My notes (from a physics course) justifies the following equality by invoking the convolution thm:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_{-\infty}^{+\infty} \chi(\omega)\vec{E}_0(\omega)e^{-i\omega t}d\omega=\int_{-\infty}^{+\infty} \chi(\tau)\vec{E}_0(\tau-t)d\tau[/tex]

From a mathematical standpoint (i.e. without reference to what the functions chi and E are), is the above some consequence of the convolution thm? The convolution thm says that [itex]\mathcal{F}(f*g)=\mathcal{F}(f)\cdot\mathcal{F}(g)[/itex] I dont see how this applies here. For one, there are two integrals mutilplied by one another on the RHS of the convolution thm. And for another, the RHS of my equation is simply the convolution of chi and E; not its fourier transform or anything. Anyone sees?

And what if I were to add that

[tex]\vec{E}_0(\omega)=\int_{-\infty}^{+\infty} \vec{E}(T)e^{i\omega T}dT[/tex] ?

Notice the 0 subscript on the LHS but not in the integral; the difference is meant to represent two dinscting functions [of course one is just the "fourier coefficients" of the other]

Final hypothesis: perhaps my professor had another "convolution thm" in mind when he wrote that?

P.S. For clarity, I have not kept track of all the 2pi's popping up now and then in the above equations.

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# Convolution and fourier transform puzzle

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