Continuity of the tangential components of electric field across an interface

[hunt_mat]

TL;DR

How much can you get via natural boundary conditions?

Recently, I've gotten interested in applying Lagrangian mechanics to everything. I thought I would apply it to electrostatics. If I choose a Lagrangian of the form
$$L=\int_{\Omega}\frac{\epsilon}{2}|\nabla V|^{2}+\rho VdA$$
For two different regions, the interfacial condition for the normal components of the electric field pop out as natural boundary conditions.

I see no way of obtaining the interfacial conditions for the tangential components of the electric field. Does anyone know how to do this?

 

[renormalize]

$$L=\int_{\Omega}\frac{\epsilon}{2}|\nabla V|^{2}+\rho VdA$$

Sorry, I have questions about how I should interpret your notation:

1. Is $dA$ a 2-D area-element or a 3-D volume-element?
2. Is your integration region $\Omega$ a 2-D area or a 3-D volume?
3. Why is $\frac{\epsilon}{2}|\nabla V|^{2}$ not multiplied by $dA$?

 

[hunt_mat]

1) dA is an area element for simplicity.
2) $\Omega$ is an area.
3) I think you're a little confused to how integrals work. $\frac{\epsilon}{2}|\nabla V|^{2}+\rho V$ is the integrand.

 

[renormalize]

1) dA is an area element for simplicity.
2) $\Omega$ is an area.

Thanks for clarifying.

3) I think you're a little confused to how integrals work. $\frac{\epsilon}{2}|\nabla V|^{2}+\rho V$ is the integrand.

Then you should use parentheses to avoid confusing your readers: $L=\int_{\Omega}\left(\frac{\epsilon}{2}|\nabla V|^{2}+\rho V\right)dA$.

 

[Vincf]

I believe that continuity of the component parallel to the interface is automatically satisfied, provided your variational formulation implicitly assumes continuity of the potential at the interface.

 

[Roberto Pavani]

I may be missing something, but in electrostatics $E_s = -\frac{\partial V}{\partial s}.$ If the potential is continuous across the interface, doesn't the continuity of the tangential field (along the $s$-direction) follow automatically?