<p>I'll assume <span class="math-container">$\mathcal{P}(k)$</span> is the &quot;dimensionless&quot; power spectrum (also known as <span class="math-container">$\Delta^2(k)$</span>), i.e. <span class="math-container">$\mathcal{P}(k)\equiv k^3/(2\pi^2) P(k)$</span>. The density variance is
<span class="math-container">$$\sigma^2=\int_0^\infty \frac{\mathrm{d}k}{k}\mathcal{P}(k).\tag{1}\label{1}$$</span></p>
<p>&quot;How much of <span class="math-container">$\sigma^2$</span> is contributed by scales around <span class="math-container">$k$</span>&quot; is ambiguous, but the most natural interpretation (to me) is &quot;how much of <span class="math-container">$\sigma^2$</span> is contributed by a logarithmic interval around <span class="math-container">$k$</span>&quot;, in which case the answer is just (proportional to) the integrand, <span class="math-container">$\mathcal{P}(k)$</span>. To see this explicitly, we can define
<span class="math-container">$$\sigma^2(k)=\int_0^k\frac{\mathrm{d}k^\prime}{k^\prime}\mathcal{P}(k^\prime)\tag{2}\label{2}$$</span>
as the variance contributed by wavenumbers smaller than <span class="math-container">$k$</span> (i.e. scales larger than <span class="math-container">$k$</span>). Then the differential variance contributed by scales near <span class="math-container">$k$</span>, per logarithmic interval in <span class="math-container">$k$</span>, is
<span class="math-container">$$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma^2(k)=\mathcal{P}(k).\tag{3}\label{3}$$</span></p>
<p>Now, you are not asking about <span class="math-container">$\sigma^2$</span> but about <span class="math-container">$\sigma$</span>. In this case there is further ambiguity to the question, because <span class="math-container">$\sigma$</span> is no longer simply a sum over the power contributed at different scales and is instead a nonlinear function thereof. In general, the answer to &quot;how much rms variance is contributed by scales near <span class="math-container">$k$</span>&quot; depends on the order in which you want to add up the power at different scales. If we add up power in order from the largest scales (smallest <span class="math-container">$k$</span>) to the smallest scales (largest <span class="math-container">$k$</span>), as in equation \eqref{2}, then the answer is
<span class="math-container">$$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma(k)
=\frac{\mathrm{d}}{\mathrm{d}\log k}[\sigma^2(k)]^{1/2}
=\frac{1}{2}[\sigma^2(k)]^{-1/2}\frac{\mathrm{d}\sigma^2(k)}{\mathrm{d}\log k}
=\frac{\mathcal{P}(k)}{2\sigma(k)}.\tag{4}\label{4}$$</span>
To add up the scales in a different order, just reorder the integral in equation \eqref{2}; equation \eqref{4} remains accurate with the redefined <span class="math-container">$\sigma(k)$</span>.</p>

CC BY-SA 4.0
2023-03-14T20:06:27.963Z
2023-03-14T20:06:27.963Z
Sten
<p>Imagine a Universe in which the matter power spectrum behaves as
<span class="math-container">$$\mathcal{P}\sim k^{0.8}$$</span></p>
<p>How much of the variance (not variance squared!) of matter is due to scales around <span class="math-container">$k=k_1=10^{2.9}\mathrm{Mpc}^{-1}$</span>,
versus scales around <span class="math-container">$k=k_2=10^{-3.0}\mathrm{Mpc}^{-1}$</span></p>
<p>To be explicit, how much is:
<span class="math-container">$$\frac{\sigma(k\sim k_1)}{\sigma(k\sim k_2)}?$$</span></p>

CC BY-SA 4.0
2023-03-14T17:07:17.580Z
I'd recommend formatting this with latex, currently, with images cutting through the text, it looks very confusing..
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-03-14T18:08:18.923Z
Please explain some of the magic numbers here.  Why 0.8?  Why 2.9?  Why -3.0?
James K
CC BY-SA 4.0
2023-03-14T18:38:40.940Z
They are random values assigned, nothing theoretical just to obtain certain value.
Alba
<universe>
How much of the variance of matter is due to scales
Alba
# How much of the variance of matter is due to scales

(Asked by: username_0 on 2023-03-14)

Imagine a Universe in which the matter power spectrum behaves as
$$\mathcal{P}\sim k^{0.8}$$

How much of the variance (not variance squared!) of matter is due to scales around $k=k\_1=10^{2.9}\mathrm{Mpc}^{-1}$,
versus scales around $k=k\_2=10^{-3.0}\mathrm{Mpc}^{-1}$

To be explicit, how much is:
$$\frac{\sigma(k\sim k\_1)}{\sigma(k\sim k\_2)}?$$

**Tags:** universe

> **Comments:**
> **username_1**: I'd recommend formatting this with latex, currently, with images cutting through the text, it looks very confusing..  
> **username_2**: Please explain some of the magic numbers here.  Why 0.8?  Why 2.9?  Why -3.0?  
> **username_0**: They are random values assigned, nothing theoretical just to obtain certain value.  

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## 1 answers

### Answer by username_3 on 2023-03-14. Score: 4

I'll assume $\mathcal{P}(k)$ is the "dimensionless" power spectrum (also known as $\Delta^2(k)$), i.e. $\mathcal{P}(k)\equiv k^3/(2\pi^2) P(k)$. The density variance is
$$\sigma^2=\int\_0^\infty \frac{\mathrm{d}k}{k}\mathcal{P}(k).\tag{1}\label{1}$$

"How much of $\sigma^2$ is contributed by scales around $k$" is ambiguous, but the most natural interpretation (to me) is "how much of $\sigma^2$ is contributed by a logarithmic interval around $k$", in which case the answer is just (proportional to) the integrand, $\mathcal{P}(k)$. To see this explicitly, we can define
$$\sigma^2(k)=\int\_0^k\frac{\mathrm{d}k^\prime}{k^\prime}\mathcal{P}(k^\prime)\tag{2}\label{2}$$
as the variance contributed by wavenumbers smaller than $k$ (i.e. scales larger than $k$). Then the differential variance contributed by scales near $k$, per logarithmic interval in $k$, is
$$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma^2(k)=\mathcal{P}(k).\tag{3}\label{3}$$

Now, you are not asking about $\sigma^2$ but about $\sigma$. In this case there is further ambiguity to the question, because $\sigma$ is no longer simply a sum over the power contributed at different scales and is instead a nonlinear function thereof. In general, the answer to "how much rms variance is contributed by scales near $k$" depends on the order in which you want to add up the power at different scales. If we add up power in order from the largest scales (smallest $k$) to the smallest scales (largest $k$), as in equation \eqref{2}, then the answer is
$$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma(k)
=\frac{\mathrm{d}}{\mathrm{d}\log k}[\sigma^2(k)]^{1/2}
=\frac{1}{2}[\sigma^2(k)]^{-1/2}\frac{\mathrm{d}\sigma^2(k)}{\mathrm{d}\log k}
=\frac{\mathcal{P}(k)}{2\sigma(k)}.\tag{4}\label{4}$$
To add up the scales in a different order, just reorder the integral in equation \eqref{2}; equation \eqref{4} remains accurate with the redefined $\sigma(k)$.


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