# How to show that Electric and Magnetic fields are transverse

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1. May 21, 2015

### leonardthecow

1. The problem statement, all variables and given/known data
This isn't necessarily a problem, but a question I have about a certain step taken in showing that the electric and magnetic fields are transverse.

In Jackson, Griffiths, and my professor's written notes, each claims the following. Considering plane wave solutions of the form $$\textbf{E}(\vec{x}, t) = Re[\vec{E_0}e^{-i(\vec{k} \cdot \vec{x} - \omega t)}] \\ \textbf{B}(\vec{x}, t) = Re[\vec{B_0}e^{-i(\vec{k} \cdot \vec{x} - \omega t)}]$$ since the Maxwell equations demand that the divergences of both E and B are zero, this in turn demands that $$\vec{k} \cdot \textbf{E} = 0 \\ \vec{k} \cdot \textbf{B} = 0.$$

2. Relevant equations

See above, plus the fact that $\vec{E_0}$ and $\vec{B_0}$ are complex functions.

3. The attempt at a solution

This has to just be my missing something stupid; I just don't see how the plane wave solutions and the Maxwell equations imply that condition (where the wave vector dotted into the E and B fields is zero). Even doing the divergence out for, say, the x component of the E field, you would have something like $$(\nabla \cdot \textbf{E})_x = \partial_x ({E_0}_xe^{-i(k_x x - \omega t)}) = \partial_x {E_0}_x - ik_x {E_0}_xe^{-i(k_x x - \omega t)}$$ which, combined with the other components would give you $$\nabla \cdot \vec{E_0} - i\vec{k} \cdot \textbf{E} = 0$$ which clearly isn't what any of the textbooks are saying is the case. Is it just that the divergence of the complex function $\vec{E_0}$ is zero? If so, why is that the case? Where am I going wrong here? Thanks!

2. May 21, 2015

### BvU

$\vec E_0$ is an amplitude, a constant for the plane waves you describe.

3. May 21, 2015

### leonardthecow

Ah okay, I buy that, thanks! Related question though; $\vec{E_0}$ is defined as $$\vec{E_0}=\textbf{A}_1 + i\textbf{A}_2,$$ where $\textbf{A}_2$ and $\textbf{A}_2$ are in $\mathbb{R}^3$. In a later proof, my professor makes the claim that $$\vec{k} \cdot \textbf{A}_1 = \vec{k} \cdot \textbf{A}_2 = 0.$$ Now, just by simple substitution into $\vec{k} \cdot \vec{E_0} = 0$ would this not imply only that $\vec{k} \cdot \textbf{A}_1 = - \vec{k} \cdot \textbf{A}_2$? I don't see why we would assume that both dot products are individually zero.

4. May 21, 2015

### BvU

Well, if $\vec k$ is real, then $\vec{k} \cdot \vec{E_0} = \vec{k} \cdot ( \textbf{A}_1 + i\textbf{A}_2) = 0 + i 0$ implies $\vec{k} \cdot \textbf{A}_1 = \vec{k} \cdot \textbf{A}_2 = 0$ and you are in business. Is $\vec k_0$ real ? why (or: why not) ?