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We know that (1) the transmission of light is defined by the wave propagation of the electromagnetic field, and that (2), in the absence of charges and currents, the magnetic field is always perpendicular to the electric field. Given this, I am told that the electric field vector $\mathbf{E}$ of length $E$ defines the wave propagation of light

$$\dfrac{\partial^2{\mathbf{E}}}{\partial{t}^2} = c^2_n \dfrac{d^2 \mathbf{E}}{dx^2},$$

where $c_n = \dfrac{c}{n}$ is the speed of light altered depending on the refractive index $n$.

Where did $\dfrac{\partial^2{\mathbf{E}}}{\partial{t}^2} = c^2_n \dfrac{d^2 \mathbf{E}}{dx^2}$ come from, and what does it represent? This equation was presented, but its significance is not clear to me.

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  • $\begingroup$ @Brick I don't see any mention of the equation $\dfrac{\partial^2{\mathbf{E}}}{\partial{t}^2} = c^2_n \dfrac{d^2 \mathbf{E}}{dx^2}$. Which part are you specifically referring to? $\endgroup$ May 18, 2021 at 14:03
  • $\begingroup$ It's equation (1) in the question that I linked. They just used vector notation, which is more general. $\endgroup$
    – Brick
    May 18, 2021 at 14:05
  • $\begingroup$ @Brick Oh, ok, that makes sense. Thanks for that! $\endgroup$ May 18, 2021 at 14:06
  • $\begingroup$ Also, your equations are wrong. The right-hand side in all places should be $\partial^2 \mathbf{E}/\partial x^2$ in what you wrote. That full derivative on the RHS is wrong. I took that to be a typo, but maybe it reflects a conceptual mistake. $\endgroup$
    – Brick
    May 18, 2021 at 14:06
  • $\begingroup$ @Brick Ohh, you mean the partial derivatives? Yes, now that you mention it, that does seem like a typo by the author of the notes. Thanks again. $\endgroup$ May 18, 2021 at 14:08

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