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There is a Fourier transform that I don't really understand in my textbook (p.218).

I have the following equation: $$\ddot{\delta} + 2H\dot{\delta} -\frac{3}{2} \Omega_m H^2 \delta = 0 $$

Then using the Fourier transform

$$\delta_{\vec{k}} = \frac{1}{V} \int \delta(\vec{r}) e^{i \vec{k} \cdot \vec{r}} d^3 r $$

Where $\delta(\vec{r})$ is the density fluctuation.

We get: $$\ddot{\delta_{\vec{k}}} + 2H\dot{\delta_{\vec{k}}} -\frac{3}{2} \Omega_m H^2 \delta_{\vec{k}} = 0 $$

The only function that does that is a gaussian function, I guess. I don't understand the process here.

Thank you

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    $\begingroup$ What happens when you multiply both sides of the first equation by $e^{i\vec k\cdot\vec r}$ and then integrate both sides over $d^3\vec r$? $\endgroup$
    – Sten
    Commented Mar 19 at 16:23
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    $\begingroup$ From a mathematical viewpoint, there are lots of (very nice, also...) functions whose Fourier transforms are themselves, in addition to (normalized) Gaussians. I'm wondering whether there might be a typo in your source, because Fourier transform converts derivatives $d/dx$ into multiplication by $ix$ (with normalization constants). Can you clarify? $\endgroup$ Commented Mar 19 at 18:14
  • $\begingroup$ @paulgarrett There are no spatial derivatives in the equation given, which is why the same equation that applies to $\delta$ applies also to its (spatial) Fourier transform. (Note that in no way is there any implication that $\delta$ and $\delta_{\vec k}$ are the same function of $\vec r$ or $\vec k$! They just have the same time dependence.) $\endgroup$
    – Sten
    Commented Mar 19 at 20:21
  • $\begingroup$ @Sten, ah, thanks, I evidently did not understand the conventions. In hindsight, I guess the "dots" are time derivatives, etc. :) $\endgroup$ Commented Mar 19 at 20:45
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    $\begingroup$ @Sten Thank you! I can't believe I didn't see it. $\endgroup$ Commented Mar 22 at 0:48

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