# Current sheets as sources of plane waves: reasoning behind boundary conditions

I am currently studying chapter 1.5 GENERAL PLANE WAVE SOLUTIONS of the textbook Microwave Engineering, fourth edition, by David Pozar. Example 1.3 CURRENT SHEETS AS SOURCES OF PLANE WAVES says the following:

An infinite sheet of surface current can be considered as a source for plane waves. If an electric surface current density $$\bar{\mathbf{J}}_s = \mathbf{J}_0 \hat{x}$$ exists on the $$z = 0$$ plane in free-space, find the resulting fields by assuming plane waves on either side of the current sheet and enforcing boundary conditions.

Solution
Since the source does not vary with $$x$$ or $$y$$, the fields will not vary with $$x$$ or $$y$$ but will propagate away from the source in the $$\pm z$$ direction. The boundary conditions to be satisfied at $$z = 0$$ are $$\hat{n} \times (\bar{E}_2 - \bar{E}_1) = \hat{z} \times (\bar{E}_2 - \bar{E}_1) = 0 \\ \hat{n} \times (\bar{H}_2 - \bar{H}_1) = \hat{z} \times (\bar{H}_2 - \bar{H}_1) = \mathbf{J}_0 \hat{x},$$ where $$\bar{E}_1$$, $$\bar{H}_1$$ are the fields for $$z < 0$$, and $$\bar{E}_2$$, $$\bar{H}_2$$ are the fields for $$z > 0$$. To satisfy the second condition, $$\bar{H}$$ must have a $$\hat{y}$$ component. Then for $$\bar{E}$$ to be orthogonal to $$\bar{H}$$ and $$\hat{z}$$, $$\bar{E}$$ must have an $$\hat{x}$$ component. Thus the fields will have the following form: $$\text{for z < 0,} \ \ \ \ \ \ \ \ \ \ \bar{E}_1 = \hat{x} A \eta_0 e^{jk_0z},$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bar{H}_1 = - \hat{y} A e^{jk_0z}$$ $$\text{for z > 0,} \ \ \ \ \ \ \ \ \ \ \bar{E}_2 = \hat{x} B \eta_0 e^{-jk_0z},$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \bar{H}_2 = \hat{y} B e^{-jk_0z},$$ where $$A$$ and $$B$$ are arbitrary amplitude constants. The first boundary condition, that $$E_x$$ is continuous at $$z = 0$$, yields $$A = B$$, while the boundary condition for $$\bar{H}$$ yields the equation $$-B - A = \mathbf{J}_0.$$ Solving for $$A$$, $$B$$ gives $$A = B = - \mathbf{J}_0 / 2,$$ which completes the solution. $$\blacksquare$$

What is the reasoning behind $$\hat{n} \times (\bar{E}_2 - \bar{E}_1) = \hat{z} \times (\bar{E}_2 - \bar{E}_1) = 0$$ and $$\hat{n} \times (\bar{H}_2 - \bar{H}_1) = \hat{z} \times (\bar{H}_2 - \bar{H}_1) = \mathbf{J}_0 \hat{x}$$? It seems like these boundary conditions were just conjured out of no where. And where did the negative sign in $$\bar{H}_1 = - \hat{y} A e^{jk_0z}$$ come from?

## EDIT

I think I figured out the reasoning behind $$\hat{n} \times (\bar{E}_2 - \bar{E}_1) = \hat{z} \times (\bar{E}_2 - \bar{E}_1) = 0$$ and $$\hat{n} \times (\bar{H}_2 - \bar{H}_1) = \hat{z} \times (\bar{H}_2 - \bar{H}_1) = \mathbf{J}_0 \hat{x}$$ (see my comments to user hyportnex's answer), but I still can't figure out where the negative sign in $$\bar{H}_1 = - \hat{y} A e^{jk_0z}$$ comes from.

The current is homogeneous on the plane of symmetry hence the difference vector between the left and right propagating $$\hat E$$ fields must be parallel with the plane's normal, that is $$\hat n \times (\hat E_1-\hat E_2)=0$$.
The other equation is a consequence of Ampere's law $$\oint_{\partial \mathcal A} \hat H \cdot d{\hat \ell} = \int_{\mathcal A} \hat J \cdot d\hat A$$ applied to a thin rectangle $${\mathcal A}$$ to be parallel with the $$xz$$ plane whose long side $$b$$ is parallel with $$\hat y$$ and its very short side $$a$$ is parallel with $$\hat z$$, and use Stokes' theorem to show the other boundary condition. Note that the enclosed current on that rectangle is $$J_0 b$$.
• With regards to what you're saying in the first part, is $\bar{E}_2 - \bar{E}_1$ the boundary between the two electric fields $\bar{E}_2$ and $\bar{E}_1$, and so $\hat{n} \times (\bar{E}_2 - \bar{E}_1)$ is the counter-clockwise circulation around that boundary, and so $\hat{n} \times (\bar{E}_2 - \bar{E}_1) = \hat{z} \times (\bar{E}_2 - \bar{E}_1) = 0$ means that that circulation is equal to $0$? Jan 18, 2023 at 17:43
• And I guess it is analogously for $\hat{n} \times (\bar{H}_2 - \bar{H}_1) = \hat{z} \times (\bar{H}_2 - \bar{H}_1) = \mathbf{J}_0 \hat{x}$, except that the anti-clockwise circulation on the boundary for the two magnetic fields $\bar{H}_2$ and $\bar{H}_1$ is equal to the electric surface current density on that boundary plane, $\bar{\mathbf{J}}_s = \mathbf{J}_0 \hat{x}$? Jan 18, 2023 at 17:47
• Also, with regards to another matter, where did the negative sign in $\bar{H}_1 = - \hat{y} A e^{jk_0z}$ come from? Jan 18, 2023 at 18:05
• @ThePointer Both $H_1$ and $H_2$ are from the same current source but on the opposite sides of it. Integrate along the rectangular loop $\mathcal A$ the $H$, from the symmetry their magnitudes are the same, so they must point in the opposite direction for otherwise the Ampere integral would be zero and not the enclosed current. Feb 14, 2023 at 1:59