Suppose a cubic waveguide, made of perfect conductor, has only two open parallel sides. And the boundary conditions in this case are that the electric field at the surface must satisfy:

$$\vec{B} \cdot \vec{n}=0,$$

and magnetic field:

$$\vec{E} \times \vec{n}=0,$$

where the $\vec{n}$ is the normal vector pointing outwards from the conductor. These two relations come from the equations:

$$\nabla \cdot \vec{B}=0,$$ $$\nabla \times \vec{E}=0.$$

The question is how to derive the other boundary condition that at the surface the electric field must satisfy:

$$\frac{\partial{E_n}}{\partial n}=0.$$

$E_n$ means the electric field along normal direction.

  • $\begingroup$ Integrate $\nabla\cdot\vec{E}=\rho$ in a cylinder whose axis is perpendicular to the surface and shrink the said cylinder to a point. $\endgroup$
    – user154997
    Sep 25 '17 at 11:38
  • $\begingroup$ Yeah, but I think after using the divergence theorem, the result only shows that the electric field along the normal direction is zero, rather than the derivative is also zero. $\endgroup$
    – Edwin
    Sep 25 '17 at 12:11
  • $\begingroup$ The clearest way to write the boundary condition is: $E_n^+ - E_n^- = \sigma$ where $\sigma$ would be a surface charge, and $E_n^\pm$ means the value just "above" and just "below" the surface. That difference $E_n^+ - E_n^-$ is what you wrote as $\partial E_n/\partial n$ I believe, a notation which can be confusing as it makes you believe a derivative is involved. This boundary condition can be demonstrated as per my previous hint. $\endgroup$
    – user154997
    Sep 25 '17 at 12:28
  • $\begingroup$ If that notation $\partial E_n/\partial n$ is contrary to what I thought a real derivative, then I think it is just an argument of continuity. Inside the perfect conductor, $E$ is identically zero everywhere, and so are any derivatives. If there are not surface charges, then by continuity, $\partial E/\partial n$ must also be zero at the interface. $\endgroup$
    – user154997
    Sep 25 '17 at 12:35
  • $\begingroup$ Well, I should make clear the meaning of $\partial E_n/\partial n$. Suppose the surface is the whole xz plane in Cartesian Coordinates. And x<0 space represents the perfect conductor and x>0 the vacuum space. By meaning of $\partial E_n/\partial n$, it is $\partial E_y {(x,y,z)}/\partial y$ when y goes to $+0$. It is kind of a limit process rather than the difference between the values at left and right. $\endgroup$
    – Edwin
    Sep 25 '17 at 13:35

This simply follows from the Gauss' law:

$$\nabla\cdot\vec E=\frac\rho{\varepsilon_0}.\tag1$$

Since we know that $\vec E\parallel\vec n$ at the boundary inside the waveguide, the divergence of $\vec E$ reduces$^\dagger$ to $$\nabla\cdot\vec E=\frac{\partial E_n}{\partial n}.\tag2$$

Since the waveguide doesn't have any charges inside, $\rho=0$ in the whole internal region including the vicinity of the conductor. Inserting this into $(1)$ simplified by $(2)$, we get our boundary condition.

$^\dagger$ This is true because $\vec E$ is analytic at the interface (with $n=0^+$), in which case $\lim\limits_{n\to0}E_m\to0$, where $m$ is the coordinate along any tangent to the interface and $n$ – along the normal, implies $\frac{\partial E_m}{\partial m}\to0$.


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