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 $$\dfrac{\partial^2{E_x}}{\partial{z}^2} + k^2 E_x = 0. \tag{1.44}$$

It then later says the following:

Plane Waves in a General Lossy Medium
Now consider the effect of a lossy medium. If the medium is conductive, with a conductivity $\sigma$, Maxwell's curl equations can be written, from (1.41a) and (1.20) as $$\nabla \times \bar{E} = - j \omega \mu \bar{H}, \tag{1.50a}$$ $$\nabla \times \bar{H} = j \omega \epsilon \bar{E} + \sigma \bar{E}. \tag{1.50b}$$ The resulting wave equation for $\bar{E}$ then becomes $$\nabla^2 \bar{E} + \omega^2 \mu \epsilon \left( 1 - j \dfrac{\sigma}{\omega \epsilon} \right) \bar{E} = 0, \tag{1.51}$$ where we see a similarity with (1.42), the wave equation for $\bar{E}$ in the lossless case. The difference is that the quantity $k^2 = \omega^2 \mu \epsilon$ of (1.42) is replaced by $\omega^2 \mu \epsilon[1 - j(\sigma/\omega \epsilon)]$ in (1.51). We then define a complex propagation constant for the medium as $$\gamma = \alpha + j \beta = j \omega \sqrt{\mu \epsilon} \sqrt{1 - j \dfrac{\sigma}{\omega \epsilon}}, \tag{1.52}$$ where $\alpha$ is the attenuation constant and $\beta$ is the phase constant. If we again assume an electric field with only an $\hat{x}$ component and uniform in $x$ and $y$, the wave equation of (1.51) reduces to $$\dfrac{\partial^2{E_x}}{\partial{z}^2} - \gamma^2 E_x = 0, \tag{1.53}$$

Notice that both these sections discuss an "electric field with only an $\hat{x}$ component" and "uniform in $x$ and $y$." This is written as $E_x$. But, beyond just being written as $E_x$, what precisely is this (mathematically)? Is this just an electric field with only an $x$-component, so that $E_x = E(x)$? But, then, if there's no $y$ component, then how does it make sense to say "uniform in $x$ and $y$"? Is the basis vector $\hat{i} = (1, 0, 0)$ included somehow, and is it somehow related to $\hat{x}$? Furthermore, a subscript often denotes the partial derivative, so is $E_x$ somehow the partial derivative of $E$ with respect to $x$? Overall, I do not think the author has phrased this well, or done enough to clarify his phrasing (using mathematics). What, precisely, using mathematics, is meant by "electric field with only an $\hat{x}$ component"?

  • $\begingroup$ "It only has an $x$ component" means that the electric field vector at each point in space only points along the $x$ direction, i.e. $\vec{E} = E_x \hat{x}$. However, the component $E_x$ (which is not indicating a derivative) can generally be a function of $x$, $y$, and $z$, i.e., $\vec{E} = E_x(x,y,z) \hat{x}$. Then, the field being uniform in $x$ and $y$ means that it doesn't depend on $x$ and $y$, i.e., $\vec{E} = E_x(z) \hat{x}$. $\endgroup$
    – march
    Commented May 15, 2022 at 4:00
  • $\begingroup$ @march Thanks for the clarification. But notice that the author excludes the $\hat{x}$, so that he only writes $E_x$, such as in $$\dfrac{\partial^2{E_x}}{\partial{z}^2} - \gamma^2 E_x = 0$$ So, if what you're saying is true, then shouldn't it be written as $$\dfrac{\partial^2{E_x \hat{x}}}{\partial{z}^2} - \gamma^2 E_x = 0$$? $\endgroup$ Commented May 15, 2022 at 4:44

1 Answer 1


A lot of three dimensional situations can be confusing.

In general an electric field might vary across all three dimensions but in this case you are told that it does not vary with position (x,y) and you are also told that there is only a x-component of the electric field.

So instead of the electric field being \begin{pmatrix} E_{\rm x}(x,y,z) \\ E_{\rm y}(x,y,z) \\ E_{\rm z}(x,y,z) \end{pmatrix}

in this example it is \begin{pmatrix} E_{\rm x}(z) \\ 0 \\ 0 \end{pmatrix}

Your equation $\dfrac{\partial^2{E_x \hat{x}}}{\partial{z}^2} - \gamma^2 E_x = 0$ is incorrect as you are subtracting a scalar $ \gamma^2 E_x$ from a vector $\dfrac{\partial^2{E_x \hat{x}}}{\partial{z}^2}$.

A correct form of the equation might be $\dfrac{\partial^2{(E_x \hat{x})}}{\partial{z}^2} - \gamma^2 (E_x\hat x) = 0$,


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