Difference between revisions of "Electromagnetic Potentials"

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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>
  
−
The field experienced by a moving charge q</math> in a dynamic electromagnetic field is:
+
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>
 
<math>\mathbf{F}_{moving,dynamic} = - \nabla \varphi_q - ∂\mathbf{A}_q/∂t</math>
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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:
 
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}_\mathbf{v}|\mathbf{a}/|\mathbf{v}|</math>
+
<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:
 
Where <math>\mathbf{a}</math> is the convective acceleration of the charge, which equals:
Line 88: Line 88:
 
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:
  
−
<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}_\mathbf{v}|\mathbf{a}/|\mathbf{v}|</math>
+
<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:
 
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}_\mathbf{v}|/|\mathbf{v}|\right]</math>
+
<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{A}_\mathbf{v}\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 force on a moving charge in a changing magnetic field becomes:
+
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}_\mathbf{v}|\mathbf{a}/|\mathbf{v}|</math>
+
<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>
  
 
{{Site map}}
 
{{Site map}}
  
 
[[Category:Function Conjunction]]
 
[[Category:Function Conjunction]]

Revision as of 02:15, 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 [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]

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