Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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
Sincerely

share|improve this question
1  
Yaay, pictures! :) Also a great question as usual, +1 –  Marek Jul 28 '11 at 15:44
2  
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

1 Answer 1

up vote 2 down vote accepted
+50

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.