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madness
Apr15-09, 08:18 AM
Consider a plane wave f = cos(kz - wt). Applying Maxwell's equation (divE=0 in vacuo) gives
kcos(kz - wt) = 0 which means that k = 0. This surely doesn't make sense?

whybother
Apr15-09, 08:31 AM
Well, mathematically,

\bigtriangledown \cdot E_0 \cos(kz - wt) = E_0 \sin(kz - wt) = 0

\Longrightarrow k = 0 or sin(kz - wt) = 0 \forall{ z, t }

EDIT: But yes, addressed below is the conceptual issue here...

nicksauce
Apr15-09, 08:33 AM
Well the reason you get that is that EM waves are transverse waves, and you have oversimplified the situation. The more general result is:
Consider a plane wave
\vec{E} = \vec{E_0}cos(\vec{k}\cdot\vec{r} - \omega t})
Then,
\nabla\cdot\vec{E} = \vec{k}\cdot\vec{E} = 0

Which shows that the wave vector is perpendicular to the field.

madness
Apr15-09, 08:55 AM
I think I see my mistake. I thought all i was doing was choosing my coordinates such that my z axis coincided with the wave vector k. The way i did it missed out the fact that k and E point in different directions.