{a}</math> where <math>\mathbf{F}</math> is the force, <math>m</math> is the mass, and <math>\mathbf{a}</math> is the acceleration. However, in special relativity, force is four-dimensional, because momentum is four-dimensional, and force is generally the rate change of momentum over time<ref>https://en.wikipedia.org/wiki/Four-force</ref>. Even light itself is known to have momentum<ref>https://en.wikipedia.org/wiki/Photon#Physical_properties</ref>.
+
If <math>q_s</math> was a loose electron embedded in another electrically-neutral wire element, let <math>\mathbf{I}_r dl</math> stand for this target current element, analogous to the source current element <math>\mathbf{I}_s dl</math>. Therefore the force on current element <math>\mathbf{I}_r dl</math> due to current element <math>\mathbf{I}_s dl</math> is:
  
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There are three dimensions are associated with space and one dimension is associated with time. Per Emmy Noether's theorem, momentum conservation is due to invariance due to spatial translations, while energy conservation is due to invariance due to time translations<ref>https://en.wikipedia.org/wiki/Noether's_theorem#Examples_2</ref>.
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:<math>\mathbf{F} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{\mathbf{I}_s \mathbf{I}_r}{c^2}\right)</math>
  
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It is known that the norm of the spatial-symmetry contribution of the four-momentum is frame dependent, while the norm of the entire four-momentum is itself invariant<ref>https://en.wikipedia.org/wiki/Four-momentum</ref>. This is a manifestation of the fact that a mass may radiate or absorb energy, and therefore the time-symmetry contribution of the four-momentum may change. The contribution to the four-force due to time-symmetry is essentially equal to velocity vector multiplied by power absorbed divided by the speed of light squared, which essentially gives velocity times the rate change of mass over time<ref>https://www.physicsforums.com/threads/force-on-a-spherically-uniform-radiator-moving-through-space.887479/</ref>.
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In '''SI units''', this becomes:
  
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The Lorentz force gives us the rate change of the momentum per charge with respect to time. However, the relationship between momentum <math>\mathbf{p}</math> and energy <math>\mathcal{E}</math> is<ref name="Akira Hirose">http://physics.usask.ca/~hirose/p812/notes/Ch10.pdf</ref>:
+
:<math>\mathbf{F} = - \frac{\mathbf{\hat{r}}}{4 \pi \epsilon r^2} \left( \frac{\mathbf{I}_s \mathbf{I}_r}{c^2}\right)</math>
  
−
:<math>\mathbf{p} = \frac{\mathbf{v}}{c^2}\mathcal{E}</math>
+
This is equivalent to:
  
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Therefore, the acceleration <math>\mathbf{a}</math> of a ''target point'' charge <math>q_t</math> due to the Lorentz force it receives is<ref name="Akira Hirose"/>:
+
:<math>\mathbf{F} = - \left( \mathbf{I}_r \dot\ \nabla \right)\mathbf{A_s}</math>
  
−
:<math>\mathbf{a} = \frac{q_t}{m\gamma}\left[\mathbf{E} + \boldsymbol{\beta}_t \times \mathbf{B} - \boldsymbol{\beta}_t \left(\boldsymbol{\beta}_t\ \dot\ \mathbf{E}\right)\right]</math>
+
This is essentially the "Marinov Force", but with restrictions. Like the Marinov Force, it exists without necessitating a magnetic field <math>\mathbf{B}</math> and so may exist in regions where the curl of <math>\mathbf{A_s}</math> is zero. However, this scenario is constricted to the case that "all charges are confined to a line" with the restriction that the current elements are electrically neutral.
  
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I (''[[User:S.H.O.|S.H.O.]] <sup>[[User_talk:S.H.O.|talk]]</sup>'') hereby make the suggestion that the first instance of <math>\mathbf{E}</math> is different than the second instance of <math>\mathbf{E}</math>. The first instance corresponds to the electric field according to the Lorentz-Maxwell model of the electric field. Under that model of the electric field, there is no effect on the longitudinal components of the electric field dependent on the chosen inertial frame. Therefore, as per the Joules-Bernoulli equation<ref>https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Transformation_of_the_fields_between_inertial_frames</ref> describing the transformation of the electric field between differing inertial frames, the following is true for the component of the electric field co-linear with the velocity:
+
===Scenario 2: Large drift velocities and negligible wire velocities===
  
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:<math>\mathbf {{E}_{\parallel}}' = \mathbf {{E}_{\parallel}}</math>
+
The magnetization currents embodied inside permanent magnets consist of electrons with very high effective velocities per the Virial theorem which relates the kinetic energy to the potential energy of the system<ref>https://en.wikipedia.org/wiki/Virial_theorem</ref>. Therefore, in the case of permanent magnets interacting with other permanent magnets, one can simply consider the effective difference of velocity of the charges comprising the effective magnetization currents and disregard the velocities of the atoms which confine these currents in place. Let <math>\mathbf{v}_{rel}</math> be the relative velocity of charge <math>q_s</math> respect to <math>q_r</math>. Therefore:
  
−
This is in contrast to the electric field model of Liénard–Wiechert, whose components co-linear with the velocity are dependent on the magnitude of the velocity, as is evident in the prior subsection titled "The Longitudinal E-field derived from the Liénard–Wiechert Potentials". I (''[[User:S.H.O.|S.H.O.]] <sup>[[User_talk:S.H.O.|talk]]</sup>'') am making a suggestion today (Wednesday March 15, 2017) that the second instance for the above ''Lorentz acceleration'' equation applies for the electric field as derived from the potentials of Alfred-Marie Liénard<ref name="Liénard"/> and Emil Wiechert<ref name="Wiechert"/>.
+
:<math>\mathbf{E} = q_s\frac{\mathbf{\hat{r}}}{r^2} \left(\frac{\mathbf{v}_{rel\bot}^2}{2c^2} - 2\frac{\mathbf{v}_{rel\parallel}^2}{2c^2} \right)</math>
  
−
To distinguish between these two electric fields, <math>\mathbf{E}_{LM}</math> will denote the electric field used in Lorentz force equation + Maxwell equations, while <math>\mathbf{E}_{LW}</math> will denote the electric fields as derived by the Liénard–Wiechert potentials. Therefore the acceleration <math>\mathbf{a}</math> becomes:
+
The electric force on <math>q_r</math> by <math>q_s</math> is therefore:
  
−
:<math>\mathbf{a} = \frac{q_t}{m\gamma}\left[\mathbf{E}_{LM} + \boldsymbol{\beta}_t \times \mathbf{B} - \boldsymbol{\beta}_t \left(\boldsymbol{\beta}_t\ \dot\ \mathbf{E}_{LW}\right)\right]</math>
+
:<math>\mathbf{F} = q_r q_s\frac{\mathbf{\hat{r}}}{r^2} \left(\frac{\mathbf{v}_{rel\bot}^2}{2c^2} - 2\frac{\mathbf{v}_{rel\parallel}^2}{2c^2} \right)</math>
  
−
The extra acceleration term <math>\mathbf{a}_{LW}</math> is therefore:
+
====Special Scenario 2.1: Forces between magnetization currents from magnets of the same grade====
  
−
<math>\mathbf{a}_{LW} = - \boldsymbol{\beta}_t \left(\boldsymbol{\beta}_t\ \dot\ \mathbf{E}\right)</math>
+
In the case where the magnets are of the same grade (e.g. N52), the effective drift speed of the magnetization currents are identical. This makes it possible to calculate the above using current elements <math>\mathbf{I}_s</math> and <math>\mathbf{I}_r</math>.
  
−
===Deriving an effective longitudinal force on a current element===
+
The differential version of this is:
  
−
Per the section titled "The Longitudinal E-field derived from the Liénard–Wiechert Potentials" there is a "quasistatic" contribution to the electric field based upon the Liénard–Wiechert potentials.
+
:<math>d^2\mathbf{F} = \frac{\mathbf{\hat{r}}}{r^2} \left(\frac{\left(\mathbf{I}_s-\mathbf{I}_r\right)_\bot^2}{2c^2} - 2\frac{\left(\mathbf{I}_s-\mathbf{I}_r\right)_\parallel^2}{2c^2} \right)</math>
  
−
:<math>\mathbf{E}(\mathbf{r}, t)_{quasistatic} = \frac{|\mathbf{A}(\mathbf{r},t)|c}{|\mathbf{r} - \mathbf{r}_s|} \mathbf{\hat{n}}</math>
+
In '''SI units''', this becomes:
  
−
Then extra acceleration term <math>\mathbf{a}_{LW}</math> from the previous section titled "Force is four-dimensional, not three dimensional" becomes:
+
:<math>d^2\mathbf{F} = \frac{\mathbf{\hat{r}}}{4 \pi \epsilon r^2} \left(\frac{\left(\mathbf{I}_s-\mathbf{I}_r\right)_\bot^2}{2c^2} - 2\frac{\left(\mathbf{I}_s-\mathbf{I}_r\right)_\parallel^2}{2c^2} \right)</math>
−
 
+
−
:<math>\mathbf{a}_{LW} = - \frac{q_t}{m\gamma} \boldsymbol{\beta}_t \left(\boldsymbol{\beta}_t\ \dot\ \mathbf{\hat{n}} \frac{|\mathbf{A}(\mathbf{r},t)|c}{|\mathbf{r} - \mathbf{r}_s|} \right)</math>
+
−
 
+
−
Because this scenario considers non-relativistic velocities, that is to say <math>{\boldsymbol \beta} \ll 1</math> or <math>|\mathbf{v}_s| \ll c</math>, we can approximate the Lorentz factor <math>\gamma</math> as <math>1</math>. Therefore:
+
−
 
+
−
:<math>\mathbf{a}_{LW} = - \frac{q_t}{m} \boldsymbol{\beta}_t \left(\boldsymbol{\beta}_t\ \dot\ \mathbf{\hat{n}} \frac{|\mathbf{A}(\mathbf{r},t)|c}{|\mathbf{r} - \mathbf{r}_s|} \right)</math>
+
−
 
+
−
If acceleration <math>\dot{\boldsymbol \beta}_s</math> of the ''source point'' charge <math>q_s</math> at position <math>\mathbf{r}_s</math> is negligible, then:
+
−
 
+
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:<math>\mathbf{a}_{LW} = - \frac{q_t}{m} \boldsymbol{\beta}_t \left(\boldsymbol{\beta}_t\ \dot\ \nabla\right) \mathbf{A}(\mathbf{r},t)c</math>
+
−
:<math>\mathbf{a}_{LW} = - \frac{q_t}{m} \boldsymbol{\beta}_t \left(\frac{\mathbf{v}_t}{c} \dot\ \nabla\right) \mathbf{A}(\mathbf{r},t)c</math>
+
−
:<math>\mathbf{a}_{LW} = - \frac{q_t}{m} \boldsymbol{\beta}_t \left(\mathbf{v}_t \dot\ \nabla\right) \mathbf{A}(\mathbf{r},t)</math>
+
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+
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The ''effective'' longitudinal force on charge <math>+q</math> of velocity <math>\mathbf{v}_{+q}</math> is:
+
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+
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:<math>\mathbf{F_{+q}}_{LW} = - (+q) \boldsymbol{\beta}_{+q} \left(\mathbf{v}_{+q} \dot\ \nabla\right) \mathbf{A}(\mathbf{r},t)</math>
+
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+
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Likewise, the ''effective'' longitudinal force on charge <math>-q</math> of velocity <math>\mathbf{v}_{-q}</math> is:
+
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+
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:<math>\mathbf{F_{-q}}_{LW} = - (-q) \boldsymbol{\beta}_{-q} \left(\mathbf{v}_{-q} \dot\ \nabla\right) \mathbf{A}(\mathbf{r},t)</math>
+
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+
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Per the above results, the effective longitudinal force is proportional to the square of the velocity of the charge <math>q</math>. In the case where <math>\mathbf{v}_{-q} > \mathbf{v}_{+q}</math> The total force on a pair of equal and opposite charges <math>+q</math> and <math>-q</math> is varies as:
+
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+
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<math>\mathbf{F_{-q}}_{LW} \propto v_{-q}^2 - v_{+q}^2</math>
+
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+
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<math>\mathbf{F_{-q}}_{LW} \propto (v_{avg}+v_{dev})^2 - (v_{avg}-v_{dev})^2</math>
+
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+
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<math>\mathbf{F_{-q}}_{LW} \propto v_{avg} v_{dev}</math>
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+
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Where:
+
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+
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* <math>v_{avg}</math> represents the "center of velocity" of the current element, considering both positive and negative charges.
+
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* <math>v_{dev}</math> is represents one-half of the drift velocity of the negative charges relative to the positive charges.
+
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+
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This result is consistent with the results of the Marinov Generator experiment conducted by Cyril W. Smith (See figure 9 of <ref name="Marinov Generator"/>). Sincerely,  ''[[User:S.H.O.|S.H.O.]] <sup>[[User_talk:S.H.O.|talk]]</sup>'' 23:41, 15 March 2017 (PDT)
+
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+
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To finalize the above results, let's consider the specific weighted average squared velocity:
+
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+
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: <math>v_{weighted}^2 = (50\%) v_{-q}^2 - (50\%) v_{+q}^2</math>
+
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: <math>v_{weighted}^2 = (1/2) v_{-q}^2 - (1/2) v_{+q}^2</math>
+
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+
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In terms of <math>v_{avg}</math> and <math>v_{dev}</math>, the result is:
+
−
 
+
−
: <math>v_{weighted}^2 = (1/2) (v_{avg}+v_{dev})^2 - (1/2) (v_{avg}-v_{dev})^2</math>
+
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: <math>v_{weighted}^2 = 2 v_{avg}v_{dev}</math>
+
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+
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Since the deviation velocity <math>v_{dev}</math> is one-half the drift velocity, let <math>v_{drift}</math> be the drift velocity. In the case that the drift velocity of current-carry charge carriers <math>q_{carrier} = -q</math> is co-linear with average velocity, the force on a current element <math>Idl</math> due to the aforementioned longitudinal force is:
+
−
 
+
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:<math>\mathbf{F}_{LW} = - q_{carrier} \mathbf{v}_{drift} \left(\boldsymbol{\beta}_{avg} \dot\ \nabla\right) \mathbf{A}(\mathbf{r},t)</math>
+
−
 
+
−
:<math>\mathbf{F}_{LW} = - I dl \left(\boldsymbol{\beta}_{avg} \dot\ \nabla\right) \mathbf{A}(\mathbf{r},t)</math>
+
−
 
+
−
In the case of the Marinov Generator, as the conductive slip ring increases speed, the charge carriers conveying the current are thinned out due to the sliding motion of the contacts, and so their quantity per length varies inversely to the speed. The drift velocity is determined by the voltage applied to the slip ring divided by the slip ring resistance and is not a function of the externally driven shaft's rotational rate. As a result, the induced voltage increases directly with the increase of the velocity of the slip ring <math>v_{+q}</math> and therefore <math>\boldsymbol{\beta}_{+q}</math>. Note comparatively minute drift velocities of the current-carrying loose electrons <math>-q</math> against the remaining charge of the slip ring <math>+q</math>.
+
−
 
+
−
This further confirms the above prior results.
+
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+
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I am more certain now than before that the M.A.K.E.R.A.R.C will work. Sincerely,  ''[[User:S.H.O.|S.H.O.]] <sup>[[User_talk:S.H.O.|talk]]</sup>'' 00:29, 16 March 2017 (PDT)
+
  
 
==A New Idea: The Makerarc==
 
==A New Idea: The Makerarc==

Revision as of 18:25, 18 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

Stefan Marinov's proposal

Stefan Marinov proposed adding the "motional-transformer induction" on charge q:[5]

[math](\mathbf{v_{wire}}\cdot\nabla)\mathbf{A}[/math]

or rather:

[math]-(\mathbf{v_{charge}}\cdot\nabla)\mathbf{A}[/math]

to the Lorentz force.

The extra term is equivalent to:[6]

[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 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, CGS 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, 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 [7][8] (for details, see the appendix of [9]),
[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 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 subscript s to it:

[math]\mathbf{E} = q_s\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_s\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 receiver charge [math]q_r[/math] located at [math]\mathbf{r}[/math] at rest in the observer's inertial frame. The inertial observer and the charge [math]q_r[/math] agree on what the electric field [math]E[/math] is, they agree that that there is no magnetic force on [math]q_r[/math], and finally, they agree on the acceleration of [math]q_r[/math].

The above equation can be broken up into two parts, one based on the transverse relative velocity of the source [math]\mathbf{v}_{s\bot}[/math], and one based on the longitudinal relative velocity of the source[math]\mathbf{v}_{s\parallel}[/math]:

[math]\mathbf{E} = q_s\frac{\mathbf{\hat{r}}}{r^2} \left(\frac{\mathbf{v}_{s\bot}^2}{2c^2} - 2\frac{\mathbf{v}_{s\parallel}^2}{2c^2} \right)[/math]

Scenario 1: All charges confined to a line

We can consider in the surrounding environment a system of charges with arbitrary position and velocity which observe an electric field [math]\mathbf{E}[/math] that is unique to each charge's rest frame.

Let's start off by considering the simplest case where all the charges are distributed along a straight line and all of their velocities are tangent to this line. Therefore, the velocity-dependent electric field experienced by a charge [math]q_r[/math] in its rest frame due a charge [math]q_s[/math] is:

[math]\mathbf{E} = - q_s\frac{\mathbf{\hat{r}}}{r^2} \frac{\left(\mathbf{v}_{s}-\mathbf{v}_{r}\right)^2}{c^2}[/math]
[math]\mathbf{E} = - q_s\frac{\mathbf{\hat{r}}}{r^2} \frac{\left(\mathbf{v}_{s \to r}\right)^2}{c^2}[/math]

Let's consider that along this line there is an element of a wire at velocity [math]\mathbf{v}_{wire}[/math] relative to charge [math]q_r[/math] carrying loose electrons with drift velocity [math]\mathbf{v}_{drift}[/math] relative to that wire. And let's say the total electrical charge of that wire [math]Q_{wire}=+Q[/math] + loose electrons [math]Q_{drift}=-Q[/math] is zero. What is the electric field created by this current element according to a receiver charge [math]q_{r}[/math] moving at velocity [math]\mathbf{v}_r[/math]?

[math]\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q \mathbf{v}_{wire \to r}^2}{c^2} + \frac{-Q \left(\mathbf{v}_{wire \to r}+\mathbf{v}_{drift}\right)^2}{c^2}\right)[/math]
[math]\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q \mathbf{v}_{wire \to r}^2}{c^2} - \frac{Q \left(\mathbf{v}_{wire \to r}+\mathbf{v}_{drift}\right)^2}{c^2}\right)[/math]
[math]\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{2 Q \mathbf{v}_{wire \to r}\mathbf{v}_{drift} + \mathbf{v}_{drift}^2}{c^2}\right)[/math]

To simplify this result, let's consider the average velocity of charges comprising the wire+current:

[math]\mathbf{v}_{rel} = \mathbf{v}_{wire \to r} + \mathbf{v}_{drift}/2[/math]

This is effectively the relative velocity of a current element [math]\mathbf{I}_s dl = Q_{drift} \mathbf{v}_{drift}[/math] with respect to receiver charge [math]q_r[/math]

From here, we can find the alternate expression for electric field at charge [math]q_{r}[/math] due to the current element [math]\mathbf{I}_s dl[/math]:

[math]\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q \left(\mathbf{v}_{avg} - \mathbf{v}_{drift}/2\right)^2}{c^2} + \frac{-Q \left(\mathbf{v}_{avg} + \mathbf{v}_{drift}/2\right)^2}{c^2}\right)[/math]
[math]\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q \left(\mathbf{v}_{avg} - \mathbf{v}_{drift}/2\right)^2}{c^2} - \frac{Q \left(\mathbf{v}_{avg} + \mathbf{v}_{drift}/2\right)^2}{c^2}\right)[/math]
[math]\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{-Q \mathbf{v}_{avg} \mathbf{v}_{drift}}{c^2}\right)[/math]
[math]\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{Q_{drift} \mathbf{v}_{avg} \mathbf{v}_{drift}}{c^2}\right)[/math]
[math]\mathbf{E} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{\mathbf{I}_s \mathbf{v}_{avg}}{c^2}\right)[/math]

Special Scenario 1.1: Forces between co-linear current elements confined to a straight line

If we wish to consider the force on charge [math]q_s[/math] due to the electric field of this current element, we simply multiply by target charge [math]q_r[/math].

[math]\mathbf{F} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{\mathbf{I}_s q_r \mathbf{v}_{avg}}{c^2}\right)[/math]

If [math]q_s[/math] was a loose electron embedded in another electrically-neutral wire element, let [math]\mathbf{I}_r dl[/math] stand for this target current element, analogous to the source current element [math]\mathbf{I}_s dl[/math]. Therefore the force on current element [math]\mathbf{I}_r dl[/math] due to current element [math]\mathbf{I}_s dl[/math] is:

[math]\mathbf{F} = - \frac{\mathbf{\hat{r}}}{r^2} \left( \frac{\mathbf{I}_s \mathbf{I}_r}{c^2}\right)[/math]

In SI units, this becomes:

[math]\mathbf{F} = - \frac{\mathbf{\hat{r}}}{4 \pi \epsilon r^2} \left( \frac{\mathbf{I}_s \mathbf{I}_r}{c^2}\right)[/math]

This is equivalent to:

[math]\mathbf{F} = - \left( \mathbf{I}_r \dot\ \nabla \right)\mathbf{A_s}[/math]

This is essentially the "Marinov Force", but with restrictions. Like the Marinov Force, it exists without necessitating a magnetic field [math]\mathbf{B}[/math] and so may exist in regions where the curl of [math]\mathbf{A_s}[/math] is zero. However, this scenario is constricted to the case that "all charges are confined to a line" with the restriction that the current elements are electrically neutral.

Scenario 2: Large drift velocities and negligible wire velocities

The magnetization currents embodied inside permanent magnets consist of electrons with very high effective velocities per the Virial theorem which relates the kinetic energy to the potential energy of the system[10]. Therefore, in the case of permanent magnets interacting with other permanent magnets, one can simply consider the effective difference of velocity of the charges comprising the effective magnetization currents and disregard the velocities of the atoms which confine these currents in place. Let [math]\mathbf{v}_{rel}[/math] be the relative velocity of charge [math]q_s[/math] respect to [math]q_r[/math]. Therefore:

[math]\mathbf{E} = q_s\frac{\mathbf{\hat{r}}}{r^2} \left(\frac{\mathbf{v}_{rel\bot}^2}{2c^2} - 2\frac{\mathbf{v}_{rel\parallel}^2}{2c^2} \right)[/math]

The electric force on [math]q_r[/math] by [math]q_s[/math] is therefore:

[math]\mathbf{F} = q_r q_s\frac{\mathbf{\hat{r}}}{r^2} \left(\frac{\mathbf{v}_{rel\bot}^2}{2c^2} - 2\frac{\mathbf{v}_{rel\parallel}^2}{2c^2} \right)[/math]

Special Scenario 2.1: Forces between magnetization currents from magnets of the same grade

In the case where the magnets are of the same grade (e.g. N52), the effective drift speed of the magnetization currents are identical. This makes it possible to calculate the above using current elements [math]\mathbf{I}_s[/math] and [math]\mathbf{I}_r[/math].

The differential version of this is:

[math]d^2\mathbf{F} = \frac{\mathbf{\hat{r}}}{r^2} \left(\frac{\left(\mathbf{I}_s-\mathbf{I}_r\right)_\bot^2}{2c^2} - 2\frac{\left(\mathbf{I}_s-\mathbf{I}_r\right)_\parallel^2}{2c^2} \right)[/math]

In SI units, this becomes:

[math]d^2\mathbf{F} = \frac{\mathbf{\hat{r}}}{4 \pi \epsilon r^2} \left(\frac{\left(\mathbf{I}_s-\mathbf{I}_r\right)_\bot^2}{2c^2} - 2\frac{\left(\mathbf{I}_s-\mathbf{I}_r\right)_\parallel^2}{2c^2} \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

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. ↑ http://redshift.vif.com/JournalFiles/Pre2001/V05NO3PDF/v05n3phi.pdf
  2. ↑ http://www.distinti.com/docs/pdx/paradox2.pdf
  3. ↑ 3.0 3.1 3.2 http://overunity.com/14691/the-marinov-generator/
  4. ↑ 4.0 4.1 http://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=13897
  5. ↑ https://archive.org/stream/thornywayoftruthpart4maririch#page/104/mode/2up/search/motional-transformer+induction
  6. ↑ http://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=13908
  7. ↑ https://docs.google.com/file/d/0B817m31MAj0wZTZjZmMwMjgtY2Y5YS00YTQ5LThjM2EtNzhjYTYzNzFlZDY0/edit?hl=en_GB&pli=1
  8. ↑ https://docs.google.com/file/d/0B817m31MAj0wMDI1YjllYjctY2NhOS00M2M2LWFlMTUtYjVmYTkyZmVlY2M2/edit?hl=en_GB
  9. ↑ http://web.archive.org/web/20170318210550/http://puhep1.princeton.edu/~mcdonald/examples/ph501/ph501lecture24.pdf
  10. ↑ https://en.wikipedia.org/wiki/Virial_theorem

See also

Site map

HQ ● Glossary ● April 2016 Presentation

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