Consider an arbritrary well-behaving function $$f: (\mathbb{R},\mathbb{R}) \rightarrow \mathbb{C}, \ \ \ (x, t) \mapsto f(x, t)$$ and it's Fourier transform, $\mathscr{F}\left\{f(x,t)\right\} = \tilde{f}(x,\omega)$. Now consider a time shift in the spatial coordinate as $f(x+vt, t)$. I'm looking for a way to describe this shift in the spectral domain, i.e. $$\mathscr{F}\left\{\tilde{f'}(x, \omega)\right\} = f(x+vt, t)$$ I couldn't really come up with a fruitful search term for this. My problem is that if one writes the Fourier transform, $$ \mathscr{F}\left\{\tilde{f'}(x, \omega)\right\} = \int_{-\infty}^{\infty} f(x+vt, t)\ e^{-i \omega t}\,dt, $$ there seems no obvious identity or coordinate transformation that lets escape integration in the first variable. Moving the transformation to the 2nd variable by $u = x+vt$, thus $t = (u-x)/v$ doesn't help, for example, as the integration still needs to be evaluated in both arguments: $dt = du/v$. I'm not sure it's even possible to determine $\tilde{f}(x,\omega)$ in such a general case, as the first argument is already involved without related information; but on the other hand, this is just a simple linear coordinate transform which is easy to manage in most cases. I would be grateful if you could point me to the right direction.
- If it's possible, what is the identity, and how can it be derived?
- If not, why? What minimum level of specification is necessary to determine $\tilde{f'}(x, \omega)$, even if not uniquely?
My goal is to find how should I transform $\tilde{f}(x,\omega)$ to get the desired coordinate shift. If it helps finding the solution, the transform would be done via Fast Fourier Transform.