# I How to write the relationship of B and E in vector form

1. Nov 14, 2016

### garylau

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2. Nov 14, 2016

### vanhees71

These are obviously the plane-wave modes of the electromagnetic field. You start, of course, from the Maxwell equations with vanishing charges and currents (in Heaviside-Lorentz units),
$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=0.$$
You can use some manipulations to see that $\vec{E}$ and $\vec{B}$ obey the homogeneous wave equation for waves with phase velocity $c$. From this to get the plane-wave modes you make the ansatz
$$\vec{E}=\vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}), \quad \omega(\vec{k})=c|\vec{k}|.$$
From the last Maxwell equation you get
$$\vec{\nabla} \cdot \vec{E} =\mathrm{i} \vec{k} \cdot \vec{E}_0 \exp(\cdots)=0, \; \Rightarrow \; \vec{k} \cdot \vec{E}_0=0,$$
i.e., the electric field is a transverse wave.

The first equation gives
$$\partial_t \vec{B}=-\mathrm{i} c \vec{k} \times \vec{E}_0 \exp(\cdots).$$
Integrating gives
$$\vec{B}=\frac{c \vec{k}}{\omega} \times \vec{E}_0 \exp(\cdots)=\hat{k} \times \vec{E}_0 \exp(\cdots)=\vec{B}_0 \exp(\cdots).$$
It's easy to check that this also solves the 2nd and the 3rd Maxwell equation. This shows that $\vec{E}_0$, $\vec{B}_0$, and $\vec{k}$ form a righthanded orthogonal set of vectors and $|\vec{E}_0|=|\vec{B}_0|$.

The additional factor of $1/c$ in your figure comes most likely from the use of SI units. Indeed we have
$$|\vec{B}_{\text{SI}}|=\sqrt{\mu_0} |\vec{B}_{\text{HL}}|=\sqrt{\mu_0} |\vec{E}_{\text{HL}}|=\sqrt{\mu_0 \epsilon_0} |\vec{E}_{\text{SI}}|=\frac{1}{c} |\vec{E}_{\text{SI}}|.$$