Electromagnetic plane waves in vacuo

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  • #1
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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?
 

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  • #2
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Well, mathematically,

[tex] \bigtriangledown \cdot E_0 \cos(kz - wt) = E_0 \sin(kz - wt) = 0 [/tex]

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

EDIT: But yes, addressed below is the conceptual issue here...
 
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  • #3
nicksauce
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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
[tex]\vec{E} = \vec{E_0}cos(\vec{k}\cdot\vec{r} - \omega t})[/tex]
Then,
[tex]\nabla\cdot\vec{E} = \vec{k}\cdot\vec{E} = 0[/tex]

Which shows that the wave vector is perpendicular to the field.
 
  • #4
801
70
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.
 

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