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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:

Plasmonic Light-Harvesting Devices over the Whole Visible Spectrum(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

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Yaay, pictures! :) Also a great question as usual, +1 – Marek Jul 28 '11 at 15:44
I could have this totally wrong, but here's a stab at a partial answer. The boundary conditions are that $E_{||}$ and $(\epsilon E)_\perp$ are both continuous. This means that it's necessary, but not sufficient, for any discontinuity $\Delta E$ in the field to be perpendicular to the boundary. But angles are conserved in conformal transformations, so I'd think that this condition would be preserved; the 90-degree angle would still be a 90-degree angle. – Ben Crowell Jul 28 '11 at 16:25
@Ben Crowell: I'm pretty sure this is correct. If you write it as an answer I'll upvote it. – genneth Jul 28 '11 at 22:13
@genneth: Thanks, but I don't think it's complete. It doesn't deal with the constraint on $(\epsilon E)_\perp$. – Ben Crowell Jul 28 '11 at 22:58
@Marek: Thank you very much :) – Robert Filter Jul 29 '11 at 12:30
up vote 2 down vote accepted

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.

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thanks for the answer! Could you please state how the transformation $x_i \rightarrow f_i(x)$ relates to the conformal factor such that your mapping of the normal component is justified? I don't see it directly - if I get you right, then $g(n,\partial \phi)\rightarrow \Omega^2 g(n^\prime, \partial f \partial \phi)$ which leaves some space for discussion because its not clear what the additional $\partial f$ will look like. Thank you again – Robert Filter Aug 2 '11 at 16:11
a) I should have written the normal $n$ gets mapped to a vector that is proportional to the new normal $n'$ if we are defining the normal as having unit length. b) What you write should be $g(n,\partial \phi)\rightarrow g(n',\partial f \partial \phi) = \Omega^2 g(n,\partial\phi)$. You don't transform both the metric and the vectors - that just gives you diffeomorphism invariance. The Jacobian $J_{ij} = \partial f_i/ \partial x_j$ is $\Omega$ times a rotation, so we can calculate $\Omega$ by calculating $ (JJ^T)_{ij}= \Omega^2\delta_{ij}$. – BebopButUnsteady Aug 2 '11 at 18:02
Thank you for your clearifications, I will award the bounty as soon as it is possible. Greets – Robert Filter Aug 3 '11 at 7:57

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