What, precisely, using mathematics, is meant by "electric field with only an $\hat{x}$ component"?

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"?

• "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., $\vec{E} = E_x(z) \hat{x}$. May 15, 2022 at 4:00
• @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 $$\dfrac{\partial^2{E_x \hat{x}}}{\partial{z}^2} - \gamma^2 E_x = 0$$? May 15, 2022 at 4:44

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 $$\dfrac{\partial^2{(E_x \hat{x})}}{\partial{z}^2} - \gamma^2 (E_x\hat x) = 0$$,