May 20, 2019 - Bispectra of the Poisson field

Definitions

Using definitions from https://arxiv.org/pdf/astro-ph/0005036.pdf

[math]\displaystyle{ \langle a_{\ell_1m_1} a_{\ell_2m_2} a_{\ell_3m_3} \rangle = \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ m_1 & m_2 & m_3 \end{pmatrix} B_{\ell_1 \ell_2 \ell_3} = \sqrt{\frac{(2\ell_1+1)(2\ell_2+1)(2\ell_3+1)}{4\pi}} \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ m_1 & m_2 & m_3 \end{pmatrix} b_{\ell_1 \ell_2 \ell_3} }[/math]

Poisson

Bispectrum

Taken from https://arxiv.org/abs/1303.3535 .

Let's have pixels of size [math]\displaystyle{ A }[/math] that are much smaller than the beam (I believe Tom has a typo in his paper, when he claims the opposite - see his condition below his (14)). We choose them so small, that the vast majority of pixels is empty and we can later avoid thinking about what happens when two sources fall into the same pixel.

Imagine we have just a single source with flux density [math]\displaystyle{ S }[/math]. The temperature map in this case is zero everywhere, only in the pixel with the source there is temperature

[math]\displaystyle{ T_\mathrm{pix} = g_\nu \times \frac{S}{A} , }[/math]

where [math]\displaystyle{ g_\nu }[/math] is conversion factor from Jy/sr to Kelvin.

The Fourier coefficient is [math]\displaystyle{ a_{\ell m} = \int dx Y^*_{\ell m}(x) T(x) = A \times Y^*_{\ell m}(x_\mathrm{pix}) \times g_\nu \frac{S}{A} = g_\nu \times S \times Y^*_{\ell m}(x_\mathrm{pix}) . }[/math]

The three point function then reads

[math]\displaystyle{ \begin{array}{lcl} \langle a_{\ell_1m_1} a_{\ell_2m_2} a_{\ell_3m_3} \rangle &=& g_\nu^3 S^3 \langle Y_{\ell_1 m_1}^*(x_\mathrm{pix}) Y_{\ell_2 m_2}^*(x_\mathrm{pix}) Y_{\ell_3 m_3}^*(x_\mathrm{pix})\rangle \\ &=& g_\nu^3 S^3 \frac{1}{4\pi} \int dx Y_{\ell_1 m_1}^*(x) Y_{\ell_2 m_2}^*(x) Y_{\ell_3 m_3}^*(x) \\ &=& g_\nu^3 S^3 \frac{1}{4\pi} \sqrt{\frac{(2\ell_1+1)(2\ell_2+1)(2\ell_3+1)}{4\pi}} \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ m_1 & m_2 & m_3 \end{pmatrix} \end{array} }[/math]

In case we have several sources, the cross terms (with the possible exception of some [math]\displaystyle{ \ell_i = 0 }[/math]) vanish, because we are as likely to have positive and negative values in the [math]\displaystyle{ Y^*_{\ell_im_i} }[/math] that is evaluated at a different pixel (and we assume it never happens two sources are at the same pixel). The three point function then reads

[math]\displaystyle{ \begin{array}{lcl} \langle a_{\ell_1m_1} a_{\ell_2m_2} a_{\ell_3m_3} \rangle &=& g_\nu^3 (S_1^3 + \dots +S_N^3) \frac{1}{4\pi} \sqrt{\frac{(2\ell_1+1)(2\ell_2+1)(2\ell_3+1)}{4\pi}} \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ m_1 & m_2 & m_3 \end{pmatrix} \end{array}, }[/math]

which corresponds to the reduced bispectrum

[math]\displaystyle{ b_{\ell_1\ell_2\ell_3} = g_\nu^3 (S_1^3 + \dots +S_N^3) \frac{1}{4\pi} . }[/math]

Assuming a continuous distribution of sources,

[math]\displaystyle{ b_{\ell_1\ell_2\ell_3} = g_\nu^3 \frac{1}{4\pi} \int dS \frac{dN}{dS} S^3 . }[/math]

The factor of [math]\displaystyle{ 4\pi }[/math] is weird, but it is not clear how to get rid of it.

After studying the literature it actually appears this factor is actually correct. See (38) of Komatsu & Spergel, where they point out their [math]\displaystyle{ \frac{dn}{dS} }[/math] is density per unit angle. This factor seems to be missing in Tom's paper, but he mentions "[math]\displaystyle{ \Delta \Omega_\mathrm{map} }[/math] correction", which may be talking about this.

Power spectrum

As for the three point function, we can look at contribution from a single source

[math]\displaystyle{ \begin{array}{lcl} \langle a_{\ell_1m_1} a^*_{\ell_2m_2}\rangle &=& g_\nu^2 S^2 \langle Y_{\ell_1 m_1}^*(x_\mathrm{pix}) Y_{\ell_2 m_2}(x_\mathrm{pix})\rangle \\ &=& g_\nu^2 S^2 \frac{1}{4\pi} \int dx Y_{\ell_1 m_1}^*(x) Y_{\ell_2 m_2}(x) \\ &=& g_\nu^2 S^2 \frac{1}{4\pi} \delta_{\ell_1 \ell_2} \delta_{m_1 m_2} \end{array} }[/math]

The powerspectrum with many sources is then [math]\displaystyle{ C_\ell = g_\nu^2 \frac{1}{4\pi} \int dS \frac{dN}{dS} S^2 . }[/math]

Skewness using top hat filters

As can easily be seen (eg Komatsu https://arxiv.org/abs/astro-ph/0005036v2), skewness of a filtered field is

[math]\displaystyle{ S_3 = \frac{1}{4\pi} \sum_{\ell_1 \ell_2 \ell_3} \frac{(2\ell_1+1)(2\ell_2+1)(2\ell_3+1)}{4\pi} \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ 0 & 0 & 0 \end{pmatrix}^2 W_{\ell_1}W_{\ell_2}W_{\ell_3} b_{\ell_1\ell_2\ell_3} , }[/math]

where [math]\displaystyle{ W_\ell }[/math] is weighting the data based on what we measure.

Using Stirling's formula

[math]\displaystyle{ n! \approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n }[/math]

and analytic form for [math]\displaystyle{ \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ 0 & 0 & 0 \end{pmatrix} }[/math] we can write down approximation (https://iopscience.iop.org/article/10.1088/0004-637X/760/1/5/pdf)

[math]\displaystyle{ \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ 0 & 0 & 0 \end{pmatrix} \approx \sqrt{\frac{2}{\pi}} \frac{(-1)^{L/2}}{[L(L-2\ell_1)(L-2\ell_2)(L-2\ell_3)]^{1/4}} , }[/math]

where [math]\displaystyle{ L = \ell_1 + \ell_2 + \ell_3 }[/math] and we get zero for odd [math]\displaystyle{ L }[/math]. This approximation is valid as long as all the elements in the denominator are sufficiently large, as they are the arguments we apply Stirling's formula for.

Restricting [math]\displaystyle{ \ell \in I = [\ell_c - \Delta \ell/2, \ell_c + \Delta \ell/2] }[/math], we have

[math]\displaystyle{ \begin{array}{lcl} S_3 &=& \frac{1}{4\pi} \sum_{ \ell_i \in I } \frac{(2\ell_1+1)(2\ell_2+1)(2\ell_3+1)}{4\pi} \begin{pmatrix} \ell_1 & \ell_2 & \ell_3\\ 0 & 0 & 0 \end{pmatrix}^2 b_{\ell_1\ell_2\ell_3} \\ &\approx& \sum_{ \ell_i \in I, \ell_1+\ell_2+\ell_3\ \mathrm{even} } \frac{\ell_1 \ell_2 \ell_3}{2\pi^2} \times \frac{2}{\pi} \frac{1}{[L(L-2\ell_1)(L-2\ell_2)(L-2\ell_3)]^{1/2}} b_{\ell_1\ell_2\ell_3} \\ &\approx& \sum_{ \ell_i \in I, \ell_1+\ell_2+\ell_3\ \mathrm{even} } \frac{\ell_c^3}{\pi^3} \times \frac{1}{[3 l_c^4]^{1/2}} b_{\ell_c\ell_c\ell_c} \\ &=& \frac{\ell_c}{\sqrt{3}\pi^3} b_{\ell_c\ell_c\ell_c} \sum_{ \ell_i \in I, \ell_1+\ell_2+\ell_3\ \mathrm{even} }1 \\ &=& \frac{\ell_c}{\sqrt{3}\pi^3} b_{\ell_c\ell_c\ell_c} \times \frac{(\Delta \ell)^3}{2} \end{array} }[/math]

Here we assumed [math]\displaystyle{ \ell_c \gg \Delta \ell }[/math], which allows us to take everything outside of the sum. On the last line we just take half of the volume, because of the condition on even [math]\displaystyle{ \ell_1 + \ell_2 + \ell_3 }[/math].