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Longitudinal and transversal parts

Helmholtz' theorem: \bgroup\color{col1}$ \mathbf{E}$\egroup and \bgroup\color{col1}$ \mathbf{B}$\egroup can be decomposed into
$\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \mathbf{E}_\Vert + \mathbf{E}_\perp,\quad \mathbf{B}= \mathbf{B}_\Vert + \mathbf{B}_\perp$ (1.5)
$\displaystyle \mathbf{\nabla \times} \mathbf{E}_\Vert$ $\displaystyle =$ $\displaystyle \mathbf{\nabla \times} \mathbf{B}_\Vert =0,\quad \div\mathbf{E}_\perp = \div\mathbf{B}_\perp = 0.$ (1.6)

This becomes clearer in Fourier space, e.g.
$\displaystyle \hat{\mathbf{E}}({\bf k})$ $\displaystyle =$ $\displaystyle \hat{\mathbf{E}}_\Vert({\bf k}) + \hat{\mathbf{E}}_\perp({\bf k})$ (1.7)
$\displaystyle \hat{\mathbf{E}}_\Vert({\bf k})$ $\displaystyle =$ $\displaystyle [{\bf k}\cdot \hat{\mathbf{E}}({\bf k}) ]{\bf k}/k^2,\quad
\hat{\...
...}_\perp({\bf k})= [{\bf k} \times \hat{\mathbf{E}}({\bf k})]\times {\bf k}/k^2.$ (1.8)



Subsections

Tobias Brandes 2005-04-26