Boundary Conditions Invariant Under Conformal Transformations in Electrostatics?

Question

in two dimensional electrostatics it is assumed that the whole physical system is translationally invariant in one direction. Here, the two-dimensional Laplace equation $$\Delta \phi(x,y) = \frac{\sigma(x,y)}{\epsilon_0}$$ holds in free space and solutions can be mapped on solutions if space is transformed conformally - the metric changes from $$g\rightarrow \Omega^2 g\ .$$

This approach has a lot of nice applications like the possibility to calculate the field around a complicated conductor or, in aerodynamics, the flow around an airfoil as discussed earlier.

Aubry et al. have in a number of papers (see e.g. Interaction between Plasmonic Nanoparticles Revisited with Transformation Optics (PRL) and Plasmonic Light-Harvesting Devices over the Whole Visible Spectrum (Nano Lett.)) used this technique to calculate the eigenfunctions of "crescend" structures and coupled cylinders:

(taken from the Nano Lett.)

I think this is a very nice approach but I am not sure why it can be done in this way. My problem is that as far as I know usually only Neumann and Dirichlet boundary conditions are invariant under conformal transformations. In electrodynamics, however, we have that both $E_t = -d\phi(\mathbf{t})$ (tangential) and $\epsilon\cdot E_n = -\epsilon\cdot d\phi(\mathbf{n})$ (normal) are continuous at a boundary where the permittivity $\epsilon$ changes.

In the papers I could not find a discussion of this issue so I might ask here:

Are the boundary conditions for the electrostatic potential $\phi$ invariant under conformal transformations?

Thank you in advance
Sincerely

Answer 1

Under a conformal transformation $f$ we have that the coordinate $x_i \rightarrow f_i(x)$, the potential $\phi \rightarrow \phi$ and therefore the derivative $\partial_i \phi \rightarrow \frac{\partial f_j}{\partial x_i} \partial_j \phi$. A tangent vector $t$ to some surface $S$ gets mapped to a tangent vector to $f(S)$ as always, and because the transformation is conformal the normal to a surface $n$ gets mapped to a vector proportional to the normal $n'$ on the surface $f(S)$ since $t\cdot n = 0$ is preserved by conformal transformation.

The usual boundary conditions $\phi(S) = \phi_0$ and that $\phi(S)$ is continuous are preserved, since $\phi$ transforms trivially. A different condition is that $\epsilon_{+}n\cdot \partial \phi_{+} - \epsilon_{-}n\cdot \partial \phi_{-} = 0$, where $\phi_{\pm}$ is the limit of $\phi$ approaching the surface from either direction and $\epsilon_{\pm}$ are constants. This gets mapped to the equation $\Omega^2\times\left(\epsilon_{+}n'\cdot \partial \phi_{+} - \epsilon_{-}n'\cdot \partial \phi_{-}\right) = 0$ so that is preserved as well (where $\Omega$ is the scale factor of $f$). The same occurs for the part of the $\partial_j \phi$ tangent to the the surface as well. A sheet of charge $\sigma$ produces the b. c. $\epsilon_{+}n\cdot \partial \phi_{+} - \epsilon_{-}n\cdot \partial \phi_{-} = \sigma$ which is preserved as well with the correct transformation $\sigma \rightarrow \Omega^2 \sigma$.

So all the usual b. c. of electrostatics work well with conformal transformations.