I'm confused after reading of a book in which the author proves mathematically that electric and magnetic fields are orthogonal to each other (for TEM mode). I'm calculating it in the same way, however something is missing in his solution. I begins from Maxwell's equations: $$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$$

Due to fact, that this is TEM mode, the electric field vector varies only with the z plane, so: $$\frac{\partial \mathbf{E}}{\partial x} = \frac{\partial \mathbf{E}}{\partial y} = 0$$

According to my calculations, the answer should be:

$$\left(\mu \frac{\partial H_x}{\partial t}-\frac{\partial E_y}{\partial z}\right)\mathbf{i} + \left(\frac{\partial E_x}{\partial z}+\mu\frac{ \partial H_y}{\partial t}\right)\mathbf{j} = 0 $$

However, in the book the answer is:

$$\left(\mu \frac{\partial H_x}{\partial t}\right)\mathbf{i} + \left(\frac{\partial E_x}{\partial z}+\mu\frac{ \partial H_y}{\partial t}\right)\mathbf{j} = 0 $$

So I'm wondering what happened to this "element" of my answer: $$\left(-\frac{\partial E_y}{\partial z}\right)\mathbf{i}$$

Is this a mistake in the book, or have I made a mistake?


1 Answer 1


If you are in a Transverse Electric Mode, only one of the components of the electric field is non-zero. Apparently the author has decided that this is the x direction, therefore $E_y = E_z = 0$. This is, of course, absolutely arbitrary, basically corresponding to choosing the orientation of your frame of reference (much in the same way as you’re choosing the $z$ direction as the direction of propagation).

Personally, I’ve always found the following coordinate-free proof to be “cleaner”. Consider the Fourier transform of your equation, if $\mathbf{k}$ is the wave vector and $\omega$ is the angular frequency, you will have that $\nabla \times \mathbf{E} = i \mathbf{k} \times \mathbf{E}$ and $\frac{\partial \mathbf{B}}{\partial t} = -i \omega \mathbf{B}$ (or with the signs switched, depends on what convention you are using). $$\mathbf{k} \times \mathbf{E} = \omega \mathbf{B}$$ From the fact that the cross product is perpendicular to both factors, you obtain that $\mathbf{E} \bot \mathbf{B}$.

  • $\begingroup$ Sorry I made mistakes many times. Corrected: The equation $\mathbf{k}\times\mathbf{E}=\omega\mathbf{B}$ does not imply transverse wave. If $\mathbf{E}_0$ such that $\nabla\times\mathbf{E}_0=\mathbf{0}$ (the form $\mathbf{E}_0=-\nabla\phi$ in mind), is oriented in the $\mathbf{k}$-direction, $\mathbf{k}\times\mathbf{E}_0=\mathbf{0}$. Thus for any solution which satisfies $\mathbf{k}\times\mathbf{E}_1=\omega\mathbf{B}_1$, the electric field $\mathbf{E}=\mathbf{E}_0+\mathbf{E}_1$ is also a solution. Such $\mathbf{E}_1+\mathbf{E}_0$ does not belong to a transverse wave. $\endgroup$
    – HEMMI
    Nov 13, 2022 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.