- #1

- 10

- 0

##

\frac{\partial{\vec{E}_t}}{\partial{z}}+i\frac{\omega}{c}\hat{e}_z\times \vec{B}_t=\vec{\nabla}_tE_z

##

- Thread starter flintbox
- Start date

- #1

- 10

- 0

##

\frac{\partial{\vec{E}_t}}{\partial{z}}+i\frac{\omega}{c}\hat{e}_z\times \vec{B}_t=\vec{\nabla}_tE_z

##

- #2

Paul Colby

Gold Member

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- #3

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Will you please clarify the notations? What does ##t## stand for?

- #4

nrqed

Science Advisor

Homework Helper

Gold Member

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It comes from taking the curl of the B field (here they are assuming a plane wave). But the equation is not completely correct, the term on the rhs does not make any sense.

##

\frac{\partial{\vec{E}_t}}{\partial{z}}+i\frac{\omega}{c}\hat{e}_z\times \vec{B}_t=\vec{\nabla}_tE_z

##

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