Difference between revisions of "Electromagnetic Potentials"

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==Comment Record==
 
==Comment Record==
  
−
Beginning with the velocity-dependent electromagnetic potential, which many alleged to be equivalent to the Lorentz force, extra force terms not seen in the Lorentz force appear as a result of taking velocity to be an explicit function of the coordinates, as per the S.R.-like Lorentz Ether theory, in contrast to Special Relativity (S.R.). So herein, the actual vector identity for the gradient of a dot is employed, resulting in mathematical consistency, as opposed to "magically" waving away the velocity-gradient terms as is usually done to impose consistency of the Electromagnetic Lagrangian with the Lorentz Force. These extra terms are gauge-dependent, and so an appropriate gauge must be selected (by Nature itself) to render these (heretical) gauge-dependent forces meaningful. Applying the Lorenz gauge would make it consistent with the finite speed of light, while applying the Coulomb gauge would imply dependence of the force on the instaneous position of the sources of electromagnetic potential. The force scales directly with the magnitude acceleration of the particle, and so the corresponding multiplier may be thought of as a modification to the "effective mass". In the case of charge whose velocity is deflecting, the "effective mass" is also a tensor mass. It is anticipated that these potentials will play a significant role in the thermal characteristics of the S.H.O. Drive, themselves being gauge-dependent. ''[[User:S.H.O.|S.H.O.]] <sup>[[User_talk:S.H.O.|talk]]</sup>'' 13:37, 28 August 2016 (PDT)
+
Beginning with the velocity-dependent electromagnetic potential, one normally may derive the Lorentz force from it. This occurs by substituting a gradient term via an approximate, rather than exact, vector identity <math>\nabla (\mathbf{A} \cdot \mathbf{v}) = \nabla_\mathbf{A}(\mathbf{A} \cdot \mathbf{v})</math>. This is "allowed" normally because velocities in Special Relativity are not an explicit function of the coordinates, a matter simply assumed to be fact.<ref>http://www.nhn.ou.edu/~gut/notes/cm/lect_09.pdf ("The triple cross product can be written in a more compact form where we use the fact that the velocity is not an explicit function of the coordinates.")</ref> However, by taking velocity to be an explicit function of the coordinates, as per the S.R.-like Lorentz Ether theory, the extra force term not seen in the Lorentz force appears, which is <math>\nabla_\mathbf{v}(\mathbf{A} \cdot \mathbf{v})</math>. ''[[User:S.H.O.|S.H.O.]] <sup>[[User_talk:S.H.O.|talk]]</sup>'' 03:18, 12 September 2016 (PDT)
 +
 
 +
So herein, the actual vector identity for the gradient of a dot is employed, resulting in mathematical consistency, as opposed to "magically" waving away the velocity-gradient terms as is usually done to impose consistency of the Electromagnetic Lagrangian with the Lorentz Force. These extra terms are gauge-dependent, and so an appropriate gauge must be selected (by Nature itself) to render these (heretical) gauge-dependent forces meaningful. Applying the Lorenz gauge would make it consistent with the finite speed of light, while applying the Coulomb gauge would imply dependence of the force on the instaneous position of the sources of electromagnetic potential. ''[[User:S.H.O.|S.H.O.]] <sup>[[User_talk:S.H.O.|talk]]</sup>'' 13:37, 28 August 2016 (PDT)
 +
 
 +
The extra force term <math>\nabla_\mathbf{v}(\mathbf{A} \cdot \mathbf{v})</math> consists of a changing effective mass-correction term (in units of kg/s) multiplied by a velocity. Such a force is not preserved under Galilean transformations, but neither is the derivative <math>\frac{d(m\mathbf{v})}{dt} = m \frac{d\mathbf{v}}{dt} + \mathbf{v} \frac{dm}{dt}</math>. The extra force term is essentially a <math>\mathbf{v} \frac{dm}{dt}</math> term and could conceivably substitute for it, unless other terms may also assume a similar role (more on this below). Such forces change the time-like component of the relativistic 4-momentum, and therefore they are related to changes of rest energy of a particle (charge) subject to potentials, as viewed by an arbitrary inertial observer. Whatever other such terms may be, after cancelling the terms on both sides, what remains on the left hand side is equal to <math>m \frac{d\mathbf{v}}{dt}</math> of the particle (charge) and the right hand side yields the standard Lorentz force on the particle (charge). Unlike the 4-momentum, which is Lorentz invariant, the 3-momentum (i.e. the set of the 3 space-like components of the 4-momentum) is not. The Lorentz force, as normally expressed, only deals with changes of the 3-momentum over time. An observer may observe the time-like component of its 4-momentum transform as it changes velocity resulting in an unaccounted for "thrust", particularly if the time-like component of the momentum <math>E/c</math> in some way depends on the ''velocity-dependent electromagnetic potentials'', which in turn would yield, in effect, a ''velocity-dependent rest mass''.
 +
 
 +
To allude this possibility, consider that in cases where there are changing charge densities due to divergent/convergent electrical currents, and yet where current densities are constant, the kinetic forces <math>m \frac{d\mathbf{v}}{dt}</math> between charges do not sum to zero. A key such example of another non-Galilean invariant terms involving the variation mass over time at a given velocity, is the hypothetical Longitudinal force density conceived by Koen J. van Vlaenderen in his 2015 paper "General Classical Electrodynamics"<ref>http://vixra.org/abs/1512.0297</ref>, which from equations (2.8) and (2.9) can be written as <math>\mathbf{f}_L(\mathbf{x}) = \mathbf{J} (- \nabla \cdot \mathbf{A})</math>. The extra force term, adapted to help generalize Whittaker's force law to cover field force densities, was intended to preserve Newton's Third Law of Motion ("For every action there is an equal and opposite reaction"). Under the Lorenz gauge condition, this equals <math>\mathbf{f}_L(\mathbf{x}) = \frac{dq}{dV} \mathbf{v} \left( \epsilon_0 \mu_0 \frac{∂\varphi}{∂t} \right) = \frac{d(E_{coulomb})}{c^2dtdV} \mathbf{v}</math>, which when integrated over volume elements <math>dV</math> gives <math>\frac{d(m_{coulomb})}{dt} \mathbf{v} = \frac{d(E_{coulomb})}{c^2dt} \mathbf{v}</math>. When this <math>\mathbf{v} \frac{dm}{dt}</math>-type term is subtracted from both sides of the full force equation (a <math>\frac{d(m \mathbf{v})}{dt}</math> type equation), we once again return to the Lorentz 3-force, which is the standard electromagnetic force of type <math>F = m \frac{d\mathbf{v}}{dt} = ma</math>. ''[[User:S.H.O.|S.H.O.]] <sup>[[User_talk:S.H.O.|talk]]</sup>'' 03:18, 12 September 2016 (PDT)
  
 
==Background==
 
==Background==
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Using Feynman subscript notation, we can separate the last term on the right into two separate terms:
 
Using Feynman subscript notation, we can separate the last term on the right into two separate terms:
  
−
: q<math>\nabla (\mathbf{v} \cdot \mathbf{A}) = q\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) + q\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A})</math>
+
: <math>q\nabla (\mathbf{v} \cdot \mathbf{A}) = q\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) + q\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A})</math>
  
 
In an article titled "A Discussion on the Magnetic Vector Potential"<ref>http://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=13908</ref>, Cyril W. Smith (Professor Ph.D of Electronic and Electrical Engineering from 1964-1989<ref>http://www.positivehealth.com/author/cyril-smith-ph-d</ref>), the last term (without the charge <math>q</math>) can be expressed as:
 
In an article titled "A Discussion on the Magnetic Vector Potential"<ref>http://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=13908</ref>, Cyril W. Smith (Professor Ph.D of Electronic and Electrical Engineering from 1964-1989<ref>http://www.positivehealth.com/author/cyril-smith-ph-d</ref>), the last term (without the charge <math>q</math>) can be expressed as:

Revision as of 03:18, 12 September 2016

The basic idea here is that the electromagnetic potentials [math]\phi[/math] and [math]A[/math] and their derivatives can be used to derive all electromagnetism.

Comment Record

Beginning with the velocity-dependent electromagnetic potential, one normally may derive the Lorentz force from it. This occurs by substituting a gradient term via an approximate, rather than exact, vector identity [math]\nabla (\mathbf{A} \cdot \mathbf{v}) = \nabla_\mathbf{A}(\mathbf{A} \cdot \mathbf{v})[/math]. This is "allowed" normally because velocities in Special Relativity are not an explicit function of the coordinates, a matter simply assumed to be fact.[1] However, by taking velocity to be an explicit function of the coordinates, as per the S.R.-like Lorentz Ether theory, the extra force term not seen in the Lorentz force appears, which is [math]\nabla_\mathbf{v}(\mathbf{A} \cdot \mathbf{v})[/math]. S.H.O. talk 03:18, 12 September 2016 (PDT)

So herein, the actual vector identity for the gradient of a dot is employed, resulting in mathematical consistency, as opposed to "magically" waving away the velocity-gradient terms as is usually done to impose consistency of the Electromagnetic Lagrangian with the Lorentz Force. These extra terms are gauge-dependent, and so an appropriate gauge must be selected (by Nature itself) to render these (heretical) gauge-dependent forces meaningful. Applying the Lorenz gauge would make it consistent with the finite speed of light, while applying the Coulomb gauge would imply dependence of the force on the instaneous position of the sources of electromagnetic potential. S.H.O. talk 13:37, 28 August 2016 (PDT)

The extra force term [math]\nabla_\mathbf{v}(\mathbf{A} \cdot \mathbf{v})[/math] consists of a changing effective mass-correction term (in units of kg/s) multiplied by a velocity. Such a force is not preserved under Galilean transformations, but neither is the derivative [math]\frac{d(m\mathbf{v})}{dt} = m \frac{d\mathbf{v}}{dt} + \mathbf{v} \frac{dm}{dt}[/math]. The extra force term is essentially a [math]\mathbf{v} \frac{dm}{dt}[/math] term and could conceivably substitute for it, unless other terms may also assume a similar role (more on this below). Such forces change the time-like component of the relativistic 4-momentum, and therefore they are related to changes of rest energy of a particle (charge) subject to potentials, as viewed by an arbitrary inertial observer. Whatever other such terms may be, after cancelling the terms on both sides, what remains on the left hand side is equal to [math]m \frac{d\mathbf{v}}{dt}[/math] of the particle (charge) and the right hand side yields the standard Lorentz force on the particle (charge). Unlike the 4-momentum, which is Lorentz invariant, the 3-momentum (i.e. the set of the 3 space-like components of the 4-momentum) is not. The Lorentz force, as normally expressed, only deals with changes of the 3-momentum over time. An observer may observe the time-like component of its 4-momentum transform as it changes velocity resulting in an unaccounted for "thrust", particularly if the time-like component of the momentum [math]E/c[/math] in some way depends on the velocity-dependent electromagnetic potentials, which in turn would yield, in effect, a velocity-dependent rest mass.

To allude this possibility, consider that in cases where there are changing charge densities due to divergent/convergent electrical currents, and yet where current densities are constant, the kinetic forces [math]m \frac{d\mathbf{v}}{dt}[/math] between charges do not sum to zero. A key such example of another non-Galilean invariant terms involving the variation mass over time at a given velocity, is the hypothetical Longitudinal force density conceived by Koen J. van Vlaenderen in his 2015 paper "General Classical Electrodynamics"[2], which from equations (2.8) and (2.9) can be written as [math]\mathbf{f}_L(\mathbf{x}) = \mathbf{J} (- \nabla \cdot \mathbf{A})[/math]. The extra force term, adapted to help generalize Whittaker's force law to cover field force densities, was intended to preserve Newton's Third Law of Motion ("For every action there is an equal and opposite reaction"). Under the Lorenz gauge condition, this equals [math]\mathbf{f}_L(\mathbf{x}) = \frac{dq}{dV} \mathbf{v} \left( \epsilon_0 \mu_0 \frac{∂\varphi}{∂t} \right) = \frac{d(E_{coulomb})}{c^2dtdV} \mathbf{v}[/math], which when integrated over volume elements [math]dV[/math] gives [math]\frac{d(m_{coulomb})}{dt} \mathbf{v} = \frac{d(E_{coulomb})}{c^2dt} \mathbf{v}[/math]. When this [math]\mathbf{v} \frac{dm}{dt}[/math]-type term is subtracted from both sides of the full force equation (a [math]\frac{d(m \mathbf{v})}{dt}[/math] type equation), we once again return to the Lorentz 3-force, which is the standard electromagnetic force of type [math]F = m \frac{d\mathbf{v}}{dt} = ma[/math]. S.H.O. talk 03:18, 12 September 2016 (PDT)

Background

According Emil John Konopinski, a nuclear scientist[3] and professor of Mathematics[4] who worked on the Manhattan Project[5], the electromagnetic fields [math]\mathbf{E}[/math] and [math]\mathbf{B}[/math] can be re-expressed in terms of the electromagnetic potentials [math]\varphi[/math] and [math]A[/math] through substitutions. From his article on "What the electromagnetic vector potential describes"[6], he presents the equation of motion for a localized point charge:

[math]d(M\mathbf{v})/dt = q \left[ \mathbf{E} + \mathbf{v} \times \mathbf{B}/c \right]\ (Gaussian\ units)[/math]:
[math]d(M\mathbf{v})/dt = q \left[ \mathbf{E} + \mathbf{v} \times \mathbf{B} \right]\ (SI\ units)[/math]

With the standard substitutions for the fields in terms of the potentials, which were taken to be:

[math]\mathbf{E} = - \nabla \varphi - \frac{∂\mathbf{A}}{c∂t}\ (Gaussian\ units)[/math]
[math]\mathbf{E} = - \nabla \varphi - \frac{∂\mathbf{A}}{∂t}\ (SI\ units)[/math]
[math]\mathbf{B} = \nabla \times \mathbf{A}\ [/math]

Konopinski determined the function relating the time derivative of the "total" momentum with the velocity-dependent potential, as evaluated by an inertial observer who sees the localized charge [math]q[/math] of mass [math]m[/math] travelling with velocity [math]\mathbf{v}[/math] which is subject to a magnetic vector potential [math]\mathbf{A}[/math] and an electric scalar potential [math]\varphi[/math].

[math]\frac{d}{dt} \left[ M\mathbf{v} + (q/c)\mathbf{A} \right] = - q \left[ \nabla \varphi - (\mathbf{v}/c) \cdot \mathbf{A} \right]\ (Gaussian\ units)[/math]
[math]\frac{d}{dt} \left[ M\mathbf{v} + q\mathbf{A} \right] = - q \left[ \nabla \varphi - \mathbf{v} \cdot \mathbf{A} \right]\ (SI\ units)[/math]

The terms on the right can be separated as follows:

[math]\frac{d}{dt} \left[ M\mathbf{v} + q\mathbf{A} \right] = - q \nabla \varphi + q \nabla (\mathbf{v} \cdot \mathbf{A})\ (SI\ units)[/math]

Using Feynman subscript notation, we can separate the last term on the right into two separate terms:

[math]q\nabla (\mathbf{v} \cdot \mathbf{A}) = q\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) + q\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A})[/math]

In an article titled "A Discussion on the Magnetic Vector Potential"[7], Cyril W. Smith (Professor Ph.D of Electronic and Electrical Engineering from 1964-1989[8]), the last term (without the charge [math]q[/math]) can be expressed as:

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

Written in this form, [math]\mathbf{a}_x[/math], [math]\mathbf{a}_y[/math], and [math]\mathbf{a}_z[/math] are the unit basis vectors for x, y, and z, respectively. However, for the purposes of the S.H.O. Drive Wiki Site, [math]\mathbf{a}[/math] will stand for the vector for acceleration, so it doesn't hurt that the standard variables for the basis vectors are really [math]\mathbf{e}_x[/math], [math]\mathbf{e}_y[/math], and [math]\mathbf{e}_z[/math].

Predictably, the values for first term on the right (without the charge [math]q[/math]) can be expressed as:

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

The sum of the first and last terms gives:

[math]\nabla (\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} + A_x \frac{∂v_x}{∂x} + A_y \frac{∂v_y}{∂x} + A_z \frac{∂v_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} + A_x \frac{∂v_x}{∂y} + A_y \frac{∂v_y}{∂y} + A_z \frac{∂v_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} + A_x \frac{∂v_x}{∂z} + A_y \frac{∂v_y}{∂z} + A_z \frac{∂v_z}{∂z} \right] \mathbf{e}_z \end{matrix}[/math]

By choosing the [math]x[/math] Cartesian axis so that it is aligned with the velocity [math]\mathbf{v}[/math] of charge [math]q[/math] such that [math]\mathbf{v}/|\mathbf{v}| = \mathbf{\hat{v}} = \mathbf{e}_x[/math], the above reduces to:

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

[math]\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A})[/math] simplifies to:

[math]\left[ \mathbf{A} \cdot \mathbf{a}/|\mathbf{v}| \right] \mathbf{\hat{v}} = \left[ A_x \frac{∂v_x}{∂x} + A_y \frac{∂v_y}{∂x} + A_z \frac{∂v_z}{∂x} \right] \mathbf{e}_x[/math]

The above may also be expressed as:

[math]\nabla_\mathbf{v}(\mathbf{A} \cdot \mathbf{v}) = (\mathbf{A} \cdot \mathbf{a})\mathbf{v} / |\mathbf{v}|^2[/math]

Draft

The field experienced by a charge [math]q[/math] viewed at rest in a static electromagnetic field is:

[math]\mathbf{F}_{rest,static} = - \nabla \varphi[/math]

The field experienced by a charge [math]q[/math] viewed at rest in a dynamic electromagnetic field is:

[math]\mathbf{F}_{rest,dynamic} = - \nabla \varphi - ∂\mathbf{A}/∂t[/math]

The field experienced by a moving charge [math]q[/math] in a dynamic electromagnetic field (ignoring dilation of proper time relative to coordinate time) is:

[math]\mathbf{F}_{moving,dynamic} = - \nabla \varphi_q - ∂\mathbf{A}_q/∂t[/math]

Where:

  • [math]\varphi_q = \varphi - \mathbf{v} \cdot A[/math] is the scalar potential experienced by the moving charge.
  • [math]∂ \mathbf{A}_q/∂t = ∂\mathbf{A}/∂t + (\mathbf{v} \cdot \nabla)\mathbf{A}[/math] is the partial time derivative of the magnetic vector potential experienced by the moving charge.

Substituting per the above, the field experienced by the moving charge [math]q[/math] is:

[math]\mathbf{F} = - \nabla (\varphi-\mathbf{v} \cdot \mathbf{A}) - ∂\mathbf{A}/∂t - (\mathbf{v} \cdot \nabla)\mathbf{A}[/math]
[math]\mathbf{F} = - \nabla \varphi + \nabla (\mathbf{v} \cdot \mathbf{A}) - ∂\mathbf{A}/∂t - (\mathbf{v} \cdot \nabla)\mathbf{A}[/math]

Using Feynman subscript notation:

[math]\nabla (\mathbf{v} \cdot \mathbf{A}) = \nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) + \nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A})[/math]
[math]\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{v} \times \nabla \times \mathbf{A} + (\mathbf{v} \cdot \nabla)\mathbf{A}[/math]
[math]\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{A} \times \nabla \times \mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}[/math]

Substuting for the curl of the vector potential and the curl of the immediate velocity field for a moving charge, we have:

[math]\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{v} \times \mathbf{B} + (\mathbf{v} \cdot \nabla)\mathbf{A}[/math]
[math]\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}[/math]

Where:

[math]\mathbf{B} = \nabla \times \mathbf{A}[/math] is the magnetic field.
[math]\mathbf{ω}_\mathbf{v} = \nabla \times \mathbf{v}[/math] is the angular rate of deflection.

Substituting per the above, the field experienced by the moving charge is:

[math]\mathbf{F} = - \nabla \varphi - ∂\mathbf{A}/∂t + \mathbf{v} \times \mathbf{B} + (\mathbf{v} \cdot \nabla)\mathbf{A} + \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v} - (\mathbf{v} \cdot \nabla)\mathbf{A}[/math]
[math]\mathbf{F} = - \nabla \varphi - ∂\mathbf{A}/∂t + \mathbf{v} \times \mathbf{B} + \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}[/math]

This field includes the field from Lorentz plus two additional terms:

[math]\mathbf{F} = \mathbf{F}_{Lorentz} + \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}[/math]
[math](\mathbf{A} \cdot \nabla)\mathbf{v}[/math] is the dot product of the magnetic vector potential with the gradient of the velocity field.

For a velocity field defined in the immediate neighborhood of a moving charge [math]q[/math] at point [math]p[/math], where the local [math](\nabla_\mathbf{v} \mathbf{A})_p[/math] is a tangent vector on [math]\mathbf{A}[/math] (the Lie derivative of [math]\mathbf{v}[/math] along [math]\mathbf{A}[/math]), the above is equivalent to:

[math](\mathbf{A} \cdot \nabla)\mathbf{v} = |\mathbf{A}|(\nabla_\mathbf{v} \mathbf{A})_p = (\mathbf{A} \cdot \mathbf{v})\mathbf{a}/|\mathbf{v}|^2[/math]

Where [math]\mathbf{a}[/math] is the convective acceleration of the charge, which equals:

[math]\mathbf{a} = (∂\mathbf{v}/∂x)|(∂\mathbf{x}/∂t)|[/math]

If the charge is taken as a point particle, the convective acceleration is the same as the acceleration.

[math]\mathbf{a} = ∂²\mathbf{x}/∂t²[/math]
[math]\mathbf{A} \times \mathbf{ω}_\mathbf{v}[/math] is the cross product of the magnetic vector potential and the angular rate of deflection.
[math]\mathbf{ω}_\mathbf{v} = (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2[/math]

When fields are static, the field experienced by a moving charge is:

[math]\mathbf{F}_{moving,static} = - \nabla \varphi + \mathbf{v} \times \mathbf{B} + \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}[/math]

So for the case of static fields, the force on an accelerating charge is:

[math]\mathbf{F}_{moving,static} = - \nabla \varphi + \mathbf{v} \times \mathbf{B} + \mathbf{A} \times (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2 + (\mathbf{A} \cdot \mathbf{v})\mathbf{a}/|\mathbf{v}|^2[/math]

While the power on an accelerating charge q subject to a static field is:

[math]P_{moving,static} = q \left[\ -\nabla \varphi \cdot \mathbf{v} + (\mathbf{v} \times B) \cdot \mathbf{v} + \mathbf{A} \times (\mathbf{v} \times \mathbf{a}) \cdot \mathbf{v}/|\mathbf{v}|^2 + (\mathbf{A} \cdot \mathbf{v})(\mathbf{a} \cdot \mathbf{v})/|\mathbf{v}|^2\ \right][/math]
[math]P_{moving,static} = q \left[\ -\nabla \varphi \cdot \mathbf{v} + \mathbf{A} \times (\hat{\mathbf{v}} \times \mathbf{a}) \cdot \hat{\mathbf{v}} + (\mathbf{A} \cdot \mathbf{\hat{v}})(\mathbf{a} \cdot \mathbf{\hat{v}})\ \right][/math]

The field on a moving charge in a changing electromagnetic field becomes:

[math]\mathbf{F}_{moving,dynamic} = - \nabla \varphi - ∂\mathbf{A}/∂t + \mathbf{v} \times \mathbf{B} + \mathbf{A} \times (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2 + (\mathbf{A} \cdot \mathbf{v})\mathbf{a}/|\mathbf{v}|^2[/math]

Summarizing the derivation of the last two terms above, we have:

[math]\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{A} \times \nabla \times \mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}[/math]
[math]\mathbf{ω}_\mathbf{v} = \nabla \times \mathbf{v}[/math] is the angular rate of deflection.
[math]\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}[/math]
[math]\mathbf{A} \times \mathbf{ω}_\mathbf{v} = \mathbf{A} \times (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2[/math]
[math](\mathbf{A} \cdot \nabla)\mathbf{v} = (\mathbf{A} \cdot \mathbf{v})\mathbf{a}/|\mathbf{v}|^2[/math]
[math]\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{A} \times (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2 + (\mathbf{A} \cdot \mathbf{v})\mathbf{a}/|\mathbf{v}|^2[/math]

A concise alternative to the above is:

[math]\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = (\mathbf{A} \cdot \mathbf{a})\mathbf{v}/|\mathbf{v}|^2 [/math]

The field on a moving charge [math]q[/math] in a changing electromagnetic field becomes:

[math]\mathbf{F}_{moving,dynamic} = - \nabla \varphi - ∂\mathbf{A}/∂t + \mathbf{v} \times \mathbf{B} + (\mathbf{A} \cdot \mathbf{a})\mathbf{v}/|\mathbf{v}|^2[/math]

The power on a moving charge [math]q[/math] in a changing electromagnetic field becomes:

[math]P_{moving,dynamic} = q \left[\ \left(-\nabla \varphi - ∂\mathbf{A}/∂t\right) \cdot \mathbf{v} + (\mathbf{A} \cdot \mathbf{a})(\mathbf{v} \cdot \mathbf{v})/|\mathbf{v}|^2\ \right][/math]
[math]P_{moving,dynamic} = q \left[\ \left(-\nabla \varphi - ∂\mathbf{A}/∂t\right) \cdot \mathbf{v} + \mathbf{A} \cdot \mathbf{a}\ \right][/math]

References

  1. ↑ http://www.nhn.ou.edu/~gut/notes/cm/lect_09.pdf ("The triple cross product can be written in a more compact form where we use the fact that the velocity is not an explicit function of the coordinates.")
  2. ↑ http://vixra.org/abs/1512.0297
  3. ↑ Emil Konopinski, 78, Atomic Bomb Scientist, New York Times
  4. ↑ Eugene Greuling at the Mathematics Genealogy Project
  5. ↑ (October 1991). "Obituary: Emil J. Konopinski". Physics Today 44 (10): 144. Digital object identifier: 10.1063/1.2810306.
  6. ↑ http://exvacuo.free.fr/div/Sciences/Dossiers/EM/ScalarEM/J%20Konopinski%20-%20What%20the%20Electromagnetic%20Vector%20Potential%20Describes%20-%20ajp_46_499_78.pdf
  7. ↑ http://www.overunityresearch.com/index.php?action=dlattach;topic=2470.0;attach=13908
  8. ↑ http://www.positivehealth.com/author/cyril-smith-ph-d

See also

Site map

HQ ● Glossary ● April 2016 Presentation

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