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In D.K.Cheng's equation 8-29 he makes the following correlation between the magnetic field intensity $\mathbf{H}$ and the electric field intensity $\mathbf{E}$ in an electromagnetic wave.

enter image description here

Where $\eta$ is the instrinsic impedance. So if I know the electric field $\mathbf{E}$, I can also find $\mathbf{H}$.

However, I'm confused how you choose a proper normal vector $\mathbf{a_n}$ when doing the cross-product in the formula.

Example

Let's say I have an electric field $\mathbf{E}=C(\mathbf{a_x}+j\mathbf{a_y}) \cdot e^{-jk_0z}$,

where $C$ is a constant, $j$ is the imaginary unit and $k_0$ is the wavenumber. Since $\mathbf{E}$ has an $x$-component and a $y$-component, a normal vector to $\mathbf{E}$ must then be $\mathbf{a_z}$, since the dot-product now results in $0$.

$$\mathbf{H} = \frac{C}{\eta} \Bigg( \begin{bmatrix}0 \\0 \\ 1 \end{bmatrix} \times \begin{bmatrix} 1 \\j\\0\end{bmatrix} \Bigg)=\frac{C}{\eta} (-j\mathbf{a_x}+\mathbf{a_y}) \cdot e^{-jk_0z}$$

And we get the following $\mathbf{H}$-field. However! another normal vector to $\mathbf{E}$ is also $-\mathbf{a_z}$, because the dot-product still results in $0$. Calculating the $\mathbf{H}$-field now gives.

$$\mathbf{H} = \frac{C}{\eta} \Bigg( \begin{bmatrix}0 \\0 \\ -1 \end{bmatrix} \times \begin{bmatrix} 1 \\j\\0\end{bmatrix} \Bigg)=\frac{C}{\eta} (j\mathbf{a_x}-\mathbf{a_y}) \cdot e^{-jk_0z}$$

As seen, we get two different $\mathbf{H}$-fields. I get they are parallel but they have opposite direction.

My question

So how do I choose the correct normal vector?

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  • $\begingroup$ en.wikipedia.org/wiki/Normal_(geometry)#Choice_of_normal $\endgroup$
    – Brick
    Commented Dec 1, 2020 at 17:48
  • $\begingroup$ Thank you @ChemicalChems ! Now I know what vector to use. $\endgroup$
    – Carl
    Commented Dec 1, 2020 at 19:15
  • $\begingroup$ @Carl I posted the comment as an answer, as after I was finished writing it, I understood it was pretty much all the information needed to solve your doubts. Consider accepting to close the question. $\endgroup$
    – ChemiCalChems
    Commented Dec 1, 2020 at 19:19

1 Answer 1

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Not only is $\mathbf{a_n}$ a normal vector, it's the direction of travel of the electromagnetic wave, which is in the direction of $\mathbf{E}\times\mathbf{H}$. This is the appropriate choice of your vector.

In fact, I would argue thinking about this vector as simply a "normal vector" is misleading, as it is fundamental to know in which direction the wave travels to understand the relative direction of the electric and magnetic fields.

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  • $\begingroup$ This logic is ultimately circular as $E \times H$ is also a cross product and the bigger issue is about consistency in how that is defined across all of the places it occurs including the cases here, curls in Maxwell's equations, etc. $\endgroup$
    – Brick
    Commented Dec 1, 2020 at 19:57
  • $\begingroup$ @Brick I don't think this question was debating why the scientific community uses the right hand rule for cross products as a general convention, I would argue it was asking which of the two vectors normal to the $\mathbf{E}-\mathbf{H}$ plane was to be chosen. Given the right hand rule is chosen as a convention, and the Maxwell equations taking the form they take using this convention, the result follows. $\endgroup$
    – ChemiCalChems
    Commented Dec 1, 2020 at 20:16
  • $\begingroup$ The right hand rule specifies which normal to take, and using that rules consistently leads to a consistent treatment throughout. Your answer implies that the physically-measurable direction of travel of the wave specifies the positive direction of the normal, and that is not true. $\endgroup$
    – Brick
    Commented Dec 1, 2020 at 20:21
  • $\begingroup$ @Brick I very clearly stated in my answer that thinking about this vector as simply a normal to the $\mathbf{E}-\mathbf{H}$ plane is misleading. This vector is the unit vector in the direction of the wave's propagation, and from Maxwell's equations, it is seen that electric and magnetic fields oscillate perpendicularly to the wave's propagation, i.e. electromagnetic waves are transverse waves. Thinking of this vector as the normal to a plane is less fundamental that thinking of the fields as oscillating perpendicularly to the wave's direction of propagation. $\endgroup$
    – ChemiCalChems
    Commented Dec 1, 2020 at 20:28
  • $\begingroup$ That's exactly what makes your answer, in my understanding of it, wrong. But I'll let it go at that. $\endgroup$
    – Brick
    Commented Dec 1, 2020 at 20:31

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