Why aren’t the electric and magnetic components of an EM wave complementary?

Question

Every visualization of an electromagnetic wave is essentially some variation of this picture:

In every one of these graphs, both the electric and magnetic components are shown as being sine waves with the same phase angle. However, it is just as frequently stated that the amplitude of a magnetic field is related to the derivative of the electric field at any given point in space and/or time. Shouldn’t it follow, then, that the magnitude of any given point along the magnetic wave should peak when the corresponding point along the electric wave is at 0? In other words, if the electric component of an EM wave were a sine wave (with no phase angle), shouldn’t the magnetic component be a cosine wave (also with no phase angle)?

Answer 1

Maxwell's equations relate spacial derivatives with time derivatives, e.g. $$\nabla\times\mathbf{B}=\mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$$ So your assertion

However, it is just as frequently stated that the amplitude of a magnetic field is related to the derivative of the electric field at any given point in space and/or time

is false.

Answer 2

Equations of the form below are solutions of the wave equation:

$\mathbf E = \mathbf E(u)$, and $\mathbf B = \mathbf B(u)$, where $u = \mathbf {k.x} – \omega t + \theta$, $\omega = |\mathbf k|c$, where $\mathbf k$ is a constant vector and $\theta$ is a constant.

The expression for $u$ means: $u = k_xx + k_yy + k_zz - ωt + θ$ and $|\mathbf k| = (k_x^2 + k_y^2 + k_z^2)^{1/2}$

They are called plane waves because for a given value of $t$ and $u$, the expression above for $u$ is an equation of a plane normal to $\mathbf k$.

Using that general equation for a plane wave, and doing the derivatives, it is possible to prove:

$$\nabla \times \mathbf E = \left(\frac{1}{\omega}\right)\left(\mathbf k \times \frac{\partial \mathbf E}{\partial t}\right) = \left(\frac{1}{\omega}\right)\frac {\partial (\mathbf k \times \mathbf E)}{\partial t}$$

According to one of the Maxwell equations: $$\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}$$

It follows that: $$ -\left(\frac{1}{\omega}\right) \mathbf k \times \mathbf E = \mathbf B$$

We conclude that $\mathbf B$ is perpendicular to $\mathbf k$ and $\mathbf E$. But also that, as $\mathbf k$ is a constant vector, and has a constant angle with $\mathbf E$, that $|\mathbf E|$ is always proportional to $|\mathbf B|$.

If we suppose $\mathbf E = \mathbf E(u)$, and $\mathbf B = \mathbf B(u+\alpha)$, the only way to get that is making $\alpha = 0$.

Answer 3

For a traveling plane wave E and B are in phase. For a standing plane wave they are 90 degrees out of phase. This follows from the Maxwell equations.

Answer 4

after some research it seems difficult for me to give a comprehensive answer. But here are the key notes:

• The displayed pictures, where $E$ and $B$ are in phase are correct for the $\textbf{far field}$ --> So look up near and far field

• The $\textbf{curl}$ of $E$ and $B$ are related to the time derivatives of $B$ and $E$, respectively. So that involves also a derivative in space.

Here the math that might interest you (from https://de.wikipedia.org/wiki/Elektromagnetische_Welle):

\begin{align} \vec{E}& = \vec{E}_0 f(\vec{k}\vec{x}-ct) \\ \vec{\nabla} \times \vec{E} = \vec{k} \times \vec{E}_0 \frac{\partial f(\vec{k}\vec{x}-ct) }{\partial (\vec{k}\vec{x})}& = -\frac{\partial\vec{B}}{\partial t} \\ \text{with} \qquad \frac{\partial f(\vec{k}\vec{x}-ct) }{\partial (\vec{k}\vec{x})} &= - \frac{\partial f(\vec{k}\vec{x}-ct) }{\partial (ct)} \\ \text{follows} \qquad \vec{B} &= \frac{1}{c} \vec{k} \times \vec{E} \end{align}