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I have a monochromatic ribbon beam with $E(x)e^{i(kz-\omega t)}$ being the electric field's amplitude. I want to show that the lowest order approximation in terms of plane waves is

$\mathbf{E}(x,z,t)=\mathbf{\epsilon} \int{d\kappa A(\kappa) e^{i(\kappa x+kz-\omega t)}}$

where $\mathbf{\epsilon}$ is the polarisation direction and $A(\kappa)$ is the Fourier transform of $E(x)$ around $\kappa=0$.

From the result I can understand/identify that the Fourier kernel is $e^{i(\kappa x)}$ but usually when you use a Fourier transformation you go from $f(x)\rightarrow F(\omega)$, from one variable to another that is, but here all of a sudden you just add one new variable in the transformed field. How is that possible?

EDIT There is a new approach to this, but there is a tiny little point that I don't get. Consider the following geometry.

$d=x_1\sin{\theta},\; \sin{\theta}=\dfrac{x_0}{r_{01}}\approx \dfrac{x_0}{z},\; d\approx \dfrac{x_0x_1}{z}$

Consider the travelling wave

$e^{ikr}=e^{ik(r_{01}-d)}$

Why $r$ becomes $(r_{01}-d)$?

I beieve that $r$ describes the optical path. It feels like the two rays have an optical difference of $d$, but I am not sure if this is the answer.

Any ideas would be more that welcomed!

Solution found here

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The Fourier transform can be for any arbitrary periodic variable, not just $t$ and $\omega$. Often times people will compute the Fourier transform in space to get estimates of a wave number spectrum, which is what your problem looks like. –  honeste_vivere Oct 3 at 18:04

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