From S.H.O.
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.
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 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} = - \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]
See also
Site map
HQ ● Glossary ● April 2016 Presentation
