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Basically, my book (Modern Optics by Robert Guenther) presents the "proof" of the interdependancy of the plane waves E and B as follow:
Suppose \vec{E} is an electric plane wave:
\vec{E} = \vec{E_0}e^{i(\omega t - \vec{k}\cdot \vec{r}+ \phi)}.
Then we find that
\frac{\partial \vec{E}}{\partial t}=i\omega \vec{E}.
And if \vec{B} is a plane wave in-phase with \vec{E}, such as
\vec{B} = \vec{B_0}e^{i(\omega t - \vec{k}\cdot \vec{r}+ \phi)},
then
\vec{\nabla}\times \vec{B} = -i\vec{k}\times \vec{B}.
And thus, given \vec{E} a plane wave, \vec{B} a plane in-phase satify the Maxwell equation
\vec{\nabla}\times \vec{B} = \mu\epsilon \frac{\partial \vec{E}}{\partial t}
under the simple condition that E_0 = cB_0 but what tells me that given \vec{E} a plane wave, this the only solution? It's this little detail that bugs me.
Suppose \vec{E} is an electric plane wave:
\vec{E} = \vec{E_0}e^{i(\omega t - \vec{k}\cdot \vec{r}+ \phi)}.
Then we find that
\frac{\partial \vec{E}}{\partial t}=i\omega \vec{E}.
And if \vec{B} is a plane wave in-phase with \vec{E}, such as
\vec{B} = \vec{B_0}e^{i(\omega t - \vec{k}\cdot \vec{r}+ \phi)},
then
\vec{\nabla}\times \vec{B} = -i\vec{k}\times \vec{B}.
And thus, given \vec{E} a plane wave, \vec{B} a plane in-phase satify the Maxwell equation
\vec{\nabla}\times \vec{B} = \mu\epsilon \frac{\partial \vec{E}}{\partial t}
under the simple condition that E_0 = cB_0 but what tells me that given \vec{E} a plane wave, this the only solution? It's this little detail that bugs me.
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