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I would like please to demonstrate the expression of Power spectrum in Cosmology :

First, I have the relative contrast:

$$\delta_{i}(\vec{x}, z) \equiv \rho_{i}(\vec{x}, z) / \bar{\rho}_{i}(z)-1\quad(1)$$

After, we decompose this relative contrast on Fourrier basis :

$$\delta_{i}(\vec{x}, z)=\int \frac{\mathrm{d}^{3} k}{(2 \pi)^{3}} \tilde{\delta}_{i}(\vec{k}, z) \exp (\mathrm{i} \vec{k} \cdot \vec{x})\quad(2)$$

and finally, how to find the following expression (3) from (1) and (2) :

$$\left\langle\tilde{\delta}_{i}(\vec{k}, z) \tilde{\delta}_{i}\left(\vec{k}^{\prime},z\right)\right\rangle=(2 \pi)^{3} \delta_{\mathrm{D}}\left(\vec{k}+\vec{k}^{\prime}\right) P_{i}(\vec{k}, z)\quad(3)$$

?

Any help is welcome.

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This is just a Fourier transform: (let $\boldsymbol{x}=\boldsymbol{r}_2-\boldsymbol{r}_1$)

$$ \begin{aligned} \langle \delta(\boldsymbol{k}_1)\delta(\boldsymbol{k}_2) \rangle&=\int\int d^3r_1d^3r_2\langle \delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2) \rangle e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\ &=\int d^3r_1 e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}\int d^3r_2\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2)\rangle e^{-i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\ &=\int d^3r_1 e^{-i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}\int d^3x\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_1+\boldsymbol{x})\rangle e^{-i\boldsymbol{k}_2\cdot(\boldsymbol{r}_1+\boldsymbol{x})}\\ &=\int e^{-i(\boldsymbol{k}_1+\boldsymbol{k}_2)\cdot\boldsymbol{r}_1}d^3r_1\int\xi(\boldsymbol{x})e^{-i\boldsymbol{k}_2\cdot\boldsymbol{x}}d^3x\\ &=(2\pi)^3\delta_D(\boldsymbol{k}_1+\boldsymbol{k}_2)P(\boldsymbol{k}_2) \end{aligned} $$

Here, $\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2\rangle)$ is two-point correlation function (2pcf) in real space. If we assume our universe is statistically homogeneous, $\langle\delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2\rangle)$ should have the form $\xi(\boldsymbol{r}_1-\boldsymbol{r}_2)$. So power spectrum is the Fourier transform of 2pcf.

In addition, if we assume our universe is statistically isotropic (not true in redshift space), 2pcf can be $\xi(|\boldsymbol{r}_1-\boldsymbol{r}_2|)$ and power spectrum can be $P(k)$.

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  • $\begingroup$ Zhao : Thanks. which trick do you use in integrals to pass from first line to second line in your demonstration : $\endgroup$ – youpilat13 Feb 12 at 19:31
  • $\begingroup$ $\begin{aligned} \langle \delta(\boldsymbol{k}_1)\delta(\boldsymbol{k}_2) \rangle&=\int\int d^3r_1d^3r_2\langle \delta(\boldsymbol{r}_1)\delta(\boldsymbol{r}_2) \rangle e^{i\boldsymbol{k}_1\cdot\boldsymbol{r}_1}e^{i\boldsymbol{k}_2\cdot\boldsymbol{r}_2}\\ &=\int e^{i(\boldsymbol{k}_1+\boldsymbol{k}_2)\cdot\boldsymbol{r}_1}d^3r_1\int\xi(\boldsymbol{r}_1-\boldsymbol{r}_2)e^{i\boldsymbol{k}_2\cdot(\boldsymbol{r}_1-\boldsymbol{r}_2)}d^3(r_1-r_2)\\ &=(2\pi)^3\delta_D(\boldsymbol{k}_1+\boldsymbol{k}_2)P(\boldsymbol{k}_2) \end{aligned}$ ? Best regards. $\endgroup$ – youpilat13 Feb 12 at 19:32
  • $\begingroup$ @youpilat13: Thanks for your comments, I fixed my Fourier notation and improved the derivation. Hope this helps! $\endgroup$ – R.Y. Zhao Feb 13 at 10:11
  • $\begingroup$ Zhao. Thank you, it is clearer. Best regards $\endgroup$ – youpilat13 Feb 13 at 11:22

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