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

  1. Nov 14, 2016 #1

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  3. Nov 14, 2016 #2

    vanhees71

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    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
    $$|\vec{B}_{\text{SI}}|=\sqrt{\mu_0} |\vec{B}_{\text{HL}}|=\sqrt{\mu_0} |\vec{E}_{\text{HL}}|=\sqrt{\mu_0 \epsilon_0} |\vec{E}_{\text{SI}}|=\frac{1}{c} |\vec{E}_{\text{SI}}|.$$
     
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