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

I'm confused about this part:

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

If the electric field only has an $\hat{x}$ component, as stated, then how does it make sense to say that it has "uniform (no variation) in the $x$ and $y$ directions"? Shouldn't there not be anything in the $y$ direction? And how does this all reduce the Helmholtz equation to $\dfrac{\partial^2{E_x}}{\partial{z}^2} + k^2 E_x = 0$? I would appreciate an explanation with mathematics that illustrates this.

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Equation 1.44 is for (an infinite) plane wave propagating in the $z$ direction. The $\mathbf E$ vector points in the $x$ direction and is assumed to have the same value for all values of $x$ and $y$. It does vary with $z$ and $t$.

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  • $\begingroup$ Oh, ok, I see what it's saying now. $\endgroup$ Commented Apr 22, 2022 at 17:29

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