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: $$\frac{\sigma(k\sim k_1)}{\sigma(k\sim k_2)}?$$

  • 1
    $\begingroup$ I'd recommend formatting this with latex, currently, with images cutting through the text, it looks very confusing.. $\endgroup$ Mar 14 at 17:07
  • $\begingroup$ Please explain some of the magic numbers here. Why 0.8? Why 2.9? Why -3.0? $\endgroup$
    – James K
    Mar 14 at 18:08
  • $\begingroup$ They are random values assigned, nothing theoretical just to obtain certain value. $\endgroup$
    – Alba
    Mar 14 at 18:38

1 Answer 1


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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