The boundary conditions in a waveguide

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.

• Integrate $\nabla\cdot\vec{E}=\rho$ in a cylinder whose axis is perpendicular to the surface and shrink the said cylinder to a point. – user154997 Sep 25 '17 at 11:38
• 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. – Edwin Sep 25 '17 at 12:11
• 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. – user154997 Sep 25 '17 at 12:28
• 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. – user154997 Sep 25 '17 at 12:35
• 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. – Edwin Sep 25 '17 at 13:35

$$\nabla\cdot\vec E=\frac\rho{\varepsilon_0}.\tag1$$
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 $$\frac{\partial E_m}{\partial m}\to0$$.