# Relation between $\textbf{E}$ and $\textbf{H}$ field for TM wave

My book (Nanoscale Energy Transport & Conversion by Gang Chen, pg. 173) considers the following scenario of a TE wave reflecting at an interface:

It says the incident $$\bf{E}$$ field is: $$\textbf{E}_{\parallel i}exp\left[-i\omega\left(t-\frac{n_1xcos\theta_i+n_1zcos\theta_i}{c_0}\right)\right]$$

and that we can obtain

$$H_i=\frac{n_1}{\mu c_0}E_{\parallel i}$$

from a Maxwell equation. How is this result obtained?

I assume we use

$$\nabla \times \textbf{E} =-\frac{\partial \textbf{B}}{\partial t}$$

but

$$\nabla \times \textbf{E} = \left( \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} \right)\hat{\textbf{y}}$$

and the the components $$E_x$$ and $$E_z$$ are unknown.

Those components are not unknown—in fact in linear media $$\mathbf E_\text i$$ is going to be perpendicular to $$\mathbf k_\text i$$ as shown in your figure.

It looks like the author has done something very strange with the actual components of their $$\mathbf k$$-vector; I would state that given the diagram you have, if the $$+x$$-direction is away from the interface and the $$+z$$-direction is otherwise in the direction of $$\mathbf k_\text i$$ as you have shown, then the right-handed $$+y$$-direction is into the plane of the figure, we have $$\mathbf k_\text i = (-\cos\theta_\text i, 0, +\sin\theta_\text i),$$ and then in somewhat more detail we have $$\mathbf E = E_0 ~\begin{bmatrix}\sin\theta_\text i\\ 0\\ \cos\theta_\text i\end{bmatrix}~\exp\left[-i\omega\left(t - \frac{-n_1 x \cos\theta_\text i + n_1 z \sin\theta_\text i}{c_0}\right)\right],$$notice the minus sign appearing by $$n_1 x$$.

Then indeed we can compute $$\nabla \times\mathbf E$$ to find $$\nabla\times\mathbf E = -i\omega \frac{n_1}{c_0} E_0\begin{bmatrix}0\\-\cos^2\theta_\text i -\sin^2\theta_\text i\\0\end{bmatrix}~\exp\left[-i\omega\left(t - \frac{-n_1 x \cos\theta_\text i + n_1 z \sin\theta_\text i}{c_0}\right)\right],$$ and of course the quantity inside the column vector there simplifies to just $$(0,-1,0).$$ The time integral effectively divides by $$-i\omega$$ leading to \begin{align}\mathbf B &= -\int dt~\nabla\times \mathbf{E}\\ &=-\frac{-i\omega}{-i\omega}\frac{E_0 n_1}{c_0} ~\begin{bmatrix}0\\-1\\0\end{bmatrix}~\exp\left[-i\omega\left(t - \frac{-n_1 x \cos\theta_\text i + n_1 z \sin\theta_\text i}{c_0}\right)\right]\\ &=\frac{E_0 n_1}{c_0} ~\begin{bmatrix}0\\1\\0\end{bmatrix}~\exp\left[-i\omega\left(t - \frac{-n_1 x \cos\theta_\text i + n_1 z \sin\theta_\text i}{c_0}\right)\right], \end{align}and then in linear media we have $$\mathbf H = \mathbf B/\mu,$$ hence $$\|\mathbf H\| = \frac{n_1}{\mu c_0}\|\mathbf E\|.$$Note that we haven't used the fact that there exists an interface yet—this is a completely general relation holding for all plane waves in linear media, not just ones that happen to be in a certain plane and happen to be heading towards an interface.

• Your B field is polarized in the negative y-direction (into the screen). Shouldn't it be pointing in the positive y-direction (out of the screen)? Commented Nov 4, 2019 at 1:07
• I'm saying this based on the figure - the B field is positive out of the screen, in the positive y-direction. Your B field seems to violate right hand rule? Commented Nov 4, 2019 at 1:34
• @Drew You are correct that the sign was wrong; I have corrected it. Commented Nov 4, 2019 at 19:56