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I'm trying to prove the equation from Griffiths' Introduction to Electrodynamics which tells us that the vector potential ${\bf A}$ is continuous across any boundary (both tangential and normal components) but its derivative somehow inherits the discontinuity already proved for ${\bf B}$, namely: $$ \frac{\partial {\bf A}_{above} }{\partial n}- \frac{\partial {\bf A}_{below}}{\partial n} = -\mu_0 \bf{K} \tag{1}$$ Now I tried a very weird approach yesterday, but it was evidently flawed and did not take me anywhere so, this time, I tried following some of the author's suggestions. Let us consider a cartesian coordinate system (${\bf \hat{i},\hat{j}, \hat{k} }, x,y,z$). Let us consider now a current sheet, namely a 2D-surface $\Sigma$ with surface current density $\bf{K}$ which lies in the $xy$-plane. Moreover, let's set the current parallel to the $x$-axis, therefore ${\bf K} = K \bf{\hat{i}}$. By doing so, the ${\bf\hat{k}}$ direction will be perpendicular to $\Sigma$ as well. Let us eventually rename ${\bf A}_{above/below} \equiv {\bf A}_{a,b}$. Now, using ${\bf B = \nabla \times A }$ and computing the cross product in the RHS I can rewrite the equation: $$ {\bf B}_{a}-{\bf B}_{b} = \mu_0 (\mathbf{K} \times \mathbf{\hat{n}}) \tag{2}$$ as $$\nabla \times(\mathbf{A}_a - \mathbf{A}_b) = -\mu_0 K \bf{\hat{j}}$$ Well, I then compute the curls in the LFH and regroup the terms according to the cartesian unit vectors, and by comparison with the RHS I finally get to the following system of three equations: $$ \begin{cases} \frac{\partial A_{a,z}}{\partial y} + \frac{\partial A_{b,y}}{\partial z} = \frac{\partial A_{b,z}}{\partial y} + \frac{\partial A_{a,y}}{\partial z} \\ \frac{\partial A_{a,y}}{\partial x} + \frac{\partial A_{b,x}}{\partial y} = \frac{\partial A_{b,y}}{\partial x} + \frac{\partial A_{a,x}}{\partial y} \\ \frac{\partial A_{a,x}}{\partial z} - \frac{\partial A_{b,x}}{\partial z} + \frac{\partial A_{b,z}}{\partial x} - \frac{\partial A_{a,z}}{\partial x} = -\mu_0 K \end{cases} \tag{3}$$ Now, if we ignore for a second the partial derivatives with respect to $x$, we can see that the equation tells us that the derivative of the components of the vector potential which are parallel to the current, with respect to the unit normal to the surface do indeed inherit a discontinuity. Now, I tried using the equation: $$ {\bf A}_{above} = {\bf A}_{below} $$ which, expanded, gives us: $$A_{a,i}-A_{b,i} = 0, i \in [x,y,z], \tag{4}$$ Now going back to $(3)$ and using $(4)$, we can see that the first two equations vanish and so does the extra term in the third equation that was causing us problems. We are left with the sole equation: $$ \frac{\partial A_{a,x}}{\partial z} - \frac{\partial A_{b,x}}{\partial z} = -\mu_0 K $$ Now, being totally honest, I can't see how this is equal to the author's result $\bf{(1)}$. Nevertheless, I cannot unsee the symmetry: in the case of $\mathbf{B}$, the component that presents the discontinuity is the one parallel to the surface but orthogonal to the current, while in the case of $\mathbf{A}$, I obtained that the component in question is the one parallel to the current. This has made me just a bit more confident in the result I got. If somebody could tell me if my procedure is correct and, in that case, if the result is indeed equivalent to Griffiths', I'd really appreciate it. Moreover, if someone wants to propose a proof of his own, well that would also be much appreciated since my methods are far from rigorous (almost primitive I'd say) and lack all of that physical intuition which make those kind of proofs much easier and which I'd like to possess myself

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Start with $$\mathbf {B} = \nabla \times \mathbf A \tag{1}\label{1}$$ then assuming the Coulomb gauge $$\nabla \cdot \mathbf A = 0 \tag{2}\label{2}$$ you get $$\nabla \times (\nabla \times \mathbf A) = \nabla (\nabla \cdot \mathbf A) - \nabla ^2 \mathbf A \\ =- \nabla ^2 \mathbf A \tag{3}\label{3}.$$ Therefore $$\nabla \times \mathbf B = - \nabla ^2 \mathbf A = \mu_0 \mathbf J \tag{4}\label{4}.$$ Eq. $\eqref{4}$ holds both above and below the interface. For a surface current that is, say, one in the $xy$ plane $$\mathbf J = \mathbf K \delta (z) \tag{5}\label{5}$$ and then $$\nabla ^2 \mathbf A = -\mu_0 \mathbf K \delta (z) \tag{6}\label{6}.$$ Now integrate form below to above over an infinitesimal segment along $\hat z$: $$\frac{\partial}{\partial z} \left(\mathbf {A_a - A_b}\right)= -\mu_0 \mathbf K \tag{7}\label{7},$$ which is what you wanted to prove.


Added details: $\mathbf A$ is continuous everywhere, and since the discontinuity is along the $\hat z$ direction, both $\frac{\partial \mathbf A}{\partial x}$ and $\frac{\partial^2 \mathbf A}{\partial x^2}$ are also continuous, similarly in the $\hat y$ direction.

Now let $\mathbf K = K\hat y$, this implies that $\nabla^2A_y=-\mu_0K\delta(z)$. Then integrating along $\hat z$ across the interface you only have to show that $\frac{\partial^2 A_y}{\partial x^2}+\frac{\partial^2 A_y}{\partial y^2}$ is zero, but that is true because of its continuity$.

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  • $\begingroup$ how did you perform the integration of the LHS of $(6)$? My approach is: $$ lim_{\epsilon \to 0^+}\int_{-\epsilon}^{+\epsilon}[(\nabla^2 A_x){bf \hat{i}}+ (\nabla^2A_y){\bf \hat{j}} + (\nabla^2A_z){\bf \hat{k}}]dz$$ The first term, for example, would result in $$ {\bf\hat{i}}\lim_{\epsilon \to 0^+}\int_{-\epsilon}^{+\epsilon} ( \partial^2_{xx}A_x(x,y,z) + \partial^2_{yy}A_x(x,y,z) + \partial^2_{zz}A_x(x,y,z) )dz$$I don't know how to treat the integration of the $x,y$ second derivatives, while the $z$ one gives: $$\hat{i}[\partial_z A_x(x,y,+\epsilon)-\partial_z A_x (x,y,-\epsilon)] $$ $\endgroup$ Aug 11, 2023 at 16:19
  • $\begingroup$ $\int \delta (z) = 1$ and $B_z$ is continuous across the surface current. $\endgroup$
    – hyportnex
    Aug 11, 2023 at 16:26
  • $\begingroup$ The delta function was ok since we're integrating on the spike. For what concerns the LHS: the continuity of the normal component of $B$ across the surface allows us to apply the FTOC only to the $z$ component. then you express $B_z$ in terms of the vector potential again? What happens to $B_x,B_y$. Those might be functions of $z$ as well right? $\endgroup$ Aug 11, 2023 at 16:35
  • $\begingroup$ I do not know what "FTOC" is but the rest to yes, and that is where the jump is. $\endgroup$
    – hyportnex
    Aug 11, 2023 at 16:41
  • $\begingroup$ could I request the formal passages made when integrating the laplacian (or the curl of the magnetic field I guess)? I already ticked the answer but I'd like to see if I'm missing something. The FTOC is the fundamental theorem of calculus, sorry I thought it was called like this in English. I have a problem with integrating the non $z$ partial derivatives $\endgroup$ Aug 11, 2023 at 17:37

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