Normal derivative of vector potential discontinuity

[physiks]

Homework Statement

In Griffiths, the following boundary condition is given without proof:
∂A_above/∂n-∂A_below/∂n=-μ₀K
for the change in the magnetic vector potential A across a surface with surface current density K, where n is the normal direction to the surface. A later problem asks for a proof of this, by using cartesian coordinates at the surface, with z perpendicular to the surface and x parallel to the current, and using the first three equations below.

Homework Equations

∇.A=0
A_above=A_below
B_above-B_below=μ₀(Kxn)
where n is a unit normal (I'm dropping all of the hats on my unit vectors).

The above information tell us
K=Kx
n=z

The Attempt at a Solution

First of all I have probably a really silly question because I know it's blatantly wrong, but why isn't this the case
B_above-B_below=μ₀(Kxz)
(∇xA_above)-(∇xA_below)=μ₀(Kxz)
∇x(A_above-A_below)=μ₀(Kxz)
0=μ₀(Kxz) because A_above=A_below

Aside from that, the solution states that because A_above=A_below all over the surface, ∂A/∂x and ∂A/∂y are also the same above and below the surface. Where does this come from, and why not ∂A/∂z? I can't get any further than this at present. Thanks for any help.

 

[Orodruin]

Regarding your "silly" question, it essentially boils down to the following: Study the function f(x)=|x|. At x=0, the left derivative is -1 and the right derivative is +1, yet the function itself is continuous. The potentials above and below describe the potential field in two different domains. You cannot first put z=0 in both and then take the limit that gives the derivatives.

Regarding the actual problem: try to think of an integration surface for which the current density is going to be relevant.

 

[physiks]

Regarding your "silly" question, it essentially boils down to the following: Study the function f(x)=|x|. At x=0, the left derivative is -1 and the right derivative is +1, yet the function itself is continuous. The potentials above and below describe the potential field in two different domains. You cannot first put z=0 in both and then take the limit that gives the derivatives.

Regarding the actual problem: try to think of an integration surface for which the current density is going to be relevant.

Aaaah, so B_above|_{on surface}=(∇xA_above)|_{on surface}, not B_above|_{on surface}=∇x(A_above|_{on surface}). Writing down where things are evaluated really clears that up...

Hmm I'm really not sure what you mean - the problem suggests the whole setup itself, as above, if that's what you mean?

 

[Orodruin]

Sorry, I missed that you were allowed to use the discontinuity in B. Can you think of some way of multiplying this equation with something else in order to just keep μ₀K?

 

[physiks]

Sorry, I missed that you were allowed to use the discontinuity in B. Can you think of some way of multiplying this equation with something else in order to just keep μ₀K?

Well I could dot both sides with (-y) then multiply by x. Not sure that would help though...

 

[Orodruin]

What operation between two vectors that give a vector as a result are you aware of?

 

[physiks]

What operation between two vectors that give a vector as a result are you aware of?

Ok, cross both sides with -z?
Well, z actually to get the negative on the RHS too?

That leaves a nasty vector triple product on the LHS which I don't like the look of - I'm hoping when I work it through it should condense down nicely?

 

[Orodruin]

Using the BAC-CAB rule (and remembering where the derivatives act), you should be able to break the LHS down to the directional derivatives and a gradient of one of the components of A. It is in order to put this gradient to zero you will need to apply that the x and y derivatives of A vanish.

Both z and -z will work, it is just a matter of an overall minus sign on both sides (which can also be obtained by crossing from the right or the left).

 

[physiks]

Using the BAC-CAB rule (and remembering where the derivatives act), you should be able to break the LHS down to the directional derivatives and a gradient of one of the components of A. It is in order to put this gradient to zero you will need to apply that the x and y derivatives of A vanish.

Both z and -z will work, it is just a matter of an overall minus sign on both sides (which can also be obtained by crossing from the right or the left).

I'll try working through that and get back if there's any issues... Why is it that the derivatives of A wrt x and y vanish though?

 

[Orodruin]

For those derivatives you do not have the same continuity problem across the surface since the limits are in different directions. The cancelation then follows from the continuity similar to what you tried in your original post.

 

[physiks]

For those derivatives you do not have the same continuity problem across the surface since the limits are in different directions. The cancelation then follows from the continuity similar to what you tried in your original post.

Do you mean this bit:
'∂A/∂x and ∂A/∂y are also the same above and below the surface. Where does this come from, and why not ∂A/∂z?'
Sorry I still don't understand why this is the case, despite your post.

 

[Orodruin]

So, for $\partial_z A_\pm$ (let me denote an arbitrary component of $\vec A$ above/below the surface at $z = 0$ by $A_\pm$) at a point where $z = \pm \varepsilon$ (where $\varepsilon > 0$) you would have
$$
\left.\partial_z A_\pm\right|_{z=\pm\varepsilon} = \lim_{a \to 0}\left(\frac{A_\pm(\pm \varepsilon+a)-A_\pm(\pm\varepsilon)}{a}\right)
$$
by definition (where $x$ and $y$ are kept fixed and thus suppressed in my notation). The difference of the derivatives above/below the surface $z = 0$ is
$$
\lim_{\varepsilon\to 0} \left(\left.\partial_z A_+\right|_{z=\varepsilon} - \left.\partial_z A_-\right|_{z=-\varepsilon} \right)
$$
The two limits (that of $\varepsilon$ and that of $a$) do not commute in general (they commute only when the $z$ derivative is continuous). However, when you instead look at the definition of the $x$ or $y$ derivative, the involved limit is going to commute with the limit of $\varepsilon \to 0$ as long as $A$ is continuous across $z = 0$.

The more intuitive way to think of it is that the potential on either side must have a limit which is the same on the surface in order to be continuous. Thus, the derivative along the surface must also be the same and the $x$ and $y$ directions are here what define "along the surface" in your coordinate system.