Function Conjunction → Electromagnetic Potentials

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 q' viewed at rest in a static electromagnetic field is:

F_rest,static = - ∇Φ

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

F_rest,dynamic = - ∇Φ - ∂A/∂t

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

F_moving,dynamic = - ∇Φ' - ∂A'/∂t

Where:

Φ' = Φ - v·A is the scalar potential experienced by the moving charge.

∂A'/∂t = ∂A/∂t + (v·∇)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:

F = - ∇(Φ-v·A) - ∂A/∂t - (v·∇)A

F = - ∇Φ + ∇(v·A) - ∂A/∂t - (v·∇)A

Using Feynman subscript notation:

∇(v·A) = ∇_v(v·A) + ∇_A(v·A)

∇_A(v·A) = v x ∇ x A + (v·∇)A

∇_v(v·A) = A x ∇ x v + (A·∇)v

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

∇_A(v·A) = v x B + (v·∇)A

∇_v(v·A) = A x ω_v + (A·∇)v

Where:

B = ∇ x A is the magnetic field.

ω_v = ∇ x v is the angular rate of deflection.

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

F = - ∇Φ - ∂A/∂t + v x B + (v·∇)A + A x ω_v + (A·∇)v - (v·∇)A

F = - ∇Φ - ∂A/∂t + v x B + A x ω_v + (A·∇)v

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

F = F_Lorentz + A x ω_v + (A·∇)v

(A·∇)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 (∇_v A)_p is a tangent vector on A (the Lie derivative of v along A), the above is equivalent to:

(A·∇)v = |A|(∇_v A)_p = |A_v|a/|v|

Where a is the convective acceleration of the charge, which equals:

a = (∂v/∂x)|(∂x/∂t)|

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

a = ∂²x/∂t²

A x ω_v is the cross product of the magnetic vector potential and the angular rate of deflection.

ω_v = (v x a)/|v|^2

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

F_moving,static = - ∇Φ + v x B + A x ω_v + (A·∇)v

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

F_moving,static = - ∇Φ + v x B + A x (v x a)/|v|^2 + |A_v|a/|v|

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