# 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. 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. 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 $$\frac{\partial E_m}{\partial m}\to0$$.