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Consider a solution to the wave equation $ \psi\left(x,t\right) $, then using Fourier transform, we can represent:

$ \psi\left(x,t\right)=\left(\frac{1}{2\pi}\right)^{2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\widetilde{\psi}\left(k,\omega\right)e^{i\left(kx+wt\right)}dkdw $

Now if we'll apply this form into the wave equation $ \frac{\partial^{2}\psi}{\partial x^{2}}-\frac{1}{c^{2}}\frac{\partial^{2}\psi}{\partial t^{2}}=0 $

We'll get:

$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\widetilde{\psi}\left(k,\omega\right)e^{i\left(kx+wt\right)}\left(-k^{2}+\frac{\omega^{2}}{c^{2}}\right)dkdw=0 $

Now according to my book, this obligates the term $ \left(-k^{2}+\frac{\omega^{2}}{c^{2}}\right) $ to be $0$.

I really cant understand why. My knowledge in Fourier transform is very low, we've just learned maybe $ 2 $ hours just getting familier with the equations and applying it to some basic physics exercise. But I want to understand in a more profound way.

As I understand (with my basic knowledge of just year of math learnings), taking a fourier transform is equvivalent to representing a vector in a vector space using orthogonal basis. Which means $ e^{i\left(kx+wt\right)} $ those are forming the orthogonal vector basis and the inner product is probably the integrals $ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} d\omega dk$.

So why does $ \left(-k^{2}+\frac{\omega^{2}}{c^{2}}\right) $ has to be $ 0 $ in order for the equation to make sense? as I see it maybe the term $ \widetilde{\psi}\left(k,\omega\right) $ can also cause everything to be $ 0 $.

If someone can explain in a simple way which dosent require profound understanding of the math behind the scenes, I'll be greatful. Thanks

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The Fourier transform is invertible. A basic requirement of invertibility is that the transform of something is zero if an only if that something is zero.

So if the integral you give is to be zero, then $$ \tilde \Psi(k,\omega)(\omega^2-c^2k^2) $$ has to be zero for every $\omega$ and $k$. Thus either $\tilde \Psi$ is identically zero or $(\omega^2-c^2k^2)$ is zero.

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  • $\begingroup$ I see, that makes sense. thanks ! $\endgroup$
    – FreeZe
    Oct 27 '20 at 12:20

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