Boundary condition for magnetic vector potential

[issacnewton]

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.

$$\frac{\partial \vec{A_2} }{\partial n}- \frac{\partial \vec{A_1} }{\partial n}=-\mu_o \vec{K}$$

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

$$\vec{B_2}-\vec{B_1}=\mu_o(\vec{K}\times \hat{n})$$

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

$$\vec{\nabla}\times \vec{A_2}-\vec{\nabla}\times \vec{A_1}=\mu_o(\vec{K}\times \hat{n})$$

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

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

$$\vec{K} = K\hat{x}$$

and

$$\hat{n}=\hat{z}$$

so that

$$\mu_o(\vec{K}\times \hat{n}) = \mu_o K(-\hat{y})$$

so I take the dot products of the equation 1 with $\hat{x},\hat{y},\hat{z}$.

There will be three equations in all.

$$\frac{\partial A_{2z} }{\partial y}+\frac{\partial A_{1y} }{\partial z}=<br /> \frac{\partial A_{2y} }{\partial z}+\frac{\partial A_{1z} }{\partial y}$$$$\frac{\partial A_{2y} }{\partial x}+\frac{\partial A_{1x} }{\partial y}=<br /> \frac{\partial A_{2x} }{\partial y}+\frac{\partial A_{1y} }{\partial x}$$

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

$$\vec{\nabla}\cdot \vec{A} = 0$$and at the boundary, A is continuous.

$$\vec{A_2}=\vec{A_1}$$

But I don't know how to use this information to prove the result I am seeking. Any guidance
will be appreciated.

 

[lepton5]

Hi, Griffith's fan too here,

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


then, eqn 1 will become eqn 2 as below:


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



at boundary condition $$\vec{A_2}=\vec{A_1}$$,

then eqn 2 will be:


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


or, another write in vector term:


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


with $$\hat{z}$$ is perpendicular to the boundary surface.

 

[issacnewton]

Hi Lepton

while deriving the following equation

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

I think you assumed that

$$\frac{\partial A_{2z} }{\partial x}=\frac{\partial A_{1z} }{\partial x}$$

why is that so ?

you could also have said,

$$\frac{\partial A_{2x} }{\partial z}=\frac{\partial A_{1x} }{\partial z}$$

because $A_{2x}=A_{1x}$

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.

$$f(x)=\lvert x \rvert$$

This is continuous at the boundary x=0, but its not differentiable. So how did you do that ?

 

[lepton5]

for normal component, using the fact $$\nabla\cdot A = 0$$, or write it in integral form,

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


or we can write it in cartesian coordinate, $$A_{2z} = A_{1z}$$.

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.

 

[issacnewton]

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.

 

[namphcar22]

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 $\frac{\partial \vec{A}}{\partial x}, \ \frac{\partial \vec{A}}{\partial y}$ 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 $\vec{A}$. If true, this would solve our problem.

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

 

[issacnewton]

hi, if you find an answer, post here...