Boundary condition for magnetic vector potential

In summary, the conversation discusses the boundary condition for the magnetic vector potential (\vec{A}) as given by equation 5.76 in Griffiths' book. The problem of proving this boundary condition is brought up and the attempt to use equation 5.74 is discussed. The conversation also touches on the use of dot products and the information that \vec{A} is continuous and has a zero divergence. The difficulty in proving the boundary condition is mentioned and a possible solution is suggested, but the mathematical justification for this solution is questioned.
  • #1
issacnewton
998
29
Hi

I am studying magnetic vector potential from Griffiths book. The eq 5.76 in his book gives
the boundary condition for the magnetic vector potential.

[tex]\frac{\partial \vec{A_2} }{\partial n}- \frac{\partial \vec{A_1} }{\partial n}=-\mu_o \vec{K} [/tex]

where n is the vector perpendicular to the boundary surface and pointing from region
1 to region 2.

There is a problem in the book asking to prove this. So here's my attempt to do it. And I couldn't do it right. The book has eq 5.74 , which is

[tex]\vec{B_2}-\vec{B_1}=\mu_o(\vec{K}\times \hat{n}) [/tex]

So I decided to use this for my purpose. Writing B in terms of the vector potential A, we have

[tex] \vec{\nabla}\times \vec{A_2}-\vec{\nabla}\times \vec{A_1}=\mu_o(\vec{K}\times \hat{n}) [/tex]

Now writing in terms of the Cartesian components explicitly and collecting x,y,z components, we have

[tex]\hat{x}\left[\frac{\partial A_{2z} }{\partial y}- \frac{\partial A_{2y} }{\partial z}-
\frac{\partial A_{1z} }{\partial y}+\frac{\partial A_{1y} }{\partial z}\right]
-\hat{y}\left[\frac{\partial A_{2z} }{\partial x}-\frac{\partial A_{2x} }{\partial z}-
\frac{\partial A_{1z} }{\partial x}+\frac{\partial A_{1x} }{\partial z}\right]
+\hat{z}\left[\frac{\partial A_{2y} }{\partial x}-\frac{\partial A_{2x} }{\partial y}
-\frac{\partial A_{1y} }{\partial x}+\frac{\partial A_{1x} }{\partial y}\right]
=\mu_o(\vec{K}\times \hat{n}) [/tex]lets call this equation 1where K is the surface current. Now I take

[tex] \vec{K} = K\hat{x} [/tex]

and

[tex] \hat{n}=\hat{z} [/tex]

so that

[tex]\mu_o(\vec{K}\times \hat{n}) = \mu_o K(-\hat{y}) [/tex]

so I take the dot products of the equation 1 with [itex]\hat{x},\hat{y},\hat{z}[/itex].

There will be three equations in all.

[tex]\frac{\partial A_{2z} }{\partial y}+\frac{\partial A_{1y} }{\partial z}=
\frac{\partial A_{2y} }{\partial z}+\frac{\partial A_{1z} }{\partial y} [/tex][tex]\frac{\partial A_{2y} }{\partial x}+\frac{\partial A_{1x} }{\partial y}=
\frac{\partial A_{2x} }{\partial y}+\frac{\partial A_{1y} }{\partial x} [/tex]

[tex]\frac{\partial A_{2z} }{\partial x}-\frac{\partial A_{2x} }{\partial z}
-\frac{\partial A_{1z} }{\partial x}+\frac{\partial A_{1x} }{\partial z}=
\mu_o K [/tex]Now after this , I am totally lost. We actually know two more things about the vector
potential.

[tex]\vec{\nabla}\cdot \vec{A} = 0 [/tex]and at the boundary, A is continuous.

[tex]\vec{A_2}=\vec{A_1} [/tex]

But I don't know how to use this information to prove the result I am seeking. Any guidance
will be appreciated.
 
Last edited:
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  • #2
Hi, Griffith's fan too here,

you see from [tex]\mu_o(\vec{K}\times \hat{n}) = \mu_o K(-\hat{y}) [/tex] automatically x and y component from eqn 1 will dissapear.


then, eqn 1 will become eqn 2 as below:


[tex]-\hat{y}\left[\frac{\partial A_{2z} }{\partial x}-\frac{\partial A_{2x} }{\partial z}-
\frac{\partial A_{1z} }{\partial x}+\frac{\partial A_{1x} }{\partial z}\right]
=\mu_o K(-\hat{y} [/tex]



at boundary condition [tex]\vec{A_2}=\vec{A_1} [/tex],

then eqn 2 will be:


[tex]\hat{y}\left[\frac{\partial A_{2x} }{\partial z}-\frac{\partial A_{1x} }{\partial z}\right]=-\mu_o K(\hat{y} [/tex]


or, another write in vector term:


[tex]\left[\frac{\partial \vec{A}_{2} }{\partial z}-\frac{\partial \vec{A}_{1} }{\partial z}\right]=-\mu_o \vec{K}[/tex]


with [tex] \hat{z} [/tex] is perpendicular to the boundary surface.
 
  • #3
Hi Lepton

while deriving the following equation

[tex]
\hat{y}\left[\frac{\partial A_{2x} }{\partial z}-\frac{\partial A_{1x} }{\partial z}\right]=-\mu_o K(\hat{y} )
[/tex]

I think you assumed that

[tex]\frac{\partial A_{2z} }{\partial x}=\frac{\partial A_{1z} }{\partial x} [/tex]

why is that so ?

you could also have said,

[tex]\frac{\partial A_{2x} }{\partial z}=\frac{\partial A_{1x} }{\partial z} [/tex]

because [itex] A_{2x}=A_{1x} [/itex]

so what reasoning did you use ?

There is also another point. A is continuous at the boundary. But how can we say that
its derivatives also continuous ? Consider the following function.

[tex]f(x)=\lvert x \rvert[/tex]

This is continuous at the boundary x=0, but its not differentiable. So how did you do that ?
 
Last edited:
  • #4
for normal component, using the fact [tex]\nabla\cdot A = 0[/tex], or write it in integral form,

[tex]\oint \vec{A}\cdot d\vec{a} = 0[/tex], we will get [tex]A_{above}^{\perp} = A_{below}^{\perp}[/tex].


or we can write it in cartesian coordinate, [tex]A_{2z} = A_{1z}[/tex].

for tangential component, Griffith also said that A is continuous but not for its derivative. There is discontinuity in derivative of A in tangential component.

I think I can safely equate for normal component and find the derivative of A in tangential component as discontinuity in boundary.
 
  • #5
Hi Lepton,

Thats what I am not convinced about. Griffiths usually doesn't give rigorous arguments. I am just trying to make sure that what he says is right from mathematical standpoint as well.
 
  • #6
I just stumbled upon this problem in Griffiths, and I, too, have the same difficulty as the OP. The problem seems to require the tangental derivatives [itex] \frac{\partial \vec{A}}{\partial x}, \ \frac{\partial \vec{A}}{\partial y} [/itex] to be continuous across the current sheet. This link http://www.physics.sfsu.edu/~lea/courses/ugrad/360notes14.PDF claims between equations 10 and 11 that continuity of the tangental derivatives follows from continuity of [itex]\vec{A}[/itex]. If true, this would solve our problem.

But in general, if a function [itex] f(x, y)[/itex] is jointly continuous in [itex]x, \ y[/itex] and differentiable in [itex]x[/itex], it does not follow that [itex]\frac{\partial f}{\partial x}[/itex] is continuous in [itex]y[/itex]. For example, the function [itex]f(x, y) = \frac{x |y|^{3/2}}{x^2 + y^2}[/itex] is continuous but [itex] \frac{\partial f}{\partial x} = \frac{y^{3/2} (y^2 - x^2)}{(x^2 + y^2)^2} [/itex] is not continuous along the line [itex]x = 0[/itex]. Thus, without extra conditions on [itex]\vec{A}[/itex] I don't see why its tangental derivatives must be continuous.
 
  • #7
hi, if you find an answer, post here...
 

1. What is a boundary condition for magnetic vector potential?

A boundary condition for magnetic vector potential is a mathematical constraint that must be satisfied at the interface between two different media in order to accurately describe the behavior of the magnetic field in that region.

2. Why is it important to consider boundary conditions for magnetic vector potential?

Boundary conditions for magnetic vector potential are important because they ensure that the magnetic field is continuous and well-behaved at the interface between different media, which is crucial for accurately modeling and predicting the behavior of electromagnetic systems.

3. What are some examples of boundary conditions for magnetic vector potential?

Some examples of boundary conditions for magnetic vector potential include the continuity of the tangential component of the magnetic field, the continuity of the normal component of the magnetic flux density, and the continuity of the normal component of the magnetic field intensity.

4. How do boundary conditions for magnetic vector potential differ from those for electric potential?

Boundary conditions for magnetic vector potential and electric potential differ in that the former are concerned with the behavior of the magnetic field at the interface between different media, while the latter are concerned with the behavior of the electric field at the interface.

5. Are there any simplified or approximate boundary conditions for magnetic vector potential?

Yes, there are simplified or approximate boundary conditions for magnetic vector potential that are often used in practical applications. These include the perfect magnetic conductor (PMC) and the perfect electric conductor (PEC) conditions, which assume that the magnetic and electric fields are completely reflected at the interface between media.

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