<p>There is a Fourier transform that I don't really understand in my <a href="https://www.cambridge.org/highereducation/books/introduction-to-cosmology/7E9E7C9C717570F1FFB3BA70F864A8FA#overview" rel="nofollow noreferrer">textbook</a> (p.218).</p>
<p>I have the following equation:
<span class="math-container">$$\ddot{\delta} + 2H\dot{\delta} -\frac{3}{2} \Omega_m H^2 \delta = 0
$$</span></p>
<p>Then using the Fourier transform</p>
<p><span class="math-container">$$\delta_{\vec{k}} = \frac{1}{V} \int \delta(\vec{r}) e^{i \vec{k} \cdot \vec{r}} d^3 r
$$</span></p>
<p>Where <span class="math-container">$\delta(\vec{r})$</span> is the density fluctuation.</p>
<p>We get:
<span class="math-container">$$\ddot{\delta_{\vec{k}}} + 2H\dot{\delta_{\vec{k}}} -\frac{3}{2} \Omega_m H^2 \delta_{\vec{k}} = 0
$$</span></p>
<p>The only function that does that is a gaussian function, I guess. I don't understand the process here.</p>
<p>Thank you</p>

CC BY-SA 4.0
2024-03-19T16:23:12.720Z
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 $d^3\vec r$?
Sten
CC BY-SA 4.0
2024-03-19T18:14:08.640Z
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?
paul garrett
CC BY-SA 4.0
2024-03-19T20:21:56.247Z
@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.)
Sten
CC BY-SA 4.0
2024-03-19T20:45:12.760Z
@Sten, ah, thanks, I evidently did not understand the conventions. In hindsight, I guess the "dots" are time derivatives, etc. :)
paul garrett
CC BY-SA 4.0
2024-03-22T00:48:44.753Z
@Sten Thank you! I can't believe I didn't see it.
merlinbluepickle
<cosmology><cosmological-inflation><density>
Fourier transform of the density fluctuation
merlinbluepickle
# Fourier transform of the density fluctuation

(Asked by: username_0 on 2024-03-19)

There is a Fourier transform that I don't really understand in my [textbook](https://www.cambridge.org/highereducation/books/introduction-to-cosmology/7E9E7C9C717570F1FFB3BA70F864A8FA#overview) (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

**Tags:** cosmology, cosmological-inflation, density
> **Comments:**
> **username_1**: 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 $d^3\vec r$?  
> **username_2**: 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?  
> **username_1**: @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.)  
> **username_2**: @Sten, ah, thanks, I evidently did not understand the conventions. In hindsight, I guess the "dots" are time derivatives, etc. :)  
> **username_0**: @Sten Thank you! I can't believe I didn't see it.  

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## 0 answers

