2
$\begingroup$

Usually when we study the CM8B we focus on the temperature anisotropies of it and we say that $\delta T/T \sim O(10^{-5})$

For the temperature anisotropies I know that the definition is $\delta T(\hat{n})=T(\hat{n})-T_0$ where $T_0$ is the average temperature.

I saw on the internet that there are maps of the CMB polarization ANISOTROPIES. I want to ask: how are those anisotropies defined and of what order are they.

$\endgroup$
  • $\begingroup$ This uses the Stokes parameters $Q$ and $U$. Are you versed in those? $\endgroup$ – user154997 Oct 22 '17 at 21:51
  • $\begingroup$ Yes, I know Q,U and V. $\endgroup$ – Saladino Oct 22 '17 at 22:24
1
$\begingroup$

The analysis of the CMBR polarisation uses the Stokes parameters. Only $Q$ and $U$ play any role: the reference value, by that I mean the equivalent of your $T_0$, is 0 here, i.e. a lack of polarisation. More precisely, $Q+iU$ and $Q-iU$ are expanded on the basis of spherical harmonics. I have the recollection there is a subtlety related to the transformation of $Q$ and $U$ under rotation. A classic paper introducing the analysis of the CMBR polarisation is [1], and I will refer you to it.

[1] Matias Zaldarriaga and Uroš Seljak. All-sky analysis of polarization in the microwave background. Phys. Rev. D, 55:1830–1840, Feb 1997.

$\endgroup$
  • $\begingroup$ I know this description(and the fact that $Q+iU$ is a 2 spin field). What I don't understand is why people talk about polarization anisotropies. I know that CMB (linear) polarization is due to Thompson scattering coming from the quadrupole part of the radiation field. So lets imagine that I have a particular direction of polarization for the scattered photon. What does it mean polarization anisotropy? $\endgroup$ – Saladino Oct 22 '17 at 23:03
  • $\begingroup$ $Q$ and $U$ are functions of the direction of observation in the sky, i.e. the unit vector $\hat{n}$ in your question. The anisotropy is their variation as $\hat{n}$ varies. What else would you have had in mind? $\endgroup$ – user154997 Oct 22 '17 at 23:13
  • $\begingroup$ Ok, so it makes sense to take a mean value and calculate two type of anisotropies $\delta U$ and $\delta Q$...Thanks. My last question is: $\delta U/U$ is of the same order as temperature anisotropies? $\endgroup$ – Saladino Oct 22 '17 at 23:35
  • $\begingroup$ That I don't remember. I will have a look at the paper tomorrow if you haven't before me! $\endgroup$ – user154997 Oct 22 '17 at 23:52
  • $\begingroup$ The article doesn't mention it..... $\endgroup$ – Saladino Oct 23 '17 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.