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I am currently studying the textbook Microwave Engineering, fourth edition, by David Pozar. Chapter 1.4 THE WAVE EQUATION AND BASIC PLANE WAVE SOLUTIONS says the following:

The Helmholtz Equation

In a source-free, linear, isotropic, homogeneous region, Maxwell's curl equations in phasor form are $$\nabla \times \bar{E} = -j \omega \mu \bar{H} \tag{1.41a}$$ $$\nabla \times \bar{H} = j \omega \epsilon \bar{E}, \tag{1.41b}$$ and constitute two equations for the unknowns, $\bar{E}$ and $\bar{H}$. As such, they can be solved for either $\bar{E}$ or $\bar{H}$. Taking the curl of (1.41a) and using (1.41b) gives $$\nabla \times \nabla \times \bar{E} = - j\omega \mu \nabla \times \bar{H} = \omega^2 \mu \epsilon \bar{E},$$ which is an equation for $\bar{E}$. This result can be simplified through the use of vector identity (B.14), $\nabla \times \nabla \times \bar{A} = \nabla (\nabla \cdot \bar{A}) - \nabla^2 \bar{A}$, which is valid for the rectangular components of an arbitrary vector $\bar{A}$. Then, $$\nabla^2 \bar{E} + \omega^2 \mu \epsilon \bar{E} = 0, \tag{1.42}$$ because $\nabla \cdot \bar{E} = 0$ in a source-free region. Equation (1.42) is the wave equation, or Helmholtz equation, for $\bar{E}$. An identical equation for $\bar{H}$ can be derived in the same manner: $$\nabla^2 \bar{H} + \omega^2 \mu \epsilon \bar{H} = 0. \tag{1.43}$$ A constant $k = \omega \sqrt{\mu \epsilon}$ is defined and called the propagation constant (also known as the phase constant, or wave number), of the medium; its units are $1/m$.

Plane Waves in a Lossless Medium

In a lossless medium, $\epsilon$ and $\mu$ are real numbers, and so $k$ is real. A basic plane wave solution to the above wave equation can be found by considering an electric field with only an $\hat{x}$ component and uniform (no variation) in the $x$ and $y$ directions. Then, $\partial/\partial{x} = \partial/\partial{y} = 0$, and the Helmholtz equation of (1.42) reduces to $$\dfrac{\partial^2{E_x}}{\partial{z}^2} + k^2 E_x = 0. \tag{1.44}$$ The two independent solutions to this equation are easily seen, by substitution, to be of the form $$E_x(z) = E^+e^{-jkz} + E^-e^{jkz}, \tag{1.45}$$ where $E^+$ and $E^-$ are arbitrary amplitude constants.
The above solution is for the time harmonic case at frequency $\omega$. In the time domain, this result is written as $$\mathcal{E}_x(z, t) = E^+ \cos(\omega t - kz) + E^- \cos(\omega t + kz), \tag{1.46}$$ where we have assumed that $E^+$ and $E^-$ are real constants.

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A complete specification of the plane wave electromagnetic field should include the magnetic field. In general, whenever $\bar{E}$ or $\bar{H}$ is known, the other field vector can be readily found by using one of Maxwell's curl equations. Thus, applying (1.41a) to the electric field of (1.45) gives $H_x = H_z = 0$, and $$H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz}), \tag{1.49}$$ where $\eta = \omega \mu / k = \sqrt{\mu/\epsilon}$ is known as the intrinsic impedance of the medium.

I'm confused by this part:

In general, whenever $\bar{E}$ or $\bar{H}$ is known, the other field vector can be readily found by using one of Maxwell's curl equations. Thus, applying (1.41a) to the electric field of (1.45) gives $H_x = H_z = 0$, and $$H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz}), \tag{1.49}$$ where $\eta = \omega \mu / k = \sqrt{\mu/\epsilon}$ is known as the intrinsic impedance of the medium.

It isn't totally clear to me how (1.41a) is "applied" to (1.45) to get $H_x = H_z = 0$ and $H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz})$, but it seems to me that they might be referring to taking the curl of (1.45). My understanding is that the curl is defined as follows:

$$\nabla \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \dfrac{\partial}{\partial{x}} & \dfrac{\partial}{\partial{y}} & \dfrac{\partial}{\partial{z}} \\ F_x & F_y & F_z \end{vmatrix} = \hat{\mathbf{i}} \left( \dfrac{\partial{F_z}}{\partial{y}} - \dfrac{\partial{F_y}}{\partial{z}} \right) - \hat{\mathbf{j}} \left( \dfrac{\partial{F_z}}{\partial{x}} - \dfrac{\partial{F_x}}{\partial{z}} \right) + \hat{\mathbf{k}} \left( \dfrac{\partial{F_y}}{\partial{x}} - \dfrac{\partial{F_x}}{\partial{y}} \right)$$

So it seems to me that we then have

$$\begin{align} \nabla \times E_x(z) &= \nabla \times E^+e^{-jkz} + \nabla \times E^-e^{jkz} \\ &= E^+ (\nabla \times e^{-jkz}) + E^- (\nabla \times e^{jkz}) \ \ \ \text{(since $E^+$ and $E^-$ are constants.)} \end{align}$$

But $e^{-jkz}$ and $e^{jkz}$ aren't fields either, so how does this work? And since we're taking the curl of $E_x$ instead of $E$, how does that work? And, finally, how do we get $H_x = H_z = 0$ and $H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz})$?


EDIT

If we apply the curl to (1.45), then we get

$$\nabla \times E_x = \hat{\mathbf{j}} \dfrac{\partial{E_x}}{\partial{z}} = (-jk E^+ e^{-jkz} + jk E^- e^{jkz}) \hat{\mathbf{j}} = -jk (E^+ e^{-jkz} - E^- e^{jkz})\hat{\mathbf{j}},$$

so it isn't clear to me how the authors got this.

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    $\begingroup$ The author considers the case of a plane wave moving in the $z$ direction, with the electric field pointing in the $x$ direction. If you take the curl of this electric field only the $d/dz$ of $E_x$ contributes, and so you arrive at $H_y$. $\endgroup$
    – M. Wind
    Apr 24, 2022 at 15:48
  • $\begingroup$ @M.Wind Thanks for the clarification. It is not clear to me where the $H_y$ comes from. I would appreciate it if you would post an answer explaining all of this. $\endgroup$ Apr 24, 2022 at 23:54
  • $\begingroup$ It appears that you are confused about the true meaning of the curl. It is a differential operator, acting on a vector field, producing another vector field. Once you have substituted the components of the initial vector field (in this case the electric field), all you have to do is perform the differentiations. In this case it is very simple, because you only have constants and phase factors. $\endgroup$
    – M. Wind
    Apr 25, 2022 at 13:36

1 Answer 1

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We know that the only component of $\mathbf{E}$ is along the $\mathbf{e_{x}}$

Therefore $$ \nabla \times \mathbf{E} = \partial_zE_x \mathbf{e_{y}} - \partial_yE_x \mathbf{e_{z}} = \partial_zE_x \mathbf{e_{y}} $$ We also have $E_x = f(z)$ Thus the only component for $\mathbf{H}$ is along the $\mathbf{e_{y}}$ or $H_y$

We have $$ \nabla \times \mathbf{E} = -j\omega \mu \mathbf{H} $$ Which given $$ \mathbf{E} = E_x(z)\mathbf{e_{x}} $$ we have $$ \frac{\partial E_x}{\partial z}\mathbf{e_{y}} = - j \omega \mu H_y\mathbf{e_{y}} $$

We have $$ H_y = \frac{-1}{j\omega \mu}\frac{\partial E_x}{\partial z} = \frac{j}{\omega \mu}\frac{\partial E_x}{\partial z} $$ If we apply $$ E_x = E^{+}\mathrm{e}^{-jkz} + E^{-}\mathrm{e}^{jkz}\\ \partial_z E_x = -jk E^{+}\mathrm{e}^{-jkz} + jkE^{-}\mathrm{e}^{jkz} = - jk \left[E^{+}\mathrm{e}^{-jkz} - E^{-}\mathrm{e}^{jkz}\right] $$ we then have $$ H_y =- \frac{j}{\omega \mu}\cdot jk \left[E^{+}\mathrm{e}^{-jkz} - E^{-}\mathrm{e}^{jkz}\right]\\ H_y =\frac{k}{\omega \mu} \left[E^{+}\mathrm{e}^{-jkz} - E^{-}\mathrm{e}^{jkz}\right]\\ H_y =\frac{1}{\eta} \left[E^{+}\mathrm{e}^{-jkz} - E^{-}\mathrm{e}^{jkz}\right] $$

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  • $\begingroup$ This doesn't explain anything beyond $$\nabla \times \mathbf{E} = \partial_zE_x \mathbf{e_{y}} - \partial_yE_x \mathbf{e_{z}} = \partial_zE_x \mathbf{e_{y}}.$$ My question was about $H_x = H_z = 0$ and $H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz})$. $\endgroup$ Apr 24, 2022 at 17:27
  • $\begingroup$ It does - since the only component of $\mathbf{H}$ is $H_y$ E.g $H_x = H_y=0$ Also this the fact that is is only dependent on $z$ gives the derivative form. $\endgroup$
    – Chinny84
    Apr 24, 2022 at 22:01
  • $\begingroup$ You are making claims ("since the only component of $\mathbf{H}$ is $H_y$ ...") without explanation. $\endgroup$ Apr 24, 2022 at 23:29
  • $\begingroup$ You have the form $\mathbf{E}$ which leads to the $H_x = H_z = 0$, in general this is not true clearly. However, based on the form of the electric field you get the above. I am not sure what else you need to know. It is not claims, but following the model outlined in eq. 1.44 $\endgroup$
    – Chinny84
    Apr 25, 2022 at 7:24
  • $\begingroup$ @ThePointer I have updated the answer, but I would suggest learning some vector manipulation and algebra. It is pretty clear the derivation. $\endgroup$
    – Chinny84
    Apr 25, 2022 at 8:48

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