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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.

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1 Answer 1

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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 $J_0 b$.

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    $\begingroup$ 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 $0$? $\endgroup$ Jan 18, 2023 at 17:43
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    $\begingroup$ 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, $\bar{\mathbf{J}}_s = \mathbf{J}_0 \hat{x}$? $\endgroup$ Jan 18, 2023 at 17:47
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    $\begingroup$ Also, with regards to another matter, where did the negative sign in $\bar{H}_1 = - \hat{y} A e^{jk_0z}$ come from? $\endgroup$ Jan 18, 2023 at 18:05
  • $\begingroup$ @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. $\endgroup$
    – hyportnex
    Feb 14, 2023 at 1:59

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