Continuity across boundaries in Electromag

[fantispug]

Homework Statement

This question is adapted from an implicit assumption in Ashcroft and Mermin question 1.5.

Consider a medium with no net charge (but possibly a net current) in which Ohm's law holds. Let an electromagnetic wave travel through the medium with angular frequency $$\omega$$. Then, by Maxwell's equations, the electric field satisfies
$$-\nabla^2 \mathbf{E} = \frac{\omega^2}{c^2} \epsilon(\omega) \mathbf{E}$$ for some appropriate function $$\epsilon(\omega)$$, where c is the speed of light.

Consider the boundary between two such regions (which may contain surface charge). Prove that $$\epsilon \mathbf{E}^{\bot}$$ is continuous across the boundary.

Homework Equations

The surface charge between two regions is given by the discontinuity of the perpendicular component of the electric field.
$$E^{\bot}_{\text{above}} - E^{\bot}_{\text{below}} = \frac{\sigma}{\epsilon_0}$$

The Attempt at a Solution

I don't think it's true. Given the perpendicular component of $$\epsilon \mathbf{E}$$ is continuous across the boundary, using the definition of $$\epsilon$$ we see
$$-\nabla^2 (E^{\bot}_{\text{above}} - E^{\bot}_{\text{below}}) = 0$$
that is
$$\nabla^2 \sigma = 0$$.

I don't see any a priori reason this should be true (given this I could run the argument backwards to prove the assertion). I think this would be the crux of the argument; and indeed I can not see another way to approach the problem.

Any help on what I'm missing?
Cheers.

 

[Jhero]

Thank you for bringing up this interesting question. I believe the issue here is that you are assuming that the surface charge is constant across the boundary, which is not necessarily true. The surface charge on each side of the boundary may be different, but the electric field should still be continuous across the boundary. This is because the electric field is determined by the distribution of charges, not just the surface charge.

To prove this, we can use the continuity equation for current, which states that the divergence of current density is equal to the rate of change of charge density. In this case, since there is no net charge in the medium, the rate of change of charge density is zero and the continuity equation reduces to:

\nabla \cdot \mathbf{J} = 0

Using Ohm's law, we can express the current density as:

\mathbf{J} = \sigma \mathbf{E}

where \sigma is the conductivity of the medium. Substituting this into the continuity equation, we get:

\nabla \cdot (\sigma \mathbf{E}) = 0

Using the product rule for the divergence, this becomes:

\sigma (\nabla \cdot \mathbf{E}) + \mathbf{E} \cdot (\nabla \sigma) = 0

Since \nabla \cdot \mathbf{E} = 0 (by Gauss's law), we can rearrange this equation to get:

\mathbf{E} \cdot (\nabla \sigma) = 0

This implies that the electric field is always perpendicular to the gradient of the conductivity. Now, going back to the original problem, we can see that since the electric field is continuous across the boundary, the gradient of the conductivity must also be continuous. This means that the conductivity itself is continuous, since the gradient of a function is continuous if the function is continuous. And since the conductivity is related to \epsilon by \sigma = \epsilon \omega, we can conclude that \epsilon is also continuous across the boundary.

I hope this helps to clarify the issue. Let me know if you have any further questions.