Difference between revisions of "Electromagnetic Potentials"

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The field experienced by a charge q' viewed at rest in a static electromagnetic field is:
 
The field experienced by a charge q' viewed at rest in a static electromagnetic field is:
  
−
F_rest,static = - ∇Φ
+
<math>F_{rest,static} = - \nabla \varphi</math>
  
 
The field experienced by a charge q' viewed at rest in a dynamic electromagnetic field is:
 
The field experienced by a charge q' viewed at rest in a dynamic electromagnetic field is:
  
−
F_rest,dynamic = - ∇Φ - ∂A/∂t
+
<math>F_{rest,dynamic} = - \nabla \varphi - ∂\mathbf{A}/∂t</math>
  
 
The field experienced by a moving charge q' in a dynamic electromagnetic field is:
 
The field experienced by a moving charge q' in a dynamic electromagnetic field is:
  
−
F_moving,dynamic = - ∇Φ' - ∂A'/∂t
+
<math>F_{moving,dynamic} = - \nabla \varphi' - ∂\mathbf{A}'/∂t</math>
  
 
Where:
 
Where:
  
−
* Φ' = Φ - v·A is the scalar potential experienced by the moving charge.
+
* <math>\varphi' = \varphi - \mathbf{v} \cdot A</math> 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.
+
* <math>∂A'/∂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 q' is:
 
Substituting per the above, the field experienced by the moving charge q' is:
  
−
F = - ∇(Φ-v·A) - ∂A/∂t - (v·∇)A
+
<math>F = - \nabla (\varphi-\mathbf{v} \cdot \mathbf{A}) - ∂\mathbf{A}/∂t - (\mathbf{v} \cdot \nabla)\mathbf{A}</math>
  
−
F = - ∇Φ + ∇(v·A) - ∂A/∂t - (v·∇)A
+
<math>F = - \nabla \varphi + \nabla (\mathbf{v} \cdot \mathbf{A}) - ∂\mathbf{A}/∂t - (\mathbf{v} \cdot \nabla)\mathbf{A}</math>
  
 
Using Feynman subscript notation:
 
Using Feynman subscript notation:
  
−
∇(v·A) = ∇_v(v·A) + ∇_A(v·A)
+
<math>\nabla (\mathbf{v} \cdot \mathbf{A}) = \nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) + \nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A})</math>
  
−
∇_A(v·A) = v x ∇ x A + (v·∇)A
+
<math>\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{v} \times \nabla \times \mathbf{A} + (\mathbf{v} \cdot \nabla)\mathbf{A}</math>
  
−
∇_v(v·A) = A x ∇ x v + (A·∇)v
+
<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:
 
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
+
<math>\nabla_\mathbf{A}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{v} \times \mathbf{B} + (\mathbf{v} \cdot \nabla)\mathbf{A}</math>
  
−
∇_v(v·A) = A x ω_v + (A·∇)v
+
<math>\nabla_\mathbf{v}(\mathbf{v} \cdot \mathbf{A}) = \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}</math>
  
 
Where:
 
Where:
  
−
B = ∇ x A is the magnetic field.
+
<math>B = \nabla \times \mathbf{A}</math> is the magnetic field.
  
−
ω_v = ∇ x v is the angular rate of deflection.
+
<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:
 
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
+
<math>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>
  
−
F = - ∇Φ - ∂A/∂t + v x B + A x ω_v + (A·∇)v
+
<math>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:
 
This field includes the field from Lorentz plus two additional terms:
  
−
F = F_Lorentz + A x ω_v + (A·∇)v
+
<math>F = F_{Lorentz} + \mathbf{A} \times \mathbf{ω}_\mathbf{v} + (\mathbf{A} \cdot \nabla)\mathbf{v}</math>
  
−
(A·∇)v is the dot product of the magnetic vector potential with the gradient of the velocity field.
+
<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 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:
+
For a velocity field defined in the immediate neighborhood of a moving charge q' at point p, where the local <math>(\nabla_\mathbf{v} \mathbf{A})_p</math> is a tangent vector on \mathbf{A} (the Lie derivative of <math>\mathbf{v}</math> along <math>\mathbf{A}</math>), the above is equivalent to:
  
−
(A·∇)v = |A|(∇_v A)_p = |A_v|a/|v|
+
<math>(\mathbf{A} \cdot \nabla)\mathbf{v} = |\mathbf{A}|(\nabla_\mathbf{v} \mathbf{A})_p = |\mathbf{a}_\mathbf{v}|\mathbf{a}/|\mathbf{v}|</math>
  
 
Where a is the convective acceleration of the charge, which equals:
 
Where a is the convective acceleration of the charge, which equals:
  
−
a = (∂v/∂x)|(∂x/∂t)|
+
<math>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.
 
If the charge is taken as a point particle, the convective acceleration is the same as the acceleration.
  
−
a = ∂²x/∂t²
+
<math>a = ∂²\mathbf{x}/∂t²</math>
  
−
A x ω_v is the cross product of the magnetic vector potential and the angular rate of deflection.
+
<math>\mathbf{A} \times \mathbf{ω}_\mathbf{v}</math> is the cross product of the magnetic vector potential and the angular rate of deflection.
  
−
ω_v = (v x a)/|v|^2
+
<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:
 
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
+
<math>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:
 
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|
+
<math>F_{moving,static} = - \nabla \varphi + \mathbf{v} \times \mathbf{B} + \mathbf{A} \times (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2 + |\mathbf{a}_\mathbf{v}|\mathbf{a}/|\mathbf{v}|</math>
  
 
While the power on an accelerating charge q subject to a static field is:
 
While the power on an accelerating charge q subject to a static field is:
  
−
P_moving,static = q (- ∇Φ·v + (v x B)·v + A x (v x a)·v/|v|^2 + (a·v)|A_v|/|v|)
+
<math>P_{moving,static} = q (- \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}_\mathbf{v}|/|\mathbf{v}|)</math>
  
−
P_moving,static = q (- ∇Φ·v + A x (v-hat x a) · v-hat + a·A_v)
+
<math>P_{moving,static} = q (- \nabla \varphi \cdot \mathbf{v} + \mathbf{A} \times (\hat{\mathbf{v}} \times \mathbf{a}) \cdot  \hat{\mathbf{v}} + \mathbf{a} \cdot A_\mathbf{v})</math>
  
 
The force on a moving charge in a changing magnetic field becomes:
 
The force on a moving charge in a changing magnetic field becomes:
  
−
F = - ∇Φ - ∂A/∂t + v x B + A x (v x a)/|v|^2 + |A_v|a/|v|
+
<math>F = - \nabla \varphi - ∂\mathbf{A}/∂t + \mathbf{v} \times \mathbf{B} + \mathbf{A} \times (\mathbf{v} \times \mathbf{a})/|\mathbf{v}|^2 + |\mathbf{a}_\mathbf{v}|\mathbf{a}/|\mathbf{v}|</math>
  
 
{{Site map}}
 
{{Site map}}
  
 
[[Category:Function Conjunction]]
 
[[Category:Function Conjunction]]

Revision as of 02:30, 27 June 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.

Draft

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

[math]F_{rest,static} = - \nabla \varphi[/math]

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

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

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

[math]F_{moving,dynamic} = - \nabla \varphi' - ∂\mathbf{A}'/∂t[/math]

Where:

  • [math]\varphi' = \varphi - \mathbf{v} \cdot A[/math] is the scalar potential experienced by the moving charge.
  • [math]∂A'/∂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 q' is:

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

[math]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]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]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]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]F = 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 q' at point p, where the local [math](\nabla_\mathbf{v} \mathbf{A})_p[/math] is a tangent vector on \mathbf{A} (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}_\mathbf{v}|\mathbf{a}/|\mathbf{v}|[/math]

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

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

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

[math]P_{moving,static} = q (- \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}_\mathbf{v}|/|\mathbf{v}|)[/math]

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

The force on a moving charge in a changing magnetic field becomes:

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

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