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Temperature Anisotropies !
 
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Muhammad Junaid



Joined: 05 Jan 2017
Posts: 7
Affiliation: University of Alberta

PostPosted: March 28 2017  Reply with quote

Hi everyone,

I just have one simple question question, I want to calculate the temperature anisotropies as a function of wave number i.e. ΔT(k) / T using the list of transfer functions output from the camb code. I just want to use my own initial power spectrum Pζ(k) of initial perturbations(comoving curvature perturbations) to calculate \langle \Delta T^2(k)/T^2 \rangle = \Delta_i^2(k) P_\zeta(k). Where Δi(k) would be some combination of transfer functions of the camb code. So, what precise combinations of transfer functions do I need here as Δi's?

Thanks in Advance!!

M Junaid
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Antony Lewis



Joined: 23 Sep 2004
Posts: 1270
Affiliation: University of Sussex

PostPosted: March 28 2017  Reply with quote

Are you sure you don't want to calculate the C power spectrum? That's easily done by modifying CAMB's power_tilt.f90 to use your own initial power spectrum (and increasing accuracy settings if needed).
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Muhammad Junaid



Joined: 05 Jan 2017
Posts: 7
Affiliation: University of Alberta

PostPosted: March 28 2017  Reply with quote

No we are working out some moments in momentum space so we don't need Cl or any angular part.
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Muhammad Junaid



Joined: 05 Jan 2017
Posts: 7
Affiliation: University of Alberta

PostPosted: March 30 2017  Reply with quote

My hunch is that the temperature anisotropy is to leading order equal to \Delta T/T \approx \frac{1}{4} \hat{\Delta}_\gamma+ 2\Phi, where Φ is Weyl potential. While, you have said in your notes that power spectrum of Weyl potential is PΦ = TΦ(k)Pζ(k) where ζ is primordial perturbations. Is the same true for other power spectra as \Delta_\gamma^2 = \hat{\Delta}_\gamma^2 P_\zeta(k). So the transfer function output from the camb code are \hat{\Delta}_\gamma^2 that don't include the primordial power spectrum Pζ(k). Please correct me if I am wrong!

M.
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Antony Lewis



Joined: 23 Sep 2004
Posts: 1270
Affiliation: University of Sussex

PostPosted: March 30 2017  Reply with quote

Transfer function outputs are not squared (and indeed not scaled by the primordial power). The temperature sources are quite complicated in general (calculated in output routine of equations.f90).
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