- #1

fog37

- 1,568

- 108

Hello,

For a linear medium, the two curl Maxwell's equations

$$\nabla \times \bf{E} = - \frac {\partial \bf{B}} {\partial t}$$

$$\nabla \times \bf{H} = \frac {\partial \bf{D}} {\partial t}$$

change to

$$\nabla \times \bf{E} = i \omega \bf{B}$$

$$\nabla \times \bf{H} = - i \omega\bf{D}$$

whose all solutions are space-dependent vector functions of the form ##\tilde{\bf{E}} (x,y,z) e^{-i \omega t}##. The last two equations actually have the complex-functions ## \tilde{\bf{E}} (x,y,z)## as solutions.

For a linear medium, the two curl Maxwell's equations

$$\nabla \times \bf{E} = - \frac {\partial \bf{B}} {\partial t}$$

$$\nabla \times \bf{H} = \frac {\partial \bf{D}} {\partial t}$$

change to

$$\nabla \times \bf{E} = i \omega \bf{B}$$

$$\nabla \times \bf{H} = - i \omega\bf{D}$$

whose all solutions are space-dependent vector functions of the form ##\tilde{\bf{E}} (x,y,z) e^{-i \omega t}##. The last two equations actually have the complex-functions ## \tilde{\bf{E}} (x,y,z)## as solutions.

- Is it correct to assume that if the field solution ##\tilde{\bf{E}} (x,y,z)## is real-valued (zero imaginary part), the overall solution field ##\tilde{\bf{E}} (x,y,z) e^{-i \omega t}## would then represent a standing wave while if ## \tilde{\bf{E}} (x,y,z)## has both nonzero ##Re## and ##Im## parts the field solution would be a traveling wave?
- The spatial complex-valued function ## \tilde{\bf{E}} (x,y,z)## is truly the three scalar functions: $$ \tilde{E}_{x} (x,y,z)$$ $$\tilde{E}_{y} (x,y,z)$$ $$ \tilde{E}_{z} (x,y,z)$$ Is it possible for any of these scalar functions to be complex-valued while the others are real-valued? What kind of field would that be? Hybrid, i.e. traveling and stationary at the same time?

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