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| | <math>\mathbf{F}_{rest,static} = - \nabla \varphi</math> | | <math>\mathbf{F}_{rest,static} = - \nabla \varphi</math> |
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| − | The field experienced by a charge <math>q</math> viewed at rest in a dynamic electromagnetic field (ignoring dilation of proper time relative to coordinate time) is: | + | The field experienced by a charge <math>q</math> viewed at rest in a dynamic electromagnetic field is: |
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| | <math>\mathbf{F}_{rest,dynamic} = - \nabla \varphi - ∂\mathbf{A}/∂t</math> | | <math>\mathbf{F}_{rest,dynamic} = - \nabla \varphi - ∂\mathbf{A}/∂t</math> |
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| − | The field experienced by a moving charge <math>q</math> in a dynamic electromagnetic field is: | + | 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: |
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| | <math>\mathbf{F}_{moving,dynamic} = - \nabla \varphi_q - ∂\mathbf{A}_q/∂t</math> | | <math>\mathbf{F}_{moving,dynamic} = - \nabla \varphi_q - ∂\mathbf{A}_q/∂t</math> |
Revision as of 21:49, 11 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.
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. S.H.O. talk 13:37, 28 August 2016 (PDT)
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]
See also
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