Quezothen
11/9/2025, 12:13:59 PM
No.16841333
>>16841292
Demonstration:
>Hoping the latex is right
Consider a vector potential field of the form:
[math]\tilde{A}_{\rm cylindrical} = \frac{1}{s} \hat{\phi}[/math]
[math]\tilde{A}_{\rm cartesian} = \frac{-y}{x^2 + y^2} \hat{x} + \frac{x}{x^2 + y^2} \hat{y}[/math]
This represents a circulating field that drops off linearly with distance from the vertical axis. Its curl is zero everywhere except along the z axis, where it is undefined.
[math]\nabla \times \vec{A} = -\frac{\partial (1/s)}{\partial z} \, \hat{s} + \frac{1}{s} \frac{\partial}{\partial s} \left( s \frac{1}{s} \right) \hat{z}[/math]
Whereas the electric singularity is a point, the magnetic singularity is a string. Here it is oriented vertically along the z axis with the field circulating around it.
The proper approach to this problem is to use Stoke's Theorem to first calculate the amount of circulation around the origin, which gives the value of magnetic flux that is present.
[math]\oint_P \vec{A} \cdot d\vec{l} = \int_S \nabla \times \vec{A} \cdot d\vec{a} = \chi[/math]
[math]\oint_P \vec{A} \cdot d\vec{l} = \oint_P \frac{1}{s} \hat{\phi} \cdot (s\, d\phi\, \hat{\phi}) = \oint_P d\phi = 2\pi[/math]
The singularity string contributes a flux of 2π for a circular path drawn around it. From Stoke's Theorem we see that the surface integral of the curl must equal this value.
[math]\int_S \nabla \times \vec{A} \cdot d\vec{a} = 2\pi[/math]
Here we can invoke the 2 dimensional Dirac delta function defined as:
[math]\delta_2(s) = \begin{cases} +\infty, & s = 0 \\ 0, & s \neq 0 \end{cases}[/math]
[math]\int_S \delta_2(s)\, d a = 1[/math]
For the surface, we may use a unit disc lying in the xy plane. Then:
[math]\int_S (\delta \times \vec{A}) \cdot d\vec{a} = \int_S (\delta \times \vec{A})_{z'} \hat{z}' \cdot \hat{z}'\, da = \int_S (\delta \times \vec{A})\, da = 2\pi[/math]
By comparing this to the delta function integral, we see that:
[math]\nabla \times \tilde{A} = 2\pi \delta_2(s)[/math]
Demonstration:
>Hoping the latex is right
Consider a vector potential field of the form:
[math]\tilde{A}_{\rm cylindrical} = \frac{1}{s} \hat{\phi}[/math]
[math]\tilde{A}_{\rm cartesian} = \frac{-y}{x^2 + y^2} \hat{x} + \frac{x}{x^2 + y^2} \hat{y}[/math]
This represents a circulating field that drops off linearly with distance from the vertical axis. Its curl is zero everywhere except along the z axis, where it is undefined.
[math]\nabla \times \vec{A} = -\frac{\partial (1/s)}{\partial z} \, \hat{s} + \frac{1}{s} \frac{\partial}{\partial s} \left( s \frac{1}{s} \right) \hat{z}[/math]
Whereas the electric singularity is a point, the magnetic singularity is a string. Here it is oriented vertically along the z axis with the field circulating around it.
The proper approach to this problem is to use Stoke's Theorem to first calculate the amount of circulation around the origin, which gives the value of magnetic flux that is present.
[math]\oint_P \vec{A} \cdot d\vec{l} = \int_S \nabla \times \vec{A} \cdot d\vec{a} = \chi[/math]
[math]\oint_P \vec{A} \cdot d\vec{l} = \oint_P \frac{1}{s} \hat{\phi} \cdot (s\, d\phi\, \hat{\phi}) = \oint_P d\phi = 2\pi[/math]
The singularity string contributes a flux of 2π for a circular path drawn around it. From Stoke's Theorem we see that the surface integral of the curl must equal this value.
[math]\int_S \nabla \times \vec{A} \cdot d\vec{a} = 2\pi[/math]
Here we can invoke the 2 dimensional Dirac delta function defined as:
[math]\delta_2(s) = \begin{cases} +\infty, & s = 0 \\ 0, & s \neq 0 \end{cases}[/math]
[math]\int_S \delta_2(s)\, d a = 1[/math]
For the surface, we may use a unit disc lying in the xy plane. Then:
[math]\int_S (\delta \times \vec{A}) \cdot d\vec{a} = \int_S (\delta \times \vec{A})_{z'} \hat{z}' \cdot \hat{z}'\, da = \int_S (\delta \times \vec{A})\, da = 2\pi[/math]
By comparing this to the delta function integral, we see that:
[math]\nabla \times \tilde{A} = 2\pi \delta_2(s)[/math]