Function Conjunction → Electromagnetic Potentials

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

Draft

The field experienced by a charge $q$ viewed at rest in a static electromagnetic field is:

$\mathbf{F}_{rest,static} = - \nabla \varphi$

The field experienced by a charge $q$ viewed at rest in a dynamic electromagnetic field is:

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

The field experienced by a moving charge $q$ in a dynamic electromagnetic field is:

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

Where:

$\varphi_q = \varphi - \mathbf{v} \cdot A$ is the scalar potential experienced by the moving charge.

$∂ \mathbf{A}_q/∂t = ∂\mathbf{A}/∂t + (\mathbf{v} \cdot \nabla)\mathbf{A}$ 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 $q'$ is:

$\mathbf{F} = - \nabla (\varphi-\mathbf{v} \cdot \mathbf{A}) - ∂\mathbf{A}/∂t - (\mathbf{v} \cdot \nabla)\mathbf{A}$

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

Using Feynman subscript notation:

$\nabla (\mathbf{v} \cdot \mathbf{A}) = \nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) + \nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A})$

$\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{v} \times \nabla \times \mathbf{A} + (\mathbf{v} \cdot \nabla)\mathbf{A}$

$\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{A} \times \nabla \times \mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}$

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

$\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{v} \times \mathbf{B} + (\mathbf{v} \cdot \nabla)\mathbf{A}$

$\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}$

Where:

$\mathbf{B} = \nabla \times \mathbf{A}$ is the magnetic field.

$\mathbf{ω}_\mathbf{v} = \nabla \times \mathbf{v}$ is the angular rate of deflection.

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

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

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

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

$\mathbf{F} = \mathbf{F}_{Lorentz} + \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}$

$(\mathbf{A} \cdot \nabla)\mathbf{v}$ 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 $q'$ at point $p$ , where the local $(\nabla_\mathbf{v} \mathbf{A})_p$ is a tangent vector on $\mathbf{A}$ (the Lie derivative of $\mathbf{v}$ along $\mathbf{A}$ ), the above is equivalent to:

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

Where $\mathbf{a}$ is the convective acceleration of the charge, which equals:

$\mathbf{a} = (∂\mathbf{v}/∂x)|(∂\mathbf{x}/∂t)|$

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

$\mathbf{a} = ∂²\mathbf{x}/∂t²$

$\mathbf{A} \times \mathbf{ω}_\mathbf{v}$ is the cross product of the magnetic vector potential and the angular rate of deflection.

$\mathbf{ω}_\mathbf{v} = (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2$

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

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

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

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

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

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

$P_{moving,static} = q \left[- \nabla \varphi \cdot \mathbf{v} + \mathbf{A} \times (\hat{\mathbf{v}} \times \mathbf{a}) \cdot \hat{\mathbf{v}} + (\mathbf{A} \cdot \mathbf{\hat{v}})(\mathbf{a} \cdot \mathbf{\hat{v}})\right]$

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

$\mathbf{F} = - \nabla \varphi - ∂\mathbf{A}/∂t + \mathbf{v} \times \mathbf{B} + \mathbf{A} \times (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2 + (\mathbf{A} \cdot \mathbf{v})\mathbf{a}/|\mathbf{v}|^2$